• # question_answer A solid sphere released from rest from the top of an inclined plane of inclination${{\theta }_{1}}$, rolls without sliding and reaches the bottom with speed ${{v}_{1}}$ and its time of descent is${{t}_{1}}$. The same sphere is then released from rest from the top of another inclined plane of inclination${{\theta }_{2}}$ but of the same height, rolls without sliding and reaches the bottom with speed ${{v}_{2}}$ and its time of descent is ${{t}_{2}}$. If ${{\theta }_{2}}>{{\theta }_{1}}$, then A) ${{v}_{2}}>{{v}_{1}};{{t}_{2}}<{{t}_{1}}$ B) ${{v}_{2}}={{v}_{1}};{{t}_{2}}<{{t}_{1}}$ C) ${{v}_{2}}<{{v}_{1}};{{t}_{2}}>{{t}_{1}}$ D) ${{v}_{2}}={{v}_{1}};{{t}_{2}}={{t}_{1}}$

 In rolling without slipping, the mechanical energy is conserved. Since both the inclined planes are of the same height${{v}_{2}}={{v}_{1}}$. The acceleration of the sphere rolling down the plane is $a=\frac{g\sin \theta }{1+\frac{I}{M{{R}^{2}}}}$ Since${{\theta }_{2}}>{{\theta }_{1}}\,\,;\,\,{{a}_{2}}>{{a}_{1}}$. Hence $\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{2}}}{\sin {{\theta }_{1}}}$ Thus${{t}_{1}}>{{t}_{2}}$. Hence the correct choice is [B].