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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)

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 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 14 Environmental Chemistry 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 14 Environmental Chemistry

1st PUC Chemistry Environmental Chemistry One Mark Questions and Answers

Question 1.
Name a source of energy which does not create pollution.
Answer:
Sun is a source of energy which does not create pollution.

Question 2.
Which acid is not present acid rain?
Answer:
Acetic acid is not present in acid rain.

Question 3.
Name two insecticides.
Answer:
DDT and BHC are insecticides.

Question 4.
Define the term pollutant.
Answer:
Pollutant is any substance which cause adverse effect on environment.

Question 5.
Why is CO more toxic than CO₂?
Answer:
Carbon monoxide reacts with haemoglobin to form carboxy-haemoglobin which does not act as oxygen carrier. Therefore, it is more harmful than CO₂.

Question 6.
Name two gases which are responsible for green house effect.
Answer:
Carbon dioxide and methane.

Question 7.
Which part of atmosphere contains ozone layer ?
Answer:
Ozone layer is present in stratosphere.

Question 8.
What is full form of BOD and COD ?
Answer:
BOD stands for Biochemical oxygen demand whereas COD stands for chemical oxygen demand.

Question 9.
What is PAN ?
Answer:
PAN is Peroxycetyl nitrate

Question 10.
What is the percentage of CO₂ in the pure dry air?
Answer:
About 0.03%

Question 11.
What is the nature of London Smog ?
Answer:
It is reducing in nature.

Question 12.
Which has caused Bhopal gas tragedy ? Give its formula.
Answer:
Methyl isocyanate gas caused Bhopal gas tragedy. Its formula is CH₃ – N = C = O

Question 13.
What are pesticides ?
Answer:
Pesticides are those chemicals which are used to destroy pests, rats and parasite fungi.

Question 14.
What is meant by deoxygenation ?
Answer:
The process of removing dissolved oxygen from water by micro-organisms in order to oxidise organic matter of sewage is called deoxygenation.

Question 15.
Define incineration.
Answer:
It is the process of converting Organic materials to CO₂ and H₂O.

Question 16.
Define green chemistry.
Answer:
It involves processes and products that reduce or eliminate the use and generation of hazardous (harmful) substances.

Question 17.
What is meant y photo-chemical smog ?
Answer:
It is highly oxidising smog having high concentration of oxidizing agent like oxides of nitrogen which absorb sunlight and form free redicals.

Question 18.
What is marine pollution ?
Answer:
Pollution of sea water due to discharge of wastes into it is called marine pollution.

Question 19.
Which main compounds are causing damage to ozone layer ?
Answer:
NO and freons.

Question 20.
Which disease ’is caused due to hole in the ozone layer and why ?
Answer:
Ultra violet rays will reach the earth after passing through the hole and cause skin cancer.

Question 21.
Name two acids present in the acid rain.
Answer:
H₂SO₄ , HNO₃ and HCl.

Question 22.
What is effect of excess of \(\mathrm{SO}_{4}^{2-}\) ion in drinking water ?
Answer:
Excess of \(\mathrm{SO}_{4}^{2-}\) in drinking water (500 ppm) may cause a laxative effect.

Question 23.
Name the different regions of the atmosphere along with their attitudes and temperature ranges.
Answer:
Troposphere (0-11 km), Stratosphere (11 – 50), Mesosphere (50 – 85 km), Thermosphere (85 – 500 km).

Question 24.
What are fungicides ?
Answer:
Those chemicals which check the growth of fungi are called fungicides.

Question 25.
Give one example of organic herbicide.
Answer:
Triazine is an example of organic herbicide.

Question 26.
Give one main reason of ozone depletion.
Answer:
Chlorofluorocarbon causes depletion of ozone layer.

Question 27.
Define primary pollutant of the air.
Answer:
Those pollutants which are directly emitted from the sources and pollute the air are called primary pollutants of the air.

Question 28.
Write the name of gas produced in Mathura refineries which can damage the great historical monument “Taj Mahal”.
Answer:
Sulphur dioxide.

Question 29.
What type of radiations are abosorbed by CO₂ in the atmosphere ?
Answer:
Infrared radiations.

Question 30.
In which season and what time of the day, there is photochemical smog ?
Answer:
In summer, in the afternoon.

1st PUC Chemistry Environmental Chemistry Two Marks Questions and Answers

Question 1.
Define environmental chemistry.
Answer:
Environment chemistry deals with study of the origin, transport, reactions, effects and fates of chemical species in the environment.

Question 2.
Explain tropospheric pollution in 100 words.
Answer:
Trophospheric Pollution: It is due to presence of undesirable solid or gaseous particles in air.
1. Gaseous Air Pollutants: These are oxides of sulphur, nitrogen and carbon, hydrogen sulphide, hydrocarbons, ozone and other oxidants.
2. Particulate Pollutants: These are dust, mist, fumes, smoke and smog.

Question 3.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
Carbon monoxide reacts with hemoglobin to form caboxy-hemoglogin which is 300 times more stable than oxy-hemoglobin complex. In blood, when concentration of carboxy-hemoglobin is about 3-4%, the oxygen carrying capacity of blood is greatly reduced. This oxygen deficiency, results into headache. Weak eye sight nervousness and cardiovascular disorder.

Question 4.
List gases which are responsible for green house effect.
Answer:
Carbon dioxide, methane, ozone, chlorofluorocarbons and water vapours in atmosphere are responsible for greenhouse effect.

Question 5.
Statues and monuments in India are affected by acid rain. How?
Answer:
Acid rain affects statues and monuments.
H₂O + CO₂ →H₂CO₃
\(\mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{CO}_{3}^{2-}\)
When pH of the rain drops below 5.6, it is called acid rain.
2SO₂ + O₂ + 2H₂O → 2H₂SO₄
3NO₂ +O₂ +2H₂O → 2HNO₃
CaCO₃ + H2SO₄ → CaSO₄ +H₂O+CO₂
(marble)
CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂

Question 6.
What is smog ? How is classical smog different from photochemical smog ?
Answer:
Smog, When smoke mixes with fog, it is called smog. It is of two types :
1. Classical smog: It occurs in cool humid climate and is the result of build-up of sulphur dioxides and particulate matter from fuel combustion. It is reducing in nature.

2. Photochemical smog : It occurs in warm, dry and sunny climate and results from the action of sunlight on the nitrogen oxides and hydrocarbons produced by automobiles and factories. It is oxidizing in nature.

Question 7.
Write down the reactions involved during the formation of photochemical smog.
Answer:
Mechanism of formation of Photochemical Smog:
\(\mathrm{NO}_{2}(\mathrm{g}) \stackrel{\mathrm{hv}}{\longrightarrow} \mathrm{NO}(\mathrm{g})+\mathrm{O}(\mathrm{g})\)
O(g) + O(g) → O₃(g)

Question 8.
What are the harmful effects of photochemical smog and how can they be controlled ?
Answer:
Effects :

• Smog causes respiratory problems like asthma attacks bronchitis, heart related disorder.
• It also causes irritation to eyes, throat and nose.
• It reduces visibility and affects road as well as traffic.
• It damages plants and other materials like electronic and electrical equipment.

Control :

• Efficient catalytic converters in the automobiles will reduce smog formation as it prevents the release of nitrogen oxides and hydrocarbons to the atmosphere.
• It can also be suppressed by certain compounds which acts as fuel radical traps. When these compounds are sprayed in the atmosphere, they generated free radicals which readily combine with free radicals responsible for formation of photochemical smog.

Question 9.
What do you mean by ozone hole ? What are its consequences ?
Answer:
Ozone hole : There is depletion of ozone over Antarctica region commonly known as a ozone hole. Its consequences are that UV light can reach the earth which can lead to ageing of skin, cataract, sunburn, skin cancer, killing of many phytoplankton, damage to fish productivity etc. It has also been observered that plant protein get easily affected by UV radiation which leads to the harmful mutation of cells.

Question 10.
What are the major causes of water pollution ? Explain.
Answer:
Causes of water pollution
1. Pathogens : The disease causing agents are called pathogens like bacteria that enter water from domestic sewage and animal excreta.

2. Organic wastes : Leaves, grass, trash etc., pollute water. Excessive growth of photoplankton within water is also source of water pollution.

Question 11.
Have you ever observed any water pollution in your area ? What measures would you suggest to control it?
Answer:
Yes, there is water pollution. Sewage water should not be mixed with drinking water. Water purifier should be used before drinking water.

Question 12.
What do you mean by Biochemical Oxygen Demand (BOD)?
Answer:
It is defined as the amount of oxygen required by bacteria to break down organic materials present in water.

Question 13.
Do you observe any soil pollution in your neighbourhood ? What efforts will you make for controlling the soil pollution ?
Answer:
Yes, there is water pollution. We can prevents soil pollution by using less pesticides, insecticides, fungicides, weedicides DDT which has been banned should not be used. Organo-phosphates and carbamates should be used instead of Aldrin and Dieldrin. Sodium chlorate and sodium arsenite are also not environment friendly. Fields should be deweeded manually.

Question 14.
What are pesticides and herbicides ? Explain giving examples.
Answer:

• Pesticides: Those which are used to kill pests, insects which are harmful for the crops, e.g., Aldrin, Dieldrin B.H.C. etc.,
• Herbicides: Those chemicals which are used to destroy weeds i.e., unwanted plants are called herbicides, e.g., sodium chlorate, sodium arsenite etc.

Question 15.
What do you mean by green chemistry? How will it help in decreasing environmental pollution?
Answer:
Green Chemistry: It is the way of thinking and is about utilizing the existing knowledge and principles of chemistry and other services to reduce the adverse effect of pollutants.
1. Avoiding use of organic solvents such as benzene, toluene, CCl₄ etc., which are highly toxic, we can save our environment.

2. Use of CO₂ instead of CFC’s for thermocol can reduce pollution or depletion of ozone layer.

Question 16.
What would have happened if the greenhouse gases were totally missing in the earth’s atmosphere ? Discuss.
Answer:
If greenhouse gases are totally missing from earth’s atmosphere then temperature of earth will decrease. Plants cannot carry out photosynthesis if CO₂ is not present. Human beings cannot survive without plants.

Question 17.
Name the oxides of nitrogen. What are the sources ?
Answer:
Nitrogen monoxide (NO), Nitrogen dioxide (NO₂), Dinitrogen trioxide (N₂O₃),
Dinitrogen tetroxide (N₂O₄), Dinitrogen pentoxide (N₂O₅)
The source of oxides of nitrogen are combustion of fossil fuel, especially, petroleum.
They are also formed by reaction of N₂ and O₂ in presence of lightening.

Question 18.
Discuss the harmful effects of oxides of sulphur.
Answer:
Oxides of sulphur are SO₂ and SO₃. They cause acid rain which damages plants, animals and buildings. SO₂ irritates the respiratory system of animals and humans. They are harmful for lungs. They are harmful for plants. They increase rate of mortality.

