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KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current.

Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

KSEEB SSLC Class 10 Science Chapter 13 Intext Questions

Text Book Part I Page No. 118

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle get deflected when brought near a bar magnet because a compass needle is in fact, a small bar magnet. The ends of the compass needle point approximately towards north and south directions.

Text Book Part I Page No. 122

Question 2.
Draw magnetic Held lines around a bar magnet.
Answer:

Field lines around a bar magnet

Question 3.
List the properties of magnetic field lines.
Answer:
The properties are:

• They travel from north pole to south pole outside the magnet and south pole to north pole inside the magnet.
• They are closed and continuous curves.
• Two magnetic field lines never intersect each other. If the lines intersect, then at the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic fields which is not possible.
• The number of field lines per unit area is the measure of the strength of magnetic field, which is maximum at poles. The magnetic field is strong, where the field lines are close together and weak where the lines are far apart.

Question 4.
Why don’t two magnetic field lines intersect each other?
Answer:
Two magnetic fields lines of force never intersect each other. If the lines intersect, then at [the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic field, which is not possible.

Text Book Part I Page No. 123, 124

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right- hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:

Since the current passes through the loop in a clockwise direction, therefore the front face of the loop will be the south pole and the back face, ie, the face touching the table will be north pole. According to right-hand rule, the direction of the magnetic field inside the loop will be pointing downward. Outside the loop, the direction of the magnetic field will be upward.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:

The figure indicates that the magnetic field is the same at all points in the solenoid. That is field is uniform inside the solenoid.

Question 3.
Choose the correct option.
The magnetic field inside a long straight solenoid-carrying current
(a) is zero.
(b) decreases as we move towards its end.
(c) increases as we move towards its end.
(d) is the same at all points.
Answer:
(d) is the same at all points.

Text Book Part I Page No. 125, 126

Question 1.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) velocity
(d) momentum.

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if

1. current in rod AB is increased;
2. a stronger horse-shoe magnet is used; and
3. length of the rod AB is increased?

Answer:

1. displacement of A is increased.
2. If a stronger horse-shoe magnet is used magnetic field is increasing.
3. current flows is more.

Question 3.
A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) upward.
Since the positively charged particle alpha particle projected towards west, so the direction of current is towards west. Now the deflection is towards north, so the force is towards north. Now hold the forefinger, centre finger and thumb of our left – hand at right angles to one another. Let us adjust the hand in such a way that our centre finger points towards west and thumb points towards north. If we look at our forefinger, it will be pointing, upward. Thus, the magnetic field is in the upward direction. So, the correct answer is (d).

Text Book Part I Page No 127

Question 1.
State Fleming’s left-hand rule.
Answer:
According to this rule, stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular. If the first finger points in the direction of the Magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Question 2.
What is the principle of an electric motor?
Answer:
A current-carrying conductor when placed in a magnetic field experiences a force. This is the principle of an electric motor.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
The split ring reverse the direction of current in the armature coil after every half rotation i.e., it acts as a commutator. The reversal of current reverses, the direction of the forces acting on the two arms of the armature after every half rotation. This allows the armature coil to rotate continuously in the same direction.

Text Book Part I Page No. 130

Question 1.
Explain different ways to induce current in a coil.
Answer:
We can induce current in a coil either by moving it in a magnetic field or by changing the magnetic around it. It is convenient in most situations to move the coil in a magnetic field.

Text Book Part I Page No. 131

Question 1.
State the principle of an electric generator.
Answer:
Based on the phenomenon of electromagnetic induction, electric generator are prepared. In an electric generator, Mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. This is the principle of an electric generator.

Question 2.
Name some sources of direct current.
Answer:
Dry cell, Battery and D.C. generator.

Question 3.
Which sources produce alternating current?
Answer:
A.C. generator and D.C. generator.

Question 4.
Choose the correct option. A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution.
Answer:
(c) half revolution.

Text Book Part I Page No. 132

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:

1. Electric fuse
2. Earthing wire.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
P = VI
Here P = 2 KW = 2000 W
V = 220

The current drawn by this electric oven is 9 A whereas the fuse in the circuit is ( only 5 A capacity. When a high current of 9 A flows through the 5 A fuse, the fuse wire will get heated too much, melt and break, the circuit. Therefore, when a 2 kW power rating electric oven is operated in a circuit having a 5 A fuse will blow off cutting off the power supply in this circuit.

Question 3.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:

1. Each appliance has a separate switch to ON/OFF the flow of current through it.
2. The use of an electric fuse prevents the electric circuit and the appliance from possible damage by stopping the flow of unduly high electric current.
3. We should not connect too many appliances to a single socket to prevent overloading.

KSEEB SSLC Class 10 Science Chapter 13 Textbook Exercises

Question 1.
Which of the following correctly describes the magnetic Held near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:
(d) The field consists of concentric circles centred on the wire.

Question 2.
The phenomenon of electro-magnetic induction is
(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor.
Answer:
(a) generator.

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electro-magnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(c) AC generator will generate a higher voltage.

Question 5.
At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false.
(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire of an electric supply.
Answer:
(a) False
(b) true
(c) true
(d) False.

Question 7.
List two methods of producing magnetic fields.
Answer:

1. Permanent magnet
2. Electromagnet.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:

One end of the solenoid behaves as a magnetic north pole, while the other behaves as the south pole. The field lines inside the solenoid are in the form of parallel straight lines. This indicates that the magnetic field is the same at the points inside the solenoid.

As shown in figure a strong magnetic field produced inside a solenoid can be used to magnetise a piece of Magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
If the direction of magnetic field and flow of electric current are mutually perpendicular then force experienced by a current-carrying conductor in a magnetic field is largest.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
According to Fleming’s left-hand rule, the magnetic field acts in the vertically downward direction.
Note that the direction of current will be opposite to that of the electron beam.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:

Principle:
An electric motor is a rotating device that converts electrical energy to mechanical energy working.

