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JEE Advanced Model Paper 4 with solutions for online practice

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Question : 13

Total: 54

A long uniform rod of length L and mass m is free to rotate in a horizontal plane about a vertical axis through its one end 0. A spring of force constant k is connected horizontally between one end (B) of the rod and a fixed wall (see figure). When therod is in equilibrium, it is parallel to the wall. When the rod is rotated slightly in the direction shown (at B) and released, the time period of resultant oscillations is T. If amplitude of these oscillations is 80 , then the maximum speed of displacedend of rod is vmax . Now

T = 2π √
m
5k
T = 2π √
m
3k
vmax = Lθ0√
3k
m
vmax =
Lθ0
2
√
5k
m

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