Question 19.
What is the §ffect of dust particles present in atmosphere and why ?
Answer:
Dust particles suspended in the atmosphere effectively reduces the amount of light rays reaching the surface of earth and thus lower the temperature of the earth. They contribute to increased fog and rain in cities. They may reduce visibility and produce blurring effect on vision.

Question 20.
Discuss the pollution caused by thermal power plants.
Answer:
The gases like CO₂, CO, SO₂, SO₃ and oxides of nitrogen are formed due to combustion of fossil fuels in thermal power plants. Fly ash is also formed by combustion of high ash fossil fuel in thermal power plants.

Question 21.
How does detergent cause water pollution?
Answer:

• Detergent are non-biodegradable and they cause water pollution.
• They inhibit oxidation of organic substances present in waste water because they form a sort of envelope around them.
• They form stable foam in rivers which extend over several hundred meters of the river water.

Question 22.
How is greenhouse effect responsible for global warming ?
Answer:
Greenhouse effect is the phenomenon in which earth’s atmosphere traps the heat from the sun and prevents it from escaping into outer space. Carbon dioxide, methane, ozone, chlorofluorocarbons and water vapours absorb infrared radiations and increase the temperature of earth which is called global warming.

1st PUC Chemistry Environmental Chemistry Four Marks Questions and Answers

Question 1.
What is environment chemistry ? Discuss is social relevance.
Answer:
Environment chemistry deals with study of the origin, transport, reactions, effects and fates of chemical species in the environment.
Mankind is faced with several types of pollution such as air pollution, water pollution, soil pollution, thermal pollution, noise pollution, metals pollution etc.
It is most essential to control various types of pollution so as to save mankind and living organisms.

Question 2.
Define an environmental pollutant. What do you understand by an environmental pollution model?
Answer:
The substances which cause adverse effect on the environment are called environmental pollutants. In an environmental pollution process, a pollutant originates from a source and gets transported by air or water or is dumped on a land by man. Some of the pollutants may be absorbed (assimilated) or chemically changed by the environment; the rest build-up to concentrations which are harmful to environment.

Question 3.
State briefly the reactions causing ozone layer depletion in the stratosphere.
Answer:
The decomposition products of CFCs destroy ozone as it is shown in the following reaction:
CF₂Cl₂ (g) + hv → Ci(g) + \(\dot{\mathrm{C}}\)F₂Cl(g)
\(\dot{\mathrm{C}}\)l(g) + O₃(g) → Cl\(\dot{\mathrm{o}}\)(g) + O₂(g)
Cl\(\dot{\mathrm{o}}\)(g) + O(g) → Cl + O₂(g)

Question 4.
Write short note on Green Chemistry.
Answer:
Green Chemistry : It involves designing and development of green chemical products and processes which do not create pollution.

• Use of CNG has reduced air pollution in Delhi.
• Development of a new method to produce ibuprofen in 99% yield avoiding the usage of large quantity of solvents and wastes associated with the traditional methods.
• Using CO₂ as blowing agent for manufacture of polystyrene foam sheet packaging material has eliminated the use of CFCs which cause ozone depletion.
• Designing of a safer marine antifouling compound ‘Sea-nine’ that degrades far more rapidly than organizations which persists in the marine environment and cause pollution problems.

Question 5.
How is photochemical smog formed ? What are its effects ? How can it be controlled?
Answer:
Photo Chemical smog : Nitrogen monoxide is formed by reaction of N₂ and O₂ at high temperature in petrol and diesel engines of cars and trucks. NO is oxidized into the air to form NO₂ which absorbs energy from sunlight and breaks up into NO and free oxygen atom.
\(\mathrm{NO}_{2}(\mathrm{g}) \stackrel{\mathrm{hv}}{\longrightarrow} \mathrm{NO}(\mathrm{g})+\mathrm{O}(\mathrm{g})\)
Oxygen atoms are’very reactive and can combine with O₂ to form O₃
O(g) + O₂ (g) → O₃ (g).
Ozone reacts with NO to form NO₂ and O₂
NO + O₃ → NO₂ + O₂
The brownish haze of photochemical smog is largely due to brown colour of NO₂.
Effects : Pungent smelling, smog produced ozone is known to be toxic. It can cause coughing, wheezing, bronchitis and irritation to mucous system.

Control of photochemical smog :

• Installation of efficient catalytic converters in the automobiles is one of the ways of reducing smog formation.
• It can be suppressed by certain compounds, which acts as free radical traps. When these compounds are sprayed in the atmosphere they generate free radicals which readily combine with free radical precursors of photochemical smog.

Question 6.
What is the cause of acid rain ? How is it harmful to the environment ?
Answer:
When rain falls through polluted air, it comes across chemicals such as gaseous oxides of sulphur, oxides of nitrogen, mists, of hydrochloric acid and phospheric acid etc., pH lowers down from 5.6 to 3.5 sometimes it becomes as low as 2.

Question 7.
Distinguish between photochemical smog and classical smog.
Answer:

Photochemical smog	Classical smog
1. it is formed by oxides of nitrogen, hydrocarbons,etc.,	1. It is formed by oxides of sulphur carbon and particulate matter from combustion
2. It is oxidizing in nature	2. It is reducing in nature
3. Photochemical smog occurs in cities having large numbers of vehicles. It is not harmful	3. It occurs in both urban and rural areas. It is harmful

Question 8.
Describe the following in brief:
1. Ozone depletion over Antarctica (do not write reaction)
2. BOD and COD
3. Eutrophication
Answer:
1. Ozone depletion over Antarctica: In summer season, nitrogen dioxide and methane react with chlorine monoxide preventing ozone depletion whereas in winter, special type of clouds called polar stratosphereic coulds are formed over Antartica. These clouds provide surface on which chlorine nitrate molecules formed gets hydrolysed to form HOCl. It also reacts with HCls to give Cl₂. When sunlight returns to Antarctica again ozone depletion starts by free radicals.

2. BOD: It is amount of oxygen required by bacteria to decompose organize wastes present in water.
COD: It is the amount of oxygen (in ppm) required to oxidize the contaminants. COD is determined by using chemical oxidizing agent K₂Cr₂O₇.

3. Eutrophication: The process in which nutrient enriched water bodies support a dense plant polulation which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.

1st PUC Chemistry Environmental Chemistry Five Marks Questions and Answers

Question 1.
Explain International standards for drinking water.
Answer:
The quality of water is of vital concern for mankind because it is directly linked with human welfare. The drinking water should be fit for human consumption having the following essential parameters for water quality:

1. It should be colourless and odourless.
2. It should be pleasant in taste.
3. It should be clear and turbidity should be less 10 ppm.
4. Its pH should be between 5.5 to 9.5.
5. The total dissolved solids should not be more than 500 ppm.
6. It should be free from disease causing microorganisms.

Question 2.
Explain causes of soil pollution.
Answer:
The main sources of soil pollution are:

1. Improper disposal of human and animal excreta, solid and liquid wastes.
2. Domestic refuse and industrial wastes dumped on the land.
3. Wastes from mining of coal and other minerals dumped on the land.
4. Fertilisers and pesticides used in agriculture.
5. Radioactive refuge from laboratories, industries and hospitals.
6. Removal of the upper fertile layer of soil.

You can Download Chapter 10 Ondu Hoo Hechige Idutini 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 10 Ondu Hoo Hechige Idutini

Ondu Hoo Hechige Idutini Questions and Answers, Notes, Summary

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Karnataka 1st PUC Chemistry Question Bank Chapter 1 Some Basic Concepts of Chemistry

1st PUC Chemistry Some Basic Concepts of Chemistry One Mark Questions and Answers

Question 1.
What is the unit for equivalent weight ?
Answer:
Equivalent weight is a mere number, it has no units. When expressed in grams it is called gram equivalent weight.

Question 2.
Define Atomic weight.
Answer:
The atomic weight of an element is the ratio of average mass of an atom of an element to \(\frac{1}{12}\) of mass of an atom of C-12 Isotope.

Question 3.
Write the mathematical form of Atomic weights.
Answer:

Question 4.
Mention the unit for equivalent weight.
Answer:
There is no unit for atomic weight, when expressed in grams it is called gram atomic weight.

Question 5.
Write the Relationship between Atomic weight and Valency weight.
Answer:

Question 6.
Define Valency.
Answer:
Valency : The combining capacity of an atom (or the number of electron present in the outer most orbital of an atom] which involves in bond formation.

Question 7.
Mention the methods to determine Equivalent weight.
Answer:

• Oxide Method
• Chloride Method
• Hydrogen Displacement method
• Inter conversion method.

Question 8.
Define Molecular weight.
Answer:
It is defined as “ The ratio of mass of 1 molecule of the substance to \(\frac{1}{12}\) mass of an atom of C-12 isotope”.

Question 9.
Write mathematical forms of Molecular weight.
Answer:

Question 10.
Define Avagadro’s Law.
Answer:
Equal volume of all gases and vapours under the same condition of temperature and pressure contains equal number of molecules.

Question 11.
Define Mole.
Answer:
1 mole represent the amount of substance which contain Avagadro number of particles.

Question 12.
What you mean by Avagadro Number ?
Answer:
The number of molecules present in 1 gm molecular weight of any substance.
OR
The number of atoms present in 1 gm atomic weght of any element. It is represented by the symbol N – 6.022 × 10²³

Question 13.
Define Molality.
Answer:
The number of moles of solute dissolved in 1 kg of the solution.

Question 14.
Define Molarity.
Answer:
Number of molecular equivalent of solute dissolved in 1 dm³ of the solution.

Question 15.
What is mole fraction?
Answer:
The ratio of number of moles of a component to the number of moles present in the solution.

Question 16.
What do you mean by Normal solution?
Answer:
A solution containing 1 gm equivalent of the substance dissolved in 1 dm³ of the solution.

Question 17.
What you mean by decinormal solution?
Answer:
A decinormal solution is the solution containing 0.1 g equivalent (\(\frac { 1 }{ 10 }\) ^th) of solute / substance dissolved in 1 dm³ of solution.

Question 18.
What are indicators ?
Answer:
Indicators are the chemical ubstance used to indicate the end point of a volumetric reaction by the change in colour.

Question 19.
Mention the types of Indicators.
Answer:

1. Internal indicator
2. External indicator
3. Self indicator

Question 20.
Define empirical formula.
Answer:
It is a formula which gives the relative number of atoms of each element in the compound.

Question 21.
Define Molecular formula.
Answer:
It is a formula which gives the actual number of atoms of each element present in the molecule of the given compound.

Question 22.
What is the relation between empirical and molecular formula ?
Answer:

Question 23.
State the law of definite proportion.
Answer:
The law states that “a chemical compound, however prepared, contains the same elements combined together in the same fixed proportion by mass”.

Question 24.
Define law of conservation of mass.
Answer:
It states that matter can neither be, created nor be destroyed.