Working:
Current in the coil ABCD enters from the source battery through conducting brush X and flow back to the battery through brush Y. Notice that the current in the Arm AB of the coil flows from A to B. In arm CD it flows from C to D that is opposite to the direction of current through arm AB on applying Fleming’s left hand rule for the direction of force on a current¬carrying conductor in a magnetic field. We find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards.

Thus the coil and the Axle O, mounted free to turn about an axis, rotate anti-clockwise at half rotation. Q makes contact with the brush X and P with brush Y. Therefore the current in the coil gets reversed and flows along the path DCBA. The reversal of current also reverses the direction of force acting on the two arms AB and CD. Thus the arm AB of the coil that was earlier pushed down, is now pushed up and the arm CD previously pushed up is pushed down. There is a continuous rotation of the coil and to the axle.
Split rings in electric motors acts as a commutator.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motor is used as an important component in electric fans, refrigerators, mixers, washing machines, computers, MP3 players etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is

1. pushed into the coil
2. withdrawn from inside the coil
3. held stationary inside the coil?

Answer:

1. There is a momentary deflection in the needle of the galvanometer.
2. Now the galvanometer is deflected towards the left showing that the current is now set up in the direction opposite to the first.
3. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero.

Question 14.
Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in coil B? Give reason.
Answer:
If the current in the coil A is changed there is a change in its magnetic field. By this electricity is induced in B. This is called Electromagnetic induction.

Question 15.
State the rule to determine the direction of a

1. magnetic field produced around a straight conductor carrying current,
2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
3. current induced in a coil due to its rotation in a magnetic field.

Answer:
(i) Right-hand thumb rule: If the current-carrying conductor is held in the right hand such that the thumb points in the direction of the current, then the direction of the curl of the fingers will be given the direction of the magnetic field.

(ii) Fleming’slefthandrule: Stretch the forefinger, the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the central finger in the direction of the current, then the thumb points in the direction of a force in the conductor.

(iii) Fleming’s right-hand rule: Stretch the thumb/ forefinger and the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, thumb in the direction conductor, then the central finger points in the direction of current induced in the conductor.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of the brushes?
Answer:

Principle: In an electric generator, mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. It is working on the principle of electromagnetic induction.

Working: When the Axle attached to the two rings is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Let us say the coil ABCD is rotated clockwise in the arrangement. By applying Fleming’s right-hand rule, the induced currents are set up in these arms along with the directions AB and CD. Thus an induced current flows in the direction ABCD. If there are a larger number of turns in the coil, the current generated in each turn adds up to give a large current through the coil. This means that the current in the external circuit flows from B₂ and B₁.

After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both the arms change, giving rise to the net induced current in the direction DCBA. The current in the external circuit now flows from B₁ to B₂. Thus after every half rotation the polarity of the current in the respective arms changes.

There are two brushes and in the electric generator, one brush is at all times in contact with the arm moving up in the field, while the other is in contact with the arm moving down. Because of these Brushes unidirectional current is produced.

Question 17.
When does an electric short circuit occur?
Answer:
Overloading can occur when the live wire and the neutral wire come into direct current (This occurs when the insulation of wires is damaged or there is a fault in the appliance) In such a situation, the current in the circuit abruptly increases. This is called short-circuiting.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
This is used as a safety measure, especially for those appliances that have a metallic body, for example, electric press, toaster, table fan, refrigerator, etc. The metallic body is connected to the earth wire which provides a low-resistance conducting path for the current. Thus earth wire ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth and the user may not get a severe electric shock.

KSEEB SSLC Class 10 Science Chapter 13 Additional Questions and Answers

1. Fill in the blanks:

Question 1.
Magnetic field is a quantity that has both …….. and ……
Answer:
Magnitude, direction.

Question 2.
An electric current through a metallic conductor produces a ……. around it.
Answer:
Magnetic field.

Question 3.
In electric motors, the …… acts as a commutator
Answer:
split ring.

Question 4.
…… current reverses its direction periodically.
Answer:
Alternating.

Question 5.
In our country the potential difference between two wires is ……
Answer:
220 V.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2

(Unless stated otherwise, take n \(=\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of ‘n’.
Solution:
Radius of hemisphere, r = 1 cm.
Volume of hemisphere, V \(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2 \pi}{3} \times(1)^{3}\)
Radius of Cone, r = 1 cm,
Height of Cone, h = 1 cm.
∴ Volume of a cone, V \(=\frac{1}{3} \pi r^{2} h\)

∴ Volume of Total cube \(=\frac{2 \pi}{3}+\frac{\pi}{3} \mathrm{cm}^{3}\)
= πcm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

solution:
Volume of cylinder, V = πr²h
= π × (1.5)² × 8
= 18π cm³.
Volume of Cylinder V \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \pi \times(1.5)^{2} \times 2\)
\(=\frac{3}{2} \pi \mathrm{cm}^{3}\)
∴ The volume of air contained in the model that Rachel made.

= 66 cm³.

Question 3.
A Gulab Jamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see figure given aside).


Solution:
Length of cylindrical Gulab jamun,
l = 5 – (1.4 + 1.4) = 2.2 cm.
Radius, r = 1.4 cm.
∴ Volume of Cylinder (1 Gulab Jamun), = Left hemisphere + Volume of Cylinder + Right hemisphere

∴ Total volume of such 45 Gulab jamun,
\(=45 \times \frac{22}{3} \times 0.28 \times 12.2\)
= 22 × 0.28 × 183
= 1127.28 cm³.
∴ Gulab Jamun contains sugar syrup upto about 30% of its volume.
∴ \(1127.28 \times \frac{30}{100}\)
= 338 cm³.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see
Figure)

Solution:
Radius of conical depression, r = 0.5 cm.
(Depth) height, h = 1.4 cm.
Volume of 1 depression which is conical, \(=\frac{1}{3} \pi r^{2} h\)

Volume of such 4 depressions
\(=4 \times \frac{11}{30} \mathrm{cm}^{3}\)
\(=\frac{44}{30} \mathrm{cm}^{3}\)
∴ Total volume of wooden pen-stand, \(=(15 \times 10 \times 3.5)-\frac{44}{30}\)
= 525 – 1.47
= 523.53 cm³

