Two wires if same material and area if cross section but with length in the ratio $5 : 3$ are streached by the same force. The ratio of work done in two cases is

5 : 8

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8 : 5

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5 : 3

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3 : 5

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Solution

The correct option is C $5 : 3$
Force in wire due to stretching, $F = \frac{Y A x}{L}$

Force is equal in both cases, $(F_{1} = F_{2})$

$\Rightarrow \frac{Y A x_{1}}{L_{1}} = \frac{Y A x_{2}}{L_{2}}$

$\Rightarrow \frac{x_{1}}{x_{2}} = \frac{L_{1}}{L_{2}} = \frac{5}{3}$

Work done in stretching a wire.

$W = F . d x = \frac{Y A x}{L} d x = \frac{1}{2} \frac{Y A}{L} x^{2} = \frac{1}{2} F . x$

$\frac{W_{1}}{W_{2}} = \frac{0.5 \times F_{1} . x_{1}}{0.5 \times F_{2} . x_{2}} = \frac{x_{1}}{x_{2}} = \frac{5}{3}$

Hence, Ratio of work done is $5 : 3$

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Two wires if same material and area if cross section but with length in the ratio 5:3 are streached by the same force. The ratio of work done in two cases is

The correct option is C 5:3Force in wire due to stretching, F=YAxL Force is equal in both cases, (F1=F2) ⇒YAx1L1=YAx2L2 ⇒x1x2=L1L2=53 Work done in stretching a wire. W=F.dx=YAxLdx=12YALx2=12F.x W1W2=0.5×F1.x10.5×F2.x2=x1x2=53 Hence, Ratio of work done is 5:3

Two wires if same material and area if cross section but with length in the ratio $5 : 3$ are streached by the same force. The ratio of work done in two cases is