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You can Download Chapter 15 Waves Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 15 Waves

1st PUC Physics Waves Textbook Questions and Answers

Question 1.
A string of mass 2.50 kg Is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse Jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Mass per unit length of the string,
µ = \(\frac{2.50 \mathrm{kg}}{20 \mathrm{m}}\) = 0.125 kg m^-1
Given tension, T = 200 N.
The speed of wave on the string is given by,
v = \(\sqrt{\frac{T}{\mu}}\) =\(\sqrt{\frac{200}{0.125}}\) = 40 ms^-1
∴ Time taken by the disturbance to reach the other end of the string:
t = \(\frac{\mathrm{s}}{\mathrm{v}}\) = \(\frac{20}{40}\) = 0.5 s

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When Is the splash heard at the top given that the speed of sound In air Is 340 m s^-1? (g= 9.8 ms^-2)
Answer:
Time taken to hear the splash at the top =Time taken by the stone to reach the pond + Time taken by the sound to travel to the top from the base of tower,
i e., T = T₁ + T₂ ……..(1)
T₁ : We know that s = u t + 1 / 2 a t²
u = 0, s = 300 m, a = g = 9.8 ms^-2
⇒ t = \(\sqrt{\frac{2 \mathrm{s}}{\mathrm{a}}}\) = \(\sqrt{\frac{2 \times 300}{9.8}}\) = 7.82 s
∴ T₁ =7.82 s
T₁ : s = vt ⇒ t = \(\frac{\mathrm{s}}{\mathrm{v}}\) = \(\frac{300}{340}\) = 0.88 s
∴ T₂ = 0.88 s
∴ T₁+ T₂ = 7.82 + 0.88 = 8.70 s

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension In the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 m s^-1?
Answer:
Mass per unit length of the wire,
µ = \(\frac{2.10 \mathrm{kg}}{12.0 \mathrm{m}}\) = 0.175 kg m^-1
The speed of wave on a string is
given by, v=\(\sqrt{\frac{T}{\mu}}\)
Given v = 343 ms^-1
∴ T = µv² =0.175 × 343²
= 20588.575 = 2.06×10⁴ N.

Question 4.
Use the formula v = \(\sqrt{\frac{rP}{\rho}}\) to explain why the speed of sound in air

1. Is Independent of pressure,
2. Increases with temperature,
3. Increases with humidity.

Answer:
Given v = \(\sqrt{\frac{\mathrm{rP}}{\rho}}\) ……….(1)
According ideal gas law, P = \(\frac{\rho \mathrm{RT}}{\mathrm{M}}\) , where
\(\boldsymbol{\rho}\) is the density, T is the temperature, M is the Molecular mass of the gas;
R – universal gas constant.
Substituting for P in (1) we get
v = \(\sqrt{\frac{\mathbf{r} \mathbf{R} \mathbf{T}}{\mathbf{M}}}\)
This shows that v is

1. Independent of pressure
2. Increases with temperature i.e. v ∝ \(\sqrt{T}\)
3. We know that the molecular mass of water (18) is less than that of N₂ (28) and O₂ (32). therefore as humidity increases, the effective molecular mass of air decreases and hence velocity increases.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + vt, i.e., y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave :
(a) (x – vt )²
(b) log [(x + Vy)/x₀]
(c) 1/(x + vt)
Answer:
The converse is not true. For any function to represent a travelling wave, an obvious requirement is that the function should be finite at all times and finite everywhere.

Function a) and b) do not satisfy this requirement and hence cannot represent a travelling wave. Only function c) satisfies the condition.

Question 6.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of

1. The reflected sound,
2.  The transmitted sound? Speed of sound in air is 340 ms^-1 and in water 1486 ms^-1.

Answer:
We know that λ = \(\frac{\mathrm{v}}{\mathrm{f}}\)
where,
λ : wavelength
v : velocity
f: frequency

1.  Reflected sound :
λ = \(\frac{340}{1000}\) = 0.34 m
(∵ v The reflected sound travels in air)

2. Transmitted sound :
λ = \(\frac{1486}{1000}\) = 1.486 m
(∵ v The transmitted sound travels in water)
Note: When sound moves from one medium to other, the frequency does not change.

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
Given
f = 4.2 × 10⁶ Hz
v = 1.7 × 10³ ms^-1
we know that, λ = \(\frac{\mathrm{v}}{\mathrm{f}}\)
\(=\frac{1.7 \times 10^{3}}{4.2 \times 10^{6}}\) = 4.05 × 10^-4 m
∴ wavelength of sound in tissue = 4.05 × 10^-4 m

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x+ π/4 ) where x and y are in cm and t in s. The positive direction of x is from left to right.

1. Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
2. What are its amplitude and frequency?
3. What is the initial phase at the origin?
4. What is the least distance between two successive crests in the wave?

Answer:

1. Travelling wave: The wave is travelling from right to left with a speed of \(\frac{36 s^{-1}}{0.018 \mathrm{cm}^{-1}}\) = 2000 cm s^-1 = 20 ms^-1
2. Amplitude: 3 Frequency : \(\frac{36}{2 \pi}\) = 5.73 Hz
3. Initial phase at origin : π/4
4. Distance between two successive crests in the wave = \(\frac{2 \pi}{0.018 \mathrm{cm}^{-1}}\) = 349.06 cm = 3.49 m.

Question 9.
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0,2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion In travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:

All the graphs have same amplitude of frequency but different initial phase. All graphs are sinusoidal.

Question 10.
For the travelling harmonic wave y(x,t) = 2 cos 2π (10t – 0.0080 x + 0.35) where x and y are In centimetre and t In seconds. Calculate the phase difference between oscillatory motion of 2 points separated by a distance of
a) 4 m
b) 0.5 m
c) λ/2
d) 3λ/2
Answer:
Given
k = 2m × 0.008 cm^-1
We know that, k = \(\frac{2 \pi}{\lambda}\) ⇒ λ = \(\frac{2 \pi}{(0.008 \times 2 \pi)}\) = 125cm
Now, (phase difference) = \(\frac{2 \pi}{\lambda}\) × (path difference)
∴ ∆Φ = \(\frac{2 \pi}{\lambda}\) ∆x

Question 11.
The transverse displacement of a string (clamped at Its both ends) is given by y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\). cos \((120 \pi \mathrm{t})\) where x and y are in m and t In s The length of the string is 1.5 m and its mass is 3.0 x10^-2 kg.
Answer the following :

1. Does the function represent a travelling wave or a stationary wave?
2. Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
3. Determine the tension in the string.

Answer:
1.  The function represents a stationary wave.

2.

∴ λ = 3 m , f = 60 Hz, v = \(\frac{120 \pi}{2 \pi} \times 3\) = 180 ms^-1
for both waves.
3.  We know that T = µ v² where µ = mass per unit length of the string
\(=\frac{3 \times 10^{-2} \mathrm{kg}}{1.5 \mathrm{m}}\) = 2 × 10^-2 kgm^-1
∴ T = 2 × 10^-2 × 180² = 648 N

Question 12.
(i) For the wave on a string described in Exercise 15.11, do all the points c n the string oscillate with the same
(a) frequency,
(b) phase,
(c) amplitude?
Explain your answers.
(ii) What Is the amplitude of a point 0.375 m away from one end?
Answer:
Given :
y (x, t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\) cos (120 πt)

i) Amplitude of wave = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)

Frequency = 60 Hz.
∴ From the above equation of wave, we see that all points on the string oscillate with same phase and frequency except the nodes (endpoints). We can also see that amplitude of the wave changes for different ‘x’.

ii) Amplitude of a point at 0.375 m
i.e., x = 0.375 m
Amplitude = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)
= 0.06 sin \(\left(\frac{2 \pi}{3} \times 0.375\right)\)
= 0.042 m

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:

1. y= 2 cos (3x) sin (10t)
2. y= 2\(\sqrt{x-v t}\)
3. y = 3 sin(5x – 0.5t) + 4cos(5x-0.5t)
4. y = cos x sin t + cos 2x sin 2t

Answer:
1.  y = 2cos (3x) sin (10 t)
Represents a stationary wave

2.  y = 2\(\sqrt{x-v t}\)
Does not represent any wave.

3.  y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
= 5 sin \(\left[5 x-0.5 t-\tan ^{-1}(3 / 4)\right]\) Represents a travelling wave,

4.  y = cos x sin t + cos 2x sin 2t
Represents superposition of 2 stationary waves.

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2kgm^-1. What is

1. The speed of a transverse wave on the string,
2. The tension in the string?

Answer:
Given :
f = 45 Hz mass of wire
= m =3.5×10^-2 kg
linear mass density
= µ = mass per unit length
= 4 × 10^-2 kg m^-1
∴ length of the wire
\(=\frac{m}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}\)
l = 0.875 m

1.  We vibrate in fundamental mode
⇒ λ = 2l

∴ λ =2×0.875 = 1.75 m
∴ Speed of wave
= v =fλ = 45×1.75
= 78.75 ms^-1

2. We know that
T = µV²
∴ Tension, T = 4×10^-2 × 78.75^-2 = 248.06 N

Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer:
For a closed organ pipe of length ‘l’, the frequency of n^th mode of vibration is
f = (2n – 1)\(\frac{v}{4 l}\) ………….(1)
Thus the (n + 1)^th mode of vibration of closed pipe is
f = (2n + 1)\(\frac{v}{4 l}\) ………….(2)
Given that the closed pipe vibrates at 340 Hz for tube length = 25.5 cm and 79.3 cm.
Let l₁ = 25.5 cm l₂ = 79.3 cm
Equating (1) and (2) we get

On solving we get n = 1.
On substituting in (1) we have
\(\frac{(2 \times 1-1) v}{4 \times\left(\frac{25.5}{100}\right)}=340\) ⇒ v = 346.8 ms^-1

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Given:
l = 100 cm = 1 m  f = 2.53 kHz

Since the rod is champed at the middle, a node is formed and since the rod is oscillating at a fundamental frequency of 2.53 kHz antinodes are formed at both ends as shown in the figure.
∴ l = λ/4 × 2 = λ/2 ⇒ λ = 2l
λ = 2 × 1 = 2 cm
We know that
v = fλ = 2.53 × 10³ × 2
= 5.06 × 10 ³ ms^-1

Question 17.
A pipe 20 cm long Is closed at one end. Which harmonic mode of the pipe 1s resonantly excited by a 430 Hz source? Will the same source be In resonance with the pipe if both ends are open? (speed of sound In air Is 340 m s^-1).
Answer:
Given :
l = 20 cm = 0.2 m
v = 340 ms^-1
f = 430 Hz
For a pipe closed at one end :
f = \(\frac{(2 n-1) v}{4 l}\) n = 1, 2, 3 ……
⇒ 430 = \(=(2 n-1) \frac{340}{4 \times 0.2}\)
⇒ n = 1
⇒ Resonance occurs only for first/fundamental mode of vibration.
For a pipe open at both ends,
f = \(\frac{\mathrm{nv}}{2 l}\) n = 1, 2, …..
⇒ 430 = \(\frac{\mathrm{n} \times 340}{2 \times 0.2}\)
⇒ n = 0.51
Since n<1, resonance does not occur.

Question 18.
Two sitar strings A and B playing the note ‘Gd’ are slightly out of tune and produce beats of frequency 6 Hz. The tension In the string A Is slightly reduced and the beat frequency Is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what Is the frequency of B?
Answer:
Let f₁ and f₂ be the frequencies of strings A and B respectively.
Given:
f₁ = 324 Hz
Beat frequency = f_b = 6 Hz
∴ f₂= f₁ ± f_b =(324 ±6) Hz
We know that T ∝ v² and v ∝ f
(∵ T=μv² and v=fλ.)
Thus f ∝ \(\sqrt{T}\) …………(1)
Decreasing the tension in string A will decrease the frequency f₁.
Since the beat frequency f_b is reduced to 3 Hz upon decreasing the tension in string A, f₂ = 318 Hz.
Note:
If f₂ = 330 Hz, then the beat frequency would have increased when the tension in string A was reduced.

Question 19.
Explain why (or how):

1. In a sound wave, a displacement node Is a pressure antinode and vice versa,
2. bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
3. A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
4. Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate In gases, and
5. The shape of a pulse gets distorted during propagation in a dispersive medium.

Answer:

1. In a sound wave, the displacement node is a point where the amplitude of oscillation is zero. But at a displacement node, the pressure changes are maximum. Hence it is also a pressure antinode. Similarly, at displacement antinode, the amplitude of oscillation is maximum whereas pressure changes are minimum. Hence displacement antinode is also a pressure node.
2. Bats emit high-frequency ultrasonic waves. These waves are reflected back from the obstacles on their path and are sensed by bats.
3. Because they emit different harmonies which can be easily differentiated by human ears.
4. This is because solids have both shear and bulk modulus of elasticity whereas gas has only bulk modulus of elasticity.
5. A sound pulse consists of waves with different wavelengths. In a dispersive medium, these waves travel with different velocities which distorts the shape of pulse.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz In still air.
1.  What Is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 m s^-1,
(b) recedes from the platform with a speed of 10 m s^-1?

2.  What Is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given, f = 400 Hz v = 340 ms^-1 v_s = 10 ms^-1 : speed of train .
1.
a) train approaches the platform

b) train recedes from the platform

2.  The speed of sound does not change, i.e., it is 340 ms^-1 for both cases.

Question 21.
A train, standing In a station-yard, blows a whistle of frequency 400 Hz In still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 ms^-1. what are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly Identical to the case when the air Is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given f = 400 Hz v = 340 ms^-1 v_m = 1.0 ms^-1 : speed of air
Since there is no relative motion between the source and the observer,
the frequency of sound for the observer = f =400 Hz.
The wind is blowing in the direction from the yard to the station, the effective speed of sound for an observer at the platform = v + v_m = 350 ms^-1
Wavelength of sound : λ= \(\frac{\left(v+v_{m}\right)}{f}\)
= \(\frac{350}{400}\) = 0.875 m
The situation is not identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms^-1 (= v₀). Because here there is a relative motion between the source and the observer.

Question 22.
A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin (0.005 x +12t + π /4)

1. what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
2. Locate the points of the string which have the same transverse displacements and velocity as the x a 1 cm point at t = 2 s, 5 s, and 11s.

Answer:
Given:
y (x, t) = 7.5 sin (0.005 x + 12t + π /4)
1. At x=1 cm and t=1s
y (1, 1)= 7.5 Sin(0.005 +12 + π /4)
= 7.5 sin (12.005 + π /4)
= 1.67 cm
Velocity of oscillation : v = \(\frac{\mathrm{d}(\mathrm{Y}(\mathrm{x}, \mathrm{t})}{\mathrm{dt}}\)
= \(\frac{\mathrm{d}}{\mathrm{dt}}\) (7.5 sin (0.005 x + 12t + π/4) dt
= 7.5 × 12 cos (0.005 x+ 12t + π/4)
At x = 1 cm and t = 1 s
v = 7.5 × 12 cos(0.005 + 12 + π/4)
= 87.75 cm s^-1
We know that velocity of wave propagation = \(\frac{\mathrm{w}}{\mathrm{k}}\)
Here w = 12 s^-1 and k = 0.005 cm^-1
∴ Velocity of wave propagation
= \(\frac{12 s^{-1}}{0.005 \mathrm{cm}^{-1}}\) = 24 ms^-1

∴ At x = 1 cm and t = 1 s velocity of oscillation is not equal to velocity of wave propagation.

2. In a wave, all the points which are separated by a distance ±λ,±2λ ……..
from x = 1 cm will have same transverse displacements and velocity. For the given
wave , λ= \(\frac{2 \pi}{0.005}\) = ±12.56 cm, +25.12
m….From x = 1 cm will have the same displacements and velocity as at x = 1 cm, t = 2s, 5s and 11 s.

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
1.  Does the pulse have a definite
(i) frequency,
(ii) wavelength,
(iii) speed of propagation?
2.  If the pulse rate is 1 after every 20 s, (that is the whistle Is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
1.  The pulse does not have definite frequency and definite wavelength but has definite speed of propagation.
2.  The frequency of the note is not 1/20 Hz or 0.05 Hz. It is only the frequency of repetition of the whistle.

Question 24.
One end of a long string of linear mass density 8.0 × 10^-3 kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is kg. The pulley end absorbs all the Incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down its transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Given : Mass per unit length of string = μ = 8×10^-3kgm^-1
f = 256 Hz A = 5 cm : amplitude Tension in the string,
T = 90 kg × 9.8 ms^-2= 882 N
velocity of the wave :
v = \(\sqrt{\frac{T}{\mu}}\)

propagation constant,
k= \(\frac{2 \pi}{\lambda}\) = 4.84 m^-1
Equation of wave y (x, t) = A sin(wt – kx)
∴ y(x, t) = 5 sin(512 π t – 4.84 x)
with x and y in centimetre and t in seconds.

Question 25.
A SONAR system fixed In a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be1450ms^-1.
Answer:
Given:
f = 40 kHz v = 1450ms^-1
Speed of enemy submarine = v₀ = 360 km/ hr
=360 × \(\frac{5}{16}\) =100 m/s
Apparent frequency of sound waves:
(source at rest and observer moving towards the source)

Now this frequency is reflected by the enemy submarine and is observed by the SONAR
∴ \(\mathrm{f}^{\prime}=\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}} \times \mathrm{f}^{\prime}\)
Note: Here the observer (SONAR) is at rest and the source is moving towards the observer at a speed of 360 km h^-1 = 100 ms^-1 (v_s)
\(\therefore \mathrm{f}^{\prime}=\frac{1450}{1450-100} \times 42.76 \mathrm{k}=45.93 \mathrm{kHz}\)

Question 26.
Earthquakes generate sound waves Inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^-1, and that of P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 minutes before the first S wave. Assuming the waves travel in a straight line, at what distance does the earthquake occur?
Answer:
Let v_p =8 km s^-1 and v_s = 4 k ms^-1
t = time gap between arrival of s and p waves = 4 min = 240 s
t = t_s – t_p = \(\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{s}}}-\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{p}}}\)
where d :
distance of earthquake centre (kms)
∴ 240 = \(\mathrm{d}\left[\frac{1}{4}-\frac{1}{8}\right]\)
∴ d = \(\frac{240}{0.25-0.125}\) = 1920 km

Question 27.
A bat Is flitting about In a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat Is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Given:
f = 40 kHz
velocity of bat = 0.03 velocity of sound i.e., v_s = 0.03 v
Apparent frequency of sound hitting the wall
\(f^{\prime}=\frac{v}{v-v_{s}} \times f=\frac{v}{v-0.02 v} \times f\)
= 1.03 \(f^{\prime}\) = 1.03 × 40 k
= 41.24 k Hz
This freuency f1 is reflected e ft tne wall and is recieved by the bat moving towards the wall
∴ \(f^{\prime \prime}=\frac{\gamma+v_{s}}{v} \times f^{\prime}=\frac{v+0.03 v}{v} \times f^{\prime}\)
= 1.01 \(f^{\prime}\) = 1.03 × 41.24 k
= 42.47 k Hz

1st PUC Physics Waves one mark Questions and Answers

Question 1.
What is meant by a wave?
Answer:
The disturbance set up in a medium is called a wave.

Question 2.
What property of the medium is essential for the propagation of mechanical wave?
Answer:
Elasticity and Inertia.

Question 3.
Which physical quantity does not change when a wave travels from one medium to another?
Answer:
Frequency

Question 4.
What is a progressive wave?
Answer:
A wave, which travels continuously in a medium in the same direction, is called a progressive wave.

Question 5.
If y = 2sin π (40t – 2x) represents a progressive wave. What is its frequency?
Answer:
20 Hz.
Solution:
Given equation is
y = 2 sin π (40t – 2x)
y =2 sin 40π (t – x/20)
Comparing this with y = a sin ω (t -x/v)
ω = 2 πf = 40π
∴ f= 20Hz

Question 6.
Two waves are represented by the equations Y₁ = a sin(ωt- kx) and Y₂ = a cos (ω t – kx). What is the phase difference between them?
Answer:
π/2 radians.
Y₁ = a sin (ω t – kx).
Y₂= a cos (ω t – kx).
= a sin(ω t – kx + π /2).
∴ Phase difference between Y₁ and Y₂ is π/2 rad.

Question 7.
What is meant by ‘Phase of a Particle’ in a wave?
Answer:
The phase of a particle at any instant represents the state of vibration of the particle at that instant.

Question 8.
What is a mechanical wave?
Answer:
A wave which requires a medium for its propagation is called a mechanical wave.

Question 9.
Give an example of a transverse wave.
Answer:
Light waves.

Question 10.
Give an example of a two-dimensional wave.
Answer:
Water waves.

Question 11.
What is a transverse wave?
Answer:
In transverse waves, particles of the medium are vibrating perpendicular to the direction of wave propagation.

Question 12.
What is the angle between the vibration of the particle of medium and direction of propagation of the wave in a transverse wave?
Answer:
90°

Question 13.
How does velocity of sound vary with pressure?
Answer:
The velocity of sound is independent of pressure.

Question 14.
How does velocity of sound vary with temperature?
Answer:
The velocity of sound in a gas is directly proportional to the square root of the absolute temperature.

Question 15.
How does velocity of sound vary with humidity?
Answer:
The velocity of sound increases with increase in humidity.

Question 16.
Why sound travels faster in moist air than in dry air?
Answer:
The density of moist air is less than that of dry air. As the velocity of sound in a gas is inversely proportional to the square root of its density, the velocity of sound in moist air is greater than that in dry air.

Question 17.
A wave has a velocity of 330 ms^-1 at one atmospheric pressure. What will be its velocity at 4 atmospheric pressure?
Answer:
330 ms^-1
[Reason: Velocity of sound in a gas is independent of pressure].

Question 18.
What are the beats?
Answer:
The rise and fall in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is known as beats.

Question 19.
What is beat frequency?
Answer:
The number of beats heard per second is called beat frequency.

Question 20.
Define the beat period.
Answer:
The time interval between two consecutive maxima or minima is called beat period. It is also equal to the reciprocal of beat frequency.

Question 21.
What is the available range of sound frequencies?
Answer:
20 Hz to 20,000 Hz.

Question 22.
By how much does the frequencies of 2 sound sources differ, if they produce 10 beats in 2 seconds.
Answer:
Number of beats per second
\(=\frac{10}{2}=5\)
∴ ∆ f = f₁ ~ f₂ = 5 Hz

Question 23.
Define reverberation.
Answer:
Reverberation is defined as the persistence of audible sound even after the source has ceased to produce the sound.

Question 24.
What is a stationary wave?
Answer:
The wave formed due to the superposition of two identical waves travelling with the same speed in opposite directions is called a stationary wave or standing wave.

Question 25.
What Is fundamental frequency?
Answer:
The vibration of a body with the lowest frequency is called the fundamental frequency.

Question 26.
What are overtones?
Answer:
Frequencies greater than fundamental frequency are called overtones.

Question 27.
What are harmonics?
Answer:
Overtones, which are an integral multiple of the fundamental frequency, are called harmonics.

Question 28.
The length of the vibrating portion of a sonometer wire Is doubled. How does its frequency change?
Answer:
Halved
[Reason: Frequency f ∝ 1//. As / is doubled f is halved].

Question 29.
Give the relation between the fundamental note and overtone in an open pipe.
Answer:
\(f^{\prime}=(n+1) f\)
f – fundamental frequency, n = 1,2, 3….
for 1^st and 2^nd  …. overtones.

Question 30.
What is the distance between a node and the neighbouring antinode of a stationary wave?
Answer:
\(\frac{\lambda}{4}\)

Question 31.
The fundamental frequency produced in a dosed pipe is 500 Hz. What Is the frequency of the first overtone?
Answer:
1500 Hz.
[Solution: Frequency of the first over tone in a closed pipe is \(f^{\prime}\) = 3f = 3 × 500 = 1500 Hz].

Question 32.
Find the distance between node and an adjacent antinode if the wavelength in 4m in a stationary wave.
Answer:
Distance between node and adjacent antinode =\(\frac{\lambda}{4}=\frac{4}{4}\) = 1 m

Question 33.
Explain why is it NOT possible to have interference between the waves produced by 2 sitars?
Answer:
Because the waves produced will not have a constant phase difference.

Question 34.
Which harmonies are present in a closed organ pipe?
Answer:
All the odd harmonies are present.

Question 35.
What will be the resultant amplitude when 2 waves y₁ = a sin ω t are superposed at any point at a particular instant?
Answer:
y = y₁ + y₂ = a sin ω t + a cos ω t
= a (sin ω t + cos ω t)
=\(\sqrt{2} \mathrm{a}(\sin (\omega \mathrm{t}+\pi / 4))\)
∴ Resultant amplitude : \(\sqrt{2}\)a

Question 36.
Write 2 characteristics of a medium which determine the speed of sound waves in the medium.
Answer:
Elasticity and inertia.

Question 37.
State the factors in which the speed of a wave travelling along a stretched Ideal string depends
Answer:
Tension and mass per unit length.

Question 38.
The fundamental frequency of oscillation of a closed pipe is 400 Hs. What will be the fundamental frequency of oscillation of open pipe of the same length?
Answer:
f_e = \(\frac{\mathrm{v}}{4 l}\) = 400 Hz
f_o = \(\frac{\mathrm{v}}{2 l}\) ⇒ f_o = 2 f_e = 800 Hz.

Question 39.
Why is it difficult some times to recognise your friend’s voice on phone?
Answer:
Because of modulation.

Question 40.
Which of the following media can pass longitudinal waves only air, water or Iron?
Answer:
Air

1st PUC Physics Waves two marks Questions and Answers

Question 1.
Explain different types of waves (based on medium)
Answer:
Waves are classified into two types:

1. Mechanical Waves:
Waves, which requires a medium for their propagation are called mechanical waves.
e.g.: Waves on the surface of water, Seismic waves (due to earthquake), Sound waves, Waves on stretched string, waves formed in an air column, shock waves.

2. Non-mechanical Waves:
Waves, which do not require a medium for their propagation are called Non-mechanical Waves.
e.g.: Light waves, heat waves, radio waves, X- rays, ultraviolet rays, Infrared rays, etc.

Question 2.
The equation of a progressive wave is y = 0.2 sin(50t-0.5x). Find the amplitude and the magnitude of the velocity, if ‘x’ and ‘y’ are in metres.
Answer:
Given equation is,
y = 0.2 sin(50t – 0.5x)
= 0.2 sin50(t – x/100)
Comparing with y = a sin ω (t – x/v)
amplitude a = 0.2m,
velocity v = 100m/s

Question 3.
State the principle of superposition. Name the phenomenon produced due to the superposition of waves.
Answer:
When two or more waves superpose, he resultant displacement of particle of the medium is equal to the vector sum of the displacements due to individual waves. Superposition of waves leads to the phenomenon of interference, diffraction, beats, and formation of stationary waves are due to the superposition of waves.

Question 4.
What is a longitudinal wave ? Give an example.
Answer:
If the particles of a medium vibrate along the direction of wave propagation then wave is called longitudinal waves,
e.g: Sound waves in air are longitudinal waves.

Question 5.
The distance between two particles is 0.1m; If the phase difference between these points is π/2 rad calculate the wavelength.
Answer:
∆= 0.1m, Φ = π/2 rad

Question 6.
How does its frequency of a tuning fork change when the prongs are

1. filed
2. waxed.

Answer:

1. When the prongs of a tuning fork are filed its frequency increases.
2. The frequency of a tuning fork decreases when the prongs are waxed.

Question 7.
What is meant by beats? What are its applications ?.
Answer:
The periodic rise and fall (Waxing and waning) in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is called beats.
The phenomenon of beats can be used.

1. To determine the unknown frequency of a tuning fork.
2. In tuning musical instruments.

Question 8.
What is the Doppler effect? Give an example.
Answer:
The apparent change in the frequency of sound due to the relative motion between the source and the observer is called the Doppler effect.
E.g.: The apparent frequency of the whistle of a train increases as it. approaches an observer on the platform and decreases when the train passes the observer.