Question 5
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Volume of water in a Cone,

Volume of lead shot \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \pi \times(0.5)^{3}\)
\(=\frac{\pi}{6} \mathrm{cm}^{3}\)
Let the number of lead balls kept in vessel \(\frac{1}{4}\) of the water flows out,
\(\Rightarrow n \times \frac{\pi}{6}=\frac{1}{4} \times \frac{200 \pi}{3}\)
\(\Rightarrow n \times \frac{\pi}{6}=\frac{100}{6} \pi\)
∴ n = 100
∴ Number of lead balls =100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. (Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14).
Solution:
Height of solid iron pole, h = 220 cm.
Base diameter of pole, d = 24 cm.
∴ Radius, r = 12 cm.
Height of cylinder, h = 60 cm.
Radius of cylinder, r= 8 cm.
Mass of the pole = ?
Volume of 1^st Cylinder, V = πr²h
= π × (12)² × 220
Volume of 2^nd cylinder, V = πr²h
= 77 × (8)² × 60
∴ Total Volume = π × (12)² × 220 + π × (8)² × 60
= {144 × 220 + 64 × 60}π
= 35520π
= 111532.8 cm³.
Mass of iron about 1 ccm is 8 gm.
∴ Mass of iron 111532.8 ……. ??

= 111.5328 × 8 kg.
= 892.26 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm. and its height is 180 cm.
Solution:
(i) Radius of cylinder GHEF, r = 60 cm.
Height, h = 120 + 60
= 180 cm.

Volume of cylinderical vessel \(=\pi r^{2} h\)
= π × (60)² × 180
(ii) Volume of hemisphere + Volume of cone

(iii) Quality of water remained in cylinder.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements and n = 3.14.
Solution:
Diameter of spherical glass vessel = 8.5 cm.
∴ r = 4.25 cm.
Length of cylindrical neck = 8 cm.
Diameter = 2 cm.
∴ Radius, r = 1 cm.
Volume of water she found = 345 cm³.

(i) Volume of Cylinder neck, V

(ii) Volume of cylindrical vessel \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \pi \times(4.25)^{3}\)
∴ Total volume of water, V

∴ Her answer is not correct.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4

Question 1.
The following distribution gives the dialy income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:

n = 50, ∴ \(\frac{n}{2}\) = 25
On a graph paper mark the following points :
(120, 12), (140, 26), (160, 34), (180, 40), (200, 50).
For the Ogive graph,

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:

To draw ogive of the less than type.
we have to join the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
From the graph, Median is 46.5 kg

n = 35 ∴ \(\frac{\mathrm{n}}{2}\) = 17.5
Class interval which has median is = (46 – 48)
l = 46, n = 35, f = 14, cf = 14, h = 2

= 46.5
∴ Median = 46.5 kg.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution, and draw its ogive.
Solution:

We can draw Ogive graph by plotting ordered pairs:
(50, 100), (55, 98), (60, 90), (65, 78). (70, 54), (75, 16).

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Solution:

(i) n = 68, ∴ \(\frac{n}{2}\) = 34
The median is in class interval (125 – 145)
l = 125, n = 68, f = 20, cf = 22, h = 20

= 125 + 12
∴ Median = 137 units.

(ii) To find out Mode:

Here, Mode is in class interval (125 – 145)
Maximum frequency, l = 125, f₁ = 20, f₀ = 13, f<sub>2/sub> = 14, h = 20

= 125 + 10.76
= 135.76
∴ Mode = 135.76 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of ‘x’ and ‘y’

Solution:
Median is 28.5
Class interval which has median is = (20 – 30)
l = 20, n = 60, f = 20, cf = 5 + a × h = 10

17 = 25 – x
∴ x = 8
5 + x + 20 + 15 + y + 5 = 60
x + y – 45 = 60
x + y = 15
y = 15 – x
y = 15 – 8
∴ y = 7
∴ x = 8, y = 7

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Solution:

n = 100 ∴ \(\frac{n}{2}\) = 50
Median in which C.I. is (35 – 40)
l = 35, n = 100, f = 33, cf = 45, h = 5

= 35 + 0.76
∴ Median = 35.76 years

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, 171.5 – 180.5)
Solution:

n = 40 ∴ \(\frac{n}{2}\) = 20
C.I. which has median is (144.5 – 153.5)
l = 144.5, n = 40, f= 12, cf = 17, h = 9

= 144.5 + 2.25
∴ Median = 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median life time of a lamp.
Solution:

n = 400 ∴ \(\frac{n}{2}=\frac{400}{2}=200\)
Class interval having median is = (3000 – 3500)
l = 3000, n = 400, f = 86, cf = 130, h = 500

= 3000 + 406.98
∴ Median = 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:

(i) n = 100, ∴ \(\frac{n}{2}\) = 50
Class interval having median is (7 – 10)
l = 7, n = 100, f = 40, cf= 36, h = 3

= 7 + 1.05
∴ Median = 8.05 Letters

(ii) Clall interval which has mode is (7 – 10)
Maximum frequency, l = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3

= 7 + 0.88
∴ Mode = 7.88

(iii) Mean(\(\overline{X}\)) : Step Deviation Method:


= 8.5 – 0.18
= 8.32
∴ Mean = 8.32
∴ (i) Median = 8.05 letters
(ii) Mode = 7.88
(iii) Mean = 8.32.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

(i) n = 30, ∴ \(\frac{\mathbf{n}}{2}\) = 15
Class interval having median is (55 – 60)
l = 55, n = 30, f = 6, cf = 13, h = 5

= 55 + 1.67
∴ Median = 56.67 kg.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds.

Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds

KSEEB SSLC Class 10 Science Chapter 4 Intext Questions

Text Book Part II Page No. 5

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂?
Answer:
In carbon dioxide molecule, the two oxygen atoms are bonded on either side with carbon atom be double bonds. These there are 2 double bonds in CO₂. Carbon shares its electrons in the formation of a double bond with one
oxygen atom and another two electrons with another oxygen atom. In this process, both the oxygen atoms and the carbon atom acquire the stable electronic configuration of the noble gas neon. The formation of CO₂ molecule is shown below.

