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Finding Significant Figures

The work done...

Question

The work done in increasing the length of a one metre long wire of cross-sectional area $1 m m^{2}$ through 1 mm will be

250 J

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10 J

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5 J

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0.1 J

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Solution

The correct option is C $0.1 J$
$W o r k d o n e$ $= \frac{1}{2} \times$ $Y o u n g^{'} s m o d u l u s$ $\times s t r a i n^{2} \times$ $v o l u m e$

$W = \frac{1}{2} \times Y \times \frac{Δ l^{2}}{L^{2}} \times A L$

$= \frac{Y A Δ L^{2}}{2 L}$

$= \frac{2 \times 10^{11} \times 10^{- 6} \times 10^{- 6}}{2 \times 1} = 0.1 J$

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Similar questions

1 m

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1 m

1 {mm}^{2}

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Y = 2 \times 10^{11} {N/m}^{2}

1 m

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Y = 2 \times 10^{11} N m^{- 2}

1 m

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(Y = 2 \times 10^{11} N / m^{2})

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