CBSE Class X

Question 2:

What is the principle of an electric motor?

Answer:

The working principle of an electric motor is based on the magnetic effect of

current. A current-carrying loop experiences a force and rotates when placed

in a magnetic field.The direction of rotation of the loop is given by the

Fleming’s left-hand rule.

Question 1:

State Fleming’s left-hand rule.

Answer:

Fleming’s left hand rule states that if we arrange the thumb, the centre finger,

and the forefinger of the left hand at right angles to each other, then the thumb

points towards

the direction of the magnetic force, the centre finger gives the direction of current,

and the forefinger points in the direction of magnetic field.

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) The direction of the magnetic field can be determined by the Fleming’s left

hand rule. According this rule, if we arrange the thumb, the centre finger, and

the forefinger of the left hand at right angles to each other, then the thumb

points towards the direction of the magnetic force, the centre finger gives the

direction of current, and the forefinger points in the direction of magnetic field.

Since the direction of positively charged alpha particle is towards west, the

direction of current will be the same i.e., towards west. Again, the direction

of magnetic force is towards north. Hence, according to Fleming’s left hand rule,

the direction of magnetic field will be upwards.

Question 2:

In Activity 13.7 (page: 230), how do we think the displacement of rod AB will

be affected if (i) current in rod AB is increased: (ii) a stronger horse-shoe

magnet is used: and (iii) length of the rod AB is increased?

Answer:

A current-carrying conductor placed in a magnetic field experiences a force. The

magnitude of force increases with the amount of current, strength of the magnetic

field, and the length of the conductor. Hence, the magnetic force exerted on rod

AB and its deflection will increase if

(i) current in rod AB is increased

(ii) a stronger horse-shoe magnet is used

(iii) length of rod AB is increased

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) and (d)

When a proton enters in a region of magnetic field, it experiences a magnetic

force. As a result of the force, the path of the proton becomes circular. Hence,

its velocity and momentum change.

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)The magnetic field inside a long, straight, current-carrying solenoid is uniform. It is

the same at all points inside the solenoid.

Question 2:

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:

The energy given to each coulomb of charge is equal to the amount of work

required to move it. The amount of work is given by the expression,

Work done = Potential difference × Charge

Where,

Charge = 1 C

Potential difference = 6 V

Work done = 6 × 1 = 6 J

Therefore, 6 J of energy is given to each coulomb of charge passing through a

battery of 6 V.

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:

Inside the loop = Pierce inside the table

Outside the loop = Appear to emerge out from the table

For downward direction of current flowing in the circular loop, the direction of

magnetic field lines will be as if they are emerging from the table outside the

loop and merging in the table inside the loop. Similarly, for upward direction

of current flowing in the circular loop, the direction of magnetic field lines will

be as if they are emerging from the table outside the loop and merging in the

table inside the loop, as shown in the given figure.

Question 3:

Why don’t two magnetic lines of force intersect each other?

Answer:

If two field lines of a magnet intersect, then at the point of intersection, the

compass needle points in two different directions. This is not possible. Hence,

two field lines do not intersect each other.

Question 2:

List the properties of magnetic lines of force.

Answer:

The properties of magnetic lines of force are as follows.

(a) Magnetic field lines emerge from the north pole.

(b) They merge at the south pole.

(c) The direction of field lines inside the magnet is from the south pole to the

north pole.

(d) Magnetic lines do not intersect with each other.

Question 1:

Draw magnetic field lines around a bar magnet.

Answer:

Magnetic field lines of a bar magnet emerge from the north pole and terminate

at the south pole. Inside the magnet, the field lines emerge from the south pole

and terminate at the north pole, as shown in the given figure.

Question 1:

Why does a compass needle get deflected when brought near a bar magnet?

Answer:

A compass needle is a small bar magnet. When it is brought near a bar magnet,

its magnetic field lines interact with that of the bar magnet. Hence, a compass

needle shows a deflection when brought near the bar magnet.

Question 19:

Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

(b) Why are the conductors of electric heating devices, such as bread-toasters and

electric irons, made of an alloy rather than a pure metal?

(c) Why is the series arrangement not used for domestic circuits?

(d) How does the resistance of a wire vary with its area of cross-section?

(e) Why are copper and aluminium wires usually employed for electricity

transmission?

Answer:

(a) The melting point and resistivity of tungsten are very high. It does not burn

readily at a high temperature. The electric lamps glow at very high temperatures.