Question 25.
State Gay Lussac’s law of combiningh volumes.
Answer:
The law sttes that ’when gaseous reactants combine to give gaseous products, they do so in volumes which bear a simple whole number ratio to one another and to the volume of producs, provided the volumes are measured under the similar conditions of temperature and pressure”.

Question 26.
What is NTP ?
Answer:
(Normal temperature and pressure or standard temperature and pressure). The temp 273K (0 °C) and pressure 101.3Kpa (or N/m²) or latm = 760 mm of Hg level are taken as STP or NTP.

Question 27.
What physical quantities are represented by the following units and what are their most common names?
(i) kg ms^-2
(ii) kg m²s^-2
(iii) dm³
Answer:
(i) Force (newton)
(ii) Work (joule)
(iii) Volume

Question 28.
Rewrite the following after required corrections : (i) The length of a rod is 10 cms, (ii) the work done by a system is 10 Joules.
Answer:
(i) The length of a rod is 10 cm (s is not used) (ii) The work done by a system is 10 joules (small letter is used in place of capital letter).

Question 29.
Classify the following substances into elements, compounds and mixtures:
(i) Milk
(ii) 22-carat gold
(iii) Iodized table salt
(iv) Diamond
(v) Smoke
(vi) Steel
(vii) Brass
(viii) Dry ice
(ix) Mercury
(x) Air
(xi) Aerated drinks
(xii) Glucose
(xiii) Petrol/Diesel/Kerosene oil
(xiv) Steam
(xv) Cloud.
Answer:
Element – (iv), (ix); Compounds – (viii), (xii), (xiv), (xv); Mixtures – (i), (ii), (iii), (v), (vi), (vii), (x), (xi), (xiii).

Question 30.
Why is air sometimes considered as a heterogeneous mixture?
Answer:
This is due to the presence of dust particles which change the phase, into
heterogeneous.

Question 31.
1 L of a gas at STP weighs 1.97 g. What is the vapour density of the gas?
Answer:
22.4 L of the gas at STP will weigh 1.97 × 22.4 = 44.1 g ie., molecular mass = 44.1.

Question 32.
Why atomic masses are the average values ?
Answer:
Most of the elements exist in different isotopes ie., atoms with different masses, eg. Cl has two isotopes with mass numbers 35 and 37 existing in the ratio 3:1. Hence, average value is taken.

Question 33.
Determine the equivalent weight of each of the following compounds assuming the formula weights of these compounds are x, y and z respectively, (i) Na₂SO₄ (ii) Na₃P0₄.12H₂O (iii) Ca₃(PO₄)₂
Answer:

Question 34.
What is the mass of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g cm^-3?
Answer:
Mass = Volumexdensity = 1000 cm³ × 13.6 g cm^-3 =13,600 g = 13.6 kg

Question 35.
Vitamin C is known to contain 1.29 × 10²⁴ hydrogen atoms. Calculate number of moles of hydrogen atoms.
Answer:

Question 36.
Define an organic compound. Give one example.
Answer:
Organic compound contains essentially carbon, hydrogen along certain elements like N, P, S, O, halogen, etc., eg: urea, sugar.

Question 37.
Define an inorganic compound. Give one example.
Answer:
Inorganic compound is made up of two or more elements containing rare percentage of carbon eg: NaCl, KC1, ZnS, etc.

Question 38.
Name the process used for refining of petroleum.
Answer:
Fractional distillation.

Question 39.
Calculate the percentage of nitrogen in NH₄. (Atomic mass of N = 14, H = 1 amu)
Answer:

Question 40.
State the law of multiple proportion.
Answer:
It states ‘whenever two elements combine to form two or more compounds, the ratio between different weights of one of the elements which combines with fixed weight of other is always simple’.

Question 41.
Calculate the number of He atoms in (i) 52 u, (ii) 52 g, (iii) 52 moles of He. Atomic weight of He is 4 u.
Answer:
(i) Atomic weight of Helium = 4, i.e. 4 u is mass of 1 atom,
52 u is mass of \(\frac { 1 }{ 4 }\) × 52 = 13 atoms.

(ii) 4 g of He contains 6.022 × 10²³ He atoms / 52 g of He contains ?
\(\frac{6.022 \times 10^{23} \times 52}{4}\) = 7.8286 × 10²³ atoms.

(iii) 1 mole of He contains 6.022 × 10²³ atoms, 52 moles of He contains 52 × 6.022 × 10²³ =3.131 × 10²³ atoms.

Question 42.
How many electrons are present in 16 g of CH₄ ?
Answer:
1 molecule of CH₄ =6 + 4 = 10 electrons, 16 g of CH₄ contains 10 × 6.022 × 10²³ = 6.022 × 10²⁴

Question 43.
Boron occurs in nature in the form of two isotopes and \(\begin{array}{l}{11} \\ {5}\end{array}\) B and \(\begin{array}{l}{10} \\ {5}\end{array}\) in ratio of 81% and 19% respectively. Calculate its average atomic mass.
Answer:

Question 44.
If 2 litres of N₂ is mixed with 2 litres of H₂ at a constant temperature and pressure. What will be the volume of NH₃ formed?
Answer:
N₂(g) + 3H₂(g) → 2NH₃(g)

Question 45.
How many atoms are present in 4 ml of NH3 at STP?
Answer:
22400 ml of NH₃ contains = 4 × 6.022 × 10²³ atoms [∵ NH₃ contains 4 atoms]. 1ml of NH₃ contains = \(\frac{4 \times 6.022 \times 10^{23}}{22400}\) = 1.07 × 10²⁰ atoms.

Question 46.
Which of these weighs most? (i) 32 g of oxygen, (ii) 2 g atom of hydrogen, (iii) 0.5 mole of Fe, (iv) 3.01 × 10²³ atoms of carbon.
Answer:
(i) 32 g of oxygen weighs most.
(ii) 2 g atom of H₂ = 2g
(iii) 0.5 mole of Fe = mole × atomic weight = 0.5 × 56 = 28 g
(iv) 6.022 × 10²³ = 1 atom of d 3.01 × 10²³ atoms of C = \(\frac { 1 }{ 2 }\) × 12 = 6 g.

Question 47.
An element has a specific heat of 0.113 cal/g °C. Calculate atomic weight of element.
Answer:
According to Dulong Petitt’s law, Atomic weight × Specific heat = 6.4 cal
Atomic weight × 0.113 cal/g = 6.4 cal
Atomic weight = \(\frac { 6.4 }{ 0.113 }\) = 56

Question 48.
Why are the atomic masses of most of the elements fractional?
Answer:
This is because atomic masses are the relative masses of atoms as compared with an atom of C-12 isotope taken as 12.

Question 49.
Match the following:

Answer:
A-(ii)
B-(iii)
C-(iv)
D-(i)

Question 50.
Match the following prefixes with their multiplies.

Answer:
(i) micro 10^-6
(ii) deca – 10
(ill) mega – 10⁶
(iv) giga – 10⁹
(v) femto – 10^-15

Question 51.
Calculate the molar mass of the following:
(i) H₂O (ii) CO₂ (iii) CH₄
Answer:
(i) H₂O = 2 × H+ 1 × O
2 × 1 + 1 × 16 = 18
(ii) CO₂ = 12 × 1 + 2 × 16 = 44
(iii) CH4 = 1 × C + 4 × H
1 × 12 + 4 × 1 =16

1st PUC Chemistry Some Basic Concepts of Chemistry Two Marks Questions and Answers

Question 1.
Define basicity of an acid with example.
Answer:
Basicity of an acid is the number of replaceable hydrogen atoms present on a molecule of an acid is called basicity of an acid.

Question 2.
Define acidity of a base with example.
Answer:
Acidity of a base is the number of a mono basic acid required to neutralize 1 molecule of a base. [The base which contain number of replaceable – OH group].

Question 3.
Define equivalent weight of acid with example.
Answer:
The number of parts by mass of the acid which contains 1 part by mass of replaceable hydrogen.

Question 4.
Calculate equivalent weight of following acid, (a) H₂C₂O₄.H₂O (b) H₂O₂O₄
Answer:

Question 5.
Define equivalent weight of base with an example.
Answer:
NaOH + HCl → NaCl + H₂O
1 equivalent of HC1 = 36.45, here 1 equivalent of HCl reacts with NaOH.

Question 6.
Write the acidity and calculate equivalent weight of following base, a) Ca(OH)₂ b) Fe(OH)₃
Answer:

Question 7.
Write the mathematical form for mole fraction.
Answer:
Mathematically;
If n₁ is the number of moles of solvent; If n₂ is the number of moles of solute

Question 8.
What is ppm (parts per million)?
Answer:
It is the mass of solute present in 1 million(10⁴) of solution.

10^-3 = 1 mg of solute /1 dm³ of solution.
∴ 1 ppm = 1 mg of solute /1 dm³ of solution.
n ppm = n mg solute / 1 dm³ of solution.

Question 9.
Define Normality.
Answer:
Number of gram equivalent of the substance dissolved in 1 dm³ of solution.

Question 10.
Calculate molarity and normality of the following solution, a) HCl b) H₂C₂O₄
Answer:

Question 11.
Write a note on internal indicator.
Answer:
Internal indicator are the substance which are added in the titration flask at the beginning. Example: Methyl orange, starch, phenolphthalein, diphenylamine.
Internal indicator are further divided according to the type of reaction taking place during titration.

(a) Acid base indicator: The choice of indicator for acid base titration depends on the nature of the acid and base used in that reaction.
Acid Base Indicator
Strong acid Strong base Methyl orange or phenolphthalein

(b) Redox indicator: Example: Diphenylamine. When titrating KMnO₄ with oxalic acid.

Question 12.
What are external indicator? Give an example.
Answer:
External indicator is added in between the titration process by taking a drop of titrated
mixture.
Example: Potassium ferrocynide is used when ferrous ammonium sulphate is titrated against K₂Cr₂O₇.

Question 13.
What is self indicator ? Give an Example.
Answer:
When in titration one of the solution acts as a self indicator. Example: KMnO₄.

Question 14.
What is gram molecular value? Explain.
Answer:
The volume occupied by one gram mole of a substance in the gaseous or is the vapour state at STP is called the gram molecular volume. The gram molecular volume of all the substance is found to be 22.4 dm ³ or 22400 cm ³. The mass in gram of 22.4 dm ³ of a gas or vapour at STP is equal to its gram molecular weight.

1 g molecule of chlorine or 1 gram molecule of carbon di-oxide or 1 gram of any other substance in the gaseous state occupies a volume of 22400 cm³ (22.4 dm³ ).

1 g molecule weight of hydrogen = 2.016 g, 2.016 g of hydrogen occupies 22.4 dm³ of volume at STP. But 1 g equivalent weight of hydrogen is 1.008 g occupies 11.2 dm³ of volume at STP or mass of 11200 cm ³ of hydrogen = its equivalent weight.

