Electricity(10th Grade > Physics ) Questions and answers for exam Preparation

10th Grade > Physics

ELECTRICITY MCQs

Total Questions : 244 | Page 2 of 25 pages

Question 11. How much electrical energy flows through a wire in 1 second when the power is 1 kW?

1400 J
1000 J
800 J
400 J

Discuss Question

Answer: Option B. -> 1000 J
:
B
We know that,

P o w e r = \frac{Energy}{time}

, the SI unit of power is watt, energy is joule and time is second.
So, 1 W = 1 joule per sec and 1 kW = 1000 joule per sec.
Therefore, 1000 Jelectrical energy flows through the wire in 1 second when the power is 1 kW.

Question 12. Ohm’s Law gives the relation between potential difference and current i.e, voltage is directly and linearly proportional to current.

True
False

Discuss Question

Answer: Option A. -> True
:
A
According to Ohm's law:

V = I \times R

; this means if R (resistance) is constant, current increases linearly as the voltage is increased.
At a constant temperature, following is the graph between voltage and current stated asOhm's law.

Ohm’s Law Gives The Relation Between Potential Difference ...

From the above graph, we can say that thecurrent varieslinearly with potential difference.

Question 13. For making a strong electromagnet, the material of the core should be:

soft iron
steel
brass
copper

Discuss Question

Answer: Option A. -> soft iron
:
A
The soft iron inside the coil makes the magnetic field stronger because it becomes a magnet itself when current is flowing through the coil.
Moreover, it loses its magnetism as soon as the current stops flowing. It forms a temporary magnet.In this way, the electromagnet can be magnetised and demagnetised by turning the electricity on and off.

Question 14. With the increase in temperature, the resistance of a pure metal ____.

increases
decreases
does not change
first increases and then decreases

Discuss Question

Answer: Option A. -> increases
:
A
With the increase in temperature, the random motion of electrons increases. As a result, the number of collisions of electrons with the positive ions increases in a metal. Hence, the resistance of a metal increases with increase in temperature.

Question 15. Five resistors are connected in circuit as shown with a 3 V battery. Find the ammeter reading.

Discuss Question

Answer: Option A. -> 1 A
:
A
Given,
Resistance values,

R 1 = R 2 = R 3 = 3 Ω

R 4 = R 5 = 0.5 Ω

Here the resistances

R 1

and

R 2

are in series. Let the equivalent resistance of this combination be

R_{12}

R_{12} = R 1 + R 2

R_{12} = 3 + 3 = 6 Ω

Now

R_{12}

and

R 3

are in parallel.
Let the equivalent resistance of thiscombination be

R_{123}

⟹ \frac{1}{R_{123}} = \frac{1}{R_{12}} + \frac{1}{R 3}

\frac{1}{R_{123}} = \frac{1}{6} + \frac{1}{3}

R_{123} = 2 Ω

Now,

R 4, R_{123}

and

R 5

are in series.
Let the equivalent resistance of this combination be R.

⟹ R = R 4 + R_{123} + R 5

R = 0.5 + 2 + 0.5 Ω

R = 3 Ω

Using Ohms law,

I = \frac{V}{R}

I = \frac{3 V}{3 Ω} = 1 A

Question 16. In how much time (in seconds) should a charge of

10 C

flow between the two terminals so as to maintain a constant current of

100 A

110 seconds
10 seconds
110 seconds
1110 seconds

Discuss Question

Answer: Option A. -> 110 seconds
:
A
Current

(I)

is the rate of flow of charge

(q)

through a cross section per unit time

(t)

I = \frac{q}{t}

Given:

q = 10 C

I = 100 A

t = \frac{q}{I}

t = \frac{10 C}{100 A}

t = \frac{1}{10} s

Question 17. If the length of a copper wire is increased by

5 %

due to stretching. What will be the percentage increase in its resistance?

12%
11.5%
20.25%
10.25%

Discuss Question

Answer: Option D. -> 10.25%
:
D
Resistance,

R = ρ \frac{L}{A}

where:

ρ

: Resistivity,

L

: Length of the wire

A

: Cross-sectional area of the wire.
The length of the copper wire is increased by

5 %

.
So, new length is:

L^{'} = L + \frac{5}{100} L = 1.05 L

Let the new cross-sectional area be

A^{'}

.
On stretching the wire, its volume will remainconstant.

⟹ L A = L^{'} A^{'}

A^{'} = \frac{A}{1.05}

New resistance will become:

R^{'} = ρ \frac{L^{'}}{A^{'}}

R^{'} = ρ (\frac{1.05 L}{A / 1.05})

R^{'} = {1.05}^{2} R = 1.1025 R

Percentage increase in its resistance is,

p = \frac{R^{'} - R}{R} \times 100

p = \frac{(1.1025 - 1) R}{R} \times 100

p = 10.25 %

Question 18. An electric bulb is rated

220 V

and

100 W .

When it is operated at

110 V,

the power consumed will be

100 W
75 W
50 W
25 W

Discuss Question

Answer: Option D. -> 25 W
:
D
Power consumed by the electric bulb,

P = \frac{V^{2}}{R}

, where

V

is the potential difference and

R

is the electrical resistance.
Given for 1st case:
Voltage

V_{1} = 220 V

and power consumed by bulb

P_{1} = 100 W,

In the 2nd case:
Voltage of the bulb,

V_{2} = 110 V

Power consumed by the bulb =

P_{2}

Since the resistance is the same as it is the same bulb

\frac{P_{1}}{P_{2}} = \frac{(V_{1})^{2}}{(V_{2})^{2}}

\frac{100}{P_{2}} = \frac{(220)^{2}}{(110)^{2}}

P_{2} = \frac{100 \times (110)^{2}}{(220)^{2}} = 25 W

Power consumed when the bulb is operated at 110V = 25W

Question 19. According to the convention, the direction of the flow of electric current is opposite to the direction of the flow of _______.

electrons
protons
neutrons
atoms

Discuss Question

Answer: Option A. -> electrons
:
A
Electric current is defined as the amount of charge flowing through a cross-section area per unit time. It's direction is conventionally taken opposite to the direction of flow of electrons.

Question 20. In parallel combination, the total resistance of the circuit becomes ___the individual resistances.

either less than or more than
equal to
more than
less than

Discuss Question

Answer: Option D. -> less than
:
D
When resistors are connected in parallel,the equivalent resistance is less than the lowest resistance in the circuit.
Proof is as follows:
As we knowwhen