A body of mass $1 k g$ is fastened to one end of a steel wire of cross-sectional area $3 \times 10^{- 6} m^{2}$ and is rotated in horizontal circle of radius $20 c m$ with a constant speed of $2 m / s$ . The elongation in the wire is $(Y = 2 \times 10^{11} N / m^{2})$

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$0.67 \times 10^{- 5} m$

$0.33 \times 10^{- 5} m$

$2 \times 10^{- 5} m$

$4 \times 10^{- 5} m$

answer is B.

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Detailed Solution

Here, $T = \frac{m v^{2}}{r} = \frac{1 \times 2^{2}}{20 \times 10^{- 2}} = \frac{4}{0.2} = \frac{40}{2} = 20 N$

Stress $= \frac{T}{A} = \frac{20}{3 \times 10^{- 6}} N / m^{2}$

$∴$ strain= $\frac{s t r e s s}{Y} = \frac{20}{3 \times 10^{- 6} \times 2 \times 10^{11}}$

Or $\frac{Δ l}{l} = \frac{20}{3 \times 10^{- 6} \times 2 \times 10^{11}}$

$∴$ $Δ l = \frac{20}{3 \times 10^{- 6} \times 2 \times 10^{11}} \times l$

$= \frac{20 \times 20 \times 10^{- 2}}{3 \times 10^{- 6} \times 2 \times 10^{11}} = \frac{2}{3} \times 10^{- 5} m$

$= 0.67 \times 10^{- 5} m$

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