Question 15.
Define the following with example, (i) Allotropy and allotropes, (ii) polymorphism, (iii) isomorphism.
Answer:
(i) The existence of an element in two or more chemically similar but physically different forms is called allotropy and the different forms are called allotropes, e.g., diamond, graphite, wood charcoal, lamp black etc., are allotropes of carbon.
(ii) The existence of a compound in different crystalline forms is called polymorphism and the different forms are called polymorphs, e.g., ZnS has two polymorphs called zinc blende and wurtzite.
(iii) The existence of different compounds with similar chemical composition in the same crystalline form is called [M₂SO₄M₂ (SO₄)₃ 24H₂O] are isomorphs.

Question 16.
Discuss smaller the quantity to be measured more precise should be the instrument.
Answer:
Quite often the uncertainty in measurement is expressed in terms of percentage by putting ± sign before it, e.g., 250 ± 1% etc. If the same instrument is used for measuring different quantities, then smaller the quantity to equal to ± 1 mg, then if we weigh 100 g on it, the result can be reported as 100 ± 0.001%. If same balance is used to weigh 10 g, the result reported will be 10 ± 0.01%, and if 1 g is weighed, the result reported will be 10 ± 0.1%. Hence, smaller the quantity to be measured, more precise should be the instrument.

Question 17.
Compute the mass of one molecule and the molecular mass of C₆H₆  (benzene). (At. mass of C = 12, H = 1 u).
Answer:
Molecular weight of C₆H₆ = 6 × 12 + 6 × 1 = 78 g,
Mass of 1 molecule = \(\frac{78}{6.002 \times 10^{23}}\) = 12.94 × 10^-23 g = 1.294 × 10^-22 g

Question 18.
An organometallic compound on analysis was found to contain, C = 64.4%, H = 5.5% and Fe = 29.9%. Determine its empirical formula. (At. mass of Fe = 56 u).
Answer:

Question 19.
4g of copper chloride on analysis was founded to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (Cl). What is the empirical formula of copper chloride? [At. mass of Cu = 63.5 u, Cl = 35.5 u].
Answer:

Question 20.
Calculate the number of grams of oxygen in 0.10 mol of Na₂CO₃.10H₂O.
Answer:
1 mole of Na₂CO₃.10H₂O contains 13 moles of oxygen atoms.
0.1 mole Na₂CO₃.10H₂O contains 13 × 0.1 = 1.3 moles of oxygen atoms.
Mass of oxygen atoms = 1.3 × 16 = 30.8 g.

Question 21.
How many grams of Cl2 are required to completely react with 0.4 g of H2 to yield HC1? Also calculate the amount of HC1 formed.
Answer:
H_2(g) + Cl_2(g) → HCl_(g) ,2 × 1 = 2 2 × 35.5 = 71
2 g of H₂ reacts with 71 g of Cl₂ to give 73 g of HCl.
0.4 g of H₂ reacts with \(\frac{71}{2}\)x 0.4 = 14.2 g of Cl₂ to give \(\frac{73 \times 0.4}{2}\) = 14.6 g of HCl.
14.2 g of Cl₂ and 14.6 g of HCl.

Question 22.
Cone. HC1 is 38% HCl by mass. What is the molarity of this solution if d = 1.19g cm^-3? What volume of cone. HC1 is required to make 1.00 L of 0.10 M HCl?
Answer:

Question 23.

Answer:

Question 24.
Calculate the volume of O₂ at STP liberated by heating 12.25 g of KClO₃. (At. wt. of K = 39, Cl = 35.5, O = 16 u).
Answer:

Question 25.
1 M solution of NaNO₃ has density 1.25 g cm ^-3. Calculate its molality. (Mol. weight of NaNO₃ = 85 g mol^-1).
Answer:
Mass of solution = Volume of solution × density of solution
Mass of solution = 1000 cm³ × 1.25 g cm^-3 = 1250 g
Mass of solute = 85 g; Mass of solvent = 1250 – 85 = 1165 g.

Question 26.
Explain how compounds differ from elements? Give two differences.
Answer:
Elements consist of only one kind of atoms.

Question 27.
Explain how mixture differs from pure substance? Give 2 difference.
Answer:
Pure substances Mixtures

Question 28.
Classify each of the following as pure substance or mixture: (a) Ethyl alcohol, (b) Oxygen, (c) Blood, (d) Carbon, (e) Steel, (f) Distilled water.
Answer:
Pure substances: (a) Ethyl alcohol, (b) Oxygen, (d) Carbon, (f) Distilled water. Mixtures: (c) Blood and (e) Steel.

Question 29.
Describe two factors that introduce uncertainty into measured figures.
Answer:
(i) Reliability of measuring instrument
(ii) Skill of the person making the measurement.

Question 30.
A 50 ml graduated cylinder has 1 ml graduation. What is maximum number of significant figures of the volumes that can be reported from this graduated cylinder?
Answer:
The maximum number of significant figures of the volumes that can be reported are 2.

Question 31.
What is the difference between accuracy and precision of measurements?
Answer:
Accuracy: It is related to closeness of signal measurement to its true value.
Precision: It refers to the closeness of the set of values obtained from identical measurements of a quantity.

Question 32.
Classify the following substances into elements, compounds, and mixture.
(1) Milk (2) Iodised table salt (3) Diamond (4) Steel (5) Dry Ice (6) Glucose (7) Petrol (8) Steam (9) Cloud (10) Aerated drinks
Answer:
Elements (3), Compounds 5,6, 8,9, Mixture 1, 2,4, 7,10

1st PUC Chemistry Some Basic Concepts of Chemistry Five Marks Questions and Answers

Question 1.
Calculate the moles of NaOH required to neutralize the solution produced by dissolving 1.1 g P₄O₆ in water. Use the following reactions:
P₄O₆ + 6H₂O → 4H₃PO₃;
2NaOH + H₃PO₃ → Na₂HPO₃ + 2H₂O
(At. mass/g mol^-1 P = 31, O = 16).
Answer:
Molecular weight of P₄O₆ =4 × 31 + 6 × 16 =124 + 96 = 220 gmol^-1

\(\frac { 1 }{ 200 }\) moles of PiOfi produce 4 × \(\frac { 1 }{ 200 }\)moles of H₃PO₃
1 mole of H₃PO₃ requires 2 moles of NaOH
\(\frac { 1 }{ 50 }\)moles of H₃PO₃ requires 2 × \(\frac { 1 }{ 50 }\) = \(\frac { 1 }{ 25 }\) = 0.04 moles of NaOH.

Question 2.
(a) A sample of NaOH weighing 0.38 g is dissolved in water and the solution is made to 50.0 cm³ in a volumetric flask. What is the molarity of the solution?
(b) State and explain law of multiple proportion.
Answer:

(b) Law of multiple proportion states whenever two elements react to form two or more compounds, the ratio between different weights of one of the elements which combine with fixed weight of another is always simple, e.g., nitrogen reacts with oxygen to form NO and NO₂. In NO, 14 g of N reacts with 16 g of O. In NO₂,14 g of N reacts with 32 g of O. The ratio between weights of oxygen which combine with fixed weight of nitrogen is 16 : 32. i.e., 1:2.

Question 3.
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Answer:
CH_4(g) + 2O_2(g) >CO_2(g) + 2H₂O₍₁₎
1 mole = 16 g of CH₄ gives 2 moles of H₂O i.e., 2 × 18 = 36 g of H₂O.

Question 4.
How many moles of methane are required to produce 22 g of CO,2(g) after combustion?
Answer:
44 g of CO₂ will be produced by 1 mole of CH₄.
22 g of CO₂ will be formed by \(\frac{1}{44}\) × 22 = 0.5 mole of CH₄.

Question 5.
(a) How many significant figures are there in 1.00 × 10⁶?
(b) One mole of sugar contains ……. oxygen atoms.
(c) Give an example of molecule in which the empirical formula is CH₂O and the ratio of molecular formula weight and empirical formula weight is 6.
Answer:
(a) 3
(b) One mole of sugar C₁₂H₂₂O₁₁ contains 11 × 6.023 × 10²³ oxygen atoms.
= 66.253 × 10 ²³ = 6.6253 × 10 ²³ atoms of oxygen
(c) C₆H₁₂O₆.

Question 6.
An organic monobasic acid was found to contain 39.5% carbon, 6.4% hydrogen and the rest oxygen. If the equivalent mass of the acid is 60, find out its molecular formula.
Answer:
Percentage of oxygen = 100 – (39.5 + 6.4) = 100 – 45.9 – 54.1

∴Empirical formula = CH₂O
Molecular mass = Eq. mass × basicity = 60 × 1 = 60
Empirical formula weight = CH20 = 12 + 2 + 16 = 30
Nows 30n = 60 => n = 2
∴Molecular formula = (Empirical formula) × n = (CH₂O)₂ = C₂H₄O₂

Question 7.
Calculate the mass of 95% pure Mn02 to produce 35.5 g of Cl₂ as per the following reaction. Mn02 + 4HC1 »MnCl2 + Cl2 + HaO.
(At. mass of Mn = 55)
Answer:

71 g of Cl₂ is produced by 87 g of MnO₂
35.5 g of Cl₂ is produced by \(\frac { 87 }{ 71 }\) × 35.5 = 43.49 g of MnO₂
Since MnO₂ is 95% pure i.e., 95 g of MnO₂ is present in 100 g of MnO₂ sample.
43.49 of pure MnO₂ will be present in \(\frac{43.49 \times 100}{95}\)= 45.8f of MnO₂ sample

Question 8.
500 ml of Na₂CO₃ solution contains 2.65 g of Na₂CO₃ (mol. mass of Na₂CO₃ = 106). If 10 ml of this solution is diluted to 1 L, what is the concentration of the resultant solution?
Answer:
Answer:

For dilution M₁V₁ = M₂V₂ ⇒ 0.05 × 10 = 1000 × M₂
M₂ = 0.0005 M (Molarity of the resultant solution).

