A sphere of radius 0.1 m and mass 8πkg is attached to the lower end of a steel wire of length 5.0 m and diameter10−3m . 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 =1.994×1011N/m2
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a
6.2m/s
b
4.5m/s
c
8.8m/s
d
7.3m/s
answer is C.
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Detailed Solution
l+ΔL+2r=5.22 ⇒ΔL=5.22−(5+0.2) = 0.02mTension T=yALe=199.4πBut here R=L+ΔL+r=5+0.02+0.1=5.12mIn circular motion at lowest point T=mg+mv2R⇒mv2R=T−mg So, 8π×V25.12=191.4π−8π×9.8V=8.8m/s