Question 9
What are the uses of the Doppler effect?
Answer:

1. Doppler effect is used in a radar system to detect the speed of automobiles and aeroplanes.
2. It is used to determine the speed submarines (Using SONAR).
3. It is used to determine the speed of stars and planets and other celestial bodies.

Question 10.
When two tuning forks A and X are sounded together produces 6 beats per second. If the frequency of A is 341 Hz. What is the frequency of X?
Answer:
f_A = 341 Hz
f=?
and f_b = 6 beats/s
f_b = f_A ~ f_x
or f = f_A ± f_b = 341 ± 6
= 335 Hz or 347 Hz

Question 11.
What are nodes and antinomies in a stationary wave?
Answer:
The points in a stationary wave where the amplitude of vibration of the particles zero are called nodes The points in a stationary wave where the amplitude of vibration of particles is maximum are called antinodes.

Question 12.
An open pipe and a closed pipe have the same fundamental frequency. Explain how their lengths are related?
Answer:
Fundamental Frequency of an open pipe f₁ = \(\frac { v }{ { 2l }_{ 1 } } \)
Fundamental Frequency of an closed pipe
f₂ = \(\frac{\mathbf{v}}{4 \ell_{2}}\)
Given f₁ = f₂ ∵ 2l₁ = 4l₂
\(\frac{\ell_{1}}{\ell_{2}}=\frac{2}{1}\) ⇒ l₁:l₂ = 2:1

Question 13.
Mention any four characteristics of a stationary wave.
Answer:

1. A stationary wave does not move in any direction.
2. There is no flow of energy.
3. All particles in a loop are in the same phase & they are in opposite phase with respect to the adjacent loop.
4. Amplitude is different for different particles.

Question 14.
The fundamental frequency produced in a closed pipe Is 100Hz. What is the frequencies of first and second overtone?
Answer:
Given, f = 100Hz
For the first overtone f_1,
i.e f₁ = 3f
f₁ = 3 × 100
= 300 Hz
For the second overtone f_2,
i.e f₂ = 5f
= 5 × 100
= 500 Hz.

Question 15.
The equation for the transverse wave on a string is y \(=4 \sin 2 \pi(t / 0.05-x / 50)\) with length expressed Iff centimetre and time In second. Calculate the wave velocity and maximum particle velocity.
Answer:
Given

Particle velocity

Question 16.
Equation of a wave travelling on a string is y = 0.1 sin(3001 – 0.01 x) cm. Here x is in centimetre and t is in seconds. Find

1. wavelength of the wave
2. Time taken by the wave to travel 1 m

Answer:

The wave takes T seconds to travel a distance of λ.
∴ Time taken to travel 1 m is

Question 17.
What is meant by RADAR and SONAR? How are long distances measured using these techniques?
Answer:

1. RADAR – Radio Detecting and Ranging.
2. SONAR – Sound Navigation and Ranging

The waves produced by the devices are sent and are reflected by the bodies and reach them back. If the speed of sound is known and the time taken for to and fro journey, the distance can be estimated.

Question 18.
If the frequency of a tuning fork is 256 Hz and speed of sound in air is 320 ms^-1. Find how far does the sound travel when the fork executes 64 vibrations.
Answer:
We know that
v = f λ f= 256 Hz v = 320 ms^-1
∴ λ = \(\frac{\mathrm{v}}{\mathrm{f}} \) = \(\frac{320}{256} m \)
Also, the distance covered in n vibrations = n λ
∴ Distance covered in 64 vibrations = \(\frac{64 \times 320}{256} \) = 80 m

1st PUC Physics Waves four/five marks Questions and Answers

Question 1.
Distinguish between longitudinal and transverse waves.
Answer:
1. The vibration of particles of the medium is along the direction of wave propagation in longitudinal waves, whereas in transverse waves, the vibration of particles of the medium is perpendicular to the direction of wave propagation.

2. The wave propagates by forming alternate compressions and rarefactions in longitudinal waves, whereas in transverse waves, the wave propagates by forming alternative crests and troughs.

3. Longitudinal waves travel through solids, gases, and liquids, whereas Transverse waves travel through solids and on 1 liquid surfaces.

4. Longitudinal waves cannot be polarised, whereas Transverse waves can be polarised.

5. Distance between two successive compressions or rarefactions is equal to the wavelength of the wave in longitudinal waves, whereas, in transverse waves, the distance between two successive crests or troughs is equal to the wavelength of the wave.

Question 2.
Define the following terms.

1. Wave amplitude
2. Wave period
3. Wave frequency
4. Wavelength
5. Wave velocity

Answer:
1. Wave amplitude (a):
The maximum displacement of a particle of the medium from its mean position is called 1 wave amplitude.

2. Wave period (T):
The time taken by a particle of the medium to complete one vibration or Wave period is the time during which one complete wave is set up in a medium.

3. Wave frequency (f):
The number of vibrations completed by a particle of the medium in one second is called wave frequency or Wave frequency is the number of waves set up in the medium in one second.

4. Wavelength (l):
The distance traveled by the wave in a time equal to its period is called wavelength.

5. Wave velocity (v):
It is the distance traveled by a wave in one second.

Question 3.
What are the characteristics of progressive wave?
Answer:

1. A progressive wave is formed due to continuous vibration of the particles of the medium.
2. The wave travels with a certain velocity.
3. There is a flow of energy in the direction of the wave.
4. No particles in the medium are at rest.
5. The amplitude of all the particles is the same.
6. Phase changes continuously from par¬ticle to particle.

Question 4.
State Newton’s formula for the velocity of sound in a gas. What is the Laplace’s correction? Explain.
Answer:
According to Newton, velocity of sound in any medium is given by v = \(\sqrt{\frac{E}{\rho}}\) where E is the modulus of elasticity and p is the density of the medium.
For gases E = B, bulk modulus
∴ v = \(\sqrt{\frac{\mathrm{B}}{\rho}}\) …………(1)
When sound waves travel through a gas alternate compressions and rarefactions are produced. At the compression region pressure increases and volume decreases and at the rarefaction region pressure decreases and volume increases. Newton assumed that these changes take place under isothermal conditions i.e., at a constant temperature.
Under isothermal condition, B = P, pressure of the gas.
∴ In (1) v= \(\sqrt{\frac{P}{\rho}}\) ………..(2)
This is Newton’s formula for velocity of sound in a gas.
For air at NTP, P = 101.3 kPa and
ρ = 1.293 kgm^-3
Substituting the values of P and ρ in . equation (1) we get v = 280m/s. This is much lower than the experimental value of 332 m/s. Thus Newton’s formula is discarded.

Laplace’s correction:
According to Laplace, in a compressed region temperature increases and in a rarefied region it decreases and these changes take place rapidly. Since air is an insulator, there is no conduction of heat. Thus changes are not isothermal but adiabatic.
Under adiabatic condition, B = γ P, where γ is the ratio of specific heats of the gas.
Substituting in equation (1), v = \(\sqrt{\frac{\gamma P}{\rho}}\)
The above equation is called Newton- Laplace’s equation
Substituting the values of P, ρ and γ in the above equation, gives the velocity of sound in air at NTP to be about 331 m/s. This is in close agreement with the experimental value.

Question 5.
Discuss the variation of velocity of sound with,

1. Pressure
2. Temperature
3. Humidity
4. Wind

Answer:
1. Effect of Pressure:
According to Boyle’s law, for the given mass of a gas, pressure (P) is inversely proportional to its volume (V) at constant temperature
\(P \propto \frac{1}{V}\)
or PV = constant if m is the mass and
ρ is the density of a gas then,
\(\mathrm{P}\left(\frac{\mathrm{m}}{\rho}\right)\) = constant
For given mass of gas \(\frac{P}{\rho}\) = constant.
∴ In the equation v = \(\sqrt{\frac{\gamma P}{\rho}}\)
γ and \(P / \rho\) are constants.
Thus velocity of sound is independent of pressure at constant temperature,

2. Effect of Temperature:
From Charle’s law, for the given mass of a gas, the volume V is directly proportional to its absolute temperature T at constant pressure.

At constant pressure, velocity of sound

Hence velocity of sound in a gas is directly proportional to the square root of the absolute temperature,

3. Effect of Humidity:
The presence of water vapour (humidity or moisture) in the air reduces the density of air.
∴ The density of dry air is greater than the moist air.
As the velocity of sound in a gas is inversely proportional to the square root of the density, the velocity of sound in moist air is greater than that in dry air. Thus, as the humidity increases, the velocity of sound also increases.

4. Effect of wind:
The velocity of sound is greater in the direction of the wind and lesser in the opposite direction. Let v be the velocity of sound waves and v_w the velocity of wind if the wind blows in the direction of sound waves, then the resultant velocity of sound is (v + v_w). If the wind blows against the velocity of sound waves, the resultant velocity of sound is (v- v_w).

Question 6.
Explain the theory of beats.
Answer:
Consider two sound waves of the same amplitude ‘a’ and slightly different frequencies f₁ and f₂ travelling in the same direction. The displacement of a particle in a time t due to the two waves is y₁ = a sin ω₁ t, and y₂ = a sin ω₂ t The resultant displacement of the particle due to the superposition of these two waves is,


Is the amplitude of the resultant wave. The intensity of resultant sound is maximum when A is maximum.
A is maximum when cos 2 π \(\left(\frac{f_{1}-f_{2}}{2}\right) t=\pm 1\)

The interval between successive maxima is \(\frac{1}{t_{1}-t_{2}}\),
The number of times intensity of sound becomes maximum per second is f_B = \(\frac{1}{\mathrm{T}_{\mathrm{B}}}\) = f₁ – f₂
Hence the beat frequency is the differ-ence between the frequencies of the two waves.

Question 7.
Derive a general expression for apparent frequency when the source moves towards the observer and observer moving away from the source.
Answer:

Consider a source S emitting sound of frequency f. Let v be the velocity of sound. Let the source move towards the observer with velocity vs and the observer move away from the source with velocity vQ. In one second the source travels a distance SS’ = v_s and wave travels a distance SP = v. In one second source emits f waves such that these waves will be contained in a length S’ P = v – v_s.
The apparent wavelength of these waves is

These waves approach the observer O with a relative velocity (v – v₀)
∴ Number of waves received by the observer in one second or apparent frequency is,

This is the general expression for apparent frequency.

Question 8.
What is Doppler’s effect? Give the expression for the apparent frequency of the note at different cases.
Answer:
The apparent change in the frequency of sound due to the relative motion between the source and the observes is called the Doppler effect.

Case (i) :
When the source moves towards the observer and observer moves away from the source.

Case (ii) :
When the source and ob¬serves move towards each other.

Case (iii) :
When the’source and ob-serves move away from each other

Case (iv):
Source moving away from the observer and the observer moving towards the source

Case (v):
When the source is in motion and the observer is at rest.
a. when the source moves towards the observer

b. when the source moves away from the observer

Case (vi):
when the source is at rest and the observer is in motion a. when the observer moves towards the source

when the observer moves away from the source

where f – real frequency
\(f^{\prime}\) – apparent frequency
v – velocity of sound
v_o – velocity of observer
v_s – velocity of source.

Question 9.
What are the beats? Define beat frequency. Explain how the frequency of a tuning fork is determined using beats.
Answer:
The periodic rise and fall (Waxing and waning) in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is called beats. The number of beats heard per second is called the beat frequency and is equal to the difference in the frequency is of the two sound waves. To determine the unknown frequency of a tuning fork

Step 1:
Consider a tuning fork A of known frequency f and another fork B of unknown frequency \(f^{\prime}\). When A and B are sounded together let m beats are heard/sec,
∴\(f^{\prime}\) = f±m

Step 2:
Let one of the prong of B is loaded with a bit of wax. The two forks are again sounded together and let m be the number of beats heard/sec.
If m’ > m, i.e. betas increases after adding wax, then real frequency of B is \(f^{\prime}\) = f – m
If nr’ < m, i.e. betas decreases or remains same after adding wax, then real frequency of B is \(f^{\prime}\) = f + m

Question 10.
Distinguish between stationary wave and progressive wave.
Answer:

1. A stationary wave is formed by the superposition of two equal progressive waves travelling in opposite directions whereas, a progressive wave is formed due to continuous vibration of the particles of the medium.
2. The wave does not travel in any direction in stationary waves whereas, in progressive waves, the wave travels with a certain velocity.
3. There is no flow of energy in stationary waves whereas, in progressive waves, there is a flow of energy.
4. Particles at the node are at rest in stationary waves whereas, in progressive waves, no particles in the medium are at rest.
5. In stationary waves, amplitude is different for different particles, whereas in progressive waves, amplitude of all the particles is the same.
6. In stationary waves, all particles in a loop are in the same phase and they are in opposite phase with respect to particles in adjacent loops, whereas in progressive waves, phase changes continuously from particle to particle.

Question 11.
Derive the equation for a stationary wave.
Answer:
The equation of two waves having the same amplitude, wavelength, and speed but propagating in opposite directions is

Where a is the amplitude, λ is the wave-length and v is the velocity of the wave. A stationary wave is formed due to the superposition of these two waves. The resultant displacement of a particle is given by,
y = y₁ + y₂

where A = 2a cos\(\frac{2 \pi}{\lambda} x\) represents the amplitude of the resultant wave.

Question 12.
What is a closed pipe? Show that the overtones ira closed pipe are odd harmonics of the fundamental.
Ans: A pipe which is open at are end and closed at the other end is called a closed pipe.

Consider a closed pipe of length l. Let v be the velocity of sound in air. The air column in a closed pipe vibrates in such. a way that always antinodes formed at the open end and node is formed at the closed end. Let f₁, f_2, and f₃ be the frequencies and l₁, l₂ art l₃ be the wavelengths of 1^st, 2^nd and 3^rd modes of vibration respectively.

For 1st mode (fundamental mode) l = \(\frac{\lambda_{1}}{4}\) or  λ₁=4l
but \(\mathrm{f}_{1}=\frac{\mathrm{V}}{\lambda_{1}} \text { or } \quad \mathrm{f}_{1}=\frac{\mathrm{V}}{4l}\) ……….(1)
For i2nd mode (1sl overtone)
\(l=\frac{3 \lambda_{2}}{4} \quad \text { or } \quad \lambda_{2}=\frac{4 l}{3}\)
but \(\mathrm{f}_{2}=\frac{\mathrm{V}}{\lambda_{2}} \text { or } \quad \mathrm{f}_{1}=\frac{3 \mathrm{V}}{4l}=3 \mathrm{f}_{1}\) ………(2)
from (1)
For 3rd mode (2nd overtone)
\(l=\frac{5 \lambda_{3}}{4} \quad \text { or } \quad \lambda_{3}=\frac{4 l}{5}\)
but \(\mathrm{f}_{3}=\frac{\mathrm{V}}{\lambda_{3}} \text { or } \quad \mathrm{f}_{3}=\frac{5 \mathrm{V}}{4l}=5 \mathrm{f}_{3}\) …………(3)
from (1)
From (1), (2) and (3)
f₁: f₂: f₃ = 1:2:3
Therefore in the case of an closed pipe, the frequencies of overtones are odd harmonics of the fundamental.

Question 13.
What is an open pipe Show that overtones In an open pipe are harmonics of the fundamental? OR Discuss the modes of vibration of air in an open pipe. OR Show that all harmonics are present in an open pipe.
Answer:
A pipe which is open at both ends is called open pipe.

Consider an open pipe of length l. Let v be the velocity of sound in air. The air column in an open pipe vibrates in such a way that always antinodes are formed at the open ends. Let f₁ f₂ and f₃ be the frequencies and λ₁, λ₂ and λ₃ be the wavelengths of the 1^st, 2^nd and 3^rd modes of vibration respectively. For first mode (fundamental mode).

From (1), (2) and (3)
f₁: f₂: f₃ = 1:2:3
Therefore in the case of an open pipe, the frequencies of overtones are simple harmonics of the fundamental.

Question 14.
Derive an expression for fundamental frequency In the case of stretched string.
Answer:
In the fundamental mode of vibration of the string, there will be an antinode in between the two nodes a the fixed points.

If l is the length of the string then

Velocity of the-wave along the string is

where T is the tension and m is the mass per unit length (linear density) of the string. Fundamental frequency of vibration of the string is,

Question 15.
Given below are some examples of wave motion. State In each case If the motion is transverse, longitudinal or a combination of both.

1. Motion of a kink In a long coil spring produced by displacing one end of the spring sideways.
2. Wave produced in a cylinder containing water by moving its piston back and forth.
3. Wave produced by a motorboat sailing in water.
4. Ultrasonic waves In air produced by a vibrating crystal.

Answer:

1. Longitudinal wave
2. Transverse wave
3. Combination of both
4. Longitudinal wave

Question 16.
What do you mean by wave motion? Discuss Its four important characteristics.
Answer:
Wave motion is a motion where the energy is transferred without shifting the material particles.
Four characteristics:

1. It is a Simple Harmonic Motion.
2. Energy is transported without material shift.
3. The velocity of waves depends on the medium (only for longitudinal waves).
4. The particles oscillate in SHM.

Question 17.
A simple harmonic wave Is ex-pressed by the equation

y and x are In centimetre and t in seconds. Calculate the following:
i) amplitude
ii) frequency
iii) wavelength
iv) wave velocity
v) phase difference between two particles separated by 17.0 cm
Answer:


= \(\frac{2 \pi}{5}\) rad.

1st PUC Physics Waves numerical problems Questions and Answers

Question 1.
A transverse wave is represented by y = 5 sin (50 πt – πx). Find the wavelength and velocity of the wave.
Solution:
Given equation is,
y = 5 sin (50 πt – πx)
Comparing with the standard transverse wave equation,

∴ Velocity of the wave, u = 50 m/s

Question 2.
A wave travelling at a speed of 200 m/s has a frequency of 1000 Hz. Its amplitude is 2 units. Write down the wave equation.
Solution:
Standard wave equation is y = a sin 2πt \(\left[\frac{t}{T}-\frac{x}{\lambda}\right]\)
Given a = 2 units.
Frequency = 1/T = 1000 Hz
Velocity, u= 200 m/s

Therefore the wave equation is

Question 3.
The distance between two particles on a string is 10 cm. Find the phase difference between these particles if the frequency of the wave is 400 Hz and speed Is 100m/s.
Solution :
If the distance between two points is ∆x, the phase difference is given by
\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\)
Here, Wavelength \(\lambda=\frac{v}{f}=\frac{100}{400}=0.25 \mathrm{m}\)
path difference \(\Delta x=10 \mathrm{cm}=0.1 \mathrm{m}\)
∴ Phase difference \(\Delta \phi=\frac{2 \pi}{0.25} \times 0.1\)
= 0.8π radians
=144°.

Question 4.
A sinusoidal wave propagating through air has a frequency of 200Hz. If the wave speed is 300m/s, how far apart are two points in the medium with a phase difference of 45° and 60°?
Solution:
v = 300m/s, f = 200Hz, Φ₁=45°, Φ₂=60°
v = f λ
∴ wavelength \(\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{300}{200}=1.5 \mathrm{m}\)
path difference = \(\frac{\lambda}{2 \pi}\) phase difference

∴ Distance between the points

Question 5.
A wave traveling along a string is described by y(x,t) = 0.00327 Sin(72x-2.72t) in which the numerical constants are in SI units (i.e.,\ 0.00327m, 72.1 rad/m and 2.72 rad/s.) Find the amplitude, wavelength, period and speed of the wave.
Solution:
The equation for a wave travelling along the +ve x direction is
y(x, t) = a sin(kx – ωt)-(1)
The given equation is
y(x, t) = 0.00327sin(72.1 x -2.72t) – (2)
comparing equation (1) and equation (2)
1.  Amplitude a = 0.00327 m

2.  Wavelength:
We have k =72.1 rad m^-1
ω = 2.72 rad s^-1

3.  period, T = \(\frac{2 \pi}{\omega}\)

\(=\frac{2 \pi}{2.72} \quad=2.31 \mathrm{sec}\)

4.  Frequency f = \(\frac{1}{T}\)
\(\frac{1}{2.31}=0.4329 \mathrm{Hz}\)

5.  Speed of the wave: from the equation
v = fλ
= 0.4329×0.0872
= 0.0377ms^-1.

Question 6.
A sinusoidal wave propagating through air has a frequency of 150HZ. If the wave speed is 200 ms^-1 how far apart are two points in the medium which have a phase difference of 45° and 150°?
Solution:
f =150Hz, v = 200ms^-1 φ₁ =45° φ₂=150°
v= fλ
wave length \(\lambda=\frac{v}{f}=\frac{200}{150}=1.333 \mathrm{m}\)
path difference =\(\frac{\lambda}{2 \pi}\) phase difference

∴ Distance between the points is

Question 7.
A wave travelling along a string is represented by the equation,
y = 0.08 Sin (5t – 3x) Where x and y are in metre and t is in second.

1. At t = 0.1 sec, find the displacement at x = 0.2m
2. At x s 0.1m, find the displacement at t = 0.4 sec.

Answer:
1.

(∵sin (- θ) = -sin (θ) and sine function is in radian. It is converted into degree by multiplying by 180/3.14)
y =-7.98x 10^-3m.

2. t = 0.4s and x = 0.1m.
y = 0.08 sin (5t – 3x) = 0.08 sin (5 × 0.4 – 3 × 0.1)
y = 0.08 sin (2 – 0.3) = 0.08 sin (1.7)
y = 0.08 sin \(\left(\frac{1.7 \times 180}{3.14}\right)\) = 0.08 sin (97.45°)
y = 0.08 cos (7.45°) = 0.07932 m
(∵sin (90 + θ) = cos (θ)).

Question 8.
A train moving at a speed of 72kmph towards a situation sounding a whistle of frequency 600 Hz. What are the apparent frequency of the whistle as heard by a man on the platform when the train

1. approaches him?
2. recedes from him? (speed of sound In air Is 340 ms^-1).

Answer:
V_s= 72m/hr = 72x 1000m/3600s,
= 20m/s, V = 340m/s, f = 600Hz

1. Apparent frequency when train approaches the observer
\(f^{\prime}=\left(\frac{V}{V-V_{s}}\right) f=\frac{340 \times 600}{340-20}=637.5 \mathrm{Hz}\)

2. Apparent frequency when train recedes from observer
\(\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{f}=\frac{340 \times 600}{340+20}=566.7 \mathrm{Hz}\)

Question 9.
Find the ratio of velocity of sound in oxygen and velocity of sound In hydrogen at S.T.P. Given the molecular weight of Oxygen is 32 and that of Hydrogen is 2.
Solution :
Let v₀ and v_H be the velocities of sound in oxygen and hydrogen respectively
Then

Where Po and pH are the densities of oxygen and Hydrogen respectively. X is the ratio of specific heats which is same for hydrogen and oxygen. But density is directly proportional to the molecular weight.

Question 10.
At what temperature will the velocity of sound in air reduces to half of its value at 0° C?
Solution :

Question 11.
Two tuning forks X and Y sounded together produce 10 beats per second. When Y is slightly loaded with wax, the number of beats reduces to 6 per second. If the frequency of X is 480 Hz, find that of Y.
Solution :
Frequency of x = 480 Hz
No. of beats with y = 10
∴ Frequency of y = 480+10 or 480-10
i.e., 490 or 470
On loading y, number of beats = 6
Frequency of y after loading = 486 or 474
∴ The frequency of y before loading can not be 470, because if it were 470 before loading, it must be less than 470 after loading.
∴ Actual frequency of y = 490 Hz.

Question 12.
Two tuning forks A and B gives 4 beats per second. The frequency of A is 510 Hz. When B Is filed 4 beats per second are again produced. Find the frequency of B before and after filing.
Solution:
Frequency of A =510 Hz.
Beats per second = 4
Therefore the frequency of B before filing,
= 510 + 4 or 510 – 4
= 514 or 506
After filing B, 4 beats are produced again the frequency of B after filing
=510 + 4 or 510 – 4
= 514 or 506
The frequency of B before filing can not be 514 Hz because if it is 514 before filing, after filing its frequency must be more than 514.
Therefore the frequency of B before filing must be 506 Hz and after filing it is 514 Hz.

Question 13.
Two sound waves of wavelength 1.34 m and 1.36 m produces 4 beats per second. Calculate the velocity of sound in the medium.
Solution :
Let v be the velocity of sound in the given medium. Then Frequency of the first wave
\(f_{1} \quad=\frac{v}{\lambda_{1}}=\frac{v}{1.34}\)
Frequency of the second wave

Velocity of sound v = \(\frac{4 \times 1.34 \times 1.36}{0.02}\)
= 364.5 ms^-1

Question 14.
A set of 65 tuning forks are arranged In the Increasing order of frequencies such that each gives 3 beats per second with the previous one. Find the frequency of the first tuning fork If the frequency of the last tuning fork Is one Octave above the first one.
Solution :
Let N be the frequency of the first tuning fork. As each tuning fork is giving 3 beats with the preceding one, and they are arranged in the ascending order of frequencies,
Frequency of the second tuning fork = N + 3
Frequency of the third T.F. = N + 6
= N + 3 × 2
Frequency of the fourth T.F = N+9
= N + 3 × 3
Similarly Frequency of the 65th T.F. = N + (65-1 ) × 3
= N + 64 × 3 = N+192 ……………….(1)
But the frequency of the last tuning fork (65th) is one octave above that of the first one.
∴ Frequency of the 65th T.F.= 2N ……………(2)
From (1) and (2) 2N = N + 192
N = 192
∴ Frequency of the first tuning fork = 192 Hz.

Question 15.
A source of ultrasonic wave is emitting ultrasonic waves of frequency 30 kHz. It is placed in a moving car. With what velocity is the car moving If the frequency appears to be 20 kHz to a stationary observer? Velocity of sound in car 340 ms^-1.
Solution:
Here the listener is at rest and the source is moving. As the apparent frequency is lesser than the actual frequency, the source is moving away from the listener.
To find the velocity with the source is moving:
The apparent frequency is given by,
\(\mathrm{f}^{\prime}=\mathrm{f} \frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \quad \therefore \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\)
Here, \(f^{\prime}\) is the apparent frequency = 20 kHz
f is the actual frequency = 30 kHz
v is the velocity of sound = 340 ms^-1
On substituting the values, \(\frac{20}{30}=\frac{340}{340+v_{s}}\)
cross multiplying, 2(340 + v_S) = 3 × 340 2 × 340 + 2v_s = 3 × 340
2 . v_s = 3 × 340 – (2 × 340) or 2 . v_s = 340
v_s =\(\frac{340}{2}\) = 170ms^-1
∴ The car is moving with a velocity of 170ms^-1 away from the listener.

Question 16.
An engine moving with a speed of 25 ms^-1 sends out a whistle at a frequency of 280 Hz. Find the apparent frequency of the whistle for a stationary observer

1. when the engine Is approaching him and
2. when it is moving away from him. The velocity of sound is 330 ms^-1.

Solution :

1. When the source is moving towards the observer,
apparent frequency \mathrm{f}^{\prime}=\mathrm{f} \frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}
Here source velocity v_s = 25 ms^-1,
velocity of sound v = 330 ms<sup.-1
frequency f = 280 Hz.
\(\therefore \mathrm{f}^{\prime}=\left(\frac{330}{330-25}\right) \times 280=302.95 \mathrm{Hz}\)

2. When the source is moving away from the observer, apparent frequency is given by,

Question 17.
The apparent frequency of the whistle of an engine changes in the ratio 6:5 as the engine passes a stationary observer. If the velocity of sound is 330ms^-1 what is the velocity of the engine?
Solution:
Let f be frequency of the sound emitted by the engine and vs be its velocity. The apparent frequency f of the whistle as the engine is approaching the observer is, \(\mathrm{f}^{\prime}=f \frac{\mathrm{V}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\)
But v = 330 ms^-1
\(\therefore \mathrm{f}^{\prime}=\mathrm{f} \frac{330}{330-\mathrm{v}_{\mathrm{s}}}\)   ………(1)
Let f” be the apparent frequency of the sound heard when the engine is moving away from the observer. Then
But v = 330 ms^-1

Cross multiplying
6 × (330 – v_s) = 5 × (330 + v_s)
330 × 6 – 6V_s = 5 × 330 + 5 v_s
330 × 6 – 330 × 5 = 5v_s + 6v_s
330 (6 – 5) = 11v_s
330 × 1 = 11 v_s
∴ v_s \(=\frac{330}{11}\) = 30 ms^-1

Question 18.
Two cars approach each other with a common speed of 20ms1. The first car sounds a horn and a passenger in the other car estimates it to be 700 Hz. The speed of sound is 332 ms^-1.

