A long steel wire of length 'L' is suspended from the ceiling of a room. A sphere of mass 'm' and radius 'r' is attached to the lower end of the wire. The height of the ceiling is (L+e). When the sphere is made to oscillate as a pendulum, its lowest point just touches the floor. The velocity of the sphere at the lowest point will be (L >> r,l and r is the radius of the wire)

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$\sqrt{\frac{π r^{2} e Y}{m} - g L}$

$\sqrt{\frac{π r^{2} Y}{m e} - g L}$

$\sqrt{\frac{π r^{2} m}{Y e} - g L}$

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answer is A.

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Detailed Solution

During oscillation, when the string is vertical, pseudo force mV²/L acting on the sphere is vertically downward so tension in the string becomes maximum.
$∴ T_{max} = m g + \frac{m V^{2}}{L}$
Elongation in the string = Height of ceiling - Length of string = (L + e) - L = e
$∴ e = \frac{T_{max \cdot L}}{A Y} = \frac{(m g + \frac{m V^{2}}{L}) L}{π r^{2} \cdot Y}$
$\Rightarrow V = \sqrt{\frac{π r^{2} e Y}{m} - g L}$