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A sphere of radius 0.1 m and mass $8 π k g$ is attached to the lower end of a steel wire of length 5.0 m and diameter $10^{- 3} m$ . The wire is suspended from 5.22 m high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. Y for steel $= 1.994 \times 10^{11} N / m^{2}$

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$6.2 m / s$

$4.5 m / s$

$8.8 m / s$

$7.3 m / s$

answer is C.

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Detailed Solution

$l + Δ L + 2 r = 5.22 \Rightarrow Δ L = 5.22 - (5 + 0.2)$ = 0.02m

Tension $T = \frac{y A}{L} e = 199.4 π$

But here $R = L + Δ L + r = 5 + 0.02 + 0.1 = 5.12 m$

In circular motion at lowest point $T = m g + \frac{m v^{2}}{R} \Rightarrow \frac{m v^{2}}{R} = T - m g$

So, $\frac{8 π \times V^{2}}{5.12} = 191.4 π - 8 π \times 9.8$

$V = 8.8 m / s$

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