1. Calculate the actual frequency of the horn
2. When the cars move away from each other, what Is the estimated frequency by the same passenger?

Answer:
1. To find true frequency f:

2.

Question 19.
An observer standing by the side of a highway estimates a drop of 20% in the pitch of the horn of a car as it crosses him. If the velocity of sound is 330 m/s, calculate the speed of the car.
Answer:
Given :
v = 330 m/s
Let f₁ and f₂ be the apparent frequencies heard ty the observer, before and after the source crossing respectively.


Solving this we get vs = 36.67 ms^-1

Question 20.
A person standing in front of a mountain at a certain distance beats a drum at regular intervals. The drum¬ming rate is gradually increased, and he finds that the echo is not heard distinctly when the rate becomes 50 per minute. He then moves nearer to the mountain by 100 meters, and finds that the echo is again not heard when the drumming rates become 60 per minute. Calculate:

1. The distance between the mountain and the initial position of the man,
2. Velocity of sound.

Answer:
Let ‘s’ be the distance between the man and the mountain and let ‘v’ be the velocity of sound.
Given :
Drumming rate = 50 per minute.
∴ Interval between successive beats
\(=\frac{60}{50}=1.2 / \mathrm{sec}\)
Time taken by the echo \(=\frac{2 \mathrm{s}}{\mathrm{v}}\)
Given that when the drumming rate is 50 per minute, echo is not heard by the man ⇒ beats overlap in time frame

Similarly when the person moves 100 m
towards the mountain, with drumming rate = 60 per minute

Substituting this in (1) we get.
v = 1000 m/s

Question 21.
An open pipe Is 30 cm long. Find the fundamental frequency of vibration. Which harmonic is excited by a tuning fork of frequency 22 kHz? velocity of sound is 340 ms^-1.
Solution:
The fundamental frequency of an open pipe is

The frequency of the n^th mode of vibration is given by

Thus the 2.2 kHz source will produce 4^th harmonic.

Question 22.
Two open pipes when sounded together produce 10 beats. If the lengths of the pipes are In the ratio of 4:5, calculate their frequencies.
Solution :
Let f₁ and f₂ be the fundamental frequencies of the two pipes and L₁ be their lengths.
Then,

Then,

Question 23.
Two tuning forks A and B gives 6 beats per second. A. is in resonance with a closed pipe of length 15 cm. and B Is In resonance with an open pipe of length 30.5 cm. Calculate the frequencies of A and B.
Solution:
Let f₁ and f2 be the frequencies of the tuning forks A and B respectively.
Then, f₁ – f₂ = 6 ………….(1)
But tuning fork A is in resonance with a closed pipe of length 15 cm
\(\therefore f_{1}=\frac{V}{4 L_{1}}=\frac{v}{4 \times 0.15}=\frac{v}{0.6}\)
Similarly tuning fork B is in resonance with an open pipe of length 30.5 cm

Frequency of the tuning fork A is
\(\mathrm{f}_{1}=\frac{\mathrm{v}}{0.6}=\frac{219.6}{0.6}=336 \mathrm{Hz}\)
Frequency of the tuning fork B is
\(\mathrm{f}_{2}=\frac{v}{0.61}=\frac{219.6}{0.61}=360 \mathrm{Hz}\)

Question 24.
A closed pipe resonates In its fundamental mode of frequency 500 Hz In air. What will be the fundamental frequency if air is replaced by hydrogen at the same temperature? Density of air = 1.20 kg/m³ and density of hydrogen = 0.089 kg/ m³.
Solution:
Let v_a be the velocity of sound in air and v_H be the velocity of sound in hydrogen. Then fundamental frequency of the pipe filled with air is,

If f is the fundamental frequency of the closed pipe, filled with hydrogen then,

Where γ is the ratio of specific heats, P is the pressure and Pa is the density of air.
Similarly, \(v_{H}=\sqrt{\frac{y P}{\rho_{H}}}\)
\(\therefore \quad \frac{v_{\mathrm{a}}}{v_{\mathrm{H}}}=\sqrt{\frac{\rho_{\mathrm{H}}}{\rho_{\mathrm{a}}}}=\sqrt{\frac{0.089}{1.20}}\) =0.2734
On substituting in equation (3),
\(\frac{500}{f}=0.2734\)
\(f=\frac{500}{0.2734}\)
=1836 Hz.

Question 25.
A string vibrates with a frequency of 200 Hz. When its length Is doubled and Its tension is altered it begins to vibrate with a frequency of 300 Hz. What Is the ratio of new tension to the original tension?
Solution:
\(f=\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
Let L be the length of the string and m be the mass per unit length and T₁ be the original tension. Then,
\(200=\frac{1}{2 L} \sqrt{\frac{T_{1}}{m}}\) …………..(1)
In the Second case length = 2L, let T₂ be the tension

Question 26.
Two tuning forks A and B when vibrated together gives 6 beats per second. The tuning fork A is in unison with an air column in a closed pipe 0.15m long vibrating In fundamental mode and tuning fork B is in unison with an air column In an open pipe 0.61 m long vibrating In first overtone. Calculate the frequencies of the tuning forks.
Solution:
Let f and f be the frequencies of tuning forks A and B respectively.

\(\frac{\omega 1}{m} f_{m}-f_{c}=6 \quad \Rightarrow \frac{f_{B}}{f(1)}=6\)
f_B = 360 Hz
∴ f_A = 6+f_B = 366 Hz.

Question 27.
Two tuning forks, when sounded together, produce 5 beats per second. A sonometer wire of length 0.24 m is in unison with one of them. If the length of the wire is increased by 0.01 m, ft is in unison with the other fork. Find the frequencies of the forks.
Solution:
Let f₁ and f₂ be the frequency of the two turning forks.
given f₁ ~ f₂ =5 ……….(1)
l₁ = 0.24m, l₂ = 0.25m.
Frequency of vibration of the wire is given by

Question 28.
A closed organ pipe of length 0.42 m and an open organ pipe, both contain air at 35° C. The frequency of the first overtone of the closed pipe is equal to the fundamental frequency of open pipe. Calculate the length of open pipe and the velocity of sound in air at 0° C. Given that closed pipe is in unison in its fundamental mode with a tuning fork of frequency 210 Hz.
Solution:
Let l₁ be the length of the closed pipe and l₂ be the length of the open pipe. l₁ = 0.42m
Let v_t be the velocity of sound in air at t° c, t = 35° c
Let f be the fundamental frequency of the closed pipe
\(t=v_{1} / 4 / 1\)
Let f be the fundamental frequency of the open pipe
\(f^{\prime}=v_{1} / 2 l_{2}\)
Given that frequency of first overtone of the closed pipe is equal to the fundamental frequency of the open pipe

Also given that closed pipe is in unison with a tuning fork of frequency 210Hz i.e.,
f = 210 Hz
∴ In (1) 210 \(=\frac{v_{t}}{4 \times 0.42}\)
v_t=840×0.42 =352 .8m /s
Velocity of sound in air at 0°C is
\(v_{0}=v_{t} \sqrt{\frac{273}{273+t}}\)
\(=3528 \sqrt{\frac{273}{273+35}}=3322 \mathrm{m} / \mathrm{s}\)

Question 29.
A stretched wire emits a note of fundamental frequency 35 Hz. When the tension Is increased by 0.5 kg. wt., the frequency of the fundamental rises to 40 Hz. Find the initial tension and the length of the wire. (Mass per unit length of the wire = 1.33 × 10^-3 kg/m).
Solution:
Fundamental frequency of vibration in a stretched string is

Question 30.
One end of a horizontal wire is fixed and the other passes over a smooth frictionless pulley and has a heavy body attached to it. The fundamental frequency is 400 Hz. When the body is totally immersed in water, the frequency drops to 350 Hz. Find the density of the body. \(\left[\rho_{\omega}=19 / \mathrm{cm}^{3}\right]\).
Answer:
Given :
f₁= 400 Hz. f₂ = 350 Hz
\(\rho_{\omega}=^{1} 9 / \mathrm{cm}^{3}\)
We know that, for a string under tension T, the frequency of oscillation
\(\mathrm{f}=\frac{\mathrm{n}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
l: length of wire, \(\mu\) : mass per unit length
∴ T₁ = mg   T₂ = mg – B
Now m: \(\mathrm{v} \rho\)   v: volume      ρ: mass density
∴ mg = \(\mathrm{v} \rho\)g     B = \(\mathrm{v} \rho\)_ωg
g = 9.8 m s^-2
Given:

Question 31.
A whistle, emitting a sound of frequency 300 Hz is tied to a string of 2 m length and is rotated with an angular velocity of 15 rad / second in the horizontal phase. Find the range of frequencies heard by the observer stationed at a large distance from the whistle.
Answer:
Given:
radius r = 2 m ;
w = 15 rad s^-1;  v = 330 ms^-1
We know that v_s = r w = 30 m s^-1

The observer will receive maximum frequency when the source is approaching (A) and minimum when its receding (B).

∴ Frequency range = 275 Hz – 330 Hz.

Question 32.
The wavelengths of 2 notes are 7/165^m and 7/167^m. Each note produces 5 beats per second with a 3^rd note of a fixed frequency. Calculate the velocity of sound in air.
Answer:
Given

Question 33.
An open pipe is suddenly closed at one end with the result that the frequency of 3^rd harmonic of closed pipe is found to be higher by 50 Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of open pipe?
Answer:
Let f0 be the fundamental frequency of the pipe of length l. Then,
\(\mathrm{f}_{0}=\frac{\mathrm{v}}{2 l}\) ………..(1)
Let f_e be the 3rd harmonic of closed pipe then,

From (3) and (4) we get f₀ = 100 Hz

Question 34.
Calculate the number of beats heard per second if there are 3 sources of frequencies (n -1), n and (n +1) Hz of equal intensity sounded together.
Answer:
Let us assume each disturbance has an amplitude ‘A’ then the resultant displacement is given by,
y = A sin 2π(n – 1)t + A sin2πnt+ A sin2π(n + 1)t
i.e. y= 2A sin 2πnt cos2πt + A sin2πnt
y= A(1 + 2cos2πt) sin2πnt
∴ Resultant amplitude: A(1 + 2 cos2πt)
Amplitude is maximum when cos2πt = 1
i.e., when 2πt = 2πk k = 0,1,2……….
i.e., when t = 0, 1, 2, 3……….
∴ Time difference between successive maxima = 1 s
Similarly, amplitude is ‘0’ when cos2πt = -1/2
i.e., when cos2πt = 2πk + 2π/3
k = 0, 1, 2………….
i.e. when t = k + 1/3
i.e., when t = 1/3, 4/3, 7/3 …….
Again the time difference between successive minima = 1 s
∴ The frequency of beats is also 1 Hz.
Thus, one beat is heard per second.

Question 35.
Two sound waves originating from the same source, travel along different paths in air and then meet at a point. If the source vibrates at a frequency of 2 kHz and one path is 41.5 cm longer than the other, what will be the nature of interference ? The speed of sound in air is 332 m/s
Answer:
We know that v = fλ
\(\therefore \lambda=\frac{v}{f}\)
Given v = 332 ms^-1
f = 2 k Hz
⇒ \(\lambda=\frac{332}{2 \times 10^{3}}=0.166 \mathrm{m}\)
We know that phase difference (∆Φ)) and path difference (∆x) are related by
\(\Delta \phi=\frac{2 \pi}{\lambda} \quad \Delta x \Rightarrow \Delta \phi=\frac{2 \pi}{0.166} \times 0.415\)
∴ \(\Delta \phi=5 \pi\)
Since phase difference is an odd multiple ot π, the interference is destructive.

Question 36.
A metallic rod of length 2 m is rigidly clamped at its midpoint. Longitudinal stationary waves are set up in the rod in such a way that there are 2 nodes on either side of the midpoint. The amplitude of an antinode is 4 × 10^-6 m. Write an equation of motion of the constituent waves in the rod. (Young’s modulus = 2×10¹¹Nm^-2 and density = 8000 kg m^-3)
Answer:
General equation of a standing wave is y = 2A sink x cos ωt where
\(\mathrm{k}=\frac{2 \pi}{\pi}\) and \(\omega=2 \pi \mathrm{f}\)
Which is obtained by adding 2 identical progressive waves travelling in opposite directions
i.e., y = y₁ + y₂ where
y₁ = A sin (kx – ωt)
y₂ = A sin (kx + ωt)
Let l be the length of the rod: l = 2 m
From the figure below, we see that
\(l=\frac{5 \lambda}{2}\)

We know that velocity of longitudinal wave is given by

∴ Equation of standing wave :
y = 2(4 × 10^-6) sin (2.5 πx) cos (12.5 × 10³π) t
Equation of constituent waves
y₁ = (4 ×10^-6) sin (2.5πx – 12.5 × 10³ πt)
y₂= (4 × 10^-6) sin (2.57πx + 12.5 × 10³ πt)

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Karnataka 1st PUC Hindi Textbook Answers Sahitya Vaibhav Chapter 22 मत घबराना

मत घबराना Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए:

प्रश्न 1.
कवि किस प्रकार आगे बढ़ने के लिए कहते हैं?
उत्तर:
कवि कदम साधकर आगे बढ़ने के लिए कहते हैं।

प्रश्न 2.
कवि किसके साथ होने की बात कहते हैं?
उत्तर:
कवि कहते हैं कि कोई साथ रहे या न रहे, परन्तु चन्दा और तारे साथ रहेंगे।

प्रश्न 3.
कवि किसे अपनी व्यथा सुनाने के लिए कहते हैं?
उत्तर:
कवि अपनी सारी व्यथा चन्दा-तारे को सच्चा साथी जानकर सुनाने के लिए कहते हैं।

प्रश्न 4.
पथ पर बार-बार क्या टकराती है?
उत्तर:
पथ पर बार-बार बाधाएँ टकराती हैं।

प्रश्न 5.
कवि बीच राह में कैसे न रुकने को कहते हैं?
उत्तर:
कवि डरकर या ललचाकर बीच राह में न रुकने को कहते हैं।

प्रश्न 6.
वीर काँटों को क्या समझता है?
उत्तर:
वीर काँटों को भी फूल समझता है।

प्रश्न 7.
वीर किससे हाथ मिलाता है?
उत्तर:
वीर विपदाओं से हाथ मिलाता है।

अतिरिक्त प्रश्नः

प्रश्न 8.
वीराने में क्या बहती रहती हैं?
उत्तर:
वीराने में नदियाँ बहती रहती हैं।

प्रश्न 9.
कवि जीवन-पथ पर क्या बढ़ाने के लिए कहते हैं?
उत्तर:
कवि जीवन-पथ पर कदम बढ़ाने के लिए कहते हैं।

प्रश्न 10.
रास्ता कौन रोक लेती हैं?
उत्तर:
रास्ता तितलियाँ रोक लेती हैं।

प्रश्न 11.
धीर-वीरता से जो बढ़ता है उसे क्या मिलता है?
उत्तर:
धीर-वीरता से जो बढ़ता है उसे मंजिल मिलती है।

प्रश्न 12.
कवि ने किसे कायर कहा है?
उत्तर:
कवि ने घबराने वालों को कायर कहा है।

प्रश्न 13.
वीर कभी क्या नहीं करते?
उत्तर:
वीर कभी बहाना नहीं करते।

II. निम्नलिखित प्रश्नों के उत्तर लिखिएः

प्रश्न 1.
‘मत घबराना’ कविता में प्रकृति को प्रेरणास्त्रोत क्यों कहा गया है?
उत्तर:
हमारे साथ कोई रहे या न रहे, परन्तु ये चन्दा और तारे हमेशा साथ रहते हैं। ये हमारा दर्द बाँट लेंगे, अपनी बात कहेंगे। ये नदियाँ और झरने हरदम हँसते-गाते आगे बढ़ने को कहेंगे। इस प्रकार कवि प्रकृति को प्रेरणा का स्रोत मानता है। प्रकृति हमारा साथ देती है और आगे बढ़ने की प्रेरणा भी देती है।

प्रश्न 2.
कवि ने जीवन की किन विशेषताओं का उल्लेख किया है?
उत्तर:
कवि कहता हैं कि जीवन में कई बाधाएँ आती हैं और वे हमें रोकती हैं। तितलियाँ भी अपने सुन्दर रंग-बिरंगे पंखों से हमें अपनी ओर आकर्षित करती हैं, रोकने की कोशिश करती है। ऐसी स्थिति में बाधाओं से न डरकर और तितलियों से न ललचाकर हमें बीच राह में नहीं रुकना चाहिए। हमें नदियों की तरह अकेले ही निडर होकर बहते रहना चाहिए। यही जीवन की विशेषताएँ हैं।

प्रश्न 3.
मंजिल किन्हें मिलती है? अपने शब्दों में लिखिए।
उत्तर:
‘मत घबराना’ कविता में कवि नवयुवकों को जीवनपथ पर हमेशा आगे बड़ने का संदेश देते हैं। प्रकृति को प्रेरणा के रूप में चित्रित करते हुए कवि ने कहा है ‘मंजिल’ उन्हीं को मिलती है जो बाधाओं को चुनौती देकर धीर वीर के समान आगे बड़ते हैं। जो काँटों को फूल समझकर विपदाओं को स्वीकार करते हैं, जो कायर नहीं है, जो बहाने बनाकर निष्क्रीय नहीं होते वही ‘मंजिल’ को हासिल कर लेते हैं।

प्रश्न 4.
‘मत घबराना’ कविता का संदेश अपने शब्दों में लिखिए।
उत्तर:
‘मत घबराना’ कविता में कवि युवकों को सन्देश देते हैं कि जीवन के पथ पर सदा आगे बढ़ते रहना चाहिए। साथ कोई हो या न हो प्रकृति माता सदा हमारे साथ रहेगी। प्रकृति से हमें प्रेरणा लेनी चाहिए। यही इस कविता का सन्देश है।

अतिरिक्त प्रश्नः

प्रश्न 5.
कवि ‘मानव’ ने नवयुवकों को किन बातों से न घबराने की सलाह दी है?
उत्तर:
कवि ‘मानव’ युवाओं को बाधाओं से नहीं घबराने की सलाह दे रहे हैं। वे कहते हैं- रास्तें में बाधाएँ तुमसे बार-बार टकराएगी लेकिन तुम डरकर बीच रास्ते ही मत रुक जाना। तुम्हें तो बस जीवन पथ पर अपने कदम बढ़ाते जाना है।

III. ससंदर्भ भाव स्पष्ट कीजिए:

प्रश्न 1.
कोई साथ न रहने पर भी
चन्दा-तारे साथ रहेंगे।
दर्द तुम्हारा बाँटेंगे वे,
तुमसे अपनी बात कहेंगे।
उत्तर:
प्रसंग : प्रस्तुत पंक्तियाँ हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘मत घबराना’ नामक कविता से ली गई हैं, जिसके रचयिता डॉ. रामनिवास ‘मानव’ हैं।
संदर्भ : कवि नवयुवकों को संदेश देते हैं कि जीवन के पथ पर तुम सदा आगे बढ़ों। प्रकृति हमें प्रेरणा देती है।
स्पष्टीकरण : कवि कहते हैं कि बाधाओं से, प्रतिकूल परिस्थितियों से घबराना नहीं चाहिए। इनसे बिना डरे निरन्तर आगे बढ़ते रहना चाहिए, कोई साथ दे या न दे। आप मंजिल तय करते समय अकेले पड़ जाएँ तब घबराना नहीं। यह चाँद, यह तारे तुम्हारे साथ हर पल रहेंगे। यह कभी भी साथ नहीं छोड़ते। प्रकृति मनुष्य का कभी भी साथ नहीं छोड़ती। मनुष्य का प्रकृति के साथ रिश्ता अटूट है। तुम उनसे बात करना। अपना कष्ट सुनाना। वे बिना परेशान हुए तुम्हारी बात को गंभीरता से सुनेंगे। वे अपनी बात तुमसे कहेंगे। उन्हें अपना सच्चा मित्र समझकर आगे बढ़ते जाओ।
विशेष : खड़ी बोली हिन्दी का प्रयोग। प्रेरणास्पद कविता।

प्रश्न 2.
मंजिल सदा उसी को मिलती
धीर-वीर जो बढ़ता जाता।
काँटों को भी फूल समझता,
विपदाओं से हाथ मिलाता।
कायर तो घबराते वे ही,
वीर न करते कभी बहाना।
उत्तर:
प्रसंग : प्रस्तुत पंक्तियाँ हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘मत घबराना’ नामक कविता से ली गई हैं जिसके रचयिता डॉ. रामनिवास ‘मानव’ हैं।
संदर्भ : कवि यहाँ नव युवकों को संदेश देते हुए कह रहे हैं कि तुम वीर हो, विपदाओं का सामना करते हुए मंजिल के पथ पर तुम आगे बढ़ते रहो। कवि ने इस कविता में प्रकृति को प्रेरणा के रूप में चित्रित किया है।
स्पष्टीकरण : कवि कहता है कि मंजिल सदा उसी को मिलती है, जो धीर-वीर बनकर आगे बढ़ता है। जो काँटों को फूल समझकर, विपत्तियों से हाथ मिलाकर आगे बढ़ता है और डरने का बहाना न बनाकर जो आगे बढ़ता है, वही जीवन में सफल होता है।

मत घबराना कवि परिचयः

बहुमुखी प्रतिभा के धनी रामनिवास ‘मानव’ जी का जन्म 2 जुलाई 1954 ई. को तिगरा जिला महेन्द्रगढ़ (हरियाणा) में हुआ। आपकी शिक्षा एम.ए. (हिन्दी), पीएच.डी. एवं डी.लिट्. तक हुई है। आपको स्नातक तथा स्नातकोत्तर कक्षाओं को पढ़ाने का तीन दशक से अधिक अनुभव है। आपको, अनेक प्रतिष्ठित सम्मान, पुरस्कार तथा मानद उपाधियों से सम्मानित किया गया है। आप संप्रति कुरुक्षेत्र विश्वविद्यालय, कुरुक्षेत्र में कार्यक्रम क्रियान्वयन केन्द्र में शोध निदेशक पद पर कार्य कर रहे हैं।
प्रमुख कृतियाँ : ‘धारा-पथ’, ‘रश्मी-रथ’, ‘साँझी है रोशनी’, ‘बोलो मेरे राम’, ‘सहमी सहमी आग, ‘हम सब हिन्दुस्तानी’ आदि|

कविका का आशय :

कवि नवयुवकों को संदेश देते हैं कि जीवन-पथ पर सदा आगे बढ़ो। प्रकृति सदा हमारे साथ है| प्रकृति हमें प्रेरणा देती है|

मत घबराना Summary in Kannada

मत घबराना Summary in English

The poet, Dr Ramniwas ‘Manav’, through this poem gives the message to youngsters, that one must always move forward in life. Nature will always support us. Nature also gives us inspiration.

In the first stanza, the poet says that one must move forward in life without fearing the difficulties or obstructions. Whether there is anyone with us or not, we must always move forward. The moon and the stars will share our difficulties and our pain. Considering them as our true companions, we must move forward.

In the second stanza, the poet says that rivers and streams laugh and sing as they move along continuously. They have no companions. Similarly, we must also go ahead alone. You are your own companion. Considering the rivers and streams as our true companions, we must move forward.

In the third stanza, the poet says that our path is filled with obstacles and difficulties. Butterflies will stop and distract us, however, one must not be afraid and one must not get distracted and must continue to walk down the middle of the path without stopping.

In the fourth stanza, the poet says that the destination is reached by only those who continue to move forward bravely and with determination, and by those who treat thorns as roses and cross over them, and by those who do not shake hands with adversity or misfortune. Cowards always feel scared and they make excuses saying that they are scared. The brave never make excuses. They only keep moving forward.

मत घबराना Summary in Hindi

1) बाधाओं से मत घबराना।
कदम साधकर बढ़ते जाना।
कोई साथ न रहने पर भी
चन्दा-तारे साथ रहेंगे।
दर्द तुम्हारा बाँटेंगे वे,
तुमसे अपनी बात कहेंगे।
सच्चा साथी जान उन्हें तुम
अपनी सारी व्यथा सुनाना।

कवि कहता है कि बाधाओं से बिना डरे आगे बढ़ते जाना चाहिए। साथ कोई हो न हो, आगे बढ़ते जाना चाहिए। चाँद-सितारे तुम्हारा दर्द बाँट लेंगे। उन्हें सच्चा साथी मानकर अपनी व्यथा सुनाएँ।

2) वीराने में नदियाँ-निर्झर,
जैसे हरदम हँसते-गाते।
आप अकेले अपने साथी,
सदा अकेले बढ़ते जाते।
सच्चा साथी जान उन्हें तुम
जीवन-पथ पर कदम बढ़ाना।

कवि कहते हैं- नदियाँ व झरने हँसते-गाते हुए बढ़ते है। उनका कोई साथी नहीं रहता। वैसे ही तुम भी अकेले जाना। तुम अपने साथी स्वयं ही हो। उनको अपना सच्चा साथी समझकर वैसे तुम भी आगे बढ़ो।

3) पथ क्या वह बाधाएँ जिसमें
बार-बार टकराती ना हों।
और तितलियाँ रोक रास्ता
राही को ललचाती ना हों।
पर तुम डरकर या ललचाकर
बीच राह में मत रुक जाना।

कवि कहते हैं- पथ में कई बाधाएँ आती हैं। तितलियाँ रोककर तुम्हें ललचाएगी, परन्तु तुम न डरते हुए और न ललचाते हुए बीच राह में न रूकते हुए आगे बढ़ना चाहिए।

4) मंजिल सदा उसी को मिलती
धीर-वीर जो बढ़ता जाता।
काँटों को भी फूल समझता,
विपदाओं से हाथ मिलाता।
कायर तो घबराते वे ही,
वीर न करते कभी बहाना।

कवि अन्त में कहते हैं कि मंजिल उन्हीं को मिलती है, जो धैर्य के साथ आगे बढ़ते हैं, जो काँटों को भी फूल समझकर पार करते हैं, जो विपत्तियों से हाथ मिलाते हैं। कायर तो घबराते हैं और वे डरने का बहाना बनाते हैं। वीर कभी बहाना नहीं बनाते। आगे बढ़ते रहते हैं।

कठिन शब्दार्थः

• बाधाएँ – रुकावट, विघ्न;
• कदम साधकर – कदम से कदम मिलाकर;
• व्यथा – दुःख, दर्द;
• ललचाना – लोभग्रस्त होना;
• विपदा – विपत्ति;
• कायर – डरपोक, बुजदिल;
• चन्दा – चाँद;
• वीराना – शून्य प्रदेश;
• हरदम – हमेशा;
• मंजिल – पड़ाव

Karnataka 1st PUC Geography Question Bank Chapter 11 Natural hazards and disasters

You can Download Chapter 11 Natural hazards and disasters Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Natural hazards and disasters One Mark Questions and Answers

Question 1.
What is Natural Hazard? (T.B Qn)
Answer:
It is a threat of naturally occurring event that will have a negative effect on people or the environment.
Ex: Earthquake, Landslide, Volcanic eruption.

Question 2.
What do you mean by Natural Disaster? (T.B Qn)
Answer:
A natural disaster is a major adverse event resulting from natural processes of the Earth e.g. Earthquakes, floods, drought and famine, cyclones, landslides, coastal erosion.