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?
(Hint – The eight atoms of sulphur are joined together in the form of a ring.)
Answer:

Text Book Part II Page No. 12

Question 1.
How many structural isomers can you draw for pentane?
Answer:
Pentane has 3 structural isomers. We can write as follows.
i) CH₃-CH₂-CH₂-CH₂-CH₃

Question 2.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

1. Catenation: It is the ability to form bonds with other atoms of carbon.
2. Tetravalency: With the valency of four, carbon is capable of bonding with four other atoms.

Question 3.
What will be the formula and electron dot structure of cyclopentane?
Answer:
Molecular formula of cyclopentane is: C₅H₁₀
Electron dot structure:

Question 4.
Draw the structures for the following compounds.

1. Ethanoic acid
2. Bromopentane*
3. Butanone
4. Hexanal.

* Are structural isomers possible for bromopentane ?
Answer:

Question 5.
How would you name the following compounds?
(i) CH₃—CH₂—Br


Answer:

1. Bromoethane
2. Methanol
3. Hexane.

Text Book Part II Page No. 15

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
Addition reaction means adding oxygen. Adding ethanol to potassium permanganate, we get ethanoic acid. Hence this reaction is called oxidation reaction.

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
Air, also contains other gases like nitrogen, carbon dioxide and few more gases apart from oxygen. When ethyne is burnt in air, it gives a sooty flame. This is due to incomplete combustion caused by the limited supply of oxygen. However, if ethyne is burnt with oxygen, it gives a clean flame with temperature 3000°C because of complete combustion. This oxy-acetylene flame is used for welding. It is not possible to attain such a high temperature without mixing oxygen. This is the reason why a mixture of ethyne and air is not used,
2HC ≡ CH + 5O₂ → 4CO₂ + 2H₂O + Heat.

Text Book Part II Page No. 18

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:
All the carboxylic acids decompose sodium hydrogen carbonate giving brisk effervescence of carbon dioxide gas whereas ethanol does not react with sodium hydrogen carbonate

Experiment:

1. Take two test tubes, label them as A and B
2. Take about 0.5 g of sodium hydrogen carbonate (NaHco₃) in each test tube
3. Add 2 ml of ethanol in test tube A and 2ml of ethanoic acid in test tube B.
4. We can observe the gas bubbles in test tube B. No such bubbles are seen in test tube A. Pass the gas produced in test tube B through lime water taken in another test tube
5. We will find that lime water turns milky It is a test for carbon dioxide.

Hence, this experiment proves that when ethanoic acid reacts with sodium hydrogen carbonate, then carbon dioxide gas is produced with an effervescence (a rapid evolution of gas bubbles). Ethanol does not react with NaHCO₃.

Question 2.
What are oxidising agents?
Answer:
Oxidising agents are the substances that gain electrons in redox reaction and whose oxidation number is reduced. Examples: KMnO₄ or K₂Cr₂O₇. They have the ability to oxidize or give their oxygen to other substances.

Text Book Part II Page No. 20

Question 1.
Would you be able to check if water is hard by using a detergent?
Answer:
Detergent gives lather both with hard and soft water, while a soap gives lather with soft water only. Thus, it is not possible to check if the water is hard by using a detergent.

Question 2.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
A soap molecule has two parts namely hydrophobic and hydrophilic. With the help of these particles, it attaches to the grease or dirt particle and forms a cluster called micelle. These micelles remain suspended as a colloid. To remove these micelles, it is necessary to agitate clothes.

KSEEB SSLC Class 10 Science Chapter 4 Textbook Exercises

Question 1.
Ethane, with the molecular formula C₂H₆ has
(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds.
Answer:
(b) 7 covalent bonds.

Question 2.
Butanone is a four-carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone.
(d) alcohol.
Answer:
(c) ketone.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH3Cl.
Answer:
Carbon has a valency of four. It shares one electron to each 3 hydrogen atoms and one more electron with chlorine.

Question 5.
Draw the electron dot structures for
(a) ethanoic acid.
(b) H₂S.
(c) propanone.
(d) F₂.
Answer:
a) Ethanoic acid

b) H₂S

c) propanone

(d) F₂

Question 6.
What is an homologous series? Explain with an example.
Answer:

A homologous series is a series of carbon compounds that have different numbers of carbon atoms but contain the same functional group. Every next member of a homologous series has a clear difference of 14 units of mass.

For example, methane, ethane, propane, etc., are all part of the alkane homologous series. The general formula of this series is C_nH_2n+2.

An example is explained with formula as below:

1. Methane, CH₄
2. Ethane, CH₃CH₃
3. Propane, CH₃CH₂CH₃
4. Butane, CH₃CH₂CH₂CH₃

It can be noticed that there is a difference of -CH₂ unit between each successive compound.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:

Ethanol and Ethanoic acid can be differentiated on the basis of their following properties by:

1. Ethanol is a liquid at room temperature with a pleasant smell. Ethanoic acid has a melting point of 17°C. Since it is below the room temperature so, it freezes during winter. Moreover, ethanoic acid has a smell like vinegar.
2. Ethanol does not react with metal carbonates while, ethanoic acid reacts with metal carbonates to form a salt, water and carbon dioxide.
   For example:
   2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ +H₂O
3. Ethanol does not react with NaOH while ethanoic acid reacts with NaOH to form sodium ethanoate and water.
   For example,
   CH₃COOH + NaOH → CH₃COONa + H₂O
4. Ethanol is oxidized to give ethanoic acid in the presence of acidified KMnO₄ while no reaction takes place with ethanoic acid in the presence of acidified KMnO₄.

Difference in physical properties:

Ethanol	Ethanoic acid
This is in liquid form at room temperature. Its melting point is 156° K.	Its melting point is 290K and hence it often freezes during winter in cold climates.
Difference in chemical properties
Ethanol will not react with metallic carbo­nates.	Ethanoic acid reacts with carbo­nates and Hydrogen carbonate and forms salts, carbon dioxide and water.