Hence, tungsten is mainly used as heating element of electric bulbs.

(b) The conductors of electric heating devices such as bread toasters and electric

irons are made of alloy because resistivity of an alloy is more than that of metals.

It produces large amount of heat.

(c) There is voltage division in series circuits. Each component of a series circuit

receives a small voltage for a large supply voltage. As a result, the amount of

current decreases and the device becomes hot. Hence, series arrangement is not

used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section

(A), i.e.,

 

(e) Copper and aluminium wires have low resistivity. They are good conductors of

electricity. Hence, they are usually employed for electricity transmission.

Question 18:

An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours.

Calculate the rate at which heat is developed in the heater

Answer:

Rate of heat produced by a device is given by the expression for power as

P = I²R

Where,

Resistance of the electric heater, R = 8 Ω

Current drawn, I = 15 A

P = (15)² × 8 = 1800 J/s

Therefore, heat is produced by the heater at the rate of 1800 J/s.

Question 17:

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer:

Energy consumed by an electrical appliance is given by the expression,

H = Pt

Where,

Power of the appliance = P

Time = t

Energy consumed by a TV set of power 250 W in 1 h = 250 × 3600 = 9 × 10⁵ J

Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 × 600

= 7.2× 10⁵ J

Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy

consumed by a toaster of power 1200 W in 10 minutes.

Question 16:

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected

in parallel to electric mains supply. What current is drawn from the line if the supply

voltage is 220 V?

Answer:

Both the bulbs are connected in parallel. Therefore, potential difference across

each of them will be 220 V, because no division of voltage occurs in a parallel

circuit.

Current drawn by the bulb of rating 100 W is given by,

 

Similarly, current drawn by the bulb of rating 100 W is given by,

 

Question 15:

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected

in parallel to electric mains supply. What current is drawn from the line if the supply

voltage is 220 V?

Answer:

Both the bulbs are connected in parallel. Therefore, potential difference across

each of them will be 220 V, because no division of voltage occurs in a parallel

circuit. Current drawn by the bulb of rating 100 W is given by,

 

Similarly, current drawn by the bulb of rating 100 W is given by,

 

Question 14:

Compare the power used in the 2 Ω resistor in each of the following circuits: (i)

a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel

with 12 Ω and 2 Ω resistors.

Answer:

(i) Potential difference, V = 6 V

1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of

the circuit, R = 1 + 2 = 3 Ω

According to Ohm’s law,

V = IR

Where,

I is the current through the circuit

 

This current will flow through each component of the circuit because there is no

division of current in series circuits. Hence, current flowing through the 2 Ω resistor

is 2A.

Power is given by the expression,

P=(I)²R = (2)² × 2 = 8 W

(ii) Potential difference, V = 4 V

12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component

of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be

4 V.

Power consumed by 2 Ω resistor is given by

 

Therefore, the power used by 2 Ω resistor is 8 W.

Question 13:

A hot plate of an electric oven connected to a 220 V line has two resistance coils

A and B, each of 24 Ω resistances, which may be used separately, in series, or in

parallel. What are the currents in the three cases?

Answer:

Supply voltage, V = 220 V

Resistance of one coil, R = 24 Ω

(i) Coils are used separately

According to Ohm’s law,

V = I₁R₁

Where,

I1 is the current flowing through the coil

 

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, R₂ = 24 Ω + 24 Ω = 48 Ω

According to Ohm’s law,

V = I₂R₂

Where,

I₂ is the current flowing through the series circuit

 

Therefore, 4.58 A current will flow through the circuit when the coils are connected

in series.

(iii) Coils are connected in parallel

Total resistance, R₃ is given as

  

According to Ohm’s law,

V = I₃R₃

Where,

I₃ is the current flowing through the circuit

 

Therefore, 18.33 A current will flow through the circuit when coils are connected

in parallel.

Question 12:

Several electric bulbs designed to be used on a 220 V electric supply line, are

rated 10 W. How many lamps can be connected in parallel with each other across

the two wires of 220 V line if the maximum allowable current is 5 A?

Answer:

Resistance R₁ of the bulb is given by the expression,

Where,

Supply voltage, V = 220 V

Maximum allowable current, I = 5 A

Rating of an electric bulb P₁ = 10 W

According to Ohm’s law,

V = I R

Where,

R is the total resistance of the circuit for x number of electric bulbs

Resistance of each electric bulb, R₁ = 4840 Ω

Therefore, 110 electric bulbs are connected in parallel.