Question 3.
Mention any two types of disasters. (T.B Qn)
Answer:
Major types of Disaster are Tectonic, Meteorological, and Topographical disasters.

Question 4.
What are floods? (T.B Qn)
Answer:
Floods are temporary inundation of large regions as a result of heavy rainfall, prolonged rain cyclones, storm surge along coast.

Question 5.
Name the most important flood prone area of India. (T.B Qn)
Answer:
The Ganga basin is the most important flood prone area of India.

Question 6.
Why are cyclones are caused in the Bay of Bengal? (T.B Qn)
Answer:
The Bay of Bengal is subject to intense heating, giving rise to humid and unstable air masses that produce cyclones.

Question 7.
What is drought? (T.B Qn)
Answer:
The term drought is applied to an extended period when there is a shortage of water availability due to inadequate precipitation, excessive rate of evaporation and over utilization of water from the reservoirs other storages, including the groundwater.

Question 8.
Which region of India is in the extreme drought prone area? (T.B Qn)
Answer:
The regions are western parts of Rajasthan, Kutch regions of Gujarat and semi-arid and arid regions of Western and North western parts of India.

Question 9.
Why does landslide occur? (T.B Qn)
Answer:
Landslides are occur by severe marine erosion of sea coast, Seismic activity, Heavy rainfall, construction of roads, railway lines, canal construction, mining and quarrying, over grazing deforestation.

Question 10.
Mention the most important avalanche prone area of India. (T.B Qn)
Answer:
The most important avalanche prone area of India are mainly Jammu and Kashmir, Himachal radish, uttarkhand, Sikkim, parts of Arunchal Pradesh etc.

Question 11.
What is an Earthquake?
Answer:
An earthquake is a sudden movement, trembling of the earth’s crust.

Question 12.
Which is the highest earthquake intensity region of India?
Answer:
Himalayan region.

Question 13.
Which is the only one active volcano in India?
Answer:
Barren Volcanic Island in the Andaman Island

Question 14.
When national flood control programme was launched?
Answer:
In 1954.

Question 15.
What is coastal erosion?
Answer:
Coastal erosion means eroding down the coastline by sea waves.

Question 16.
What is mean by avalanche?
Answer:
Avalanches are a hurtling mass of snow, ice and rock debris descending a mountain side.

1st PUC Geography Natural hazards and disasters Two Marks Questions And Answers

Question 1.
Name the two most important seismic zones of India. (T.B Qn)
Answer:
Zone V: This is the most severe seismic zone and is referred as very high damage Risk zone. The areas are Northeastern states, parts of Jammu Kashmir, Uttarkhand, and Bihar and Kutch region.

Zone IV: This zone is second in severity zone. Northern regions of Jammu and Kashmir, Himachal Pradesh, parts of Bihar, UP, Gujarat, West Bengal.

Question 2.
Mention any four factors that cause floods. (T.B Qn)
Answer:
Floods are caused by both natural and man-made factors. They are:
(a) Natural factors

• Continuous rainfall for a long period
• Cyclones
• Obstruction on flow of river water.

(b) Man made factors:

• Deforestation
• Unscientific Agricultural practice
• Urbanization

Question 3.
State two important flood prone areas of the country. (T.B Qn)
Answer:
The Ganga Basin: The badly affected states of the Ganga basin area Uttar Pradesh, Bihar and West Bengal.
The Brahmaputra basin: the Brahmaputra along with its tributaries floods the areas of Assam and North West Bengal regions.

Question 4.
Name any four factors that cause drought and famine. (T.B Qn)
Answer:
The main causes for the occurrence of drought and famine are reduction in annual rainfall, long period scarcity of surface and underground water, scarcity of stored water, excess utilization of freshwater. Overgrazing, deforestation. Improper agricultural practice, mining.

Question 5.
Mention any four consequences of natural hazard and disasters. (T.B Qn)
Answer:
The most important consequences of natural disasters are los of human life and property, animal wealth, destruction of vegetation etc. Natural disasters create fear, anguish and trauma in the human beings leading to various physical, biological and psychological changes. Natural disasters affect population, its distribution and density. It affects on agriculture, cropping pattern, industries, transport and communication, public health, water supply.

Question 6.
What is an earthquake? What are the main causes of an earthquake?
Answer:
An earthquake is a sudden release of energy accumulated in rocks causing the ground to . tremble or shake.
The main causes of earthquake are natural and man – made factors.

1. Tectonic forces
2. Volcanic activity
3. Landslides and Landslips
4. Collapse of underground cave roofs

Question 7.
How can drought prevented in India?
Answer:
We can reduce the intensity and impact of drought through individual and collective actions:

• Community based rainwater harvesting structures should b constructed.
• Watershed programmes should be increased
• Through plantation programmes, forest cover should be increased.
• Encouraging farmers to grow drought-resistant crops.

Question 8.
Mention the important regions of land slides in India.
Answer:
There are three important region.

1. Himalayan zone
2. Western Ghats
3. Southern Plateau.

Question 9.
Mention the different types of drought in India.
Answer:
Drought is a weather hazard, uncertainty of monsoon rainfall, deficient rainfall. So India is more frequently affected by droughts.

India droughts are classified into four types:

1. Meteorological Drought
2. Hydrological Drought
3. Agricultural Drought
4. Ecological Drought

1st PUC Geography Natural hazards and disasters Five Marks Questions And Answers

Question 1.
Explain the major seismic zones of India. (T.B Qn)
Answer:
Zone V: This is the most severe seismic (intensity above 7 in Richter scale) seismic zone and is referred as Very High Damage risk zone. The areas are. Northeastern states, parts of Jammu Kashmir, Uttarkhand, and Bihar and Kutch region.

Zone IV: This zone is second in severity (intensity between 5 and 7 in R.S) to zone VG. This is referred to as High Damage Risk zone. Northern regions of Jammu and Kashmir, Himachal Pradesh, Parts of Bihar, UP, Gujarat, West Bengal lie in this region zone. Northern regions of Jammu and Kashmir, Himachal Pradesh, parts of Bihar, UP, Gujarat, West Bengal.

Zone III: This is termed as Moderate Damage (very strong) Risk zone (intensity between 3 and 5 in R.S). The areas are Gujarat, Madya Pradesh, Rajasthan, Chhattisgarh, Odisha, Maharashtra, Northern Karnataka, Andhra Pradesh, West coastal region etc.

Zone II: This zone is referred to as low Damage (strong) Risk Zone (intensity 2 to 3 R.S). The areas are Rajasthan, Madhya Pradesh, Parts of Karnataka, Andhra Pradesh, Odisha etc.

Zone I: This zone is termed as Very Low Damage (Slight-tremor) Risk Zone. The left out parts of India and Deccan Plateau region.

Question 2.
Briefly explain the distribution of flood prone areas of India. (T.B Qn)
Answer:
a. The Ganga basin: The badly affected states of the Ganga basin are U.P, Bihar and West Bengal. Besides the Ganga River, Sarada, Gandak and Ghagra cause flood in Eastern part of U.P. The Yamuna is famous for flooding Haryana, U.P and Delhi. Bihar experiences massive and dangerous flood every year by the Kosi. Rivers like the Mahanadi, Bhagirathi and Damodar also cause floods.

b. The Brahmaputra basin: The Brahmaputra along with its tributaries floods the areas of Assam and North West Bengal regions.

c. The Central India and Peninsular river basin: In odisha spilling over of river banks by the Mahanadi, Baitarnika and Brahmani causes havoc. Southern and central India experiences floods caused by the Narmada, Godavari, Tapti and Krishna during heavy rainfall. Cyclonic storms in the deltaic regions of the Godavari, Mahanadi and the Krishna flood the coastal regions of Andhra Pradesh.

Question 3.
Explain the major drought prone areas of India. (T.B Qn)
Answer:
On the basis of severity of droughts, India can be divided into three drought prone areas.

a. The Extreme drought prone areas: This is the most important drought prone areas of the country which has been recording continuous drought for many years. The regions are western parts of Rajasthan, Kutch regions of Gujarat and semi-arid regions of Western and North western parts of India.

b. The Severe drought prone areas: This is the second important drought prone areas of the county. The eastern parts of Rajasthan, western parts of Madhya Pradesh, Parts of Maharashtra, interior parts of Andhra Pradesh. North and northeastern parts of Karnataka and Tamil nadu.

c. The Moderate drought prone areas: This region is mainly found in regions of U.P, parts of Gujarat, Maharashtra, Jharkhand, Tamil Nadu and interior parts of Karnataka.

Karnataka 1st PUC Geography Question Bank Chapter 5 Atmosphere

You can Download Chapter 5 Atmosphere Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Atmosphere One Mark Questions and Answers

Question 1.
Define Atmosphere? (T. B. Qn )
Answer:
The thin layer of gaseous matter encircling the earth as an envelope or a blanket is known as Atmosphere

Question 2.
What is the percentage of Nitrogen in the Atmosphere? (T. B. Qn )
Answer:
78% of Nitrogen in the Atmosphere.

Question 3.
Why is Carbon dioxide important in the Atmosphere? (T. B. Qn )
Answer:
It is needed for plants. It traps heat and acts as an insulating agent, that is why the Earth is warm during night.

Question 4.
Which layer of Atmosphere is called ‘Weather manufacturer? (T. B. Qn )
Answer:
Troposphere is called ‘Weather manufacturer’.

Question 5.
In which layer do you find Aurora? (T. B. Qn )
Answer:
Thermosphere.

Question 6.
What is Stratopause? (T. B. Qn )
Answer:
Stratopause is the boundary between Stratosphere and Mesosphere.

Question 7.
Which is the coldest layer in the Atmosphere? (T. B. Qn )
Answer:
Mesosphere extends to a height of 80km from the surface of the Earth. It is the coldest layer of the atmosphere.

Question 8.
Which layer is known as Radio layer?
Answer:
Ionosphere is known as radio layer, because it reflects sound and radio waves back to the earth’s surface.

Question 9.
What is tropopause?
Answer:
The upper part of troposphere is known as “tropopause”.

Question 10.
What is stratosphere?
Answer:
The layer just above the troposphere is called stratosphere. It extends, up to a height of about 50km.

Question 11.
Which is the lowest layer of the atmosphere?
Answer:
Troposphere

Question 12.
What is stratopause?
Answer:
The upper limit of the stratosphere is known as “stratopause”

Question 13.
What is Ozonosphere?
Answer:
The lower portion of the stratosphere having a greater concentration of ozone is called “Ozonosphere”. It lies between the heights of 30 to 60km.

Question 14.
What is Mesosphere?
Answer:
A layer above the stratosphere, which extends between 50 to 80km, is known as the mesosphere.

Question 15.
What is Mesopause?
Answer:
A thin layer of air separating mesosphere from other upper layers is named as ‘Mesopause’.

Question 16.
What is ionosphere?
Answer:
The thin layer extends from the ozonosphere form an altitude of about 80 to 640 km above the earth surface.

Question 17.
Which is the isothermal zone?
Answer:
Stratosphere is also known as isothermal zone because in this zone the temperature is almost uniformly distributed.

Question 18.
Which layer is known zone of radio waves?
Answer:
Ionosphere is the zone of radio waves because it reflects sound and radio waves back to the earth’s surface.

Question 19.
Which is the upper most layer of the atmosphere?
Answer:
Exosphere.

Question 20.
In which layer Ozonosphere exists?
Answer:
Stratosphere

Question 21.
What is Exosphere?
Answer:
The outer-most layer of the earth’s atmosphere lies between 640 and 1000km.

Question 22.
Define Aerosols.
Answer:
The dust and other microscopic solid particles of the atmosphere are known as ‘Aerosols.’

Question 23.
What is condensation?
Answer:
It is a process of Conduct the heat from warmer to colder regions.

Question 24.
Which layer is known as ‘Magnetosphere.’?
Answer:
Exosphere is also known as‘Magnetosphere’.

Question 25.
Which layer absorbs ultraviolet rays of the sun?
Answer:
Ozonosphere

Question 26.
Which gas is known as green house gas?
Answer:
Carbon Dioxide

Question 26.
Which gas absorbs ultraviolet rays from the sun?
Answer:
Ozone

Question 27.
Why sky appears blue color?
Answer:
It is because of selective scattering of solar radiation by dust particles. (Atmosphere)

Question 28.
What is Radiation?
Answer:
The process of transfer of heat from one body to the other without the material medium is known as ‘Radiation’.

Question 29.
Define Insolation. (T. B. Qn )
Answer:
The small proportion of solar radiation which reaches the earth is called insolation.

Question 30.
What is convection? (T. B. Qn )
Answer:
The transfer of heat through the movement of amass of substances from one place to another is called convection.

Question 31.
What is Thermometer? Mention the types of Thermometer?
Answer:
It is an instrument used to measure the atmospheric temperature. Two types of thermometers are a. Centigrade thermometer and b. Fahrenheit thermometer.

Question 32.
What is Radiation?
Answer:
It s the process by which a body emits radiant energy. Energy received from the sun in the form of heat.

Question 33.
What is conduction?
Answer:
It is a process by which heat is transferred directly though matter fro point of high temperature. Heat passes from warmer to colder substances, as long as temperature difference exists.

Question 34.
What is the surface temperature of the sun?
Answer:
The surface temperature of the sun is 6000°C or 11,000°F.

Question 35.
What is normal Lapse rate?
Answer:
Temperature decreases with increasing height at the rate of 6.5’C per room. This rate of decrease of temperature is called normal lapse rate (NLR).

Question 36.
What is Inversion of temperature?
Answer:
The temperature increases with an increase in the height. This state of affairs is called inversion of temperature.

Question 37.
What is isotherm?
Answer:
The lines drawn on maps joining places having the same air temperature are called Isotherms.

Question 38.
Which place records highest temperature in the world?
Answer:
Libya (Aziziya) is a part of Sahara desert records highest temperature in the world.

Question 39.
What is Radiation?
Answer:
The process of transfer of heat from one body to the other without the material medium is known as ‘Radiation’.

Question 40.
Which place records lowest temperature in the world?
Answer:
Verkhoyansk in N. E. Siberia

Question 41.
What is range of temperature?
Answer:
The difference between the highest and lowest amount of temperature in the particular duration is known as ‘range of temperature.’

Question 42.
What is thermograph?
Answer:
A continuous heat-recording thermometer is called Thermograph.

Question 43.
What is Terrestrial radiation?
Answer:
The earth surface after receiving the energy from solar radiation radiates energy in the form of long waves known as terrestrial radiation.

Question 44.
What is Advection?
Answer:
Temperature is also transferred by large scale movement of air. When warm air moves to cold regions temperature of the air increases and is contrast cold air reduces heat. This phenomenon is called advection.

Question 45.
What is the average distance between sun and earth?
Answer:
150millionkm.

Question 46.
What is Photosphere?
Answer:
The outer surface of the sun is known as Photosphere

Question 47.
What are Aurora borealis?
Answer:
It is a phenomenon of coloured light’s seen in the sky in the northern hemisphers 66l/2° North to 90°North

Question 48.
Mention the average atmospheric pressure of the earth. (T. B. Qn )
Answer:
The average atmospheric pressure at the sea level is 1013.25mb.

Question 49.
How many pressure belts are there in the globe? (T. B. Qn )
Answer:
There are seven pressure belts on the globe.

Question 50.
Define Doldrums. (T. B. Qn )
Answer:
Equatorial belt of low pressure experiences calm conditions and variable winds so it is known as the belt of doldrums.

Question 51.
Where do we find ‘Horse Latitudes*? (T. B. Qn )
Answer:
Subtropical high pressure belt is known as ‘horse Latitudes’.

Question 52.
What is Atmospheric Pressure?
Answer:
The weight of air on a unit area is called air pressure or atmospheric pressure.

Question 53.
Mention the types of Atmospheric Pressure?
Answer:
Low Pressure and High Pressure.

Question 54.
What is pressure gradient?
Answer:
The rate at which the atmospheric pressure changes horizontally is called pressure gradient.

Question 55.
What is Isobar?
Answer:
Lines drawn on a map joining regions of equal air pressure are called isobars.

Question 56.
What is Barometer?
Answer:
The atmospheric pressure is measure by means of an instrument known as ‘Barometer’.

Question 57.
Mention the types of Barometer?
Answer:
Mercury Barometer and Aneroid barometer.

Question 58.
What is Barograph?
Answer:
It is self recording instrument which continuously records the atmospheric pressure.

Question 59.
What is Altimeter?
Answer:
It is an instrument measures the pressure of atmosphere and denotes corresponding height of the place from the mean sea level.

Question 60.
Which zone is known as ‘Doldrums’?
Answer:
Equatorial low pressure region is known as ‘Doldrums.

Question 61.
Which region is also known as Horse latitudes?
Answer:
Subtropical high pressure belt is known as Horse latitudes.

Question 62.
Define Milibar?
Answer:
A unit of pressure used by meteorologists when preparing their weather charts: 1,000 millibars= labor 29.53inches (750mm) as indicated on a barometer.

Question 63.
What is Aneroid Barometer?
Answer:
A device to measure air pressure that uses an aneroid, which is a sealed, flexible metal bellow with some air removed that expands and contracts with air pressure changes.

Question 64.
What is Dew?
Answer:
Water that has condensed onto objects near the ground when their temperatures have fallen below the dew point of the surface air.

Question 65.
What do you mean by Convection?
Answer:
The upward flow of air that has been heated by contact with the Earth’s surface. As it warms, it expands and rises. Cold air takes its place at the surface, and is in turn healed and caused to rise.

Question 66.
What do you mean by Torrid Zone?
Answer:
The climate zone lying between the topic of Cancer and the tropic of Capricorn, where the weather is almost always hot and the sun shines.

Question 67.
What is Charles’s Law?
Answer:
With constant pressure, the temperature of an ideal gas is inversely proportional to the density of the gas.

Question 68.
Name the instrument used to measure speed of the wind. (T. B. Qn)
Answer:
‘Anemometer’ is used to measure the speed of winds.

Question 69.
What is ITCZ? (T. B. Qn )
Answer:
This region is the converging zone of trade winds known as ‘Inter-tropical convergent zone’.

Question 70.
Where do we see ‘Roaring Forties’? (T. B. Qn )
Answer:
‘Roaring Forties’ found around 40°south latitude.

Question 71.
Why are Tropical cyclones more dangerous? (T. B. Qn )
Answer:
Tropical cyclones more dangerous because they cause heavy rainfall with high velocity winds.

Question 72.
What are winds?
Answer:
The horizontal movement of air over the earth’s surface is called wind.

Question 73.
Name any two types of winds?
Answer:
Planetary winds and Periodic winds

Question 74.
Where is Hurricanes exists?
Answer:
U.S.A

Question 75.
Where is a typhoon existing?
Answer:
China

Question 76.
Where is Will-willy blows?
Answer:
Australia

Question 77.
Where we can find Tornadoes?
Answer:
Southern and eastern USA.

Question 78.
In which climate zone, days and night are equal throughout year?
Answer:
Equatorial type of climate.

Question 79.
What is Air current?
Answer:
The vertical or near vertical moving air is called air current.

Question 80.
What are the local winds?
Answer:
The regular pattern of planetary and seasonal winds is affected with the local disturbances. Local differences in temperature and pressure leads to the development of movement of winds are called ‘Local winds’.

Question 82.
What type of winds are the monsoons?
Answer:
The monsoons winds are Periodic types of winds. They are extensive and also well developed among the seasonal winds.

Question 83.
What are the names given to the antitrade winds in the Southern Hemisphere?
Answer:
Antitrade winds in the southern hemisphere are known as Westerlies.

Question 84.
What is meant by ‘Triple point temperature’?
Answer:
The gaseous, liquid and solid states of water are at equilibrium under standard atmospheric pressure.

Question 85.
What is wind vane?
Answer:
An instrument widely used for measuring wind direction is called ‘wind vane’ or ‘weather cock’.

Question 86.
Which winds are called Tropical Easterlies.
Answer:
Trade winds.

Question 87.
What are Westerlies?
Answer:
The winds blowing from the sub-tropical high pressure belts towards the sub-polar low pressure belts are called Westerlies.

Question 88.
What are ‘Anti trade winds?
Answer:
The winds blow in the opposite direction of the trade winds and were known as “Anti trade winds”.

Question 89.
What is Latent Heat?
Answer:
Energy stored when water evaporates into vapour or ice melts into liquid. It is released as heat when water vapour condenses or water freezes.

Question 90.
What is Rainshadow region? (T. B. Qn )
Answer:
In the leeward side of the mountain, as the wind begins to descend, temperature steadily increases resulting in dry air by forming ‘Rain Shadow Region’.

Question 91.
Which instrument is used to measure the amount of Rainfall? (T. B. Qn )
Answer:
Rain gauge is used to measure the amount of Rainfall.

Question 92.
Mention any two factors-that determine the climate of a place. (T. B. Qn )
Answer:
Latitude, distance from the sea, altitude, ocean currents.

Question 93.
What is Isohygric line?
Answer:
The line joining the places of equal amount of water vapor is called ‘Isohygric’ line.

Question 94.
What is precipitation?
Answer:
Precipitation is the falling of water in liquid or solid form to the earth’s surface.

Question 95.
What are the three types of precipitation
Answer:
Rain, snow and Hail stone are the three types of precipitation.

Question 96.
What is convectional rainfall?
Answer:
The rainfall caused with the rising of air upwards due to high temperature is known as convectional rainfall.

Question 97.
Define Hailstone?
Answer:
Hailstone is solid frozen raindrops. The rain droplets when they enter into cold layers of the atmosphere at a higher level get condensed.

Question 98.
Define snow?
Ans.
When the precipitation takes place at below freezing point temperature, it is in the form of ice crystals. Humidity of the air is condensed into hexagonal ice crystal and reach earth’s surface. It is known as‘snow’.

Question 99.
What is Convectional Rainfall?
Answer:
The rainfall caused with the rising of air upwards due to high temperature is known as convectional rainfall.

Question 100.
Define Isohyets?
Answer:
These are the imaginary lines drawn to join the places with the same amount of rainfall.

Question 101.
What is cloud?
Answer:
When the condensation of water vapour takes place on dust particles on a large scale at higher altitudes, clouds are formed clouds consists of tiny droplets of water which float in the air and are carried by the winds.

Question 102.
What is rain gauge?
Answer:
The amount of rainfall is measured by means of an instrument called rain gauge.

Question 103.
What is Saturation?
Answer:
The air has maximum humidity which it can withhold at the temperature is called saturation.

Question 104.
What is Condensation?
Answer:
Condensation is a reverse process of evaporation. The transformation of gaseous form of water vapour into solid or liquid form is called condensation. This is caused by the loss of heat.

Question 105.
Describe Haze?
Answer:
Acloud of dust, smoke, salt or other particles that reduces visibility close to the Earth’s surface. A haze is said to exist when visibility is less than 1.25 miles but more than 0.6 mile (1km).

Question 106.
Mention any two factors that determine the climate of a place. (T. B. Qn )
Answer:
Latitude, distance from the sea, altitude, ocean currents.

Question 107.
What is weather?
Answer:
The atmospheric condition of a place at a given time is known as weather.

Question 108.
What is climate?
Answer:
The average weather condition of a place for a long period like 30-33years is known as climate.

Question 109.
What is Meteorology?
Answer:
The scientific study of the atmosphere and its physical processes is called ‘Meteorology’.

Question 110.
What is climatology?
Answer:
The scientific study of climatic condition is known as climatology.

Question 111.
What are the elements of weather?
Answer:
Temperature, pressure, winds humidity and precipitation sunshine are the element of weather.

Question 112.
What is Continental climate?
Answer:
The amount of rainfall also decreases from coastal region to the interior Similarly the range of temperature is maximum in the interior regions making it a continental climate.

1st PUC Geography Atmosphere Two Marks Questions and Answers

Question 1.
Mention any two components of Atmosphere. (T. B. Qn )
Answer:
Various gases, water vapour and dust particles

Question 2.
Name any four important gases present in the atmosphere. (T. B. Qn )
Answer:
Nitrogen (78%), Oxygen 21%, Argon, Carbon dioxide, Hydrogen, Neon.

Question 3.
Why troposphere is called ‘Region of Mixing’? (T. B. Qn )
Answer:
Troposphere literally means “the region of mixing”. It is derived from the Greek word ‘tropos’ means mixing All changes in weather condition take place in this layer. Temperature, pressure, winds, clouds, thunder, lightning, rainbow and precipitation are common in this layer.

Question 4.
What is the role of Ionosphere in the Atmosphere? (T. B. Qn )
Answer:
These electrically charged particles are known as “Ions” and hence this layer is known as Ionosphere. Radio waves transmitted from the Earth are reflected back to the Earth by this layer. It helps in Radar, Navigator communication. The ionosphere protects us from meteors.

Question 5.
What is the significance of dust particles in the atmosphere?
Answer:
Dust particles are significant I the atmosphere from meteorological standpoint. It acts as a “hygroscopic nuclei” around which water vapor condenses to produce clouds. They also reflect * insolation and besides this smoke and fog are caused due to presence of dust particles.

Question 6.
What is troposphere?
Answer:
The lowermost layer of the atmosphere is known as troposphere and is the most important layer because almost all of the weather phenomena (fog, cloud, dew, frost, rainfall. Hailstorm, storms, cloud, lightning etc) occur in this layer. The word troposphere literally means Zone or region of mixing.

Question 7.
What is Ionosphere?
Answer:
The thin layer extends from the ozonosphere form an altitude of about 80 to 640 km above the earth surface. The process of ionization takes place in this zone, resulting in a dense concentration of positive ions.

Question 8.
Mention the salient features of troposphere?
Answer:
The important features of the troposphere are:

• Hydrological cycle
• Lapse rate
• Clouds
• Gaseous mass.

Question 9.
Mention the features of Inversion of Temperature. (T. B. Qn).
Answer:
It is a process of temperature increases with increasing height in troposphere. This feature is common during winter season, less cloudiness, slow movement of winds, and clear sky in the mountain valley.

Question 10.
Why are isotherms more irregular in the northern Hemisphere than in southern hemisphere?
Answer:
The isotherms are more irregular in Northern Hemisphere because of the distribution pattern of land and water and their differential heating.

Question 11.
What is temperate zone?
Answer:
It extends from 23 XA° to 66 !4° on both hemispheres. The sun is never vertically over head in this zone. It is characterized with warm summers and server winters. There is a general decrease of temperature towards poles and the severity and duration of the winter season also increase towards poles. The ancient Greeks have described this zone as an intermediary zone experiencing both winters and summers

Question 12.
State the difference between tropical and Frigid zone
Answer:
Sun is the chief source of energy. The heat energy radiated by the sun in all directions is called solar radiation.
The amount of radiation reaching the upper layers of the atmosphere is about 1.94 calories per cm2. As it varies little, is known as‘Solar constant

Question 13.
What is the inversion of Temperature?
Answer:
Under certain conditions air temperature increases with increasing height in the troposphere. This is known as inversion of temperature.

Question 14.
What is Albedo?
Answer:
The propagation of solar radiation falling on a non-luminouS body which the latter reflects is usually expressed as percentage. The albedo of the earth is approximately 40% of the solar radiation which is reflected back into space.

Question 15.
Why do Pressure belts shift? (T. B. Qn )
Answer:
The main cause of the pressure belts is the heat from the Sun, the pressure belts follow the annual apparent migration of the sun to the north in summer and to the south in winter. This is known as the shifting of the pressure belts.

Question 16.
What is Doldrums?
Answer:
The doldrums is the equatorial belt of calms and variable wind lying over the equatorial through of low pressure roughly between 5° N to S latitudes. This belt of calm lies between the two trade wind belts. In the late afternoon. There is strong convection which brings about heavy. thunderstorm. Because, of prevailing calm conditions the atmosphere in the doldrums is hot, oppressive and sticky.