Question 9.
2 L of a solution is prepared by dissolving 0.02 mole of NaBr and 0.02 mole of Na₂SO₄.1 L of mixture X + excess of AgNO₃ → Y (pale yellow ppt.).
1 L of mixture X + excess of BaCl2 → Z (white ppt.).
What is the ratio of number of moles of Y and Z?
Answer:
AgNO₃ reacts with Br ions to form AgBr as pale yellow ppt. as per the following reaction

2 L of solution will contain 0.02 mole of NaBr or 0.02 mole of Br^–.
=> 1 L of solution will contain 0.01 mole of Br^–
∴ as per equation (i) number of moles of AgBr i.e., Y formed is 0.01 mole
BaCl₂ reacts with SO₄^2- ions present in the solution as per the following reaction

2L of solution contain 0.02 mole of Na₂SO₄ or 0.02 mole of SO₄^2-.
⇒ 1 L of solution will contain 0.01 mole of SO₄^2-.
∴ as per equation (ii) number of moles of BaSO₄ i.e., Z formed is 0.01 mole
Ratio of Y and Z = 0.01: 0.01 = 1:1

Question 10.
How many grams of copper will get replaced in 2 L of CuSO₄ solution having molarity 1.50 M, if it is made to react with 54 g of aluminium? (At. mass of Cu = 63.5 and A1 = 27.0)
Answer:
The reaction with CuSO₄ solution and aluminium may be represented as
3CUSO₄ + 2Al → Al₂(SO₄)₂ +3CU
(3 × 159.5)g = 478.5g (2 × 27)g = 54g
Calculation of wt. of CuSO₄ in 2 L of CuSO₄ solution having 1.50 molarity

478.5 g of CuSO₄ reacts with 54 g of aluminium
Also, 159.5 g of CuSO₄ has 63.5 g of Cu
1 g of CuSO₄ has \(\frac{63.5}{159.5}\) g of Cu
478.5 g of CuSO₄ has \(\frac{63.5}{159.5}\) × 478.5g of Cu = 190.5g of Cu

Question 11.
0.7 g of Na₂CO₃.xH₂O dissolved to make 100 ml of solution. If 20 ml of this solution is required to completely neutralize 19.8 ml of 0.1 M HCl, then what is the value of x?
Answer:
Calculation of molarity of sodium carbonate solution
n₁M₁V₁ = n₂M₂V₂ (n factor for sodium carbonate is 2)
2 × 20 × M₁ =19.8 × 0.1 ⇒ M₁ =0.0495

The value of x i.e., number of water molecules of crystallization is 2.

Question 12.
Calculate the weight of CaO that can be obtained by heating 200 kg of limestone which is 95% pure. (At. mass of Ca = 40, C = 12 and O = 16).
Answer:

100 kg of CaCO₃ gives 56 kg of CaO.
Since lime stone is 95% pure; amount of CaCO₃ present in 200 kg of impure sample = 190 kg CaCO₃ 100 kg of CaCO₃ gives 56 kg of CaO
1 kg of CaCO₃ gives \(\frac { 56 }{ 100 }\)kg of CaO
190 kg of CaCO₃ give \(\frac { 56 }{ 100 }\) × 190 = 106.4 kg of CaO.

Question 13.
20 g of sample containing Ba(OH)₂ is dissolved in 10 ml of 0.5 M HC1 solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of Ba(OH)₂ in the sample. (Mol. wt. of Ba(OH)₂ = 171)
Answer:
Calculation of volume of HCl used in titration between NaOH and HCl.
V_NaOH = 10 ml, M_NaOH = 0.2M ,M_HCl = 0.5M, V_HCl = ?
M_NaOH × V_NaOH=M_HCl × V_HCl V_HCl = \(\frac{10 \times 0.2}{0.5}\) = 4ml
This is the volume of HCl left unused when excess of HC1 is added to Ba(OH)₂ solution.
Total volume of HCl added = 10 ml.
Volume of HCl used to react with Ba(OH)₂ = 10 – 4 = 6 ml
∴HCl react with Ba(OH)₂ as per the equation
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
6 ml of 0.5 M HCl has reacted with Ba(OH)₂ => Number of moles of HCl reacted,

Observing the molar ratio of HC1 and Ba(OH)₂,
Moles of Ba(OH)₂ reacted = \(\frac { 1 }{ 2 }\) × moles of HCl reacted = \(\frac { 1 }{ 2 }\) × O.003 = 0.0015 moles
Weight of Ba(OH)₂ reacted = no. of moles × mol. wt. = 0.0015× 171 = 0.2565 g

1st PUC Chemistry Some Basic Concepts of Chemistry Numerical Problems and Answers

Question 1.
How many significant figures are there in each of the following numbers?
(i) 6.005, (ii) 6.002 × 10²³, (iii) 8000, (iv) 0.0025, (v) π, (vi) the sum 18.5 + 0.4235, (vii) the product 14 × 6.345.
Answer:
(i) Four because the zeros between the non-zero digits are significant figures.
(ii) Four because only the first term gives the significant figures and exponential term is not considered.
(iii) Four. However, if expressed in scientific notation as 8 × 10³ , it will have only one significant figure, as 8.0 × 10³, 8.00 × 10³ or 8.000 x 10³, it will have 2, 3 or 4 significant figures.
(iv) Two because the zeros on the left of the first non-zero digit are not significant.
(v) As π = \(\frac { 22 }{ 7 }\) = 3.1428571…., hence it has infinite number of significant figures.
(vi) Three, because the reported sum will be only upto one decimal place i.e., 18.9.
(vii) Two, because the number with least number of significant figures involved in the calculation (i.e., 14) has two significant figures.

Question 2.
Express the following to four significant figures: (i) 6.45372, (ii) 48.38250, (iii) 70000, (iv) 2.65986 × 10³, (v) 0.004687.
Answer:
(i) 6.454, (ii) 48.38, (iii) 7.000 × 104, (iv) 2.660 × 10³, (v) 0.004687.

Question 3.
A sample of nickel weighs 6.5425 g and has a density of 8.8 g cm³ the volume? Report the answer to correct decimal place.
answer:

The result should have two significant figures because the least precise term (8.8) has two significant figures.

Question 4.
Express the result of the following calculation to the appropriate number of \(\frac{3.24 \times 0.08666}{5.006}\)
Answer:
\(\frac{3.24 \times 0.08666}{5.006}\)= 0.0560883 (Actual result)
As 3.24 has least number of significant figures, viz., 3 , the result should contain 3 significant figures only. Hence, the result will be reported as 0.0561 (after rounding off).

Question 5.
The mass of precious stones is at pressed in terms of carat. Given that 1 carat = 3.168 grains and 1 gram – 15.4 grains, calculate the total mass of the ring in grams and kilograms which contains 0.500 carat diamond and 7.00 gram gold.
Answer:
The unit conversion factors to be used will be;

Question 6.
Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M₃O₄, find that of the second.
Answer:
In the first oxide, oxygen = 27.6, metal = 100 – 27.6 = 72.4 parts by mass.
As the formula of the oxide is M₃O₄, this means 72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass.
In the second oxide, oxygen = 30.0 parts by mass and metal = 100 – 30 = 70 parts by mass.
But 72.4 parts by mass of metal = 3 atoms of metal.
∴ 70 parts by mass of metal =\(\frac{3}{72.4}\) × 70 atoms of metal = 2.90 atoms of metal.
Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen.
∴ 4.35 atoms of oxygen.
Hence, ratio of M: O in the second oxide = 2.90 : 4.35 = 1:1.5 = 2:3.
∴ Formula of the metal oxide is M₂O₃.

Question 7.
What is the equivalent weight of KH(IO₃)₂ as an oxidant in presence of 4.0 (N) HCl when IC becomes the reduced form? (K = 39.0,1 = 127.0).
Answer:
In KH(IO₃)₂, IO₃ is present as IO₃^–. Oxidation state of I will be

Question 8.
Calculate the percentage of the naturally occurring isotopes ³⁵C1 and ³⁷Cl that accounts for the atomic mass of chlorine taken as 35.45.
Answer:
Suppose ³⁵Cl present = x%. Then ³⁷Cl present = (100 -x)%
∴ Average atomic mass = \frac{x \times 35+(100-x) \times 37}{100} = 35.45 (Given)
or 35x + 3700-37x = 3545 or 2x = 155 or x = 77.5%
This ³⁵Cl = 77.5% and ³⁷Cl = (100 – 77.5) = 22.5%.

Question 9.
Convert 22.4 L into cubic metres.
Answer:

Question 10.
Calculate the molar mass of water if it contains 50% heavy water (D₂O).
Answer:
As water contains 50% D₂O, this means that it contains \(\frac{1}{2} mole of H20 and \frac{1}{2}\) mole
of D₂O. Mass of \(\frac{1}{2}\) mole of D₂O = \(\frac{1}{2}\) × (2 × 2 + 16) = 10 g.
Hence, molar mass of the given sample of water = 9 + 10 = 19 g mol^-1.×

Question 11.
60 cc of oxygen was added to 24 cc of carbon monoxide and the mixture ignited. Calculate the volume of oxygen used up and the volume of carbon dioxide formed.
Answer:

∴ volume of oxygen used up is 12 cc. and the volume of carbon dioxide formed is 24 cc.

Question 12.
200 cm³ of carbon monoxide is mixed with 200 cm³ of oxygen at room temperature and ignited. Calculate the vol of CO₂ formed on cooling to room temperature. What other gas if any may also be present.
Answer:

Volume of carbon dioxide formed is 200 cm³ and the other gas present is 100 cm³ of oxygen.

Question 13.
Calculate the volume of oxygen required to burn completely a mixture of 22.4 dm³ of CH₄ and 11.2 dm³ of H₂ (all volumes measured at STP) [1 dm³ = 1 litre].
Answer:

Total oxygen required = 44.8 dm³ +5.6 dm³ =50.4 dm³ Pits].
∴ the number of moles in 7 g of nitrogen is 0.25 moles.

Question 14.
Calculate the mass of 50 cc of CO at STP (C = 12, O = 16).
Answer:
[1 mole = 1 g mol. wt. and occupies 22.4 lit. at STP]
g mol. wt. of carbon monoxide = 12 + 16 = 28 g
28 g of CO occupies 22400 cc at STP
How many grams of CO will occupy 50 cc at STP = \(\frac{28 \times 50}{22400}\) = 0.0625 g

Question 15.
Calculate the volume at STP occupied by 6.023 × 10²² molecules of a gas X.
Answer:
[1 mole of any substance contains 6.023 × 10²³ number of molecules [Avogadro’s number]].
(a) 6.022 × 10²³ molecules occupied 22.42 cm³ (Its) = \(\frac{22.4 \times 6.023 \times 10^{22}}{6.023 \times 10^{23}}\) = 2.24 lits

Question 16.
Calculate the number of molecules in 1 kg of sodium chloride. [Na = 23, Cl = 35.5].
Answer:
(a) 1 mol wt. of NaCl = 58.5 g contains 6.022 × 10²³
(b) Contains molecules = \(\frac{1000}{58.5}\) × 6.023 × 10²³ = 102.96 × 10²³ molecules.

