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) \[{{v}_{2}}={{v}_{1}};{{t}_{2}}={{t}_{1}}\]
Correct Answer: B
Solution :
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]. |
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