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
A soap molecule has two ends. One end is hydrophilic and another end is hydrophobic. When soap is dissolved in water and clothes are put in the soapy solution, soap molecules converge in a typical manner to make a structure is called micelle. The hydrophobic ends of different molecules surround a particle of grease and make the micelle, which is a spherical structure.

In this, the hydrophilic end is outside the sphere and hydrophobic end is towards the centre of the sphere. This is why micelle formation takes place when soap is added to water. Since ethanol is not as polar as soap, micelles will not be formed in other solvents such as ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon in all its allotropic forms, burns in oxygen to give carbondioxide along with the release of heat and light. Most carbon compounds also release a large amount of heat and light on burning. Hence carbon and its compounds are used as fuels for most applications.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
Hard water often contains salts of calcium and magnesium. Soap molecules react with the salts of calcium and magnesium and form a precipitate. This precipitate begins floating as an off-white layer over water. This layer is called scum. Soaps lose their cleansing property in hard water because of the formation of scum.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue)?
Answer:
Soap is basic in nature, hence red litmus changes to blue. Blue litmus is seen blue only.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:

Hydrogenation is a reaction between hydrogen and other compounds in the presence of the desired catalyst. Hydrogenation is used for reducing saturated hydrocarbons. Hydrogenation is an addition reaction. For example: When ethane is heated with the catalyst, nickel, it is reduced to ethane.

Industrial application:

1. In the petrochemical industry, hydrogenation is used to convert alkenes into alkanes (paraffin) and cyclo-alkanes.
2. It is also used to prepare vegetable cooking fat from vegetable oils.

Question 13.
Which of the following hydro-carbons undergo addition reactions:
C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄.
Answer:
C₂H₆ and C₂H₂ are unsaturated Hydrocarbons. Hence these undergo addition reactions.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
Butter contains saturated fats. Therefore, it cannot be hydrogenated. On the other hand, oil has unsaturated fats. That is why it can be hydrogenated to saturated fats (solids).

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:

Soap are molecules in which the two ends have differing properties. One is hydrophilic, that is, it interacts with water, while the other end is hydrophobic, that is, it interacts with hydrocarbons. In the clusters of molecules in which the hydrophobic tails are on the surface of the cluster. This formation is called micelle.

Since the oily dirt will be collected in the centre of the micelle. The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. Thus the dirt suspended in the micelles is also easily rinsed away.

KSEEB SSLC Class 10 Science Chapter 4 Additional Questions and Answers

Question 1.
Write the electron dot formula of O₂.
Answer:

Question 2.
What is substitution Reaction? Give an example.
Answer:
If one type of atom or a group of atoms takes the place of another, it is called substitution reaction.
Eg: CH₄ + Cl₂ CH₃Cl + HCl (in the presence of sunlight)

Question 3.
Name 2 commercially important compounds.
Answer:
Ethanol and ethanoic acid.

Question 4.
Give an example for Esterification reaction.
Answer:

Question 5.
Write one use of ester.
Answer:
Esters are used in making perfumes and as flourishing agents.

Question 6.
What are detergents?
Answer:
Detergents are generally sodium salts of sulphonic acids or ammonium salts with chlorides or bromides etc.

Question 7.
Where is Ethanol used?
Answer:
It is used in medicines such as tincture iodine, cough syrups, and many tonics.

Question 8.
What is vinegar? Mention one of its use.
Answer:
5 to 8% solution of acetic acid in water is called vinegar. It is used as a preservative in pickles. We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency ?
Solution:
(i) Here the maximum class interval is 35 – 45. This class interval has mode
l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10

∴ Mode = 36.8 years

(ii) Mean (\(\overline{X}\)) : Step Deviation Method:

Here, a = 30

= 30 + 5.37
= 35.37
∴ Mode = 36.8
∴ Mean = 35.37
Hence Age group 36.8 patients are admitted more in number.
Mean age group patients are 35.57

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components:
Solution:
Class interval which has mode = 60 – 80
Here, l = 60, f₁ = 61, f₀ = 52, f₂ = 38, h = 20

∴ Mode = 65.63 Hours

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also. And the mean monthly expenditure :
Solution:

Solution:
(i) Mode is in class interval 1500 – 2000
Here, l = 1500, f₁ = 40, f₀ = 24, f₂ = 33, h = 500

(ii) Mean (\(\overline{X}\)): Step Deviation Method:

Here, a = 2750


= 2750 – 87.50
= 2662.5
∴ Modal monthly expenditure = 1847.8
∴ Mean Monthly expenditure = Rs. 2662.5

Question 4.
The following distribution gives the state- wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
(i) Class interval which has mode 30 – 35
Here, l = 30, f₁ = 10, f₀ = 9, f₂ = 3, h = 5

∴ Mode = 30.6
(ii) Mean (\(\overline{X}\)) : Step Deviation Method :

By Step Deviation Method:

= 32.5 – 3.3
= 29.2
∴ Mode = 30.6
∴ Mean = 29.2
We conclude that more number of states has teacher-student ratio about 30.6.
Mean of Age ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Solution:
Class interval which has mode is = 4000 – 5000
Maximum frequency is
Here, l = 4000, f₁ = 18, f₀ = 4, f₂ = 9, h = 1000

= 4000 + 608.7
= 4608.7
∴ Mode = 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Solution:
Class interval which has mode is 40- 50
Maximum frequency is 20.
Here, l = 40, f₁ = 20, f₀ = 12, f₂ = 11, h = 10

= 40 + 4.7
= 44.7
∴ Mode = 44.7 cars.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution.

Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution

KSEEB SSLC Class 10 Science Chapter 9 Intext Questions

Text Book Part II Page No. 53

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B have higher percentage hence, it is likely to have arisen earlier. In asexual reproduction, there would be only very minor differences generated due to small inaccuracies in DNA copying, so trait B, which exists in 60% of the same population may get inherited earlier while trait A which exists in 10% of the population may be originated late due to variations.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Variation is a process which occurs sometimes at cellular level of reproduction. It cause small changes in the genotype natural selection selects the individuals having useful variations which ensure their survival in the prevailing conditions of environment, variant individuals that can withstand or cope with prevailing environment will survive letter and will increase in number through reproduction.

