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Q.

A long steel wire of length 'L' and radius ‘r’ is suspended from the ceiling of a room. A sphere of mass 'm' and radius ' $r^{1}$ ' is attached to the lower end of the wire. the height of the ceiling is $(L + 2 r^{1} + l)$ . 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 ( $L > > (r^{1}$ ,l ) )

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a

$\sqrt{\frac{y π r l}{m}}$

b

None

c

$\sqrt{\frac{π r^{2} y l}{m} - L g}$

d

$\sqrt{\frac{π r^{2} y}{m l} - L g}$

answer is A.

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Detailed Solution

$T = \frac{y A e}{L} A t l o w e s t p o i n t, T e n s i o n T = \frac{m v^{2}}{L} + m g$

$\frac{v^{2}}{L} = \frac{y A ℓ}{m L} - g$

$v = \sqrt{(\frac{y {πr}^{2} ℓ}{m} - g L)}$

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A long steel wire of length 'L' and radius ‘r’ is suspended from the ceiling of a room. A sphere of mass 'm' and radius 'r1 ' is attached to the lower end of the wire. the height of the ceiling is (L+2r1+l) . 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>>(r1 ,l ) )