When the load on a wire is slowly increased from $3$ to $5 k g w t$ , the elongation increases from $0.61$ to $1.02 m m$ . The work done during the extension of wire is

0.16 J

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0.016 J

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1.6 J

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16 J

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Solution

The correct option is C $0.016 J$
Work done = Change in elastic potential energy
Which is given by $1 / 2 \times S t r e s s \times S t r a i n \times V o l u m e$
Thus, work $= \frac{1}{2} \times \frac{3 g}{A} \times \frac{0.61 \times 10^{- 3}}{L} \times A L - \frac{1}{2} \times \frac{5 g}{A} \times \frac{1.02 \times 10^{- 3}}{L} \times A L$

$= \frac{1}{2} \times (1.83 - 5.10) \times 10^{- 2}$

$= 1.635 \times 10^{- 2}$ , which is nearly equal to $0.016 J$

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When the load on a wire is slowly increased from 3 to 5kgwt, the elongation increases from 0.61 to 1.02mm. The work done during the extension of wire is

The correct option is C 0.016JWork done = Change in elastic potential energyWhich is given by 1/2×Stress×Strain×VolumeThus, work =12×3gA×0.61×10−3L×AL−12×5gA×1.02×10−3L×AL =12×(1.83−5.10)×10−2 =1.635×10−2, which is nearly equal to 0.016J

When the load on a wire is slowly increased from $3$ to $5 k g w t$ , the elongation increases from $0.61$ to $1.02 m m$ . The work done during the extension of wire is