• # question_answer A uniform rod of mass m, length S is placed over a smooth horizontal surface along y-axis and is at rest as shown in figure- An impulsive force F is applied for a small time At along positive x-direction at end A of the rod. The x-coordinate of end A of the rod when the rod becomes parallel to x-axis for the first time is (initially the coordinate of centre of mass of the rod is (0,0)): A) $\frac{\pi \ell }{12}$ B) $\frac{\ell }{2}\left( 1+\frac{\pi }{12} \right)$ C) $\frac{\ell }{2}\left( 1-\frac{\pi }{6} \right)$ D) $\frac{\ell }{2}\left( 1+\frac{\pi }{6} \right)$

 As torque = change om angular momentum $\therefore F,\,\,\Delta t=mv$                              (Linear)... (1) and $\left( F.\frac{\ell }{2} \right)\Delta t=\frac{m{{\ell }^{2}}}{12}.\omega$                  (angular)...(2) Dividing:            (1) and (2) $2=\frac{12v}{\omega \ell }\Rightarrow \,\,\,\,\,\,\,\,\,\omega =\frac{6v}{\ell }$ Using : S = ut: Displacement of COM is: $\frac{\pi }{2}=\omega t=\left( \frac{6v}{\ell } \right)t$ and       $x=vt$ Dividing: $\frac{2x}{\pi }=\frac{\ell }{6}$           $\Rightarrow \,\,\,\,x=\frac{\pi \ell }{12}$$\Rightarrow$            Coordination of A will be $\left[ \frac{\pi \ell }{12}+\frac{\ell }{2},0 \right]$ Hence [D].