Question 17.
State the relation between temperature and atmospheric pressure?
Answer:
There is a close relationship between temperature and pressure conditions of the air. Higher the temperature, lower will be the pressure and vice-versa. When the air is heated, the molecules expand and the air becomes light. The tropical region received almost perpendicular rays of the sun. As a result, the air is heated and rises upwards in the form of convectional currents and has become low pressure belt.

Question 18.
State the vertical distribution of pressure?
Answer:
Atmospheric pressure is highest at the sea level and decreasing with increasing of height. It is because the lower layer of the atmosphere has the weight of the entire atmosphere extending over several thousand kilometers. The lowest layer of the atmosphere, troposphere which extends up to 10-16km has 75 percent of the total atmosphere. The atmosphere near the earth’s surface up to an height of only 1km contains 1/10 of the total height. At the sea level atmospheric pressure is 1013.25mb

Question 19.
Name the two important Trade winds. (T. B. Qn )
Answer:
North East Trade winds: In the northern hemisphere, they blow from north-east to south west. South East trade winds: In the southern hemisphere, they blow from south east to North West.

Question 20.
How is Sea breeze formed? (T. B. Qn )
Answer:
During the day, the land gets heated faster than the sea. So the air gets heated and the rise upwards to produce a low pressure region. At the same time the pressure at sea sis comparatively high. The warm are of the land, being light, rises upwards allowing the air from sea to enter in. Such incoming air from the sea is called ‘Sea Breeze’.

Question 21.
Mention the features of a Cyclone. (T. B. Qn )
Answer:
Cyclone is a small low pressure area in the center surrounded by high pressure. The winds blow spirally towards the low pressure a area and form convergence of winds. In the northern hemisphere the direction of cyclonic winds is anti clock-wise and in the Southern hemisphere it is clockwise.

Question 22.
What are the planetary winds?
Answer:
The winds which blow from high pressure belts to the low pressure belts in the same direction
throughout the year are called planetary winds.

Question 23.
What is Hurricanes?
Answer:
Tropical cyclones in the Gulf of Mexico, the Caribbean sea and western pacific ocean are known as Hurricanes. To qualify as a hurricane, a storm must produce winds over 119Km/hr. It causes severe damages to life and property.

Question 24.
Define typhoons?
Answer:
Tropical cyclones that originate in the China Sea are known as typhoons. These winds are also very violent and destructive.

Question 25.
What do you mean by tornadoes?
Answer:
These are storms which form vey suddenly on land. The highest wind speeds on Earth occur .in tornadoes, sometimes reaching 500km/hr. They are most common and most violent in the U.S.A.

Question 26.
What are Anti-cyclones?
Answer:
An anti-cyclone is a high pressure system with winds blowing clockwise in the Northern hemisphere and anticlockwise in the Southern Hemisphere. An anti-cyclone may form wherever air sinks, the condition that creates high pressure. Winds begin to blow outwards form the high pressure area.

Question 27.
What is Coriolis Effect?
Answer:
The apparent curving motion of anything, such as wind, caused by Earth’s rotation. It was first described in 1835 by French scientist Gustavo- gasped Coriolis.

Question 28.
What is Cyclone?
Answer:
A region of low atmospheric pressure in which winds spiral inwards towards the centre of lowest pressure. In the Northern Hemisphere, the winds blow anti-clockwise: in the Southern hemisphere clockwise. In temperate regions cyclones are called depressions: in tropical regions they are more violent, and are called typhoons or hurricanes.

Question 29.
What are the seasonal winds?
Answer:
The winds that change their direction regularly in different seasons are called seasonal or periodic winds.

Question 30.
What is a trade wind?
Answer:
The winds that blow in the tropics from the sub-tropical high pressure belts towards the equatorial low pressure belt are called “trade winds”. They found approximately between 8’ and 30’ latitudes on both sides, blowing from the east. They are also known as “Tropical Easterlies”.

Question 31.
What is an antitrade wind?
Answer:
The winds blowing from the sub-tropical high pressure belts towards the sub-polar low pressure belts are known as sub-polar low pressure belts are known as anti-trade winds.

Question 32.
What are planetary winds?
Answer: The winds blowing regularly from the high pressure to low pressure belts are known as planetary winds. The velocity and extent of these are however affected with the seasonal migration of the pressure belts

Question 33.
Briefly describe the monsoon winds.
Answer:
The monsoons winds are typical example of seasonal are period winds. They are extensive and also well developed among the seasonal winds. The word ‘Monsoon’ is derived from an Arabic word ‘Mousim’ means seasonal. The Arabs have noticed the character of periodical reversal of winds on the Arabian sea and called them monsoons.

Question 34.
State the different names of the tropical cyclones.
Answer:
Tornado in the U.S.A, Simoon in Africa and Arabia, Sirocco in Italy and Sicily, Hurricane in West Indies, Typhoon in Japan and China and Willy-willies in Asia.

Question 35.
Stat the difference between winds and air currents.
Answer:
The distribution of atmospheric pressure on the earth’s surfaces most uneven. The horizontal movement of air, parallel to the earth’s surface is knows as ‘winds’.
The vertical movement is called “air current”. The wind and currents are interrelated and forms a system of air circulation through out world and play an important role in the distribution of temperature and humidity in the atmosphere.

Question 36.
What are the three types of planetary winds?
Answer:
There are three major types of planetary winds. They are Trade winds, Anti trade winds and Polar winds.

Question 37.
What are Storm Surge?
Answer:
Quickly rising ocean water levels associated with hurricanes that can cause widespread flooding.

Question 38.
What do you mean by El Nino?
Answer:
Linked ocean and atmospheric events that have worldwide effects, characterized by warming of water in the topical pacific from around the International Date Line to the coast of Peru.

Question 39.
Distinguish between air currents and wind?
Answer:
Air flows from high pressure to low pressure due to horizontal differences. This horizontal movement of air is called wind whereas the vertical movement of air is called air currents.

Question 40.
What is Ferrell’s law?
Answer:
The result of the earth’s rotation, a body moving in any direction over surface of the earth will tend to be deflected to the right in the northern hemisphere and to the left in the southern hemisphere. This law is known as “Ferrell’s law”. It mainly concerned with the movement of air.

Question 41.
What are land and sea breeze?
Answer:
The flow of air from land to sea is called land breeze and the flow of air from sea to land is called Sea breeze.

Question 42.
How is Mountain rainfall caused? (T. B. Qn )
Answer:
During this rain the moisture laden winds are forced to ascend over the mountains in their path. As the wind rises, it expands and looses temperature. This results in condensation, leading to rainfall. This rain fall is found in the windward side of the mountain and is heavy.

Question 43.
Mention the forms of condensation?
Answer:
The water produced by condensation may take several forms. It may remain suspended in the air in the form of mist, for or cloud, or it may be deposited on the solid objects on the ground in the form of rain, snow or hailstone. So, there are many forms of condensation. The chief forms of condensation are dew, frost, mist, fog, cloud, rain, snow and hailstone.

Question 44.
What is cyclonic rainfall?
Answer:
Cyclonic rain is associated with a depression or a low pressure. This type of rainfall takes place where air masses of different temperatures and humidity meet. The cyclonic rainfall is most common in the temperate region. The rainfall caused with a cyclone or a depression is known as cyclonic rainfall.

Question 45.
What is rain shadow region?
Answer:
When the ascending air may reach the mountain summit and descends On the leeward side, it is warmed by compression during its descent, and thus becomes drier and drier. The part of land over which this dry wind blows, it known as the rain shadow area. This influence of the mountain barrier is called the rain shadow effect.

Question 46.
State the difference between windward and leeward side?
Answer:
Orographic rainfall occurs due to ascent of air over mountain slopes. Orographic rain is heavy on the windward slope of the mountain. The slope of the mountain facing the wind is called “Wind ward side” while the opposite slope is called “leeward side” After crossing over the mountain, the air begin to descend along the leeward slope. It gets heated and becomes dry.

Question 47.
Mention the causes of rainfall?
Answer:
The cooling of saturated air mass is an important cause for rainfall. The cooling process takes place in different ways. Major causes are:

• The warm and moist air rises upward by vertical convection currents.
• Air is also forced to ascend over mountain ranges.
• The warm air rises over cold air.
• There should be sufficient humidity in the air.
• Condensation should be taken place.

Question 48.
What is Orographic Rainfall?
Answer:
It is the most common and widespread for of rainfall in the world. It is also known as mountain rain or relief rainfall. The moisture laden winds are forced to ascend over the mountains in their path. As the wind rise, it expands and looses temperature adiabatically. As a result, condensation takes place leading to rainfall. The amount of rainfall is heavy and concentrated in the windward side of the mountain.

However highest amount of rainfall is received at the crest of the mountain. In the leeward side of the mountain, the winds begin to descend along the slopes with compression. As the wind move along the slopes the temperature steadily increases. Hence the leeward side remains dry.

Question 49.
What are hailstones?
Answer:
Hailstones are solid frozen raindrops. The rain droplets when they enters into cold layers of the atmosphere at a higher level gets condensed, grew in size and finally reach earth’s surface. Hailstones are normally associated with

Question 50.
What is Hydrological cycle?
Answer:
The circulation of water in the Earth’s environment between lithosphere, hydrosphere and atmosphere is called hydrological cycle.

Question 51.
Mention any two factors that determine the climate of a place.
Answer:
Latitude, Distance from the sea, Altitude, Prevailing winds, Direction of the mountains, ocean currents are the affecting factors of climate.

Question 52.
State the difference between continental and marine climate?
Answer:
In the costal regions temperature condition is modified by the oceans to large extent and thus experiences warm summers and mild winters. The amount of rainfall also decreases from coastal region to the interior. Similarly the range of temperature is maximum in the interior regions making it a continental climate.

Question 53.
State any two differences between Weather and climate?
Answer:
Weather is the average condition of the atmosphere at a particular place for short period. But climate is the average condition of the atmosphere over an area over a long period of time. Weather is described as sultry, cloudy, stormy, chilly, fine or mild etc. Climate is described as hot, cold, dry, humid and wet.

1st PUC Geography Atmosphere Five Marks Questions and Answers

Question 1.
Explain the structure of the atmosphere. (T. B. Qn )
Answer:
The distribution of temperature is not uniform at different height of the atmosphere. Along with the variation of temperature there are unique features at different heights. Based on these characteristics atmosphere is divided into four parallel zones.

1. Troposphere: It is the lowest layer of the atmosphere and less closer to the earth. The word ‘Tropos’ means ‘turn’ It extends up to 18km at the equator and 8km at poles. Thus the average height is about 10-12km. The important feature of the troposphere are:

Hydrological cycle: It is confined to troposphere. The water evaporates and raised up, formation of clouds takes place. Later it is precipitated in various forms like rain, snow and hailstone. These processes are known as evaporation, condensation and precipitation.

Lapse rate: In this layer the temperature decreases at the rate of 6,5’ Celsius per every 1000 meters of height which is known as ‘lapse rate’.

Clouds: It is characterized with formation of clouds, thunder storms and lighting.
Gaseous Mass: The troposphere has about 75 percent of the total gaseous mass, The upper part of troposphere is known ‘Tropopause’.

2. Stratosphere: It lies above the tropo-sphere and extends up to 50km from the earth. The temperature is also most uniformly distributed. Hence it also known as isothermal zone. At a height of 22kms. There is a thin layer of ozone which absorbs ultraviolet rays of the sun. So it is called as ozOnosphre. The name staratopause is given to the upper part of the stratosphere.

3. Mesosphere: It extends from 50 to 80kms. It is an intermediary zone between the lower and upper layers of the atmosphere. A thin layer of air separating mesosphere from the other upper layers in named as‘Mesopause’.

4. Ionosphere: It extends from 90 to 500km. It consists of atoms of air ionized due to intensive temperature. So it is also known as‘Ionosphere’or Thermosphere. The radio waves of different length are reflected back from this layer.

5. Exosphere: The region beyond the Thermosphere is called Exosphere. It extends to about 1,000 km and the gravity of the Earth s too weak in this layer. Magnetosphere is found above this layer. Atmospheric layer in between 500-700kms is known as Exosphere and the atmosphere lying beyond is called‘Magnetosphere’.

Question 2.
Explain the composition of the atmosphere.
Answer:
The atmosphere is a mechanical mixture of several gases. It also contains a number of solid and liquid particles called‘Aerosols’. Some of the gases may be permanent. They remain in fixed proportion but others may vary in quantity from one place to another and from time to time. The different constituents of the atmosphere are gases, water vapor and dust particles.

Gases: The dry air that is around us is a mixture of various gases. The main component gases of dry air are listed in the table. Both nitrogen and oxygen together account for about 99% of the atmosphere. The remaining 1% to make up of other gases. Nitrogen is most plentiful of all the gases in the atmosphere. It is an important constituent of many organic compounds. It is needed to dilute the air and regulated combustion by diluting oxygen. All living organism need oxygen. For respiration. Life is not possible without it. It is essential for combustion. About 21% of the gases in the atmosphere consist of oxygen.

Carbon dioxide is found in small percentage in dry air. It used by green plants for photosynthesis. It absorbs solar energy and earth radiation and then emits a part of it towards the earth. The amount of carbon dioxide is increasing every year which can agent temperature. Another important gas is ozone, which is similar to oxygen molecule. It absorbs most of the ultra-violet rays from the sun and protects us from excessive heat.

Water vapour: This is the gaseous form of water. It is largely formed from the evaporation of water from the water bodies on the earth and transpiration from plants and soils.

Water vapor is one of the most variable gases of the atmosphere, which is representing in the lower layers of the atmosphere. It is capable of absorbing solar energy as the well as energy radiated form the earth. The condensation of water vapour is responsible for several forms ofthe precipitation. Eg: rain, snow, etc. The amount of water vapor mainly depends on temperature. So it. varies from place to place and from time to time. It decreases from the equator to the poles.

Dust particles: The atmosphere is capable of holding solid particles, suspended in the lower layers of the atmosphere. They consist of dust, salt particles, pollen, smoke and soot, volcanic ash etc. These dust particles are very important form the absorption and scattering of some of the solar energy.

Water vapor is one of the most variable gases of the atmosphere, which is representing in the lower layers of the atmosphere. It is capable of absorbing solar energy as the well as energy radiated form the earth. The condensation of water vapour is responsible for several forms ofthe precipitation. Eg: rain, snow, etc. The amount of water vapor mainly depends on temperature. So it.varies from place to place and from time to time. It decreases from the equator to the poles.

Dust particles: The atmosphere is capable of holding solid particles, suspended in the lower layers of the atmosphere. They consist of dust, salt particles, pollen, smoke and soot, volcanic ash etc. These dust particles are very important form the absorption and scattering of some of the solar energy.

Question 3.
Explain the characteristic features of Atmosphere.
Answer:
Atmosphere has certain characteristics features. They are

• Atmosphere is the second layer of the earth.
• Like lithosphere and hydrosphere, the atmosphere also is an integral part of the planet.
• Atmosphere is a gaseous layer extending upto a height of about 960 kms., but it has no
  distinct outer limit.
• It is attached to the earth’s surface by gravitational force.
• The atmosphere is denser near the earth’s surface, and becomes thinner away from the earth’s surface.

Question 4.
Briefly explain the factors affecting the distribution of temperature. (T. B. Qn )
Answer:
The distribution of temperature on the surface of the earth is not uniform. It varies from. region to region due to various factors. The various factors affecting the distribution of atmospheric temperature are:

a. Latitude or distance from the equator: Places close to the equator have higher temperature and are warmer than places awaylfom the equator This is because the Sun rays reach the Earth after passing rays reach the Earth after passing through the layers of the atmosphere. In the low latitudes the Sun rays are direct and have to travel a lesser extent through the atmosphere. Hence, the heat of these rays is more intense. But in high latitudes the Sun rays are slanting and have to passes through a greater extent of atmosphere.

b. Altitude: Temperature decreases with altitude. This is because the heat absorbing elements are found in lower altitude. So the places near the Earth’s surface are warmer than places higher up. This is because air near the surface is denser and contains gases like carbon dioxide, water vapour and other gases. Temperature decreases with increase in height at an average rate of l°C/165m or 6.4°C/1000m.

c. Distance from the sea: this factor also influence on the distribution of temperature and differential heating of land and water. Land gets heated faster compared to water. Water takes longer time to get heated and to cool than land. Hence during the day when the land gets heated quickly, water takes longer time and remains cool. Therefore, during the day time a land gets more heat than the surrounding water bodies.

d. Ocean currents: It increase or decrease the temperature of the Earth’s surface. Warm ocean currents along the coast make the coastal areas warmer and cold currents reduce the temperature and cool the coastal areas.’ Warm currents can be noticed on the eastern margins of the continents in the middle latitude, while .it is the concurrents flow at the western margins of the continents. Gulf stream a warm currents increases the temperature in the eastern coast of U.S.A and California bold current decreases the temperature of the western coast of U.S.A.

e. Winds: Winds that blow from the lower latitudes are warm and make the places warmer. On the other hand, winds that blow from the higher latitudes are cold and make the places cooler. Winds that blow from the sea bring plenty of rain especially if they are warm winds. While off shore winds hardly bring any rain.

f. Clouds: During the day clouds prevent Insolation from reaching the Earth’s surface. Clouds also prevent three escape of terrestrial’s radiation during the night. Clear sky Permits insolation readily during the day time and allow the rapid escape of terrestrial radiation during the night.

Question 5.
What is solar radiation? Explain the factors controlling solar radiation.
Answer:
The solar radiation is in the form of short waves. These waves strike the outer surface of the atmosphere. About 53% of the energy is lost through scattering by gas molecules, reflection by clouds and absorption by water vapor and other particles. Only about 47 percent of the energy reaches the surface.

Controlling factors: Insolation is not uniform on all parts of the earth’s surface. It differs from one region to another and also from one season to another. The important factors determining the amount of solar radiation are:

a. The angle of incidence of the sun’s rays: The sun rays do not strike the earth surface at an. equal angle. They are almost perpendicular as well as concentrated near the equator and slanting as well as spread over a large area at the poles.

b. Atmosphere: The amount of insolation also depends on the atmosphere through which it passes. The amount of cloud cover, its thickness,, dust, water vapour absorbs temperature.

c. Duration of Day: The duration of sunshine or day varies with the seasons and latitudes. Longer the duration of sunshine, more will be the insulation and vise versa, d. Distance between the earth and the sun: The earth revolves around the sun on its orbit. It is nearest to the sun on January 3 and far away on July Thence the solar radiation is more in January than in the month of July.

Question 6.
What is Inversion of temperature and explain the conditions of inversion of temperature? ‘
Answer:
Some times air temperature increases with increasing height in troposphere. This is known as ‘Inversion of temperature’ the suitable conditions for inversion of temperature is as follows-

(a) Long Winter Nights: The loss of heat by terrestrial radiation exceeds the amount of insolation received from the sun during long winter nights.

(b) Cloudless and clear sky: A clear sky causes rapid loss of hear through terrestrial radiation.

(c) Dry air: The dry air present near earth’s surface is not capable of absorbing much of the heat. Hence the temperature at lower layer does not rise.

(d) Calm atmosphere: When there is no-movement of air or it is very little, there is no transfer and mixing of hear in the lower layers of the atmosphere.

(e) Snow covered surface: When the ground is covered by snow, and there is a reflection of incoming solar radiation the ground heats very little. Therefore the air close to the ground undergoes rapid cooling and temperature inversion takes place.

There can be two types of inversion of temperature:

1. High-Altitude inversion: The high altitude inversion occurs mainly due to the frontal convergence, when a swarm air-mass is forced from the ground surface by the underlying of a cold air mass at a cold front. Alternatively, a similar inversion can be created when a warm air-mass overrides a colder one along with a warm front.

2. surface inversion: A surface inversion is much more localized and is often dependent on the terrain. It frequently occurs during winter anticyclonic weather when during calm cloudless nights thee is a rapid heat loss from the ground by radiation.

Question 7.
Explain the process of Heating and cooling of atmosphere?
Answer:
Atmospheres do not get its temperature directly by the short waves of the sun but through the long waves of the earth. The atmosphere is transparent to the solar radiation which is in the form of short waves. It allows them to reach the earth.
The four processes of heating and cooling of atmosphere are radiation, conduction, convection and advection.

Radiation: Radiation is the process of heating an object by the transmission of heat waves. The earth surface gets temperature through solar radiation. It radiates energy in the form of long waves which is absorbed by the atmosphere.
Conduction: Transfer of heat by molecular activity is known as Conduction. The atmospheric layer closer to the surface gets heat in this process with contact.
Convection: Transfer of heat through the movement of mass is called Convection. The lower layer of the atmosphere gets temperature and become lighter. So it moves upwards. The dense air in the upper layer being dense descends downwards. With this cyclic movement temperature is distributed in the lower layers of the atmosphere.
Advection: Temperature is also transferred by large scale movement of air. When warm air moves of cold regions temperature of the air increases and is contrast cold reduces heat. This phenomena is called advection.

Question 7.
What are the isotherms? State their Characteristics?
Answer:
The line drawn on maps joining places having the same air temperature are called Isotherms. They are drawn after reducing the temperature to mean sea level to eliminate the variation caused by altitude. The horizontal distribution of temperature is shown with the help of isotherms.

• Isotherms run east-west and are parallel to the latitudes.
• Isotherms take sudden deviation near coasts indicating contrasting temperature conditions between land and water bodies.
• Spacing of isotherms indicates latitudinal thermal gradient or steepness of temperature change or intensity of temperature variation.
• There is a seasonal difference in spacing of isotherms. Generally they are closer in the month of January while they are widely apart in the month of July.
• The isothermal lines have east-west trend following the latitudes.
• The ocean currents also influence and deviate the trend of isotherms.

Question 8.
Explain the temperature zones.
Answer:
The perpendicular rays of the sun strike different latitudes in different seasons. As a result there is marked variations in the amount of solar radiation in different parts of the world. On the basis of solar radiation, the surface of the earth is divided into three zones. They are

(a) Torrid Zone: It is the hottest zone on the earth’s surface. It extends from equator to 23/2° degrees in both the hemispheres. The summers are hot and winter is mild and warm. Summer temperatures are above 27°C and the temperature even during the winter season is more than 18°C. It is thus characterized with lower annual range of temperature. The torrid zone is described by ancient Greeks as ‘Winterless tropics’ as the winter season is less pronounced.

(b) Temperate zone: It extends from 23 14° to 66 !4° on both hemispheres. The sun is never vertically over head in this zone. It is characterized with warm summers and server winters. There is a general decrease of temperature towards poles and the severity and duration of the winter season also increase towards poles. The ancient Greeks have described this zone as an intermediary zone experiencing both winters and summers.

(c) Frigid Zone: Beyond 66 V2 on both hemispheres the zone extends up to 90°. The temperature lies below freezing point for most part of the year. During the summer season snow melts partially. The temperatures never rise above 10°C even during the summers. Summers are.mild and .short, winters are long and severe. The poles remains permanently ice covered. So this polar belt is described by the ancient Greeks as summer less frigid zone’.

Question 9.
Describe the major pressure belts of the world with a neat diagram. (T. B. Qn )
Answer:
The distribution of pressure is not equal on the earth’s surface. It changes from palace to place and time to time on the basis of air temperature and rotation of the earth. Any area in the atmosphere where air pressure is higher than in the surrounding areas is called “ High pressure”/ Thee are 4 high pressure belts and 3 low pressure belts on the earth’s surface.

Equatorial Low pressure belt: This belt lies between latitudes 5° N and 5° S. The Sun’s rays are almost vertical on the equator throughout the year. As a result, the temperature is uniformly high and pressure is low throughout the year. It is also known as “Doldrums”. The air gets warm and rises upward. Horizontal movement of air is absent and convectional currents occur. This is the zone of convergence of the trade winds.

Sub tropical high pressure belts: The air ascended in the form of convectional currents from the equatorial region partly descends in the between 30 to 40’ latitudes in both the hemispheres. The descending air has thus formed two high pressure zones known as subtropics high pressure belts. It is the zone from which trade and anti-trade winds originate. This belt is also known as “ horse altitudes’. It is dry and quite stable. The name horse latitude is given by the ancient sailors who used to transport horses on ships. Due to absence of strong winds, some times the ship could not move with horses. Hence sailors used to dump horses to make the ship move forward.

Sub Polar low pressure belts: In between polar high pressure knd sub-tropical high pressure belt, the sum-tropical low pressure belts are situated. They lies in between 60’ to 70’ latitudes in both the hemispheres. They are formed with spinning action of rotation of the earth and also uprising air as an effect of incoming cold polar winds.

Polar high pressure belts: The Polar Regions are characterized with low temperature. The air raised at the equator descends around the poles causing high pressure belts. The cold polar winds blow outward from this zone.

Question 10.
What are the characteristics of Isobar?
Answer:
The distribution of pressure over the earth’s surface or part of it is shown by means of isobars.
The term isobar means line ofjoining the places of equal pressure.
Important characteristics of isobars are:

• If the isobars are far apart, it indicates gentle pressure gradient as well as calm weather characterized with light surface winds.
• The isobars usually follows east-west trend.
• The isobars deviate in their trend at the point of their entrance from sea to land as well as from land to sea.
• The isobars are almost parallel to latitudes in the southern hemisphere due to predominance of oceans.
• If the isobars are close to earth other it represents steep isobaric gradient. It also indicates more atmospheric disturbances as well as strong winds.

Question 11.
What factors influences on the atmospheric pressure?
Answer:
Atmospheric pressure is influenced by various factors. Most important among them are:

a. Temperature: There is a close relationship between temperature and pressure conditions of the air. Higher the temperature, lower will be the pressure and vice-versa. When the air is heated, the molecules expand and the air becomes light. The tropical region receives almost perpendicular rays of the sun. As a result, the air is heated and raised upwards in the form of convectional currents and has become a low pressure belt.

b. Altitude: The air pressure is associated with altitude or height. Much of the atmospheric air is concentrated in the lower layers of the atmosphere. With the increase of height pressure decreases but not always at constant level. On the basis of this principle “Altimeter’ is invented. This instrument measures the pressure of Atmosphere and denotes corresponding height of the place from the mean sea level, c. Water Vapor: The amount of water vapour in the air also influence on the atmospheric pressure. Water vapour is lighter than air. So more the vapour lighter will be the air. The dry air is heavier than vapour laden. It is because of this reason the air with more amount of vapour becomes lighter and raised upwards leading to condensation and precipitation.

a) Rotation of the Earth: The spinning of the earth on. its axis also causes variation in the atmospheric pressure. The spinning action also causes spinning of air mass and the cold air near the poles is rarefred to form low pressure belts.

Question 12.
Explain the characteristics of the winds.
Answer:
Winds have certain characteristic features. They are.

• Winds blow parallel to the earth’s surface.
• Winds are named after the direction from which they blow. If they blow from the west, they are called westerlies and if they blow from the easterlies.
• Winds follow ferrel’s law. As per the ferrel’s law, winds get deflected to the right of their path I the northern hemisphere, and to the left of their path in the southern hemisphere
• The movement of the wind is due to pressure. That is the movement of the wind is due to the differences in air pressure between two places. Wind always moves from a high pressure area to a low pressure area.
• The speed and velocity of the wind will depend upon the pressure gradient. The steeper the pressure gradient, the grater is the velocity of the wind.
• Natural or artificial barriers like mountains, forests, buildings, etc. cause friction and check the force of the wind.

Question 13.
What are the seasonal winds? Explain the origin of monsoon winds of India.
Answer:
The winds which change their direction with the changer in the season are called periodic or seasonal winds. Monsoon winds are an example of seasonal winds. Period winds, seasonal winds or monsoons winds are the result of vast temperature differences between the hot summer and cold winter, where there are large land masses surrounded by water bodies.

The monsoon winds are typical example of seasonal or periodic winds. They are extensive and also well developed among the seasonal winds. The word ‘Monsoon’ is derived from an Arabic’ word ‘Mousim means seasonal. The Arabs have noticed the character of periodical reversal of winds on the Arabian sea and called them monsoons. The same word is now applied to the seasonal winds of the world.

