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Physics MCQs for NEET — Practice Questions with Answers

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To break a wire, a force of $10^{6} N / m^{2}$ is required. If the density of the material is $3 \times 10^{3} k g / m^{3}$ , then the length of the wire which will break by its own weight will be -

34 m

30 m

300 m

3 m

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Explanation

(a) $L = \frac{P}{d g} = \frac{10^{6}}{3 \times 10^{3} \times 10} = \frac{100}{3} = 34 m$

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If the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm will be

V/25

V/5

25V

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Explanation

(d) $U = \frac{1}{2} (\frac{Y A}{L}) l^{2} ∴ U \propto l^{2} \frac{U_{2}}{U_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} = {(\frac{10}{2})}^{2} = 25 \Rightarrow U_{2} = 25 U_{1}$

i.e. potential energy of the spring will be 25 V

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The work done in stretching an elastic wire per unit volume is or strain energy in a stretched string is

Stress $\times$ Strain

$\frac{1}{2} \times$ Stress $\times$ Strain

$2 \times s t r a i n \times s t r e s s$

Stress/Strain

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Explanation

(b) $s t r e s s e n e r g y = \frac{1}{2} s t r e s s \times s t r a i n$

work done by a force on a wire

$W = \frac{1}{2} (y A \frac{∆ L}{L}) \times ∆ L = \frac{1}{2} (y \frac{∆ L}{L}) (\frac{∆ L}{L}) (A L) = \frac{1}{2} s t r e s s \times s t r a i n \times v o l u m e \Rightarrow \frac{W}{v o l u m e} = \frac{1}{2} s t r e s s \times s t r a i n$

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Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be

1 : 2

1 : 4

2 : 1

1 : 1

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Explanation

(a)

$W = \frac{1}{2} F l ∴ W \propto l (F i s c o n s \tan t) ∴ \frac{W_{1}}{W_{2}} = \frac{l_{1}}{l_{2}} = \frac{l}{2 l} = \frac{1}{2}$

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A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

Zero

0.05 joule

100 joule

500 joule

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Explanation

(b)

$W = \frac{1}{2} \times F \times l = \frac{1}{2} m g l = \frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0.05 J$

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If the force constant of a wire is K, the work done in increasing the length of the wire by l is

$K l / 2$

$K l$

$K l^{2} / 2$

$K l^{2}$

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Explanation

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When strain is produced in a body within elastic limit, its internal energy

Remains constant

Decreases

Increases

None of the above

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Explanation

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When shearing force is applied on a body, then the elastic potential energy is stored in it. On removing the force, this energy

Converts into kinetic energy

Converts into heat energy

Remains as potential energy

None of the above

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Explanation

(b)

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A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in length is 1.0 mm, the increase in energy of the wire will be

0.01 J

0.02 J

0.04 J

1.00 J

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Explanation

(a) $I n c r e a s e i n e n e r g y = \frac{1}{2} \times 20 \times 1 \times 10^{- 3} = 0.01 J$

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The ratio of Young's modulus of the material of two wires is 2 : 3. If the same stress is applied on both, then the ratio of elastic energy per unit volume will be-

3 : 2

2 : 3

3 : 4

4 : 3

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Explanation

(a)

$E n e r g y p e r u n i t v o l u m e = \frac{{(s t r e s s)}^{2}}{2 Y} \frac{E_{1}}{E_{2}} = \frac{Y_{2}}{Y_{1}} (S t r e s s i s c o n s \tan t) ∴ \frac{E_{1}}{E_{2}} = \frac{3}{2}$

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