question_answer

The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle \[\theta \] without slipping and slipping down the incline without rolling is [AIPMT 2014]

A)  5 : 7            

B)  2 : 3

C)  2 : 5            

D)  7 : 5

Correct Answer: A

Solution :

A solid sphere rolling without slipping down an inclined plane In this case,  \[{{a}_{1}}=\frac{g\,\sin \theta }{1+\frac{{{k}^{2}}}{{{R}^{2}}}}=\frac{g\,sin\theta }{1+\frac{(2/5){{R}^{2}}}{{{R}^{2}}}}\] \[\left[ \therefore \text{for}\,\text{solid}\,\text{sphere},{{K}^{2}}=\frac{2}{5}{{R}^{2}} \right]\] \[=\frac{g\sin \theta }{7/5}\] \[\Rightarrow \] \[{{a}_{1}}=\frac{5}{7}g\,\sin \theta \] For a sphere slipping down an inclined plane \[\Rightarrow \]          \[{{a}_{2}}=g\sin \theta \] \[\Rightarrow \]\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5/7g\,\sin \theta }{g\,\sin \theta }\] \[\Rightarrow \]\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{7}\]