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 B $0.016 J$
Work done $= \frac{1}{2} \times s t r e t c h i n g f o r c e \times e x t e n s i o n$ .
Therefore, net work done in increasing the length from $0.164 m m$ to $1.02 m m$ is
$W_{2} - W_{1} = \frac{1}{2} \times 5 \times 9.8 \times 1.02 \times 10^{- 3} - \frac{1}{2} \times 3 \times 9.8 \times 0.61 \times 10^{- 3} = 0.016 J$ .

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

The correct option is B 0.016JWork done =12×stretching force×extension.Therefore, net work done in increasing the length from 0.164 mm to 1.02 mm isW2−W1=12×5×9.8×1.02×10−3−12×3×9.8×0.61×10−3=0.016 J.

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