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 |
\[\Rightarrow \] \[{{a}_{2}}=g\sin \theta \] |
\[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5/7g\,\sin \theta }{g\,\sin \theta }\] |
\[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{7}\] |
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