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A uniform cylindrical wire is subjected to a longitudinal tensile stress of $5 \times 10^{7} {Nm}^{- 2}$ . The Young's modulus of the material of the wire is $2 \times 10^{11} {Nm}^{- 2}$ . The volume change in the wire is 0.02%. The factional change in the radius is

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$0.25 \times 10^{- 4}$

$0.5 \times 10^{- 4}$

$1.0 \times 10^{- 4}$

$1.5 \times 10^{- 4}$

answer is A.

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Detailed Solution

$Y = \frac{stress}{\frac{∆ L}{L}} or \frac{∆ L}{L} = \frac{stress}{Y} = \frac{5 \times 10^{7}}{2 \times 10^{11}} = 2.5 \times 10^{- 4}$

Now, V = ${πr}^{2} L$

$\frac{∆ V}{V} = \frac{π ∆ (r^{2} L)}{{πr}^{2} L}$

or $\frac{∆ V}{V} = \frac{r^{2} ∆ L + L \times 2 r ∆ r}{r^{2} L}$

or $\frac{∆ V}{V} = \frac{∆ L}{L} + 2 \frac{∆ r}{r}$

or $2 \frac{∆ r}{r} = \frac{∆ V}{V} - \frac{∆ L}{L}$

or $2 \frac{∆ r}{r} = \frac{0.02}{100} - 2.5 \times 10^{- 4}$

or $\frac{∆ r}{r} = 1 \times 10^{- 4} - \frac{2.5}{2} \times 10^{- 4} = - 0.25 \times 10^{- 4}$

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