A block of weight $10 N$ is fastened to one end of a wire of cross-sectional area $3 m m^{2}$ and is rotated in a vertical circle of radius $20 c m$ . The speed of the block at the bottom of the circle is $2 m / s$ . Find the elongation of the wire when the block is at the bottom of the circle. Young’s modulus of the material of the wire $= 2.0 \times 10^{11} N / m^{2}$ .

0.002 c m

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0.003 c m

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0.001 c m

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0.004 c m

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Solution

The correct option is C $0.001 c m$

$W = 10 N, m = 1 k g, A = 3 \times 10^{- 6} m^{2} R = l = 0.2 m, V = 2 m s^{- 1}, Y = 2 \times 10^{11}, Δ l = ?$
$Δ l = \frac{Stress \times l}{Y}$
$T - W = \frac{m \times v^{2}}{R}$
$T = W + \frac{m \times v^{2}}{R}$
$T = 10 + \frac{1 \times 2^{2}}{0.2}$
$T = 30 N$
$Δ l = \frac{T \times l}{A \times Y}$
$Δ l = \frac{30 \times 0.2}{3 \times 10^{- 6} \times 2 \times 10^{11}}$
$Δ l = 10^{- 5} m$

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The correct option is C 0.001 cm W=10 N,m=1 kg,A=3×10−6 m2R=l=0.2 m,V=2 ms−1,Y=2×1011,Δl=? Δl=Stress×lY T−W=m×v2R T=W+m×v2R T=10+1×220.2 T=30 N Δl=T×lA×Y Δl=30×0.23×10−6×2×1011 Δl=10−5 m