Text Book Part II Page No. 57

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
The trait which appears in all the members of F1 generation and also in 75% numbers of F2 generation obtained by self fertilisation of F1 generation is dominant character.
The trait which does not appear in F1 generation but after selffertilisation of F1 generation, reappears in 25% of F2 generation is known as recessive.

Question 2.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel thought that if two different characteristics, rather than just one are bred with each other. What do the progeny of a tall plant with round seeds and a short plant with wrinkled seeds look like? They are all tall and have round seeds. Tallness and round seeds are thus dominant traits, But what happens when these F₁ progeny are used to generate F₂ progeny by self pollination? A Mendelian experiment will find that some F₂ progeny are tall plants with round seeds and some were short plants with wrinkled seeds. However there would also be some F₂ progeny that showed new combinations. Some of them would be tall, but have wrinkled seeds, while others would be short, but have round seeds, you can see as to how new combinations of traits are formed in F₂ offspring when factors controlling for seed shape and seed colour recombine to form zygote leading to form F₂ offspring. Thus the tall/short trait and the round seed/wrinkled seed trair are independently inherited.

Question 3.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
No this information is not sufficient to determine which of the traits: blood group A or O is dominant. Blood group A can be genotypically AA or AO. Hence, the information is incomplete.

Question 4.
How is the sex of the child determined in human beings?
Answer:

Most human chromosomes have a maternal and a paternal copy, and we have 22 such pairs. But one pair, called the sex chromosomes, is odd in not always being a perfect pair. Women have a perfect pair of sex chromosomes, both called x. But men have a mismatched pair in which one is a normal-sized x, while the other is a short one called y. So women are xx, while men are xy. As fig. shows half the children will be boys and half will be girls. And children will inherit an x chromosome from their mother regardless of whether they are boys of girls. Thus the sex of the children will be determined by what they inherit from their father. A child who inherits an x chromosome, from her rather will be a girl, and one who inherits a y chromosome from him will be a boy.

Text Book Part II Page No. 60

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:

1.  Natural selection,
2.  Genetic drift.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Traits acquired physically, emotionally are not a change which affects the genotype of an individual so it does not get inherited for coming generations and also acquired trait involves change in non-reproductive tissues which cannot be passed on to germ cells or the Progeny. Therefore, these traits cannot be inherited.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
The small number of tigers indicates that tiger variants are having many challenges and not capable to adopt the existing environment and may extinct soon. The small number of members in a population of tigers may cause small number of variations, which are essential for the survival of the species. A deadly disease, or calamity may be fatal to alhthe tigers. Since, tigers are among the top consumers in our ecosystem hence, their presence in our surrounding is a must.

Text Book Part II Page No. 61

Question 1.
What factors could lead to the rise of a new species?
Answer:
Genetic drift and Natural selection are the two factors which lead to the rise of a new species.

Question 2.
Will geographical isolation be a major factor in the speciation of a self pollinating plant species? Why or why not?
Answer:
No, geographical isolation cannot prevent speciation in this case, since the plants are self-pollinating, which means that the pollens are transferred from the anther of one flower to the stigma of the same flower or of another flower of the same plant. Geographical isolation can prevent the transfer of pollens among different plants only and not pollination.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, because geographical isolation does not affect much in asexually reproducing organisms. Asexually reproducing organisms pass on the parent DNA to offsprings that leaves no chance of speciation. However, geographical isolation works as a major factor in cross pollinated species. As it would result in pollinated species and accumulation of variation in the two geographically separated population.

Text Book Part II Page No. 66

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
Birds and reptiles are great example of two close species. Feathers in some ancient reptiles, as fossils indicate, they were evolved to provide insulation in cold weather. However, they cannot fly with these feathers, later on birds adapted the feathers to flight. This means that birds are very closely related to reptiles, since dinosaurs were reptile.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
The wing of a butterfly and the wing of a bat are similar in function i.e., flying. They look similar because of common use for flying, but their origins are different. Since, they perform similar function, they are analogous organs and not homologous.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
Usually when organisms die, their bodies will decompose and be lost. But every once in a while, the body or at least some parts may be in an environment that does not let it decompose completely. If a dead insect gets caught in hot-mud, for example, it will not decompose quickly, and the mud will eventually harden and retain the impression of the body parts of the insect. All such preserved traces of living organisms are called fossils.
Fossils explain about the extinct species every existed.

Text Book Part II Page No. 68

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
A species is a group of organisms that are capable of interbreeding to produce a fertile offspring. Skin colour, looks and size are all variety of features present in human beings. These features are genetic but also environmentally controlled. Various human races are formed based on these features. All human races have more than enough similarities to be classified as same species. Therefore, all human beings are a single species as humans of different colour, size and looks are capable of reproduction and can produce a fertile off spring.

Because all humans are a single species.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:

Evolution cannot always be equated with progress or better body designs. Evolution simply creates more complex body designs. However, this does not mean that the simple body designs are inefficient. In fact, bacteria having a simple body design are still the most cosmopolitan organisms found on earth. They can survive in hot springs, deep sea and even freezing environment.

Therefore, bacteria, spiders, fish and chimpanzees are all different branches of evolution.

KSEEB SSLC Class 10 Science Chapter 9 Textbook Exercises

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW.

Question 2.
An example of homologous organs is
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) all of the above.

Question 3.
In evolutionary terms, we have more in common with
(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
This information is not complete. We cannot say anything about whether the light eye colour trait is dominant or recessive only one generation is there.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:

Classification shows how clearly organisms are closely related with respect do evolution. As we know that each organism has descended from its ancestral type with some modification.

Classification involves grouping of organism based on similarities in internal and external structure or evolutionary history.

Two species are more closely related, if they have more characteristics in common. Different organisms would have common features if they are inherited from a common ancestor. And, if two species are more closely related, then it means they have a more recent ancestor. With subsequent generations, the variations make organisms more different than their ancestors. This discussion clearly proves that we classify organisms according to their resemblance which is similar to creating an evolutionary tree.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Homologous organs are those organs of different organisms which have the same basic structural design and origin but perform different functions. Example: the forelimbs of humans and the wings of birds look different externally but their skeletal structure is similar.

