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The Young's modulus of brass and steel are $10 \times 10^{10} {Nm}^{- 1} and 2 \times 10^{11} {Nm}^{- 2}$ respectivley. A brass wire and a steel wire of the same length are extended by 1 mm under the same force. The radii of the brass and steel wire are $R_{B} and R_{s} respectively . Then$

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$R_{s} = \frac{R_{B}}{\sqrt{2}}$

$R_{s} = \frac{R_{B}}{4}$

$R_{s} = \sqrt{2} R_{B}$

$R_{s} = 4 R_{B}$

answer is B.

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Detailed Solution

$Y = \frac{F}{{πR}^{2}} \times \frac{l}{∆ l}$

F, l and $∆ l are constants .$

$R^{2} \infty \frac{1}{Y}$

$\frac{{R_{s}}^{2}}{{R_{B}}^{2}} =$ $\frac{Y_{B}}{Y_{s}}$ = $\frac{10^{11}}{2 \times 10^{11}} = \frac{1}{2}$

or $\frac{R_{s}}{R_{B}}$ = $\frac{1}{\sqrt{2}}$ or $R_{s} = \frac{R_{S}}{\sqrt{2}}$

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