Question 17.
Calculate the atomicity of a gas X if 1 g of X occupies 11,200 cc at STP [At. Wt. of X= 1].
Answer:

Question 18.
0. 48 g of a gas forms 100 cm³ of vapours at STP. Calculate the gram molecular wt. of the gas.
Answer:
100 cm³ of the gas weighs 0.48 grams at STP
22,400 cm³ weight Mol. wt. at STP.
∴ \(\frac{22400}{100}\) × 0.48 = 107.52

Question 19.
Calculate the weight of a substance X which in gaseous form occupies 10 litres at 27 °C and 700 mm pressure. The molecular Weight of X is 60.
Answer:
Convert the volume to STP using the gas equation.
Initial conditions: P₁ = 700 mm of Hg, V₁ = 10 litres, T₁ = 27 + 273 = 300 K
Final conditions (at STP): P₀ = 760 mm of Hg, V₀ = X litres, T₀ = 273 K

Question 20.
Calculate the normality of oxalic acid solution containing 0.895 g crystals in 250 cm³ of its solution.
Answer:
Molecular formula of oxalic acid crystals = H₂C₂O₄.2H₂O = 126

Question 21.
25 cm³ of ferrous ammonium sulphate solution require 20 cm³ of 0.1 N potassium dichromate solution. Calculate the amount of ferrous ammonium sulphate crystals dissolved in 250 cm³ of the solution. (Given equivalent)
Answer:

Question 22.
What should be the normality of a solution prepared by diluting 250 ml of 0.4 NH₂SO₄ with 1000 ml of water?
Answer:
Total volume of the diluted solution = 250 + 1000 = 1250 ml

Question 23.
20 cm³ of a solution of oxalic acid requires 25 cm³ of 0.2 N potassium . permanganate to react completely (a) Calculate the normality of oxalic acid solution, (b) What volume of this oxalic acid solution when made up to 250cm³ gives 0.2 N solution?
Answer:

Normality of oxalic acid solution =200. Volume of oxalic acid solution required =200 cm³

Question 24.
200 cm³ of a solution of a dibasic acid contains 1.512 g of the acid and the normality of the solution is 0.12. Calculate (i) the equivalent mass and (ii) the molecular mass of the acid.
Answer:
(i) Mass of the acid in one dm³ of the solution = 1.512 × 5 = 7.56 g.

(ii) Molecular mass of the acid = eq. mass × basicity = 63 × 2 = 126.

Question 25.
25 cm³ a of a solution of sodium hydroxide required 29 cm³ of \(\frac { n }{ 10 }\) solution of oxalic acid for neutralization. Find the normality of sodium hydroxide solution and its amount dissolved in 500 cm³ of the solution.
Answer:

Substituting we have 28 × 0.1 = 25 × N₂
∴N₂ = \(\frac{29 \times 0.1}{25}\) = 0.116
∴ Normality of sodium hydroxide solution = 0.116
We know that W = N × E for 100 ml
i.e., Mass per dm³ = Normality × eq. nlass = 0.116 × 40 g
∴ Mass of sodium hydroxide in 500 cm 3 of the solution = \(\frac{0.116 \times 40}{2}\) = 2.32 g.

Question 26.
Calculate the volume of concentrated nitric acid of normality 14 required to prepare 1 dm³ of \frac{N}{10} nitric acid.
Answer:

Question 27.
Calculate the mass of hydrochloric acid in 200 cm³ of 0.2 N solution of it. What volume of this acid solution will react exactly with 25 cm3 of 0.14 N solution of sodium hydroxide?
Answer:
Normality of hydrochloric acid = 0.2; Eq. mass of hydrochloric acid = 36.5
Mass of hydrochloric acid in one dm³ of the solution is given by W = N × E = 0.2 × 36.5 g

Question 28.
0. 99 g of an acid was dissolved in water and the solution made up to 200 cm³, 20 cm³ of this solution required 15 cm³ of 0.105 N sodium hydroxide solution for complete neutralization. Find the equivalent mass of the acid.
Answer:

Substituting we have 20 × N₁ =15 × 0.105
∴ N₁ = \(\frac{15 \times 0.105}{20}\) = 0.07875
∴ Normality of acid solution = 0.07875
Mass of the acid in one dm 3 of the solution – 0.99 × 5 g = 4.95 g
Eq. mass of the acid is given by E = \(\frac{W}{N}=\frac{4.95}{0.07875}\) = 62.85

Question 29.
2.15 g of impure oxalic acid crystals were dissolved in water and the solution made up to 250 cm³ of this solution required 20 cm³ of 0.16 N sodium hydroxide solution for complete neutalization. What is the percentage purity of oxalic acid crystals?
Answer:

Substituting we have 25 × N₁ =20 × 0.16
∴ N₁ = \(\frac{20 \times 0.16}{25}\) = 0.128
∴ Normality of oxalic acid solution = 0.128
Eq. mass of oxalic acid crystals H₂C₂O₄.2H₂O = 63.
Mass of pure oxalic acid in one dm 3 is given by W = N × E = 0.128 × 63
Mass of pure oxalic in 250 cm3 = \(\frac{0.128 \times 63}{4}\) = 2.016 g
Mass of impure oxalic acid crystals dissolved in 250 cm³ = 2.15 g .
% percentage purity of the given oxalic acid crystals = \(\frac{2.016 \times 100}{2.15}\) = 93.76

Question 30.
A solution of sodium hydroxide contains 4.8 g of the substance per dm3, (a) What volume of the solution is required to prepare 500 cm3 of decinormal solution. (b) What volume of water has to be added to one dm3 of the solution in order to make it exactly decinormal ?
Answer:

Question 31.
Find the Molarity of hydrochloric acid containing 31.5% of hydrochloric acid. Its specific gravity is 1.16.
Answer:
Mass of 1000 cm³ of the acid = 1000 × 1.16 = 1160 g
Mass of hydrochloric acid in 100 g of the solution = 3.15 g

Question 32.
100 cm3 of concentrated hydrochloric acid which is 11 M are diluted to 275 cm³. If 2.5 g of limestone are required to neutralize 10 cm³ of this diluted – acid, what is the percentage of calcium carbonate in the sample of limestone?
Answer:

Substituting we get, 100 × 11 = 275 × M₂

∴ Normality of the diluted acid = 4
1000 cm³ of the acid solution contain 4 gram
∴ 10 cm³ of the diluted acid contain = \(\frac{4 \times 10}{1000}\) i.e., 0.4 gram
2.5 g of limestone contain 0.04 gram of calcium carbonate i.e., they contain 0.04 × 100 g = 4 g of calcium carbonate.
(Molar mass of calcium carbonate =100)
∴ percentage of calcium carbonate in the sample of limestone = \(\frac{4 \times 100}{2.5}\) = 16.0 .

Question 33.
25.0 cm³ of an acid required exactly 20.5 cm³ of deci molar base for complete neutralization. What is the normality of the acid?
Answer:

Question 34.
Exactly 20.0 cm³ of nitric acid neutralized 28.4 cm³ of 0.25 M NaOH. What is the molarity of the acid?
Answer:

Question 35.
18.5 cm³ of oxalic acid was completely neutralized by 20.0 cm³, 0.125 N base. Calculate the (a) normality (b) molarity and (c) mass of oxalic acid crystals in 1 dm³ of solution.
Answer:

Mass of oxalic acid crystals in 1 dm³ of solution = normality × equivalent mass
= 0.1351 N × 63.0 (equivalent mass of oxalic acid crystals = 63.0) = 8.513 g.

Karnataka 1st PUC Geography Question Bank Chapter 8 India

You can Download Chapter 8 India 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 India One Mark Questions and Answers

Question 1.
State the geographical location of India. (T.B.Qn)
Answer:
India extends between 8°4’ N to 37°6’ N Latitude and 68° 7’ E to 97°23 E longitude.

Question 2.
Name the southernmost and northernmost points of main land of India. (T.B.Qn)
Answer:
The northern tip of India is recognized as ‘Indira Col’ in Jammu & Kashmir while, the southern tip (main land) is ‘Kanyakumari’ or ‘Cape Camorin’ in Tamilnadu.

Question 3.
In which Island of India, is ‘Indira Point’ situated? (T.B.Qn)
Answer:
Indira point located at Great Nicobar Island.

Question 4.
What is the total geographical area of India? (T.B.Qn)
Answer:
Total geographical area of India is 32, 87,263 Sq.km.

Question 5.
Mention the International boundary between India and China. (T.B.Qn)
Answer:
McMahon line is the international boundary between India and China.

Question 6.
Which water bodies separate India and Sri lanka?
Answer:
Palk Strait and Gulf of Mannar

Question 7.
What percent of the world’s land area is with India?
Answer:
It is 2.4% of the world area.

Question 8.
What is the length of the land boundary of India?
Answer:
About 15,200 kilometers.

Question 9.
Name the largest island and smallest island in India.
Answer:
Middle Andaman is the largest one and Lakshadweep islands are the smallest islands.

Question 10.
Name the largest and smallest states in India.
Answer:
Rajasthan is the largest state in India and Goa is the smallest one.

Question 11.
Why Jabalpur is called as geographical centre of India?
Answer:
The Tropic of cancer and 80° E longitude lines meets at Jabalpur of Madhya Pradesh. It is known as geographical centre of India.

Question 12.
Name the state of India which has longest coastal line.
Answer:
Gujarat state has longest coastal line in India.

Question 13.
Which line of longitude is called Indian Standard Time?
Answer:
82°30′ (82!/2°) East longitude is called Indian Standard Time.

Question 14.
Name the type of Climate prevails in India.
Answer:
India has “Tropical Monsoon type” of climate.

Question 15.
Name the continent where India is located.
Answer:
Asia continent.

Question 16.
Mention the channel which separates Andaman and Nicobar islands.
Answer:
10° N channel separates Andaman and Nicobar Islands.

Question 17.
What is the total length of coastal line of India?
Answer:
The main land of the country has a coastline of 6100km. Including island the total length of the coastal of the country is about 7516.

Question 18.
How many states and Union Territories are there in India?
Answer:
India is divided into twenty eight states and seven Union Territories. Among them Delhi is the National Capital Territory.

Question 19.
What is the percentage of India’s population in the world?
Answer:
The percentage of India’s population in the world is 7.4%.

Question 20.
Which latitude divides the Indian Sub-continent in two halves?
Answer:
Latitude of 23 1/2° N divides the Indian Sub-continent.

1st PUC Geography India Two Marks Questions and Answers

Question 1.
Write the latitudinal and longitudinal extent of India. (T.B.Qn)
Answer:
The latitudinal extension is 8°4’ N to 37°6’ N and the longitudinal extent is 68° 7’ E to 97°23 E .The latitudinal and longitudinal extent of India is around 30° The country stretches to 3214km from North to South and 2933 km from West to East.

Question 2.
Name the water bodies that surround India. (T.B.Qn)
Answer:
India is a peninsula, located at the north tip of the Indian Ocean. It is bordered by the Arabian Sea in the west, Indian Ocean in the south and Bay of Bengal in the east.

Question 3.
Which latitude and longitude passes in the centre of the Country? (T.B.Qn)
Answer:

• The Tropic of Caner 23 1/2° N latitude passes through the middle of India and divides the country into almost two equal halves.
• Indian Standard Time- 82 1/2° E longitude passes through the middle of India (through Allahabad) is recognized as standard longitude of the country two keep standard time.

Question 4.
What is the growth rate of population and total population of India according to 2011 census? (T.B.Qn)
Answer:
According to 2011 census the total population of the country was 121.6 croreor 1216 million or 1.21 billions, which accounts for about 17.45% ofthe total world’s population.

