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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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The Young's modulus of a wire is Y. If the energy per unit volume is E, then the strain will be

$\sqrt{\frac{2 E}{Y}}$

$\sqrt{2 E Y}$

$E Y$

$\frac{E}{Y}$

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Explanation

(a)

$E n e r g y p e r u n i t v o l u m e = \frac{1}{2} \times Y \times {(s t r a i n)}^{2} ∴ s t r a i n = \sqrt{\frac{2 E}{Y}}$

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If a spring extends by x on loading, then the energy stored by the spring is (if T is tension in the spring and k is spring constant)

$\frac{T^{2}}{2 x}$

$\frac{T^{2}}{2 k}$

$\frac{2 x}{T^{2}}$

$\frac{2 T^{2}}{k}$

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Explanation

(b) $U = \frac{F^{2}}{2 K} = \frac{T^{2}}{2 K}$

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A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. It is stretched by an amount x. The work done is -

$\frac{Y x A}{2 L}$

$\frac{Y x^{2} A}{L}$

$\frac{Y x^{2} A}{2 L}$

$\frac{2 Y x^{2} A}{L}$

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Explanation

(c)

$W = \frac{1}{2} \times \frac{A Y}{L} \times {(x)}^{2} W h e r e x i s d e f o r m a t i o n$

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A stretched rubber has

Increased kinetic energy

Increased potential energy

Decreased kinetic energy

Decreased potential energy

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Explanation

(b)

Due to stretching, potential energy is stored in the rubber band which is released on releasing the rubber. Thus, potential energy is increased.

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When load of 5kg is hung on a wire then extension of 3m takes place, then work done will be

75 joule

60 joule

50 joule

100 joule

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Explanation

(a) $W = \frac{1}{2} F l = \frac{1}{2} \times M g \times l = \frac{1}{2} \times 5 \times 10 \times 3 = 75 J$

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