• # question_answer A simple pendulum with a bob of mass 'm' oscillates from A to C and back to A such that PB is H. If the acceleration due to gravity is 'g', then the velocity of the bob as it passes through B is        A) zero B) $2\,gH$ C) $mgH$ D) $\sqrt{2gH}$

 [d] At B, the velocity is maximum. Taking vertical downward motion of body from A to B, we have, $u=0,\,\,a=g,\,\,s=H$ and $v=?$ ${{v}^{2}}={{u}^{2}}+2\,as;$ or ${{v}^{2}}=0+2gH$ or $v=\sqrt{2g\,H}$