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A rod of mass m and length L , lying horizontally ils free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude LF acts on the rod at a distance of L/4 frm the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.

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Correct Answer - B::C
rod of mass m and lenth L, lying horzontally is free to rotate about a verticla axis passing through its centre. A force F is acting perpendicular to the road at a dance `L/4` from the centre.
Therefore Torque about the centre due to this force
`tau=Fxxr=FL/4`
This torque will produce an angular acceleration a therefore
`tau_c=I_cxxalopha`
`rarr tau_c=(mL^2)/12xxa`
`(I_cof a rod = (mL^2)/12)`
`rarr FL/4=(mL^2)/12xxalpha` ltbr.gt `rarr alpha=(3F)/(mL)`
therefore `theta=1/2alphat^2` (initially at rest)
`rarr theta=1/2xx((3F)/(mL))t^2`

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