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WBJEE 2014 Physics Solved Paper
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© examsnet.com
Question : 57
Total: 60
A thin rod AB is held horizontally so that it can freely rotate in a vertical plane about the end A as shown in the figure.The potential energy of the rod when it hangs vertically is taken to be zero. The end B of the rod is released from restfrom a horizontal position. At the instant the rod makes an angle θ with the horizontal.
the speed of end B is proportional to
√
s
i
n
θ
the potential energy is proportional to (1 - cosθ)
the angular acceleration is proportional to cos θ
the torque about A remains the same as its initial value
Validate
Solution:
Loss in Potential Energy = gain in Kinetic Energy, mg
L
2
sinθ =
1
2
l
ω
2
,
ω α
√
s
i
n
θ
, v α
√
s
i
n
θ
U = mgh = mg
L
2
(1 - sinθ)
∵ τ = I α ⇒ mg ×
L
2
cosθ =
m
l
3
2
× α . α ∝ COS θ
© examsnet.com
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