Analogus organs are those organs of different organism which have the different basic structural design and origin but have similar functions. Example: the wings of birds and bat.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Dog has 11 genes A….T. It acquires one chromosome from his father or mother. As per genetics Dog may be black or brown. They have 25% of BB and 50% Bb and 25% bb of genes. It is as follows.

	B	b
B	BB	Bb
b	Bb	bb

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:

The fossils are the remains of organisms that once existed on earth.

Fossil provide us evidence about

• The organisms that lived long ago , their structure etc.
• Evolutionary development of particular species i.e., line of their development with time and other environmental factors.
• Connecting links between groups. For example, feathers present in reptiles means that birds are very closely related to reptiles.
• Which organisms evolved earlier and which later.
• Development of complex body designs from the simple body designs.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
J.B.S. Haldane, a British scientist (who became a citizen of India later), suggested in 1929 that life must have developed from the simple in organic molecules which were present on earth soon after it was formed. How did these organic molecules arise? An answer was suggested by the experiment conducted by Stanley L. Miller and Harold C Urey in 1953.

They assembled an atmosphere similar to that thought to exist on early earth (this had molecules like ammonia, methane) and Hydrogen sulphide, but no oxygen) over water. This was maintained at a temperature just below 100° c and sparks were passed through the mixture of gases to stimulate lighting. At the end of a week, 15% of the carbon (from methane) had been converted to simple compounds of carbon including amino acids which make up protein molecules. This is the evidence we have for the life from inanimate matter.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexusal reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:

Sexual reproduction causes more viable variations due to the following reasons:

Two different types of gametes meet to form new individuals which have better possibilities of combinations of traits. Due to the presence of multiple traits, error in copying of DNA are highly significant. At the time of gamete formation, random aggregation of paternal and maternal chromosome occurs. Exchange of genetic material may occur between , homologous chromosomes during formation of gametes. Due to sexual reproduction over generation after generation and selection by nature created wide diversity. In case of asexual reproduction, only the very small changes due to inaccuracies in DNA copying pass on the progeny. Thus, offsprings of asexual reproduction are more or less genetically similar to their parents. So, it can be concluded that evolution in sexually reproducing organisms proceeds at a faster pace than in asexually reproducing organisms.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Each cell will have two copies of each chromosome, one each from the male and female parents. Every germ cell will take one chromosome from each pair and these may be of either maternal or paternal origin. When two germ cells combine, they will restore the normal number of chromosomes in the progeny, ensuring the stability of the DNA of the species. Hence there is equal genetic contribution of male and female parents ensured in the progeny.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, we agree with the statement that only variations that confer an advantage to an individual organism will survive in a population. Genetic variations are part of natural selection. Genetic variations are one form of variation. Selection is done with respect to the environment. Every variation do not have a survival in the environment in which they exist. The chances of surviving depend upon adaptation by surrounding and the nature of variations. Selection of variants by environmental factors forms the basis for revolutionary process.

KSEEB SSLC Class 10 Science Chapter 9 Additional Questions and Answers

Question 1.
How do we know how old the fossils are?
Answer:
There are two components to this estimation one is relative. If we dig into the earth and start finding fossils, it is reasonable to suppose that the fossils we find closer to the surface are more recent than the fossils we find in deeper layers. The second way of dating fossils is by detecting the ratios of different isotopes of the same element in the fossil material. It would be interesting to find out exactly how this method works!

Question 2.
What is Evolution?
Answer:
Evolution is simply the generation of diversity and the shaping of diversity by environmental selection.

Question 3.
What is scientific name of man?
Answer:
Home sapients.

Question 4.
What does evolution of human beings indicate?
Answer:
It indicates that all of us belong to a single species that evolved in Africa and spread across the world in stage.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1

(Unless stated otherwise, take π \(=\frac{22}{7}\))

Question 1.
2 cubes each of volume 64cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
If measurement of each side of each cube be l cm, then

(l)³ \(=\sqrt[3]{64}\)
∴ l = 4 cm.
Length of cubes if both joined together,
Length, l = 4 + 4 = 8 cm.
breadth, b = 4 cm.
height, h = 4 cm.
Surface area of the cuboid,
= 2lb + 2lh + 2bh
= 2(8 × 4) + 2(8 × 4) + 2(4 × 4)
= 2 × 32 + 2 × 32 + 2 × 16 = 64 + 64 + 32
= 160 sq.cm.

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
(i) Diameter of a hollo hemisphere = 14 cm.
∴ Radius of a hollow hemisphere, r
r = 7 cm.
Inner curved surface of a hemisphere = 2π²


= 14 × 22
= 308 sq. cm.
(ii) Radius of hollow cylinder, r = 7 cm.
Height, h = 13 – 7 = 6 cm.
∴ Inner Surface area of the Cylinder,
= 2πrh
\(=2 \times \frac{22}{7} \times 7 \times 6\)
= 44 × 6
= 264 sq.cm.
∴ Inner surface area of the vessel = Area of hemisphere + Area of cylinder
= 308 + 264
= 572 sq. cm.

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
(i) Radius of base of circular toy,
r = 3.5 cm.
∴ Surface Area of hemisphere, = 2πr²


= 77 sq.cm
(ii) Let slant height of cone is l cm. then

∴ Curved surface area of a cone = πrl


= 137.5 sq.cm.
∴ Total surface area of a toy = Surface area of hemisphere + Curved surface area of a cone
= 77.0 + 137.5
= 214.5 sq.cm.

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Solution:
(i) A cubical block of side 7 cm is surmounted by a hemisphere.
∴ Diameter of hemisphere is = 7 cm.
Radius of hemisphere, r = 3.5 cm.
∴ Surface area of hemisphere = Curved surface area – Area of base of hemisphere


= 38.5
(ii) The curved surface area of square,
=6 × l²
= 6 × (7)²
= 6 × 49
= 294 sq. cm.
∴ Total surface area of cubical block, = Curved surface area of cubical block + Surface area of hemisphere.
= 294 + 38.5
=332.5sq.cm.