Question 5.
Name the international boundaries between India & Pakistan and India & Afghanistan. (T.B.Qn)
Answer:

• The Radcliff line – India and Pakistan (2910km) by Sir Cyril Radcliff.
• The Durand line – India and Afghanistan (80km) by Mortimer Durand.

Question 6.
Why has India selected a Standard Meridian of India with an odd value of 82° 30’ E?
Answer:
On the International basis the globe has been divided into 24 time zones (each of 15 longitudes). In every zone local time of the middle longitude (divided by 7° 30’) is taken as standard time of the entire zone. Because 82 1/2° E is well divisible by 7° 30’ A standard adopted by almost all the countries of the world while they selected a Standard meridian for their respective countries.

Question 7.
Name the states through which Tropic of cancer passes.
Answer:
The tropic of Caner 23 1/2° N latitude passes through Gujarat, Rajasthan, Madhya Pradesh, Chhattisgarh, Jharkhand, and West Bengal, Tripura, and Mizoram states.

Question 8.
IIow India is a peninsular country?
Answer:
India is a peninsular country because it has water bodies on its three sides. The Indian Ocean lies in the South, Arabian Sea lies in the west and Bay of Bengal lies in the east.

Question 9.
Why is Indian sub-continent so called?
Answer:

• India and the adjoining countries are considered to be a sub-continent as it is comprised of all the characteristics of a continent.
• Indian sub-continent encompasses vast areas of diverse landmasses. Indian sub-continent is comprised of lofty mountains, fertile plains, desert and plateau.
• There is also great vastness and diversities in terms of climate, natural vegetation, wildlife and other resources.
• Also, the vivid characteristics of culture and tradition among the people make it a sub continent.

Question 10.
State the Neighbouring countries of India.
Answer:
Pakistan and Afghanistan in the North-west, China, Nepal and Bhutan in the north, Myanmar in the East, Bangladesh in the North-East, Sri Lanka and Maldives are in the south oceanic zone.

1st PUC Geography India Five Marks Questions and Answers

Question 1.
Explain the location, size and frontiers of India. (T.B.Qn)
Answer:
Location: The main land oflndia extends between 8°4’ N to 37°6’N latitude and 68° 7’ Eto 97°23 E longitude. The latitudinal and longitudinal extent oflndia is around 30° The country stretches to 3214 km from North to South and 2933 km from West to East. The northern tip of India is recognized as ‘Indira Col’ in Jammu & Kashmir while, the southern tip (main land) is ‘Kanyakumari’ or ‘Cape Camorin’ in Tamilnadu. In the same way the western and eastern tips of the country are ‘Rann of Kutch’ in Gujarat and ‘Luhit’ in Arunachal Pradesh respectively.

The territorial limit oflndia extends up to 6° 45’ N latitude. ‘Indira point’ situated at this latitude in Great Nicobar Islands. As a peninsular country India has both land and water frontiers. The total length of land frontier of the country is 15,200 km. The mainland of the country has a coast line of 6,100km including the islands. The total length of the coast line of the country is about 7516km. The territorial water extends into the sea to a distance of 12 nautical miles (22.2km) from the coastal baseline.

India is a peninsula, located at the north tip of the Indian Ocean. It is bordered by the Arabian Sea in the west, Indian Ocean in the south and Bay of Bengal in the east and covered by land in the north – China, Nepal, Bhutan etc.

The Tropic of Caner 23 1/2° N latitude passes through the middle of lndia and divides the country into almost two equal halves. Indian Standard Time – 82 1/2° E longitude passes through the middle of India (through Allahabad) is recognized as standard longitude of the country two keep standard time.

Size: India is the 7th largest country in the world next to Russia, Canada, China, USA, Brazil and Australia. It has a total geographical area of 32, 87,263 sq.km. This constitutes about 2.4% of the total land area of the Earth. India is the second most populous country in the world next to China. According to 2011 census the total population of the country was 121.6 crore which accounts for about 17.45% of the total world’s population. India has 28 states, 6 union territories and one national capital region (New Delhi).

Frontiers: India has 15,200km long land frontier extending from west to east running from Gujarat in the west to West Bengal in the east. The Himalayas for a natural boundary in the north, between India and China. Similarly, Thar Desert in the west & northwest and eastern hills acts as boundary between India & Pakistan and India & Myanmar respectively. India share land frontier with seven countries, they are Pakistan and Afghanistan to the northwest, China, Nepal and Bhutan to the north and Bangladesh and Myanmar to the-East.

The important international boundary lines demarcated between India and neighbouring countries are:

The Durand line- India and Afghanistan (80Km) by Mortimer Durand The Me Mahon line- India and China (PRC) (3488Km) by Henry Me Mahon.
The Radcliff line – India and Pakistan (2910km) by Sir Cyril Radcliff.
India and Bangladesh (4097km).
Sri Lanka, an island country, situated to the southeast, is separated by Palk Strait and Gulf of Mannar.

You can Download Chapter 29 दोपहर का भोजन Questions and Answers Pdf, Notes, Summary, 1st PUC Hindi Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Hindi Textbook Answers Sahitya Vaibhav Chapter 29 दोपहर का भोजन

दोपहर का भोजन Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिएः

प्रश्न 1.
रामचंद्र कितने वर्ष का था?
उत्तर:
रामचंद्र लगभग इक्कीस वर्ष का था।

प्रश्न 2.
प्रूफरीडरी का काम कौन सीख रहा था?
उत्तर:
प्रूफरीडरी का काम रामचन्द्र सीख रहा था।

प्रश्न 3.
सिद्धेश्वरी के मँझले लड़के का नाम लिखिए।
उत्तर:
सिद्धेश्वरी के मँझले लड़के का नाम मोहन था।

प्रश्न 4.
सिद्धेश्वरी के छोटे लड़के की उम्र कितनी थी?
उत्तर:
सिद्धेश्वरी के छोटे लड़के की उम्र छः साल थी।

प्रश्न 5.
मुंशी चंद्रिका प्रसाद कितने साल के लगते थे?
उत्तर:
मुंशी चंद्रिका प्रसाद लगभग पचास-पचपन साल के लगते थे।

प्रश्न 6.
किसकी शादी तय हो गई थी?
उत्तर:
गंगाशरण बाबू की लड़की की शादी तय हो गयी थी।

प्रश्न 7.
मुंशी जी की तबीयत किससे ऊब गई थी?
उत्तर:
मुंशी जी की तबीयत अन्न और नमकीन से ऊब गई थी।

प्रश्न 8.
मुंशी जी की छंटनी किस विभाग से हो गई थी?
उत्तर:
मुंशी जी की छंटनी मकान-किराया नियंत्रण विभाग की क्लर्की से हो गई थी।

प्रश्न 9.
‘दोपहर का भोजन’ कहानी के कहानीकार कौन हैं?
उत्तर:
‘दोपहर का भोजन’ कहानी के कहानीकार अमरकांत जी हैं।

अतिरिक्त प्रश्नः

प्रश्न 10.
रामचंद्र की पढ़ाई कहाँ तक हुई थी?
उत्तर:
रामचंद्र की इंटर तक पढ़ाई हुई थी।

प्रश्न 11.
सिद्धेश्वरी के छोटे बेटे का नाम क्या था?
उत्तर:
सिद्धेश्वरी के छोटे बेटे का नाम प्रमोद था।

प्रश्न 12.
मोहन किसकी तैयारी कर रहा था?
उत्तर:
मोहन हाईस्कूल का प्राइवेट इम्तहान देने की तैयारी कर रहा था।

प्रश्न 13.
प्रमोद ने क्या खाने के लिए ज़िद पकड़ी थी?
उत्तर:
प्रमोद ने रेवड़ी खाने की जिद पकड़ ली थी।

प्रश्न 14.
मुंशीजी, रामचन्द्र और मोहन के खाने के बाद थाली में कितनी रोटियाँ बची थी?
उत्तर:
मुंशीजी, रामचन्द्र और मोहन के खाने के बाद थाली में केवल एक रोटी बची थी।

प्रश्न 15.
सिद्धेश्वरी किसे बड़ा होशियार कहती है?
उत्तर:
सिद्धेश्वरी रामचंद्र को बड़ा होशियार कहती है।

II. निम्नलिखित प्रश्नों के उत्तर लिखिएः

प्रश्न 1.
सिद्धेश्वरी के परिवार का संक्षिप्त परिचय दीजिए।
उत्तर:
सिद्धेश्वरी का परिवार निर्धन था। उनके तीन बेटे थे। बड़ा बेटा रामचंद्र इक्कीस साल का था। दुबला, पतला, गोरे रंग का लड़का था जिसकी बड़ी बड़ी आँखें थी। वह किसी अखबार में प्रूफ-रीडरी का काम करता था, स्वभाव से गंभीर था। मंझला लड़का मोहन अट्ठारह वर्ष का था, हाईस्कूल का प्राइवेट इम्तेहान देने की तैय्यारी में लगा था, पढ़ाई में उसकी रुचि नहीं थी। छोटा बेटा प्रमोद छह वर्ष का था जो बहुत कमजोर और बीमार रहता था। उसके पति मुंशि चंद्रिका प्रसाद 45 साल की उम्र में, पारिवारिक समस्याओं के कारण 50 या 55 साल के लगते थे। डेढ़ महीने पूर्व मकान-किराया नियंत्रण विभाग की क्लर्की से उनकी छंटनी हो गई थी।

प्रश्न 2.
बीमार प्रमोद की हालत कैसी थी?
उत्तर:
छोटा बेटा प्रमोद बीमार पड़ा है। उम्र छः वर्ष है। अध-टूटे खटोले पर नंग-धडंग पड़ा हुआ है। इतना क्षीण हो गया है कि उसके शरीर की हड्डियाँ स्पष्ट दिखाई देने लगी हैं। उसके हाथपैर बासी ककड़ियों की तरह सूखे हुए बेजान हो गये हैं। पेट हंडिया की तरह फूला हुआ है। मुँह पर मक्खियाँ भिनभिना रही थीं। माँ सिद्धेश्वरी ने उसके चेहरे पर एक फटा, गंदा ब्लाउज डाल दिया। इस दयनीय स्थिति में प्रमोद बीमार था।

प्रश्न 3.
रामचंद्र का परिचय दीजिए।
उत्तर:
रामचंद्र मुंशी चंद्रिका प्रसाद और सिद्धेश्वरी का बड़ा लड़का है। उसके बाल अस्त-व्यस्त है, मुँह लाल है। उसके फटे-पुराने जूतों पर गर्द जमी हुई है। उसकी उम्र करीब 21 वर्ष है। दुबलापतला, लंबा, गोरा रंग, बड़ी-बड़ी आँखें व ओंठों पर झुरियाँ। वह एक स्थानीय समाचार-पत्र के ऑफिस में प्रूफरीडरी का काम सीखता है। इंटर पास किया है।

