On end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of `37^@` with the horizontal. as shown in figure. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. `(sin 37^@=3/5)`.
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The correct Answer is:
C
From work energy theorem (W=DKE) `W_("gravity")+W_("spring")=(1)/(2)mv_(1)^(2)-(1)/(2)mv_(1)^(2)` `mgh+((1)/(2)kx_(1)^(2)-(1)/(2)kx_(2)^(2))=(1)/(2)mv_(2)^(2)` `(1)/(2)(10)(3)+(1)/(2)(170)(1)^(2)-(1)/(2)(170)(0)^(2)` `=(1)/(2)((1)/(2))v^(2)-(1)/(2)((1)/(2))(0)^(2)rArrv=20m//s`
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