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the cube. Determine the surface area of the remaining solid.
Solution:
Let, Length of each edge of cubical block = Diameter of hemisphere = l
∴ Radius of hemisphere, \(r=\frac{l}{2} unit\).

Total surface area of newly formed cube = Curved surface area of cube – Upper part of Cube + Area of hemisphere which is depressed

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Figure given below). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

(i) Surface area of cylindrical capsule, = 2πrh
\(=2 \pi \times\left(\frac{5}{2}\right) \times 9\)
= 45π mm²,
(ii) Surface area of both hemispheres,

= 25π mm²
∴ Total surface area of capsule, = 45π + 25π mm²
= 70π mm²
\(=70 \times \frac{22}{7}\)
= 220 mm²

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1m and 4m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m². (note that the base of the tent will not be covered with canvas.)
Solution:
(i) Radius of base of cylinder, r= 2 mm.
Height of base of cylinder, h= 2.1 m.
∴ Curved Surface area of cylinder = 2πrh
= 2π × 2 × (2.1)
\(=4 \times \frac{22}{7} \times 2.1\)
= 26.4 m².

(ii) Radius of conical tent,
r = 2 m.
length, l = 2.8 m.
∴ Curved Surface area of cylinder = πrl
\(=\frac{22}{7} \times 2 \times 2.8\)
= 17.6 m².
∴ Total area of the tent = 26.4 + 17.6
= 44m²
Cost of 1 aq.m. of canvas is Rs. 500.
Cost of 44 sq. m. canvas … ??
500 × 44
= Rs. 22,000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:
Height of the cylinder, h = 2.4 m.
Diameter of the cylinder, d = 1.4 m.
∴ Radius, r = 0.7 m.
(i) Outer Curved surface area of cylinder = 2πrh

\(=2 \times \frac{22}{7} \times(0.7) \times 2.4\)
= 44. × 0.24
= 10.56 cm².
(ii) Area of base of = πr²

= 1.54 cm².
(iii) Inner surface area of cylinder, = πrl

= 5.5 cm²
∴ Total surface area of newly formed cube = 10.56 + 1.54 + 5.5 = 17.6 cm².
∴ Nearest value is 18 sq.cm.

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm. and its base is of radius 3.5 cm. find the total surface area of the article.

Solution:
Radius of base of cylinder, r= 3.5 cm
Height, h = 10 cm.

Total area of article, = Curved Surface area of Cylinder + 2 × Area of Hemisphere
= 2πrh + 2 × 2πr²

= 374 sq.cm.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why ?
Solution:
Average by Direct Method :

Question 2.
Consider the following distribution of daily wages of 30 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Step deviation method :


∴ Average = 545.20.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Solution:

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method

Solution:


∴ Mean heartbeats per minute = 75.9

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangeos kept in a packing box. Which method of finding the mean did you choose ?
Solution:
Step Deviation Method :

∴ Mean Number of mangoes in the boxes = 57.19

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.

Solution:
By Step deviation method:


∴ Mean Daily expenditure for food = Rs. 211.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution:
By step Deviation method:

∴ Mean Concentration of SO₂ = 0.099 ppm.

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:
Mean Deviation method:

Here, a = 17

∴ Mean number of days a student was absent = 12.48 Days.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:
Step Deviation Method:

By Step Deviation Method

∴ Mean Literacy rate = 69.43%

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4

(Use π = \(\frac{22}{7}\) unless stated otherwise.)

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Height of drinking glass, h = 14 cm.
Diameter of circular ends
D = 4 cm ∴ r = 2 cm.
d = 2 cm ∴ r = 1 cm.
Capacity of glass = ?

Volume of glass = Radius with 2 cm + Volume of frustum with radius 1 cm.

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its ciruclar ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Slant height of a i frustum of a cone, l = 4 cm.
Circumference of its circular ends = 18 cm. and 6 cm

2πr =18 cm.
∴ R = \(\frac{9}{\pi}\) cm.
2πr = 6
∴ r = \(\frac{3}{\pi}\) cm.
∴ Lateral surface area of a frustum of a Cone

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure).

If its radius on the open side is 10 cm, radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it.
Solution:
Radius on the open side of the cap, R = 10 cm.
Radius of upper end, r = 4 cm.
Slant height l = 15 cm

R = 10 cm, r = 4 cm, l = 15 cm
∴ Curved surface = π × l × (R + r)
= π × 15 × (10 + 4)
= \(\frac{22}{7}\) × 15 × 14
= 660 cm²
∴ Area of the cap at the end

∴ Area of material used for making it.

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm2. (Take π = 3.14).
Solution:

R = 20 cm
r = 8 cm
h = 20 cm
∴ Slant height,

∴ Volume of metallic sheet,

Quantity of milk in the container
= \(\frac{10449.82}{1000}\)
Cost of 1 litre of milk is Rs. 20,
Cost of 10.45 litres of milk ……. ??
∴ 20 × 10.45 = Rs. 209.
Cost of metal sheet = π(R + r) + πr²
= π {20 × (20 + 8) + (8)²}
= 3.14 × 624
= 1959.36 cm².
∴ Cost of preparing metallic container: For 100 cm² Rs. 8
∴ For 1959.36 cm² ……?
= 8 × \(\frac{1159.36}{100}\)
= Rs. 156.75.

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) cm., find the length of the wire.
Solution:
Height of a cone, h = 20 cm.
Vertical angle = 60°

Diameter of frustum wire = \(\frac{1}{16}\) cm.
i) In ⊥∆AQC, ∠Q = 90°

ii) In ⊥∆PRA, ∠R = 90°

Height of frustum, h = 10 cm.
∴ Volume of metallic frustum, ABCD,

Let the length of wire which has diameter \(\frac{1}{16}\) cm be ‘x’ cm

∴ Length of wire = 7964.4 m

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4, drop a comment below and we will get back to you at the earliest.