प्रश्न 4.
मँझले लड़के मोहन के रूप-रंग और स्वभाव के बारे में लिखिए।
उत्तर:
मोहन सिद्धेश्वरी का मँझला लड़का था। उम्र अठारह वर्ष थी और वह इस साल हाईस्कूल का प्राईवेट इम्तहान देने की तैयारी कर रहा था। वह न मालूम कब से घर से गायब था और सिद्धेश्वरी को स्वयं पता नहीं था कि वह कहाँ गया है। वह कुछ साँवला था और उसकी आँखे छोटी थीं। उसके चेहरे पर चेचक के दाग थे। वह अपने भाई रामचंद्र की तरह दुबला-पतला था किन्तु उतना लंबा न था। वह अपने उम्र की अपेक्षा कहीं अधिक गंभीर और उदास दिखाई पड़ रहा था।

प्रश्न 5.
सिद्धेश्वरी की आँखों से आँसू क्यों टपकने लगे?
उत्तर:
घर की आर्थिक स्थिति ठीक नहीं थी। पूरा परिवार संकटग्रस्त था। जैसे-तैसे सात रोटियाँ बनाई गई। बड़े बेटे को, पति को, मंझले बेटे को दो-दो रोटियाँ परोस दी गईं। अंत में सिर्फ एक रोटी बची थी। पति की जूठी थाली में बची चने की तरकारी के साथ बनी जली रोटी रखने जा रही थी कि उसका ध्यान छोटे बेटे प्रमोद की ओर गया। रोटी के दो टुकड़े किए और एक अलग से रख दिया। फिर खाने बैठ गई। घर कि इस स्थिति से वह दुखी हुई और उसकी आँखों से आँसू टपकने लगी।

प्रश्न 6.
मुंशी चंद्रिका प्रसाद की लाचारी का वर्णन कीजिए।
उत्तर:
मुंशी चंद्रिका प्रसाद की डेढ़ महीने पूर्व मकान-किराया नियंत्रण विभाग की क्लर्की से छंटनी हो गयी थी। दूसरी कहीं भी नौकरी नहीं मिली थी। घर में बीमार लड़का। बड़े लड़के को कहीं अच्छी नौकरी नहीं। मँझला प्राईवेट ही हाईस्कूल परीक्षा के लिए बैठा है। दो वक्त का भोजन भी नहीं मिल पाता। इनके फटे-पुराने गर्द लगे जूते, पत्नी की गंदी साड़ी, गंदगी से भरा मक्खियाँ से भिनभिनाता घर, बच्चों का खाने-पीने के लिए तरसना आदि…. इस लाचारी के सामने वे बेबस थे।

अतिरिक्त प्रश्नः

प्रश्न 7.
कड़ी धूप में आये रामचंद्र की स्थिति का वर्णन कीजिए।
उत्तर:
रामचन्द्र कड़ी धूप से आकर धम-से चौकी पर बैठ गया और फिर वहीं बेजान सा लेट गया। उसका मुँह लाल तथा चढ़ा हुआ था। उसके बाल अस्त-व्यस्त थे और उसके फटे-पुराने जूतों पर धूल जमी हुई थी।

प्रश्न 8.
‘दोपहर का भोजन’ कहानी के आधार पर सिद्धेश्वरी की स्थिति का परिचय दीजिए।
उत्तर:
सिद्धेश्वरी का परिवार घोर आर्थिक संकट से गुजर रहा है। घर में खाने के लाले पड़े हुए हैं। किसी को भी भरपेट भोजन नसीब नहीं है। इसके बावजूद सिद्धेश्वरी हिम्मत नहीं हारती है। वह घर की आर्थिक स्थिति का जिक्र कर किसी को दुःखी नहीं करना चाहती। वह अपने बच्चों व पति की हिम्मत बढ़ाती है। स्वयं कम खाकर बाकी परिवार को भोजन कराना अपना कर्तव्य समझती है। सिद्धेश्वरी जीवटता, त्याग और साहस की प्रतिमूर्ति है।

दोपहर का भोजन लेखक परिचयः

हिन्दी के प्रख्यात कथा-शिल्पी अमरकांत का जन्म 1925 ई. में बलिया (उ.प्र.) में हुआ। प्रबुद्ध और सुलझे हुए कहानीकारों में आपका नाम शीर्ष स्थान पर है। आपने ‘मनोरमा’ पाक्षिक पत्रिका का संपादन करते हुए अपनी सूझ-बूझ और कथा-चेतना का परिचय दिया। आपको 2009 ई. में ज्ञानपीठ पुरस्कार से विभूषित किया गया।
प्रमुख कृतियाँ : ‘जिंदगी और जोंक’, ‘देश के लोग’, ‘मौत का नगर’, ‘सूखा पत्ता’, ‘दीवार’ और ‘आँगन’ आदि।

कहानी का आशयः

एक निर्धन परिवार के रहन-सहन का यथार्थ प्रस्तुतीकरण इस कहानी का मूल स्वर है। घर का खर्च चलाना और भोजन तक का नसीब न हो पाना ही कहानी के रूप में प्रस्तुत करता है। अमरकांत ने प्रतीकात्मकता का प्रयोग करते हुए निर्धनता के उस पक्ष को उकेरा है जो इस देश की लगभग 80 प्रतिशत जनता के घर-घर में व्याप्त है। बातों-बातों में बहुत सी बातें ऐसी भी यहाँ हो गई हैं कि मन भर जाता है। कहानी का अंत बहुत ही मार्मिक है।
मध्यवर्गीय परिवार के संघर्षमय जीवन से साक्षात्कार कराने के उद्देश्य से प्रस्तुत कहानी चयनित है।

दोपहर का भोजन Summary in Hindi

अमरकांत हिन्दी कहानीकारों में अग्रगण्य हैं। इनकी कहानियों में मध्यवर्गीय परिवारों का जीवन चित्रित हुआ है। प्रस्तुत कहानी में भी एक गरीब निम्न मध्यवर्गीय परिवार की भूख का मार्मिक चित्रण किया गया है। चंद्रिका प्रसाद एक सरकारी-विभाग में क्लर्की करता था। छंटनी के कारण वह नौकरी से निकाल दिया गया। उसके तीन बेटे थे – रामचन्द्र, मोहन और प्रमोद। उसकी पत्नी का नाम सिद्धेश्वरी था।

चंद्रिका प्रसाद नौकरी की तलाश में सुबह से शाम तक घूमा करता था। परन्तु नौकरी मिलने के लक्षण नहीं दिखे। बड़ा लड़का रामचन्द्र प्रूफरीडरी सीख रहा था। उसे वेतन के रूप में कुछ नहीं मिलता था। दूसरा बेटा मोहन प्राइवेट रूप से हाईस्कूल की परीक्षा देना चाहता था। वह बेकार इधर-उधर घूमा करता था। तीसरा सबसे छोटा था प्रमोद।

चंद्रिका प्रसाद की स्थिति बुरी थी। उसकी पत्नी सिद्धेश्वरी दुःख की मूर्ति बनी हुई थी। वह रोटियाँ और दाल बनाकर पति व बेटों की राह देख रही थी। दुपहर का समय था। गर्मी झुलसा देनेवाली थी। बड़ा बेटा रामचंद्र भोजन के लिए बैठा।

सिद्धेश्वरी ने थाली में दो रोटियाँ, दाल और चने की तरकारी रखी थी। वह अनमने भाव से रोटियाँ खाकर उठ गया। उसने बाबूजी और भाइयों के बारे में पूछा, तो माँ ने अपने बेटे को झूठ-मूठ जवाब दिया। रामचंद्र कुछ बोला नहीं। मंझला बेटा मोहन आया। वह थाली के सामने बैठकर खाने लगा। वह मुश्किल से जली रोटियाँ खा पाया।

चंद्रिका प्रसाद भी देरी से आया। वह बहुत ही उदास था। सिद्धेश्वरी ने पति का हौसला बढ़ाया। वह खाना खाकर खटोले पर सोचते हुए सो गया। सिद्धेश्वरी की भूख लगभग मिट गई थी। वह थाली के सामने बैठ गई। एक ही रोंटी बची थी, उसने उस रोटी के दो टुकड़े किए। एक टुकड़ा छोटे बेटे प्रमोद के लिए रख दिया। वह रो-रोकर सो गया था। उसने रोटी के टुकड़े को मुँह में रखते ही, उसकी आँखों में से आँसू टपकने लगे।

दोपहर का भोजन Summary in Kannada

दोपहर का भोजन Summary in English

Amarkant is one of the foremost Hindi story writers. In his stories, one can get a glimpse into the lives of middle-class families. In this story, we can see a touching description of the hunger of a poor middle-class family.

Chandrika Prasad was a clerk in a government office. Due to retrenchment, he was removed from his job. He had three sons – Ramachandra, Mohan and Pramod. His wife was Siddeshwari.

Chandrika Prasad would go around from morning to evening in search of a job. However, it did not seem as if he would find a job. His eldest son, Ramachandra, was studying to become a proofreader. He did not even earn a stipend. The second son, Mohan, wanted to give his high school exams as a private student. He would waste his time roaming around here and there. The third and youngest son was Pramod.

Chandrika Prasad was in a very bad condition. His wife, Siddeshwari, was like an embodiment of sorrow. She had made rotis and dal and was waiting for the return of her husband and sons. It was afternoon. It was scorching hot outside. The oldest son sat down to eat.

Siddeshwari placed two rotis, dal and chana curry in his plate. He ate his food with a disturbed mind and got up. When he asked about his father and brothers, his mother made up a lie. Ramachandra did not reply. The second son, Mohan, arrived. He sat down and began to eat. With great difficulty, he ate the burnt rotis.

Chandrika Prasad arrived quite late. He was very dejected. Siddeshwari tried to cheer up her husband. He ate his lunch and then stretched out on the cot, thinking about something, and fell asleep. Siddeshwari had more or less lost her appetite. She sat down in front of the plate. Only one roti was left and so, she cut it into two parts. One half she saved for her youngest son Pramod, who had cried a lot before falling asleep. As soon as Siddeshwari put a piece of roti into her mouth, tears began to roll down her cheeks.

कठिन शब्दार्थः

• गगरा – घड़ा;
• ओसारा – बरामदा;
• खटोला – खटिया, चारपाई;
• गर्द – धूल;
• पीढ़ा – लकड़ी का छोटा आसन;
• पनियाई – पानी जैसी;
• चेचक – शीतल रोग;
• जुगाली – पशुओं का निगले हुए चारे को गले से थोड़ा-थोड़ा निकालकर दाँत से चबाने की क्रिया;
• अव्वल – प्रथमतः;
• उन्माद – पागल;
• बर्राक – पूर्णतः, अभ्यस्त;
• बीनना – चुनना;
• कनखी – तिरछी नज़र;
• छटा – धूर्त;
• छिपुली – छोटी थाली;
• अलगनी – कपडे टाँगने की रस्सी।