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 \[\Rightarrow \] \[{{a}_{2}}=g\sin \theta \] \[\Rightarrow \]\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5/7g\,\sin \theta }{g\,\sin \theta }\] \[\Rightarrow \]\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{7}\]You need to login to perform this action.
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