Question 14.
Explain the land and sea breezes with a neat diagram?
Answer:
The winds blowing alternatively during day and night from the sea and the land near the coasts are known as ‘Sea Breeze’ and ‘Land Breeze’ respectively. These are the best developed local winds near the coastal regions. They affect only a narrow strip of land along the coast. During the day time, the land gets heated more quickly than the adjacent sea. So the air is heated and raised upwards to produce a low pressure region.

The pressure at sea is comparatively high. The warm air of the land being light rises up allowing the air from sea to enter in. Such incoming air from the sea is called‘Sea breeze’. The sea breezes continue during the day time. At night the land looses its temperature quickly due to rapid radiation and a high pressure is developed. As the seawater is still retaining temperature the air is lighter and rises upward

Question 15.
What are the cyclones? Explain the types of cyclones.
Answer:
Cyclone is a small low pressure area in the centre surrounded by high pressure. The winds blow spirally towards the low pressure area and form convergence of winds. In the northern hemisphere the direction of cyclonic winds is anti-clockwise and in the Southern hemisphere it is clockwise. The cyclones are classified into two types.
(i) Tropical Cyclones: The origin of tropical cyclones is much related to intensive They cause heavy rainfall with high-velocity winds. Tropical cyclones are highly dangerous and devastating.

(ii) Temperate Cyclones: In the temperate region, cyclones are produced by the meeting of warm air mass of tropical region and cold air mass of the polar region. The tropical. air mass is lighter and it is pushed up by the advancing dense cold air mass. The process of mixing of these two air masses takes palace in the form of cyclones. They are ‘ associated with heavy rainfall.

Question 16.
Explain the anticyclones.
Answer:
Winds blowing spirally outward from centers of high pressure are known as anti-cyclones. They are just the reverse of cyclones. Their movement is clockwise in the Northern Hemisphere and anti clockwise direction in Southern Hemisphere. They are more common in the sub¬tropical high pressure belts and absent in the equatorial belt. They are circular in shape, larger in size and extent and their track is highly variable. The weather is generally associated with anticyclones is find and dry.

The anticyclones are divided into two groups such as permanent and temporary anticyclones. Permanent anti-cyclones are mainly centre in the temperate regions where the meeting of tropical and temperate air mass takes place. In between the two temperate cyclones there will be anti-cyclones. In the interior of continents the high pressure centers are developed during the winter season. It is due to lower amount of temperature

Question 17.
What is Rainfall? Explain the types of rainfall with neat diagram. (T. B. Qn )
Answer:
Rainfall is the natural process of condensation through which gaseous form of water is converted into liquid water droplets. It occurs due to cooling of saturated air mass, warm and moist air mass rising upward, warm air rising over cold air, sufficient humidity in the air and condensation. Rain fall is classified on the basis of condition and mechanism of upward rise of air and its cooling. Therefore, there are three types rainfall. They are explained as follows:

1. Convectional Rainfall: Earth surface is heated by the solar radiation. Due to this the warm moisture-laden air becomes light and ascends upwards vertically and quickly. The warm and moist air cools below dew point and condensation takes place rapidly, and dense clouds are formed. This led to the heavy rain fall with thunder and lighting. For convectional rainfall, there is need for local heating which leads to excessive evaporation.

There should not be any strong winds to lesser the heat such a condition of great heat, excessive evaporation and stagnant air is found in the equatorial regions through out the year As such, the convectional rainfall occurs in the daily afternoon in the equatorial region. In other topical countries this type of rainfall occurs only during summer, if sufficient moisture for local evaporation is available.

2. Orographic rainfall: When the moisture laden air from the sea is obstructed by a mountain, it is forced to move up the slope. As it moves up, it expands. Expansions lead to fall in temperature and the air cools. When the cooling takes places below the dew. point, condensation results and clouds are formed. When there is heavy condensation, the droplets of water in the clouds join to find bigger drops and there is rainfall.

Orographic rains are caused by the relief of the land. This type of rainfall is common in region where the mountain ranges are parallel and close to the sea and winds blow on shore. Behind the leeward side of the mountain, there is an area which receives very little rainfall and is called the rain shadow. Orographic rain is important in the monsoon land in summer, when the wind blows from the sea to the land. Thunderstorms also accompany this type of rainfall. It may be noted that the bulk of rainfall received by most parts of the world is of this type.

3. Cyclonic rainfall: The cyclonic rainfall is most common in the temperate region. The rainfall caused with a cyclone or depression is known as cyclonic rainfall. The winds take a circular movement in the regions where warm and cold air masses meet.

Question 18.
Explain the factors affecting Rainfall.
Answer:
There are various factors affecting the rainfall. They are as follows:

Latitude: Rainfall is influenced by the latitude. If the latitude is low, temperature will be high. Water vapour in the air depends upon the temperature. Rainfall depends upon the amount of water vapour in the air. Evaporating is highest in the equator and least at the poles. Therefore rainfall is more in the equator and less or nil at the poles.

Distance from the sea: Winds blowing from the sea bring rainfall over the coastal lands. But as these winds go further, they become drier. Therefore they do not bring much rainfall in the continents.

Mountains: There is heavy rainfall in the windward side of the mountains because the mountains obstruct the winds carrying moisture. But there is little rainfall on the leeward side of the mountains. This is because, by the time the winds cross the mountains, they lose much of their moisture and become dry.

Type and direction of winds: Winds blowing from the sea contain moisture, and therefore bring rain to the land over which they blow. But winds blowing from the land is dry, and therefore, they do not cause rains. Similarly winds blowing from the cold regions to hot regions cause little rainfall, while the winds blowing from hot regions to cold regions cause more rainfall.

Vegetation: Rainfall is more in a region where there is thick vegetation, i.e., forests, because the forests make the air cool. That is why, Assam and mainland areas get more rainfall than the desert and the malnad areas.

Question 19.
Explain the factors determining the climate?
Answer:
Climate is the average condition of the atmosphere over an area over long period of time. The changes in the atmospheric conditions or weather elements are caused by several factors of climate. The most important factors of climate are:

a. Latitude: The Latitudinal location is of prime importance as the temperature and humidity decrease from equator to the poles. But the distribution of pressure is contrasting as it decrease from polar regions to equator. Even the amount of rainfall generally decreases from equator to the poles. It indicates the relationship between latitude and climate, Son the isobars, Isotherms and Isohyets run east to west parallel to latitudes. It is clear that the climatic condition is normally almost similar and identical over the places on the same latitude.

b. Distance from the sea: The physical characteristics of land and water are quite different as they are in solid and liquid states. The land masses absorb temperature and loose it quickly than the water bodies. So the distribution of temperature and pressure conditions in different seasons of a year is contrasting. The interior land masses experience hot summers and cold winters.

Thus extremities prevail in the climatic condition. On the other hand in the coastal regions temperature condition is modified by the oceans to a large extent and thus experiences warm summers and mild winters. The amount of the interior. Similarly the range of temperature is maximum in the interior regions making it a continental climate. The coastal regions on the other hand, being much influenced by the oceans are described as Maritime climate or Oceanic climate.

c. Altitude: Climate is influenced by attitude in many ways. Attitude has a great effect on the distribution of heat, pressure, winds and precipitation. Temperature and pressure decreases with attitude. The wind is forced to rise, when they are obstructed by mountain ranges, and they give rainfall to the windward side. Mountain can obstruct the passage of cold or hot winds. In this way, attitude causes c changes in the atmospheric conditions.

d. Ocean currents: The warm ocean currents keep the coastal regions warm, while cold currents lessen the temperature of the costal areas. For instance: North West Europe coast and Western coast of America are washed by the warm Gulf stream and cold Labrador currents respectively. The moisture laden winds blowing over the warm currents also caused changes in the atmospheric conditions.

e. Slope of the land: The slope of the land also causes changes in atmospheric conditions. The slope that faces the ocean is warm. But the slope that does not face the sun is cold. For instance, the southern slopes of the Himalayas are warmer than the northern slopes.

f. Soil: Soil caused changes in atmospheric conditions. Rocky and sandy soil, which is dry, heats and cools faster. The soil which is wet and can retain moisture heats and cools slowly. Dark- colored soil absorbs heat faster than the light coloured soil.

g. Volcanic Activity: At the time of volcanic explosion, carbon dioxide, dust particles and ash are thrown out into the atmosphere, the carbon dioxide layer absorbs both solar and terrestrial energy, it affects the temperature and pressure of a regions.

Question 20.
Explain the differences between weather and climate.
Answer:
The atmospheric condition of a place at a given time is known as weather.
The average weather condition of a place for a long period like 30-33years is known as climate.
The main differences between weather and climate are:

1st PUC Geography Atmosphere Ten Marks Questions and Answers

Question 1.
Explain the planetary winds with the help of diagram. (T. B. Qn )
Answer:
Winds which blow from high pressure belts to the low pressure belts in the same direction through out the year are called planetary winds, permanent winds or prevailing winds. Characteristics: The chief characteristics of planetary winds:

a. They are connected with the pressure belts. So they blow from the high pressure belt or area to the low pressure belt or area.

b. They are regular through out the year.
c. They deflect to their right in the northern hemisphere and to their left in the southern hemisphere.

d. As a result of the shifting of the pressure belt northward in summer and southward in winter, the planetary winds also shift northwards in summer and southwards in winter. They are permanent and blow over vast areas of the globe. These winds include trade winds, westerlies and polar winds.

1. Trade winds: The winds that blow in the tropics form the sub-tropical high pressure belts towards the equatorial low pressure belt are called “Trade winds”. They found approximately between 8° and 30° latitudes on both sides blowing from the east. They are also known as “Tropical Easterlies”.

The trade winds due to the law of deflection blow from the north-east in the northern hemisphere and from south-east in the southern hemisphere. Trade winds blow from the cooler sub-tropical areas to the hotter area, hence they do not bring rain. However, when they blow over the open sea they gather moisture and bring heavy rainfall to the east coast of the continents.

2. Anti trade winds: The wind blowing from sub-tropical high pressure belts towards sub-polar low pressure belts are known as anti-trade winds. They are south-west to north east in the northern hemisphere and Norwest to south-east in the southern hemisphere.

Hence they are called ‘Westerlies”. They prevail largely between 40° and 650 north and south of the equator. They blow from the north-west in the southern hemisphere and south – west in the northern hemisphere. As they blow from hotter areas to colder areas they bring rain through the year.

3. Polar winds: Winds blowing from polar high pressure belts toward the sub-polar low pressure belts are known as polar winds. They blow from the north east in the northern hemisphere and south east in the southern hemisphere. As they blow form the snow cov4red areas they are very cold winds. They are constant in the southern hemisphere.

Karnataka 1st PUC Geography Question Bank Chapter 10 Climate, Soil and Forest

You can Download Chapter 10 Climate, Soil and Forest Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Climate, Soil and Forest One Mark Questions and Answers

Question 1.
What type of climate is found in India? (T.B Qn)
Answer:
India has “Tropical Monsoon” type of climate.

Question 2.
Define Monsoon. (T.B Qn)
Answer:
The word ‘Monsoon’ is derived from Arabic word ‘Mousim’ meaning season.

Question 3.
Mention the place which records high range of Temperature. (T.B Qn)
Answer:
In summer the western Rajasthan records more’than 55°C of temperature.

Question 4.
Which is the driest season in India? (T.B Qn)
Answer:
The summer or hot weather Season from March to End of May.

Question 5.
Name the region which receives ‘Monsoon outburst’. (T.B Qn)
Answer:
The.Malabar Coast of Kerala receives ‘Monsoon outburst’.

Question 6.
Which is called ‘Mawsynram of South India’? (T.B Qn)
Answer:
Agumbe of Karnataka is called ‘Mawsynram of South India.’

Question 7.
Why are cyclones formed during North East Monsoon season? (T.B Qn)
Answer:
In this season due to pressure variation between the Bay of Bengal and main land of India variable winds-cyclones and anti -cyclones originate in the Bay of Bengal.

Question 8.
What is mean by ‘Burst of Monsoon’?
Answer:
The sudden violent onset of rainfall during the period of ‘Monsoon’ is called the ‘Burst of Monsoon’.

Question 9.
Name any two local winds which blow in India in the summer season.
Answer:
‘Loo and Kalabaisakhi are the two local winds which blow in India in the summer season.

Question 10.
Which wind is responsible for the rainfall experienced over the greater part of India?
Answer:
A South-west monsoon winds is responsible for the rainfall experienced over the greater part of India.

Question 11.
What is the average annual rainfall of India?
Answer:
The average annual rainfall is 118 cm.

Question 12.
Name the place in Southern India which receives highest rainfall from the summer monsoon.
Answer:
Mahabaleswar receives the highest rainfall in South India from the summer monsoon.

Question 13.
What is Kalabaisaki?
Answer:
The local rainfall of summer season in West-Bengal is called ‘Kalabaisaki’.

Question 14.
In which state the South-west monsoon wind enters first.
Answer:
The south west monsoon enters first to the Malabar Coast of Kerala.

Question 15.
Define Pedology. (T.B Qn)
Answer:
The scientific study of soil is known as ‘pedology’.

Question 16.
Name the soil which covers vast area of the country. (T.B Qn)
Answer:
Alluvial soil

Question 17.
Why Black soil is called Regur soil? (T.B Qn)
Answer:
This soil derived from the weathered basalt rock. This soil holds water form long period and become hard whenever it is dry.

Question 18.
Where do we see Laterite soil? (T.B Qn)
Answer:
Laterite soil found in Western Ghats, parts of Eastern Ghats and North eastern hills of India.

Question 19.
What is Humus?
Answer:
Decomposed organic material found in the soil is called Humus.

Question 20.
State the type of soil that is found in the heavy rainfall regions?
Answer:
The laterite soils re found in the heavy rainfall regions.

Question 21.
Which soil is suitable for cotton cultivation?
Answer:
The black soil is suitable for cotton crop.

Question 22.
Which is highest fertile soil?
Answer:
The mountain soil (Forest soil) is fertile soil

Question 23.
Where is Green Gold? (T.B Qn)
Answer:
The forest and their resources are useful to man in various forms. Therefore, they are called ‘Green Gold’.

Question 24.
Mention the average forest cover of the country. (T.B Qn)
Answer:
The average forest cover of the country is 22.50%.

Question 25.
Which forest has high economic value trees? (T.B Qn)
Answer:
Monsoon forest has high economic value trees.

Question 26.
Where do we find Dehang Debaqg Biosphere reserve? (T.B Qn)
Answer:
Arunachal Pradesh.

1st PUC Geography Climate, Soil and Forest Two Marks Questions And Answers

Question 1.
Why India is called ‘Meteorological Unit’? (T.B Qn)
Answer:
Monsoons are the periodic winds in which there is reversal of wind direction periodically. On account of the variability in climatic conditions, seasonally and regionally, India is called ‘Meteorological Unit’.

Question 2.
Mention any two convectional rainfall of India. (T.B Qn)
Answer:
They are “Mango Showers” in Kerala, “cherry Blossoms” in Karnataka and “Kalabiashaki” in West Bengal and Assam.

Question 3.
Write the significance of Monsoon. (T.B Qn)
Answer:
The climatic conditions of the country are greatly influenced by monsoon winds. The winds blow in a particular direction west monsoon winds blow from south west to north east, while north east, while north east monsoon winds blow from northeast to south west.

Question 4.
Mention the annual average rainfall of different seasons in India.
Answer:
The season-wise distribution of rain fall is

• The south-west monsoon seasons – 75%
• The summer season – 10%
• The winter season – 2%
• The retreating monsoon seasons – 13%

Question 5.
Write a short note on Retreating Monsoon?
Answer:
The season of Retreating Monsoon is the period of Transition. During the period of transition low pressure of the north-west shifts to the Bay of Bengal. It results in the formation of cyclones over the Bay. These cyclones cause havoc on the coasts of Orissa and Andhra Pradesh.

Question 6.
Why does India have a monsoon type of climate?
Answer:
India has monsoon climate because there is a seasonal reversal in the wind system in India. During summer winds blow from sea to land and During winter winds blow from land to sea.

Question 7.
What is meant by the word ‘Monsoon’?
Answer:
The word ‘Monsoon’ is derived form the Arabic word ‘Mausam’ which means season. Hence, the word ‘Monsoon’ implies the seasonal reversal in the wind pattern over the year. It reveals the rhythm of season and changes in direction of winds. There is also ca change in the distribution pattern of rainfall and temperature with the change of seasons. The monsoon winds move six months from sea to land and another six months from land to sea,

Question 8.
Mention major factors affecting the climate of our country.
Answer:

• Location and Relief.
• Latitude
• Altitude
• Pressure and winds.

Question 9.
What are the branches of south-west monsoon winds?
Answer:
The south-west monsoons are divided into two branches. They are:

1. The Arabian Sea branch and
2. Bay of Bengal branch.

Question 10.
What is annual range of temperature? Explain it by giving one example.
Answer:
The difference between the maximum average temperature and minimum average temperature of a place over twelve months is known as annual range of temperature.

Ex: The max. average temperature at Jodhpur is 33.9°C and min. Average temperature is 14.9°C. Hence the annual range of temperature at Jodhpur is 19°C.

Question 11.
Mention the importance of Red soil. (T.B Qn)
Answer:
This soil is formed by the weathered granite rocks. It is red in colour and rich in ferrous content. Red soil covers the second largest area in the country. Largest part of peninsular region is covered with red soil. Tamil Nadu has the largest distribution of this soil in the country. Rice, Ragi, Jowar, Groundnut oil seeds are the main crops cultivated in this soil.

Question 12.
Name any four factors that affect soil erosion. (T.B Qn)
Answer:
High Temperature, Rainfall wind and waves are the natural agents. Deforestation, over grazing, shifting cultivation, unscientific methods of agriculture cause soil erosion.

Question 13.
State four best measures in the conservation of soil. (T.B Qn)
Answer:
Afforestation, control of over grazing, contours ploughing. Terrace farming, Erection of bunds, construction of check dams, crop rotation, and control of shifting Cultivation are the best measures in the soil conservation.

Question 14.
State the characteristics of Laterite soils.
Answer:
The important characteristics of Laterite soils are: Laterite soils are red in colour. They are rich in Iron and Aluminum, but poor in potash, Lime, Nitrogen and Phosphoric Acid. They are less retentive of moisture. They are poor in fertility. But they respond very well to manuring. So, with the help of manuring they can be used for the cultivation of plantation crops, such as Tea, Coffee, Spices, Rubber, etc.

Question 15.
Name the states which have the highest and the lowest forest areas in the country. (T.B Qn)
Answer:
Madhya Pradesh (44.8%) is the highest and Haryana state 2.6% is the lowest forest areas in the country.

Question 16.
Write the salient features of Evergreen forest. (T.B Qn)
Answer:
These forests are found in the regions of heavy rainfall (above 250 cm) and high temperature (above 27° C) Tall umbrella shaped trees with dense assemblage is a prominent feature of this forest. The evergreen forests always look green because, various species of trees are found here and they shed leaves in different seasons.

Question 17.
What is Mangrove forest? Why has it become important in the recent years?
(T.B Qn)
Answer:
Mangroves are trees and shrubs that grow in saline coastal habitats in the tropics and subtropics in India. The trees in these forests are hard, durable and are used in boat making and as fuel, in the recent years mangrove vegetation is being grown in the coastal areas to control effects of tidal waves and coastal erosion.

Question 18.
Mention any four measures of conservation of forest. (T.B Qn)
Answer:
Protection and preservation of forest is known as conservation. The important measures of conservation of forest are:

• Forest fires, pests and diseases should be controlled through the scientific methods.
• Encroachers of forest area should be severely punished.
• Forest education, research and training should be expanded through programmes like vanamahotsava, social forestry, and reforestation.

Industrial and mining activities in the forest regions should be compensated by reforestation.

1st PUC Geography Climate, Soil and Forest Five Marks Questions And Answers

Question 1.
What is Climate? Explain the factors that determine the climate of India. (T.B Qn)
Answer:
The average weather condition of place for a long period like 30-33 years in known in known as climate. India’s climate is said to be “Tropical Monson”.

The main factors are monsoon winds.

(i) Location: The northern part of India lies in sub-tropical and temperate zone and the part lying to the south of the tropic of cancer come under tropical zone. The tropical zone being nearer to the Equator, experiences high temperature throughout the year, with small daily and annual range. Tropic of Caner 23 1/2° N latitude passes through the centre of the country. So India is situated both in the tropical and temperate region.

(ii) Mountain Ranges: The lofty Himalayan Mountains have prevented the cold winds of central Asia, and keep India warm. They are also greatly responsible for the monsoon rains in the country.

(iii) Distribution of Land and Water: India is bounded by the Arabian Sea in the west and Bay of Bengal in the east, Indian Ocean in the south. These adjoining seas have influenced the climate of the country considerably. They influence the rainfall of the coastal region. Even the cyclones which originate from these seas regularly affect the weather condition.

(iv) The relief features of India also affect the temperature, air pressure, direction and speed of wind, the amount and distribution of rainfall. The windward side of Western Ghats and north east received high rainfall from June to September.

(v) Monsoon winds: The climatic conditions of other country are greatly influenced by monsoon winds. The winds blow in a particular direction during one season, but get reversed during the other season.

Question 2.
Explain the South West Monsoon season with the help of map. (T.B Qn)
Answer:
The south-west monsoon winds as starts in June and ends in mid-September. It is also known as advancing monsoon season or rainy season. During this season, India gets more than 75% of its annual rainfall and more than 90% of the country’s area receives downpour. It is the prime season for Kharif crops.

In the middle of June the direct rays of the Sun fall on tropic of caner due to shift in the position of the Sun from Equator towards northern hemisphere. Therefore, there is an increase in temperature from south to north. The temperature in the main land of India and nearby land masses is high compared to water bodies of the Indian Ocean.

a. The Arabian Sea branch: The Arabian Sea branch of the south-west monsoon strikes the western coast of India in Kerala on the 1st June. Arabian sea winds by carrying more moisture blow along the western coast of India and cause heavy rainfall in the western part of Western Ghats due to obstruction. These winds behave like sea breeze and cause continuous rainfall I the wind ward side of the Western Ghats tHl they lose their moisture.

Agumbe of Karnataka receives the highest rainfall during this season. This regions coming under southeast monsoon winds receive good rainfall wherever they get obstruction by hills and plateaus.

b. The Bay of Bengal branch blow from water bodies towards the Indian mainland due to variation in pressure. These winds carry moisture form the Bay of Bengl and blow along eastern coast and finally reach north eastern hills. In its path, whenever this wind receives obstruction, they cause good rainfall. The eastern part of Eastern Ghats and north astern hills receive heavy rainfall. These winds after crossing eastern coast merge the Arabian sea winds.

The Arabian Sea and the Bay of Bengal winds, after merging, blow towards north eastern regions of India. The shape of the Himalayan Mountains and northeastern hills greatly obstruct these winds. Therefore the Meghalaya plateau region, particularly Nokrek areas of Mawsynram and cheerapunji, receive very high rainfall. This place is popularly called Rainiest or wettest place on the Earth.

The southwest monsoon after crossing northeastern region blow towards east. Since the Himalayas obstruct these winds they have to take westerly direction and blow along the foothills of Himalayas. The shift in the direct6 sun rays from Tropic of Cancer towards Equator results in the gradual disappearance of southwest monsoons. Indian economy depends on the Monsoons to a large extent.

Question 3.
Briefly explain the characteristics features of winter and summer. (T.B Qn)
Answer:
The winter season (December to February): It is also called cold weather season. In this season direct rays of the Sun fall on Tropic of Capricorn. The temperature in the country is not uniform from north to south. Regions lying to the north of tropic of cancer record low temperature compared to regions in the south. There is a general decrease in temperature from south to north.

December is the beginning of cold weather season and it extends up to February. The annual average temperature is around 18° C. In the northern parts of the plains temperature falls below 5°C. January is the coldest month in the year. Jammu and Kashmir, Punjab, Haryana, UP and parts of Bihar record very low temperature with snow storms. Though the rainfall is small, in some parts on North India it is beneficial for Rabi crops. Annual rainfall in this season is around 2%.

The Summer Season (March to May): The summer season is also known as hot weather season. It begins in March and continues up to May. During this season there is gradual increase n temperature from south north due to shifting of Sun rays from Tropic of Capricorn towards the Equator. In this period south Indian states- Tamilnadu, Andhra Pradesh, Karnataka, and Kerala record high temperature.

Some part of Andhra Pradesh and Karnataka record more than 40° C of temperature. Sri Ganganagar of Rajasthan has recorded the highest temperature of above 52° C. The average temperature of the country will be around 24° C. In this season some parts of India receive convectional rainfall. During this season the country receives 10% of the annual rainfall.

Question 4.
Give details of North East Monsoon season. (T.B Qn)
Answer:
This season is also called North East Monsoon Season. It starts in the middle of September and extends up to middle of December. On September 23rd the direct rays of the sun falls on Equator. Therefore, there is a change in temperature and pressure in the land and water bodies. In this period the Indian sub-continent. The high pressure formed in the northern part of Bay of Bengal results in movement of wind from northeastern part of India towards southwestern region.

These winds blow along the eastern coast of India and Bay of Bengal. In this season due to pressure variation between the Bay of Bengal and main land of India variable winds cyclones take birth in this season and cause great damage in the eastern coast of India. The coastal areas of Tamilnadu, Andhra Pradesh, Odisha and West Bengal come under the frequent effect of cyclones. Some cyclones recorded in the last few years are Bola, Nargis, Nisha, Laila, jal, Neelam etc.

Question 5.
What is Soil? Explain the major types of soils. (T.B Qn)
Answer:
Soil is the minute or finer rock particles found on the surface of the Earth. It is formed naturally, due to the weathering of rocks, under the influence of climate.

The main types of soil in India are:

1. Alluvial soil: This soil is formed by depositional work of rivers and they are mainly found in the flood plains and deltas. Alluvial soil covers largest geographical are in the country. They are mainly distributed in the river plains of the Ganga, Brahmaputra and the Indus. Uttar Pradesh has the largest area under alluvial soil. It is also found in the deltas of east flowing rivers. Alluvial soils are classified into two types.

• Bhangar: Older alluvium, coarse and pebble like in nature, found at the lower depths of the plain.
• Khadar: New alluvium, finer in nature, found in the low lying flood plains and rich in fertility

2. Black soil: The black soils covered more area in peninsular plateau. This soil is also called ‘Cotton soil’ or “Regur soil”. It is derived from the weathered basalt rocks. This soil holds water from long period and become hard whenever it is dry. It is light-black to dark-black in colour. Maharashtra and Gujarat Madhya Pradesh, Karnataka, Andhra Pradesh and Tamilnadu. Black soils are good for Cotton, Sugarcane, Tobacco, Pulses, Millets, Citrus fruits, etc.

3. Red soil: This soil is formed by the weathered granite rocks. It is red in colour and rich in ferrous content. Red soil covers the second largest area in the country. Largest parts of peninsular region are covered with red soil. TamilNadu has the largest distribution of this soil in the country. Rice, Ragi, Jowar, Groundnut, Tobacco, Millets are the major crops cultivated in this soil.

4. Laterite soil: The hot and humid tropical regions of India are rich in laterite soil. This soil is derived from the fragmentation and disintegration of rocks in the mountain ranges. It is mainly found in the Western Ghats, parts of Eastern Ghats and Northeastern hills of India. Plantation crops like Tea, coffee, Rubber, Cashew nut are cultivated in this soil.

5. Desert soil: This soil is also called arid soil. They are mainly found in the desert and semi-desert regions of Western and North western parts of India. This soil has the least water holding capacity and humus content. Generally it is not suitable for cultivation of crops. This soil is mainly found in Rajasthan, parts of Gujarat and Haryana. With water facility crops like Bajra, Pulses and Guar ar cultivated in this soil.

6. Mountain Soil: The Himalayan mountain valleys and hill slopes are covered with Mountain or Forest soil. It is found in the mountain slopes of Jammu and Kashmir, Himachal Pradesh, Utarkhand regions, Crops like Tea, Almond, saffron are cultivated in this soil.

Question 6.
Explain soil erosion and conservation of soil. (T.B Qn)
Answer:
The removal or wearing away of the top soil by various natural agents and man-made factors is called ‘Soil Erosion’. High temperature, Rainfall wind and waves are the natural agents and deforestation, over grazing, shifting cultivation, improper and unscientific methods of agriculture are human activities cause soil erosion. In the hilly regions rainfall and temperature cause more soil erosion. In coastal area sea waves and in desert winds is the dominant factor s in the soil erosion process.

The prevention of soil erosion as well as the protection and maintenance of the Fertility of the soil. The important measures followed in the Conservation of soil are:

• Afforestation
• Control of overgrazing
• Contour ploughing
• Terrace Farming
• Erection of bunds
• Construction of check dams
• Crop rotation
• Strip Farming
• Mulching
• Literacy and education programmes on soil conservation.

Question 7.
Describe the major types of forest in India.(T.B Qn)
Answer:
The peninsular region of India has the largest forest cover with around 57% of the total forest area.

According to geo-climatic conditions, forests are classified into:

a. Evergreen Forests: These forests are found in the regions of heavy rainfall and high temperature. Tall umbrella shaped trees with dense assemblage is a prominent feature of this forest. The eve4rgree forest always looks green because various species of trees are found here and they shed leaves in different seasons.

The hardwood trees, rose wood, white cedar, toon, gurjan, chaplash, ebony, Mahogany, canes, bamboo, shisham etc. These are found in North-east India, Western Ghats, Andaman and Nicobar islands, parts of Assam and some areas of Himalayan foot hills.

b. The Deciduous forests: The deciduous forest covers a wide range of rainfall regimes. The trees of these forests seasonally shed their leaves. The Indian deciduous forest is found in a range of landscapes from the plains to the hills. These forests provide shelter to most endangered wild life in the country, such as the Tiger, Asian Elephant, Bison, Gaur etc. The deciduous forest are two types

(i) Moist Deciduous forests: The moist deciduous forests are found in wet regions, receiving annual rainfall between 100cm to 200cam and temperature of 25° C to 30° C. The trees of these forests shed their leaves during spring and early summer. They are found on the eastern slopes of the Western Ghats, Chota Nagpur Plateau, the siwaliksetc.

(ii) The Dry Deciduous Forests: The dry deciduous forest are found I the areas where annual rainfall is between 50cm to 150 cm and temperature of 25° C to 30° C. Sal is the most significant tree found in this forest. Varieties of acacia and bamboo are also fund here. These forests are found in areas of central Deccan plateau, South-east of Rajasthan, Punjab, Haryana and parts of Uttar Pradesh and Madhya Pradesh.

(iii) The mountain forests: As the name indicates these forests are confined to the Himalayan region, where the temperature is less compared to other parts of the country. The trees in this forest are cone shape with needle like leaves. The important trees are oak, fir, pin e spruce, silver fir, deodhar, devdar, juniper, picea chestnut etc. They provide softwood for making country boats, packing materials and sport articles.

c. The Desert forests: These forests are found in the areas of very low rainfall. Thorny bushes, shrubs, dry grass, acacia, cacti and babul are the important vegetation found in these forests. The Indian wild date known as ‘Khejurs”, is common in the deserts. They have spine leaves, long roots and thick fleshy stems in which they store water to survive during the long drought. These vegetations are found in Rajasthan, Gujarat, Punjab and Haryana.

d. The Mangrove Forests: These forests occur along the river deltas (Ganga, Mahanadi. Godavari and Krishna) of eastern coast and also concentrated in the coastal areas of Katchch, Kathiawar, and Gulf of Khambar. The mangrove forests in the Ganga delta are called Sunder bans because, they have extensive growth of Sundari trees. The trees in these forests are hard, durable and are used in boat making and as fuel. In the recent years mangrove vegetation is being grown I the coastal areas to control effects of tidal waves and coastal erosion.

Question 8.
Briefly explain the importance of forests. (T.B Qn)
Answer:
Forests are the one of the important natural resources. They provide various benefits to mankind and environment.

The important benefits are:

1. Forests supply fresh air, food and fodder.
2. Forests are the rain bearers, help in causing good rainfall.
3. They control soil erosion and desertification.
4. Forest provides various products like bamboo, timber, resin, lac, gum cane, fuel, wood etc.
5. They provide medicinal trees and plants used in ayurvedic medicines Eg.Neem tree. Basil, Brahmi etc.
6. They provide shelter to various birds and animals.
7. They absorb much of the rainwater and control floods and safeguards against drought.
8. They act as wind breakers and protect the agricultural crops.
9. The forest soils are rich in humus and thereby maintain the fertility of the soil.
10. They provide raw materials to paper, match box, plywood and sports articles industries and they provide pastures for grazing animals.

Question 9.
Explain the important measures of conservation of forest. (T.B Qn)
Answer:
The conservation of forest is concerned with proper utilization of forest, protection from destructive influences, misuses of forests etc.

The important measures of conservation of forest are:

1. Careless felling of tree, over-grazing and shifting cultivation should be avoided. Afforestation should be practiced.
2. Forest fires, pests and diseases should be controlled through the scientific methods.
3. Encroachers of forest area should be severely punished.
4. Forest education, research and training should be expanded through programmes like vanamahotsava, social forestry, and reforestation.
5. Industrial and mining activities in the forest regions should be compensated by reforestation.
6. Development of Green belts in the urban areas.
7. Plantation of trees along the roads, railway lines, river, canal banks, tanks and ponds.
8. Use of fuel wood, wood-charcoal by the tribal people must be prohibited.
9. Government should promote intensive tree planting programmes in urban centers.
10. Massive awareness about the aesthetic of forests should be created through mass media, workshops, live programmes etc.

Question 10.
What are Biosphere reserves? Mention the important biosphere reserves of India. (T.B Qn)
Answer:
A biosphere Reserve is a unique and representative ecosystem of terrestrial and coastal areas .The regions surrounding the biosphere reserves would be utilized for the research and experimentation in developing forest and other products.

The Man and the Biosphere Programme (MAB) of UNESCO was established in 1971 to promote interdisciplinary approaches to management, research and education in ecosystem conservation and sustainable use of natural resources. Eight of the eighteen biosphere reserves are a part of the world network of Biosphere reserves, based on the UNESCO man and the Biosphere Programme list.

The objectives of Biosphere reserves:

• Conservation of biodiversity and ecosystem.
• Association of environment with development.
• International network for research and monitoring.

Sl.No	Name of the Biosphere reserve	State	Estd.Year
1.	Nilgiri Biosphere Reserve	Tamilnadu, Kerala, Karnataka	2000
2.	Gulf of Mannar Biosphere Reserve	Tamil Nadu	2001
3.	Sunder bans Biosphere Reserve	West Bengal	2001
4.	Nanda Devi Biosphere Reserve	Uttarkhand	2004
5.	Nokrek Biosphere Reserve	Meghalaya	2009
6.	Panchmarhi Biosphere Reserve	Madhya Pradesh	2009
7.	Simlipal Biosphere reserve	Odisha	2008
8.	Achanakmar-Amarkantak	Chhattisgarh, Jharkhand	2012

You can Download Chapter 12 Omme Nagutteve Questions and Answers Pdf, Notes, Summary, 2nd PUC Kannada Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Kannada Textbook Answers Sahitya Sampada Chapter 12 Omme Nagutteve

Omme Nagutteve Questions and Answers, Notes, Summary

Karnataka 1st PUC Geography Question Bank Chapter 9 Physiography

You can Download Chapter 9 Physiography Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Physiography One Mark Questions and Answers

Question 1.
What is the other name to the Himalayas?
Answer:
The Himalayas means‘abode of snow’. Or Young fold mountains’.

Question 2.
Which mountain range is called ‘backbone of Asia’?
Answer:
Karakorum range is called backbone of Asia.

Question 3.
Name the longest and the largest glacier of India.
Answer:
Siachen (70Km) is the longest and largest glacier of India.

Question 4.
What is the other name to Outer Himalayas?
Answer:
Siwahk are called as outer Himalayas.

Question 5.
Name the largest Doon of India.
Answer:
Dehradun is largest Dun of India.

Question 6.
In which regional Himalayas Jclep la pass is found?
Answer:
The Central or Sikkim Himalayas.

Question 7.
What is Terai plain?
Answer:
It is a marshy land wide spread in the regions of excess dampness, thick forests, rich wild life etc. It is found to the south of Bhabar with wide marshy tract, where streams reappear to the surface.

Question 8.
Mention the highest peak of Peninsular plateau.
Answer:
The highest peak of Peninsular plateau is Anaimudi (2695m) situated in Annamalai hills of Kerala.

Question 9.
Which region of India is called ‘Ruhr of India’?
Answer:
The Chotanagpur plateau is called‘Ruhr of India’.

Question 10.
Where do Western Ghats and Eastern Ghats meet?
Answer:
The Western Ghats and Eastern Ghats meet in Nilgiri Hills

Question 11.
Name the longest coastal plains of India.
Answer:
The longest coastal plains of India from Gujarath (Rann of Kutch) in the west to West Bengal In the east, (61 OOKm)

Question 12.
State the location of the Great Indian Desert.
Answer:
The Great Indian Desert is located at the Western part of the Aravali Range.

Question 13.
Name one important river of the Great Indian Desert.
Answer:
The Luni is the important river among of the Great Indian Desert.

Question 14.
Name the Salt water lake in the Thar Desert.
Answer:
Sambhar Lake is the salt water lake in the Thar Desert.

Question 15.
Name the plateau which lies between the Western and Eastern Ghats of South India.
Answer:
The Deccan plateau lies between the Western and Eastern Ghats of South India.

Question 16.
Which is the highest peak of India?
Answer:
K2 or Mount Godwin (8611m) is the highest peak of India or Second highest peak of the world.

Question 17.
Which is the highest mountain peak in South India?
Answer:
Anai Mudi (2,695m) in Annamalai hills of Kerala is the highest peak in south India.

Question 18.
Which is the oldest and largest physiographical division of India?
Answer:
The peninsular plateau (16 lakh sq.km) is the oldest and largest physical division of India.

Question 19.
Which is the highest peak of Eastern Ghats?
Answer:
Mahendragiri in Orissa (1501m.) is the highest peak of Eastern Ghats.

Question 20.
Which island of India is formed with the volcanic activities?
Answer:
The Andaman and Nicobar Islands are formed with volcanic activities.

Question 21.
Which island of India is formed by Corals?
Answer:
The Lakshadweep islands are formed by corals.

Question 22.
Why are the Himalayan Rivers perennial?
Answer:
Most of the Himalayan Rivers originate from the glaciers and they get water from the rainfall as well as from the glaciers.

Question 23.
From which mountain pass does the river Sutlej enter India?
Answer:
Shipki-la-pass the river Sutlej enters to India.

Question 24.
Which is the longest and the largest tributary of Ganga?
Answer:
The Yamuna is the longest (1380) and largest tributary of the Ganga.

Question 25.
Which is the largest and longest river of South India?
Answer:
Godavari is the longest (1465km) and largest river of South India

Question 26.
What is Do-ab-region?
Answer:
The region or plain lying between two rivers. Ex: Ganga and Yamuna river is called as Do-ab region.

Question 27.
Which river makes Kapiladhara water fall?
Answer:
Narmada river.

Question 28.
Which river is called Dakishina Ganga?
Answer:
Kaveri (Cauvery-805 Km)

Question 29.
Which is the Asia’s First hydroelectricity generated station?
Answer:
The first hydro electric project of Asia was started on the river Kaveri in 1902atShivanasamudra (Shimsa).

Question 30.
Where do river Tungabhadra and Krishna meet?
Answer:
The Tufigabhadra and Krishna meets at Alampur near Kurnool in Andhra Pradesh.

Question 31.
Mention the major stream of river Ganga.
Answer:
The two major streams are Alakananda and Bhagirathi.

Question 32.
What is Hydel Power?
Answer:
The power generated from water is called the hydel power.

Question 33.
Which rives of India flow in rift valleys?
Answer:
Narmada and Tapi.

Question 34.
Which river is called ‘Sorrow river of Orissa’?
Answer:
Mahanadi river is called‘sarrow river of Orissa’.

Question 35.
What is a Lake?
Answer:
A water body, completely surrounded by land is known as a lake.

Question 36.
Name some important fresh water lakes of India.
Answer:
Dal Lake, Bhimtal, Nainiral, Loktak and barapani.

Question 37.
Why do many peninsular rivers have straight and linear courses?
Answer:
Because of hard rock bed and lack of silt and sand in their courses. They do not form meanders.

Question 38.
Which city is located on the water divide between the Indus and Ganga river systems?
Answer:
Amritsar located on the water divide between the Indus and Ganga river systems.

Question 39.
Which place is the confluence of rivers Alkananda and Bhagirathi?
Answer:
Devaprayag.

Question 40.
Which is the largest freshwater lake in India?
Answer:
Wulur lake is the largest freshwater lake in India.

Question 41.
Which river is called ‘sorrow river of West Bengal’?
Answer:
Damodar River is called Sorrow of West Bengal.

Question 42.
Which river is called ‘National River’?
Answer:
Ganga River is called National River.

1st PUC Geography Physiography Two Mark Questions And Answers

Question 1.
Name any four tributaries of river Indus.
Answer:
The Sutlej, Ravi, Jhelum, Chenab and the Beas are the major tributaries of Indus river.

Question 2.
Mention any four west flowing rivers of Peninsular India.
Answer:
The Luni, Sabarmati, Tapi, Kali, Sharavathi, Netravati, Peryiar are the major west flowing rivers.

Question 3.
What are the salient features of River regime?
Answer:
The pattern of the seasonal flow of water in a river is called its regime. It is the variability in its discharge throughout the course of a year in response to precipitation, temperature and drainage basin characteristics. The pattern of flow of water in the Himalayan river is different fro the peninsular river due to difference in climate. The Himalayan Rivers are perennial and the regime of the peninsular rivers is seasonal as they are dependent on monsoon rains.

Question 4.
What is the necessity of Inter-linking of Rivers?
Answer:
The distribution of rainfall in India is highly uneven and seasonal. The Himalayan rivers are perennial while the peninsular rivers are seasonal. During rainy season, much of the water is lost in floods and wasteful flow into the sea. But in other seasons there is scarcity of water. Even in India some parts gets more rainfall and some other parts get very low rainfall. The problems of floods and drought can be minimized through the inter-river linkages or through national water grid, under which water from one river basin can be transferred to another river basin for optimum utilization. ’

Question 5.
Mention the back water lakes of East Coast of India.
Answer:
Pulicat Lake (TN), Kolleru (AP) Chilka (Orissa) are the important back water lakes of India.

Question 6.
What are riverine islands?
Answer:
In the lower course of the river, due to gentle slope, the velocity of the river decreases and it involves into depositional work leading to the formation of rivierine islands. For example, Majuli in the Brahmaputra.

Question 7.
Mention any two ranges of Trans Himalayas.
Answer:
Karakoram range, Ladakh range and Zaskar range.

Question 8.
Mention any two hill stations of the Himalaya
Answer:
The important hills stations are Shimla, Mussorie, Raniket, Nainital, Almora, Chakrata, Darjeling etc.

Question 9.
Distinguish between Bhangar and Khadar plains.
Answer:
Bhangar Plains: It is an Old Alluvium. It contains the Kankar nodules with calcium carbonates and it is less fertile.
Khadar Plains: It is a new alluvium. It does not contain the kankar nodules and it is very fertile.

Question 10.
Name any two Ghats of the Western Ghats.
Answer:
Thalghat, Bhjorghat, Palghat, Agumbe ghat, Shiradighat, Charmadighat are the major Ghats of the Western Ghats.

Question 11.
Which coastal plains are found in Karnataka and Tamil Nadu?
Answer:
The Malabar Coast extends from Mangalore to Kanyakumari, Sand dunes, lagoons and backwaters are the important features of this coast.

Question 12.
State the difference between Lakshadweep and Andaman and Nicobar islands.
Answer:
Lakshadweep Islands: These islands are located to close to the Malabar Coast of Kerala. These are composed of small coral islands. They are small in size as compared to the Andaman and Nicobar islands.

Andaman and Nicobar islands: These are located in the Bay of Bengal. These are bigger in size and are more numerous and scattered. These are islands are an elevated portion of the submarine mountains.

Question 13.
What is ‘Duns’? Mention with examples.
Answer:
In the Siwaliks many flat bottomed valleys are there, they are known as duns. The important duns are Dehradun, Kotadun, and Patili, Chaukambadun in Uttaranchal and Udampur and Kotli in Jammu and Kashmir.

Question 14.
Mention the regional divisions of Himalayas
Answer:
The major regional divisions of Himalayas are:

1. Punjab Himalayas (Sindhu-Sutlej)
2. Kumaon Himalayas (Sutlej-Kali)
3. Nepal Himalayas (Kali-Tista)
4. Assam Himalayas (Tista-Brahmaputra)

Question 15.
Mention the Central Plateaus of India?
Answer:
The central plateaus are Malwa plateau, Bundhel Khand plateau, Bhagelkhand, ChotaNagpur and Ranchi plateau.

Question 16.
Which are the important valleys passes in Himalayas?
Answer:
Zojila and Burzil in Kashmir, Shipkila and Bralapcha la in Himachal Pradesh. Thang la, niti and lipu lekh in uttarPradesh,jelep laandNiithu lain Sikkim.

Question 17.
Write any four characteristics of the Indian desert.
Answer:

• It lies towards the western margins of the Aravalli hills.
• It is an undulating sandy plain covered with sand dunes.
• This region receives very low rainfall below 150mm per year.
• It has arid climate with a low vegetation cover.
• Streams appear during the rainy season.
• Luni is the only river in this region
• The Barchans cover larger areas but longitudinal dunes become more prominent near the Indo-Pakistan border.

1st PUC Geography Physiography Five Mark Questions And Answers

Question 1.
Name the important physical divisions of India. Explain the Himalayas.
Answer:
India is characterized by great diversity in its physical features. On the basis of physiography, the country is divided in to four major physical divisions. They are:

1. The Northern Mountains
2. The Northern Plains
3. The Peninsular Plateau
4. The Coastal Plains and Islands

The Himalayas: This is loftiest and snow covered mountains in the world. The area occupied by the Himalayas was earlier a part of ‘Tethys Sea’. The formation of this mountain is by tectonic forces of Gondawana land Angara land masses. It is situated to the north of the Indus and Ganga and the Brahmaputra plains.. The slopes of the Himalayas are gentle towards the north and steep towards south.

The Himalayas have distinct characteristics of high relief, snow covered peaks, complex geographical structures, parallel separated by deep valleys and rich temperate vegetation.The Himalayas are classified into three parallel ranges based on altitude and latitude.

The Great Himalayas or Himadri The lesser Himalayas or Himachal The Outer Himalayas or Siwaliks.

a. The Great Himalayas or Himadri: These are the inner most loftiest and continuous ranges of mountains. The average height of the Great Himalayas is 6200 m and the width varies between 120 and 190 km. The important peaks of great Himalayas in India are, Kanchenjunga-8598m in Sikkim, Nanga Prabat-8126m, Nandadevi, Badrinath, Karmet, Trishuletc.

b. The lesser Himalayas or Himachal: These ranges are also known as Inner Himalayas or Himachal ranges. It is situated between great Himalayas inn the north and Outer Himalayas or Siwaliks in the south. Its average height is around 1500-4500m and the width is about 60 to 80 km. These are very rugged and complex ranges due to erosion by rivers. The important ranges in Lesser Himalayas are Pirpanjal, Dhaul Dhar and nag- tiba etc. The important Hill stations are Shimla, Musooire, Ranikeht, Nainital, Almora, Chakrata, Darjeeling etc. Kulu valley, Kangra valley, Spiti valley are the famous valleys of Himachal.

c. The Outer Himalayas or Siwaliks: These are the outer most ranges situated to the south of Lesser Himalayas, known as Siwaliks. The Siwaliks extend from Jammu & Kashmir in the North West to Arunachal Pradesh in east. The average height of this range is around 600-1500m and its width varies between 15-5Qklm. The siwaliks are formed from the sediments brought down by the rivers of lesser, and Greater Himalayas.

There are flat floored structure valleys between Siwaliks and Lesser Himalayas, Known as Siwaliks. The Siwaliks extend from Jammu&Kashmir in the North West to Arunchal Pradesh in east.

Question 2.
Briefly explain the Regional Himalayas.
Answer:
The Himalayas are also classified into Regional and Longitudinal divisions. They are:
The Kashmir Himalayas The Himachal Himalayas The Kumaun Himalayas The Central or Sikkim Himalayas The Eastern Himalayas.

a. The Kashmir Himalayas: They are spread over in Jammu and Kashmir for about 700sqkm. The important parallel ranges in the Kashmir Himalaya are Karakoram, Ladak, Zaskar and Pirpanjal. They are characterized by high snow covered peaks, largest number of glaciers, deep valleys and High Mountain passes. The north-eastern part of the Kashmir Himalayas is a cold region and it lies between the Grater Himalayas and the Karakorum ranges. A special feature of the Kashmir valley is the Karewas. The important mountain passes are Banihal, zoji-la, Chang-la, Khardung-la etc.

b. The Himachal Himalayas: It is found in Himachal Pradesh and parts of Punjab, comprising of all the three ranges. The beautiful valleys of Kullu, Kangra, Lahul and Spiti known for orchards and scenic beauty are found here. Shipkila, Rohtang, bara- lacha la are the famous mountain passes and Kullu manali, shimla, Dalhousie, Chama etc.

c. The Kumaun Himalayas: This section extends from Sutlej to kali river valleys and has distance of320kms. The pilgrimage centers like Badrinath and Gangothri are located in this section of Himalayas.

d. The Central or Sikkim Himalayas: This section extends from kali to Tista and has a distance of about 800kms. It is also called as Nepal Himalaya. Mount Everest is located in this 3sectino of Himalaya. This section is further divided for the study into Sikkim, Darjeeling and Bhutan Himalayas.

e. The Eastern Himalayas: This range extends from Tista to Brahmaputra valley covering the states of Assam and Arunachal Pradesh. The width is about 730 kms Naga and Patkaibhum hills are located in this section. This region is very important for tea cultivation.

Question 3.
Describe the significance of Northern Plains.
Answer:

• The Northern Plain plays a very significant role in the life of the people and economy of the country.
• The Northern plains have high concentration of population 45% of India’s population.
• They are helpful for agro-based industries and urbanization.
• The northern plains have fertile soil, uniform surface and perennial rivers-suitable for agriculture.
• The plains have encouraged the development of transport and communication.
• The rivers in the plain help in the development of inland water transportation.
• It has rich underground water, useful for irrigation and other activities.
• It has cultural and traditional importance.
• They have great social, religious and political significance.

Question 4.
Peninsular Plateau is the largest physical divisions of India. Explain its features.
Answer:
The Peninsular Plateau is the largest and oldest physiographic division of India. It lies to the south of the Northern Great Plains and covers and area of about 161akh sq km. The elevation of this upland varies from 600 to 90m. This is in inverted triangle shape, with wide base lying in the north and the apex formed in the south, with tilt towards south eat.

It is bounded by the Aravallis in the North West, Bundelkhand plateau in the north, Rajmahal hills in the north east, the Western Ghats in the west and the Eastern Ghats in the east. The highest peak of Peninsular plateau is Anaimudi (2695 m) situated in Annamali hills of Kerala.

On the basis of relief features the peninsular plateau is divided into two main divisions. They are, The Central high lands: This is a smaller region of peninsular plateau situated to the north of the Narmada river. It is slightly tilted towards north. It include the Aravallies, the Malwa plateau, the Vindhya range, the Bundelkhand, the Baghelkhand and Chotanagpur plateau and Rajmahal hills.

The Aravallies runs from north east to south west for about 8900 km between Delhi to Gujarat. It is one of the oldest folded mountains of the world. Its highest peak is Guru Shikar (1722m). It separates Rajasthan- upland and agricultural region. The Aravallis are composed of quatizetes, gneisses and schists.

Rivers like the Luni, Sabarmati and the mahi flow from Aravalli ranges. The Malwa plateau is bordered by the Aravallis in the north and the vindyan range in the south. This plateau has to drainage systems i) Narmada and Mahi towards the Arabian Sea ii) Chambal, Sind, Betwa and Ken towards the Bay of Bengal. ’

The vindyan range extends in ease west direction for about 1050km. The kaimur hills lies in the east of Vindhya range and the Maikala range forms a link between the Vindhya and Satpura ranges.

The Deccan Plateau: this is a triangular plateau situated to the south of the river Tapi or Tapti. The Deccan trap is the crystalline core of the lava effusions forming this plateau are believed to have occurred through a fissure volcano and this region is considered a lava shield. It occupies the areas of Maharashtra, Karnataka, Andhra Pradesh and parts of Chhattisgarh, Odisha and Tamil Nadu.

Eastern Ghats: They form eastern boundary of the Deccan Plateau. They are “separated by the river valleys. The Eastern Ghats stretch to 800 km from Mahanadi valley in the north to Nilgiri hills in the south. Its average height is around 600m.Nallamalla, Kallamalai, B.R.Hills and M.M.Hilis are the important hills of Eastern Ghats.

The important peaks are Aramakonda, Singaraju, Nimalgiri, Mahendra giri etc. Aramakonda is considered as the highest peak of the Eastern Ghats. These zones are rich in Iron ore, Manganese ore, Limestone, Coal, Mica etc.

Western Ghats: It is also known as Sahyadris. They are almost continuous mountain system running parallel to west coast for about 1600km., from north-west to south direction. The Western Ghats meet the Eastern Ghats in Nilgiri hills.

The Western Ghats form a watershed of the peninsular rivers. Important rivers like the Godavari, Krishna, Kaveri, Sharavati, Periyar etc, rise in this zone. They are sources of hydro¬electricity. They are covered with dense evergreen and monsoon forest and rich bio-diversity zones.

Question 5.
Briefly explain the Coastal plains of India.
Answer:
This is the region all along the Indian coastline, lying between the coast and the mountain ranges of the peninsular plateau. India has 6100 km from Gujarat in the west to West Bengal in the east. The average width is 10-25 kms. The coastal plain of India is divided into two parts.

The West Coastal plains: It is extends between the Arabian Sea and the Western Ghats. It is narrower than the east coastal plains, stretching to a length of about 1400km and width of 10 to 80km from the Rann of katchchh to Kanyakumari. The west coastal plains have Gujarat, Konkan, Karnataka and Malabar Coasts.

The Gujarat Coast comprises of Rann of Kachchh and Cambay coasts. It is formed by the alluvial deposits of Sabarmati, Mahi, Luni and other small streams. Gujarat has the longest coastline in India Kandla and okha are famous sea ports and along is the biggest ship breaking center. It produces highest salt in the country.

Konkan Coast lies to the south of Gujarat coast and extends line which provides suitable site for natural seaports. Eg: Mumbai, Navasheva, Marmagoa, Karwar, New Mangalore etc., this coast records highest coastal erosion. It is very rich in Petroleum and natural gas. Karnataka coast: it is a part of Konkan coast.

It extends from karwar in the north to Mangalore in the south. It is the narrowest part of west coastal plains. Karwar and New Mangalore are important ports in this belt. Sea Bird, the naval base near Karwar is the largest naval base in India.

The Malabar Coast extends from Mangalore to Kanyakumari, Sand dunes, lagoons and backwaters are the important features of this coast. Cochin or Kochi is the biggest seaport in this coast. Backwaters of Kerala facilitate navigation and tourists enjoy traveling though small country boats. The first south west monsoon rainfall is received in this coast.

East coastal Plains: It lies between the Ea’stern Ghats and the Bay of Bengal stretching from the delta of Hooghly in the North to Kanyakumari in the south. Compared to the west coastal plains the east coastal plains are broader.

The Tatkal Coast: It is the coastal plain of Orissa state. It extend for about 400kmms, north from Subarnarekha river to south of the Rushikulya river. It has a chilka lake, which is the largest salt water lake in India, Para deep is the important horbour located here.

Coromandel Coast: The southern part of east coast is known as the Coromandel Coast. It gets more rainfall from the north east monsoons and it is highly affected by cyclones. The oldest harbor Chennai located here.

Question 6.
Describe the important features of Islands and Indian desert.
Answer:
India has a total of 247 islands. Ofthese204 are in the Bay of Bengal and the remaining 43are in the Arabian Sea. The islands of the Bay of Bengal are called Andaman and Nicobar islands, which are largely tectonic and volcanic in origin. In Andaman there are four groups of islands – North Andaman, Middle Andaman, South Andaman and Little Andaman, Port Blair the capital of Andaman and Nicobar Group Island is situated in South Andaman islands. Barren and Narcondam are famous volcanic islands in this group.

Nicobar are three groups of islands – Car Nicobar, Little Nicobar and Great Nicobar. The Andaman and Nicobar islands have warm tropical climate and receive heavy rainfall during monsoon seasons and they have thick forest and rich wildlife.

The islands of the Arabian Sea are called Lakshadweep Islands. These islands are very close top Kerala. These are coral in origin and are surrounded by fringing reefs. Kavarati is the capital of Lakshadweep islands. Minicoy and Amnidivi are the important groups in Lakshadweep.

Indian Desert: It lies to the west of the Aravallis. This desert is formed by the work of wind and climatic extremities. The total area of the desert is around 1, 75,000 sq.km. Rajasthan, parts of Gujarat, Punjab and Haryana come under Thar Desert.

The desert proper or central region of the desert is called ‘Marustali’. The atmospheric condition I the desert is extreme. During summer temperature exceeds 50° C and in winter it comes down to 10° C and below. Ganganagar of Rajasthan has recorded more than 54° C of temperature. The rainfall in the desert is very low. Roylee, a place in North Rajasthan, recorded the lowest rainfall in the country (8cm per year). Indian desert comprises mainly of sand dunes. There are a few salt lakes in the desert like Sambhar, Tal, Katu and it has thin vegetation.

Question 7.
Compare the North Indian rivers with South Indian Rivers.
Answer:

Question 8.
Why does River water dispute arise? Mention the important disputes and proposed measures.
Answer:
Water dispute means any dispute or difference between two or more state governments with respect to the use, distribution or control of the waters of, or in, any inter State River or river valley. Water, being the most precious resource is required for domestic, irrigation and industrial purposes. Most of the Indian rivers flow across more than one state. Each of the state of the river tries to obtain the maximum quantity of water. This has resulted in many water disputes in the country.

Some of the important inter state water disputes water disputes in India are:

International water disputes:

In the recent years the rivers flowing across more than two countries are also creating trouble between the neighbouring countries. At present river water dispute has become a global phenomenon. In the developing countries like India the inter-state dispute must be resolved quickly so that water resources could be utilized and harnesses properly for basic need and economic development. One of the measures could be to declare all the major rivers as national property and national schemes under Central assistance should be launched for the development of total command area of the concerned states.

Interlinking of Rivers are played a significant role to solve the water dispute. The distribution of rainfall in India is highly uneven and seasonal. The problems of floods and drought can be minimized through the inter-river linkages or through national water grid, under which water from one river basin can be transferred to another river basin for optimum utilization.

The important inter-linking projects proposed are:

1. The Ganga-Kaveri link canal connecting the basins of the son, Narmada, Tapti, Godavari, Krishna and Pennar.
2. The Brahmaputra- Ganga link canal passing through Bangladesh.
3. The Narmada canal passing through Gujarat and Rajasthan.
4. The link canals between the rivers of the Western Ghats towards the east.
5. The canal from the Chambal to central Rajasthan.

Question 9.
Briefly explain the importance of Inter-linking of Rivers in India.
Answer:
The distribution of rainfall in India is highly uneven and seasonal. The Himalayan rivers are perennial while the peninsular rivers are seasonal. During rainy season, much of the water is lost in floods and wasteful flow into the sea. But in other seasons there is scarcity of water.

Even in India some parts get more rainfall and some other parts get very low rainfall. Consequently there are floods in one region and drought and famine in other regions in the country. The problems of floods and drought can be minimized through the inter-river linkages or through national water grid, under which water from one river basin can be transferred to another river basin for optimum utilization.

The inter-link would consist of two parts, a northern Himalayan River Development component and a southern peninsular river development component. The northern component would consist of series of dams built along the Ganga and Brahmaputra rivers in India, for the purposes of storage, canals would be built to transfer surplus water from the astern tributaries of the Ganga to the west. The Brahmaputra and its tributaries would be linked with the Ganga and the Ganga with the Mahanadi river. This part of the project would provide additional irrigation and generate electricity.

Question 10.
Differentiate between East flowing rivers and West flowing rivers.
Answer:

1st PUC Geography Physiography Ten Mark Questions And Answers

Question 1.
Explain the river system of India with suitable maps. (T.B.Qn)
Answer:
On the basis of origin and flow the river system of India can be broadly divided into two groups. They are.

1. The Himalayan Rivers or North Indian Rivers
2. The Peninsular Rivers or South Indian rivers.

A. The Himalayan Rivers: These rivers take birth in Himalayan Mountains by glaciers and flows throughout the year (perennial). There are three main river systems in the Himalayan rivers. They are the Indus, the Ganga and the Brahmaputra.

1. The Indus river system: The Indus is one of the most important river systems of India. It rises near Mt. Kailash (6714m), has a length of 2880km, of which 709 km lies in India. It flows through narrow gorges between Ladakh and Zaskar ranges in the North West direction in Jammu & Kashmir. It is one of the oldest river systems of the world. Major part of its course and catchment area are in Pakistan. The main tributaries are Jhelum, Chenab, Ravi, Beas and Sutluj.

2. The Ganga: The Ganga is the longest (2500Km) and the largest river system of the country. It is generally called, the ‘National river’ of India. The Ganga has two head streams-the Bhagirathi and the Alakananda. The Bhagirathi takes it is birth in Gangothri and Alakananda rises near Badrinath in Garhwal Himalayas. These two meet at Devaprayag, and continue to flow as the Ganga, after flowing across the Himalayas; the Ganga enters the Great Plains at haridwar. From Haridwar it flows towards south an south east up to Mirzapur.

It continues to flow eastwards in the Gangetic plains of Bihar and West Bengal and enters Bangladesh, where it joins the Brahmaputra and become padma, and finally flows into Bay of Bengal. The important tributaries of Ganga are Ramganga, Ghagra, Gandak, Gomati, Bagmati, Kosi, Yamuna, Chambal, Betwa, sone, ken, damodar etc.

3. The Brahmaputra river system: It rises at Manasarovar lake (chanmyandung). In Tibet it is known as Tsangpo. It enters Aruncal Pradesh and is known as Brahmaputra. It joins Ganga at Golunde (Bangladesh). The total length it flows is 2900km. and only 885km. in India. In Bangladesh it is called Meghana.

B. Peninsular Rivers: The peninsular plateau of India has the largest network of river systems in the country. Most of the south Indian rivers rise in the Western Ghats and central highland regions. On the basis of the direction of flow the rivers are grouped into two types.

1. The East flowing rivers.
2. The west flowing rivers.

B. 1 The East flowing rivers: These rivers rise in Peninsular region, flow in eastern direction and Finally join the Bay of Bengal. The important east flowing rivers are:

The Mahanadi: It rises in sihawa or simhava region of Chattisharh and is the most important river of Odisha and Chattishgarh. The river flows to a length of 885 km and joins the Bay of Bengal ear Cuttack. The main tributaries of Mahanadi are Seonath, hasdeo, Mand and Jonk. The Hirakud, Naraj, Tikarapara dams are built across this river.

The Godavari: it is the longest and largest river of Peninsular India. It rises at Triambakeshwar in Nasik district of Maharashtra. It flows though Maharashtra and Andhra Pradesh to length of 1465km and joins the Bay of Bengal near Kakinada. The main tributaries of Godavari are the Puma, Penganga, Pranhita, Sabri, Indravathi and Manjra.

The Krishna: The Krishna is the second longest and largest east flowing river of peninsular India. It rises near Mahabaleshwar in Maharashtra, flows to a length of 140-0 km before joining the Bay of Bengal near Divi point. The koyna, Yerla, Panchaganga, Dudhganga, Bhima, Ghataprabha, malaprabha, Tungabhadra and the Musi are the main tributaries.

The Kaveri: The kaveri is a sacred river like the Ganga. It rises Talcauvery region in the Brahmagiri range of Coorg district in Karnataka state. If flows for a length of 805 kms before falling into the Bay of Bengal near Kaveripattinam. It drains an area of 87,900 sq.kms. Its main tributaries are Arkavathi, Hemavathi, Harangi, Lokapvani, Shimsa, Lakshmanathirtha, Kabini, Suvarnavathi, Bhavani, and Amaravathi.

B.2 West flowing rivers: These rivers rise in the peninsular region, flow in western direction and join the Arabian Sea. These are short and swift rivers are the luni, sabrmati, mahi, Narmada, Tapi, Mandovi, Zuari, kali, sharavvathi, Gangavati, Bedthi, Netravathi, and Periyar etc.

You can Download Chapter 13 Hydrocarbons Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 13 Hydrocarbons

1st PUC Chemistry Hydrocarbons One Mark Questions and Answers

Question 1.
What are hydrocarbons ?
Answer:
Organic compounds containing only carbon and hydrogen are called hydrocarbons.

Question 2.
What are alkanes ?
Answer:
Alkanes are saturated aliphatic hydrocarbons containing carbon to carbon single bonds.

Question 3.
Give the general formula of alkanes.
Answer:
The general formula of alkanes is C_nH_2n+2

Question 4.
What are saturated hydrocarbons ?
Answer:
The hydrocarbons containing only carbon – carbon single bonds saturated hydrocarbons.

Question 5.
What is the order of reactivity of halogens towards alkanes.
Answer:
The reactivity of halogens towards alkanes varies as Flourine > Chlorine > Bromine > Iodine i.e., F > Cl > Br > I.

Question 6.
Why alkanes are called parafins ?
Answer:
Because of low reactivity.

Question 7.
What is the nature ofC-C bond in alkanes?
Answer:
These are sigma (σ) bonds.

Question 8.
What type of structural isomerism is shown by alkanes?
Answer:
Chain Isomerism.

Question 9.
What is Photochemical reaction ?
Answer:
Reaction carried in the presence of light.

Question 10.
Which class of organic compounds are potential carcinogen?
Answer:
Polynuclear aromatic hydrocarbons.

Question 11.
What is huckel Rule ?
Answer:
It states that a compound is said to be aromatic. If it has (4n +2) p electrons which are de-localised where n = 0,1, 2, 3, ………

Question 12.
Name the scientist who first isolated benzene.
Answer:
Faraday isolated benzene in 1825.

Question 14.
What is decarboxylation ?
Answer:
The process of removing carbon dioxide from sodium salts of acid with the help of soda lime.

Question 15.
What is electrophilic subsitution reaction ?
Answer:
Those reaction in which weaker electrophile is replaced by stronger electrophile are called electrophilic substitution reaction.

Question 16.
Why is addition reaction of bromine to benzene difficult?
Answer:
Due to delocalisation of π electrons. It does not have pure double bond.

Question 17.
What are arenas? Give general formula of Monocyclic arenas.
Answer:
Arenes are aromatic hydrocarbons. General formula of arenas is C_nH_2n-6

Question 18.
What are unsaturated hydrocarbons?
Answer:
The hydrocarbons which contain atleast one carbon to carbon double bond or triple bond are called unsaturated hydrocarbons.

Question 19.
What type of isomerism is shown by alkenes ?
Answer:
Chain isomerism.

Question 20.
What are alkenes ?
Answer:
Alkenes are unsaturated aliphatic hydrocarbons containing one carbon to carbon double bond.

Question 21.
Give the general formula of alkenes.
Answer:
The general formula of alkenes is C_nH_2n.

Question 21.
Why are alkenes and alkenes called unsaturated hydrocarbons?
Answer:
Due to presence of C = C & C ≡ C bonds respectively.

Question 22.
What is the state of hybridisation of carbon atoms in alkenes ?
Answer:
Sp²

Question 23.
A compound decolourise yellow colour of bromine. What does it show ?
Answer:
It is unsaturated in nature.

Question 24.
What is Lindlar’s catalyst ?
Answer:
Palladium supported over calcium carbonate and deactivated with quinoline is called Lindlar’s Catalyst.

Question 25.
Name the hydrocarbon which contains acidic hydrogen.
Answer:
Ethylene (HC ≡ CH).

Question 26.
Give one hydrocarbon which delcolourises alk KMnO₄.
Answer:
Ethene (CH₂ = CH₂)

Question 27.
Name the reagent which brings about dehydrohalogination.
Answer:
Alcoholic potash.

Question 28.
What is the major product formed when propene reacts with hydrogen bromide ?
Answer:
2 – bromopropane.

Question 29.
Name the product formed when HBr reacts with 2 – methyl propene.
Answer:

Question 30.
What are alkynes?
Answer:
Alkynes are unsaturated aliphatic hydrocarbons containing one carbon to carbon triple bond.

Question 31.
Give the general formula of alkynes.
Answer:
The general formula of alkynes is C_nH_2n-2

Question 32.
What is the hybridisation of carbon atoms in acetylene ?
Answer:
sp

Question 33.
What is a vicinal dihalide ?
Answer:
A compound with halogen atoms on adjacent carbon atom.

Question 34. Give the example of vicinal dihalide.
Answer:

Question 35.
Give an example of geminal dihalide.
Answer:
CH₃ – CHCl₂ (1,1 – dichlorethane)

Question 36.
Give an example of alkyne.
Answer:
Acetylene CH ≡ CH

Question 37.
Why are alkenes called olefins ?
Answer:
Because they are oily forming compounds i.e., most of the oils are unsaturated.

Question 38.
Give an addition reaction of alkynes.
Answer:
CH ≡ CH + HBr → CH₂ ≡ CH – Br (Vinyl Bromide)

Question 39.
Give the resonance structures of benzene.
Answer:

Question 40.
Name the electrophiles in the following reactions

1. Chlorination
2. Nitration
3. Sulphonation
4. Friedel Craft’s methylation

Answer:

1. Chlorination – Cl⁺
2. Nitration – \(\mathrm{NO}_{2}^{+}\)
3. Sulphonation – SO₃ .
4. Friedel Crafts methylation – \(\mathrm{CH}_{3}^{+}\)

Question 41.
Write the Geometrical isomers of CHBr = CHBr
Answer:

Question 42.
Which catalyst is most effective even in the polymerization of ethane to polythene ?
Answer:
Ziegler – Nata Catalyst

Question 43.
What is the compostion of Zeigler Natta Catalyst?
Answer:
Triethylaluminium and titanium tetracholride inert solvent.

Question 44.
Name one Carcinogenic compound.
Answer:
Anthracene of Benzanthracene.

Question 45.
What is the number of a and n bonds in N ≡ C-CH = CH- C ≡ N?
Answer:
There are 7 σ bonds and 5 π bonds.

Question 46.
Name the compound formed when heptane is subjected to aromatisation.
Answer:
Toulene.

Question 47.
What is the cause of Geometrical Isomerism ?
Answer:
The restricted rotation around carbon to carbon double bond

Question 48.
What is cracking of hydrocarbon ?
Answer:
Cracking is a process of breaking higher hydrocarbons by heating.

Question 49.
What is the trade name of Benzene Hexacarbon ?
Answer:
Gammaxane.

1st PUC Chemistry Hydrocarbons Two Marks Questions and Answers

Question 1.
What is the action of ethane with bromine in carbon tetra chloride?
Answer:
Ethene reacts with bromine in carbon tetrachloride to give 1, 2 – dibromethane. Orange colour of bromine is discharged.

Question 2.
What is the action of hydrogen halide on alkene?
Answer:
Alkene reacts with hydrogen halide to form alkyl halide. The addition is governed by Markownikoff’s rule.

Question 3.
How do you obtain an alkene from alkyne ?
Answer:
Alkynes react with hydrogen in the presence of (Palladium supported over calcium carbonate and deactivated with quinoline) Lindlar’s catalyst, to give alkene.

Question 4.
How do you convert ethene to ethane?
Answer:
When a mixture of ethene and hydrogen is passed over heated nickel catalyst at 200°C, ethane is obtained.
\(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{H}_{2} \frac{\mathrm{Ni}}{200^{\circ} \mathrm{C}} \mathrm{CH}_{3}-\mathrm{CH}_{3}\)

Question 5.
How is propene obtained from propyl chloride ?
Answer:
When propyl chloride is heated with alcoholic potash propene is obtained.
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Cl}+\mathrm{KOH}_{(\mathrm{alc})} \stackrel{\Delta}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}\).

Question 6.
What is Wurtz reaction ? Give example.
Answer:
When alkyl halides are heated with sodium metal in ether medium higher alkanes are formed. This reaction is known as Wurtz reaction and employed for the synthesis of higher alkanes containing even number of carbon atoms.

Question 7.
How is Methane prepared from sodium acetate ?
Answer:
Decarboxylation : When sodium salt of carboxylic acid is heated with soda lime (mixture of sodium hydroxide and calcium oxide), an alkane containing one carbon atom less than parent carboxylic acid is formed. This reaction is called decarboxylation.

Question 8.
How is alkane prepared from Kolbe’s electrolytic method ?
Answer:
A concentrated sodium salt of a carboxylic acid is electrolysed. The alkane is formed at the anode as one of the products. This process is called Kolb’e electrolysis.

Question 9.
What is Pyrolysis ? Give example.
Answer:
The decomposition of higher alkane into a mixture of lower alkanes, alkenes, etc by the application of heat is called pyrolysis or cracking.

Question 10.
What is Aromatisation ? Give example.
Answer:
Hexane or higher akanes when heated in presence of vanadium pentoxide (V₂O₅), molybdenium oxide (MO₂O₃) or chromium oxide (Cr₂O₃) supported on alumina (Al₂O₃) at 800 K and 10-20 atm pressure give benzene or its alkyl derivatives with the liberation of hydrogen. This dehydrogenation process which involves cyclis^ion of alkanes is known as aromatization.

Question 11.
What is isomerisation ? Give example.
Answer:
When unbrached alkanes are heated with anhydrous aluminium chloride and hydrogen chloride, isomeric branched alkanes are formed. This process is called isomerisation.

Question 12.
What is Geometrical isomerism ?
Answer:
Compounds having same molecular formula and structural formula but differ in the spatial arrangement of atoms or group of atoms due to restricted rotation of carbon carbon double bond.

Question 13.
What are the conditions for Geometrical isomerism ?
Answer:

• The molecule should contain double bond.
• Each carbon atom joined by the double bond, must be attached to two different groups.

Question 14.
What is Ozonalysis ? Give example.
Answer:
The ozonolysis of alkenes involves addition of ozone molecule to carbon-carbon double bond of alkenes to form ozonide. This follows hydrolysis of the ozonoide by H₂O/Zn to yield corresponding aldehydes and / or ketones. The addition of ozone to alkene followed by hydrolysis is known as ozonolysis.

Question 15.
How is ethene prepared ?
Answer:
When 1, 2 – dihaloalkanes (vicinal dihalides) are heated with zinc metal in presence of methanol corresponding alkenes are formed. This reaction is called dehalogenation.
\(\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{Zn} \frac{\mathrm{CH}_{3} \mathrm{OH}}{\text { heat }} \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{ZnBr}_{2}\)

Question 16.
What are the uses of LDPE and HDPE ?
Answer:
LDPE is used for

• making transparent films for packaging garments and food,
• as an insultor.

HDPE is used for

• making crates and large boxes
• production of carry bags, containers, house hold articles and toys.

Question 17.
What happens when Ammonical solution of silver Nitrate treated with Acetylene ?
Answer:
Acetylene when passed through an ammonical solution of silver nitrate (Toallen’s reagent) white solid of silver acetylide is separated.

Question 18.
Give four examples for isomerism in Arenes.
Answer:

Question 19.
Give three examples for fused polycyclic arene
Answer:

Question 20.
State Markownkoff s rule.
Answer: When an unsymmetrical molecule adds to an unsymmetrical alkene, the negative part of the adding molecule adds to the carbon atom involved in the double bond, containing the least number of hydrogen atoms. Thus according to Markonwkoff’s rule the addition takes place as follows :

1st PUC Chemistry Hydrocarbons Four/Five Mark Questions and Answers

Question 1.
Explain the action of chlorine on methane in the presence of sunlight.
Answer:
Methane reacts with chlorine in the presence of diffused sunlight to give methyl chloride, methylene chloride, chloroform and carbon tetrachloride.

Question 2.
Explain the mechanism of chlorination of methane, according to Free Radical.
Answer:
Mechanism of chlorination of methane involves three types.
1. Initiation : Chlorine absorbs energy and undergoes homolysis to give chlorine free radicals.
\(\mathrm{Cl}-\mathrm{Cl} \stackrel{\mathrm{hy}}{\longrightarrow} 2 \mathrm{Cl}^{\bullet}\)

Step 2 : Propagation : Chlorine free radical reacts with methane to give methyl free radical.
\(\mathrm{Cl}^{\bullet}+\mathrm{CH}_{4} \longrightarrow \mathrm{CH}_{3}^{\bullet}+\mathrm{HCl}\)
The methyl free radical reacts with chlorine to form methyl chloride and chlorine free radical.

Step 3 : Termination : Free radials combine to form stable products.

Question 3.
Explain the mechanism of addition of hydrogen bromide to propene.
Answer:
The mechanism of addition of hydrogen bromide to propene takes place in three steps. Step 1: Hydrogen bromide dissociates into H⁺ and Br-
H-Br → H⁺+Br^–

Step 2 : The electrophile H+ attacks propene to form carbocation.

of the two carbocations (I) and (II), carbocations (I) is more stable and is formed more readily.

Step 3 : The nucleophile Br attacks the carbocation (I) to give 2 – bromopropane.

Question 4.
How is the manufacture of LDPE & HDPE ?
Answer:
Low density polythene is manufactured by compressing ethane under very high pressure (1500-2000 atmospheres) at about 475 K in the presence of a small amount of oxygen as catalyst.

High density polythene is obtained by passing ethene under pressure of 6 atmospheres at about 340 K into heptane, an inert solvent containing Zielgler-Natta catalyst (triethyl) aluminium and titanium tetrachloride)

Question 5.
Give four uses of aceylene.
Answer:

1. Oxyacetylene flame is used for cutting and welding of metals.
2. Acetylene and its derivatives are widely used in synthetic organic chemistry for synthesis of cis and trans-alkenes, methyl ketones, etc.
3. Acetylene is used as illuminant in hawker’s lamp and in light houses.
4. Acetylene is used for ripening of fruits and vegetables.
5. Acetylene is used for manufacture of ethyl alcohol, acetaldehyde, acetic acid, vinyl plastics, synthetic rubbers such as Buna-N and synthetic fibres such as Orlan.

Question 6.
Give 4 uses of ethene.
Answer:

1. Lower members of the family are used as fuels and illuminants.
2. Alkenes and substituted alkenes upon polymerization form a number of useful polymers such as polythene, PVC, Teflon, orlon, etc.,
3. Ethene is employed for preparation of ethyl alcohol and ethylene glycol (anti-freeze).
4. Ethylene is used for artificial ripening of green fruits.
5. Ethylene also used in oxygen-ethylene flame for cutting and welding of metals.

Question 7.
Give four uses of methane.
Answer:

1. Methane in the form of natural gas (CNG) is used for running scooters, cars, buses, etc., LPG (Mixture of butane and isobutene) is used as a fuel in homes as well as in industry.
2. Methane is used to make carbon black which is used in the manufacture of printing inks, paints and automobile tyres.
3. Cataltic oxidation of alkanes gives alcohols, aldehydes an carboxylic acids.
4. Higher alkanes in form of gasoline, kerosene oil, diesel, lubricating oils and paraffin wax are widely used.
5. Methane is used for manufacture halogen containing compounds such as CH₂Cl₂ , CHCl₃ , CCl₄ etc. which are used as solvents both in laboratory and industry.

Question 8.
Give the structural elucidation of benzene.
Answer:
1. From elemental analysis and molecular mass determination, the molecular formula of benzene is found to be C₆H₆.

2. On the basis of molecular formula benzene is highly unsaturated compound.

3. Benzene does not show alkene properties. Benzene is quite stable. It does not decolourise cold aqueous solution of potassium paramanganate.

4. Benzene shows cyclic structure,
(a) Benzene contains three double bonds. Benzene adds three moles of hydrogen in presence of nickel catalyst to form cyclohexane.

\(\mathrm{C}_{6} \mathrm{H}_{6}+3 \mathrm{H}_{2} \frac{180^{\circ} \mathrm{C}}{\mathrm{Ni}} \mathrm{C}_{6} \mathrm{H}_{12}\)

(b) All the six hydrogen atoms in benzene are identical – benzene reacts with bromine in presence of FeBr3 catalyst to form monobromobenzene.

\(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \stackrel{\mathrm{FeBr}_{3}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr}\)

5. Kekule proposed benzene is a planar ring structure with alternate single and double bonds.

6. To overcome this drawback, Kekule suggested that benzene is a mixture of two forms, which are in rapid oscillation.

7. It is found that the bond length in benzene is same for all C-C bonds (0.139 nm). This lies between C-C single bond (0.154 nm) and C-C double bond length (0.134 nm).

8. Resonance structure: Resonance hybrid is more stable than structure 1 and 2. The stability of benzene due to resonance is so great that n bonds of the molecule will normally resist breaking. This explain the stability of benzene.

Question 9.
Explain the mechanism of halogenation or chlorination of benzene.
Answer:
Halogenation: Benzene reacts with chlorine in the presence of FeCl₃ or AlCl₃ to form chlorobenzene.

Machanism : This involves the following steps.

Step 1: Generation of eiectrohile Cl – Cl + FeCl₃ → Cl⁺ + \(\mathrm{FeCl}_{4}^{-}\)
Step 2 : The electrophile Cl⁺ attacks benzene ring to form a carbon cation which is resonance stabilised.

Step 3 : Loss of a proton to give chlorobenzene. The proton is removed by FeCU.

Question 10.
Explain the mechanism of nitration of benzene.
Answer:
Nitration benzene reacts with a mixture of concentrated nitric acid and concentrated sulphuric acid at 50°C to form nitrobenzene.

Mechanism : This involves the following steps.
Step 1: Generation of electrophile nitronium ion \(\mathrm{NO}_{2}^{+}\)
\(\mathrm{HNO}_{3}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NO}_{2}^{+}+\mathrm{H}_{3} \mathrm{O}^{+}+2 \mathrm{HSO}_{4}^{-}\)

Step 2 : The electrophile NO₂ attacks the benzene ring to form a carbocation which is resonance stabilized.

Step 3 : Loss of a proton to give nitrobenzene. The proton is removed by \(\mathrm{HSO}_{4}^{-}\)

Question 11.
Explain the mechanism of sulphonation of benzene?
Answer:
Sulphonation : Benzene reacts with concentrated sulphuric acid at 80°C to form benzene sulphoric acid.

Mechanism : This involves the following steps.

Step 1: Generation of electrophile SO₃.
\(2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons \mathrm{SO}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}\)

Step 2 : The electrophile SO₃ attacks the benzene ring to form a carbocation

Step 3: Loss of a proton to from a sulphonate ion. The proton is removed by \(\mathrm{HSO}_{4}^{-}\)

Step 4 : Addition of proton to give benzene sulphonic acid.

Question 12.
Explain the mechanism of Friedel craft alkylation of benzene.
Answer:
Friedel – Craft alkylation reaction Benzene reacts with methyl chloride in presence of anhydrous aluminium chloride to form toluene. This is called Friedel-Crafts alkylation reaction.

The mechanism of Friedel Craft’s alkylation reaction involves the following steps: