A) \[\sqrt{1-\frac{1}{{{n}^{2}}}}\]
B) \[1-\frac{1}{{{n}^{2}}}\]
C) \[\frac{1}{2-{{n}^{2}}}\]
D) \[\sqrt{\frac{1}{1-{{n}^{2}}}}\]
Correct Answer: B
Solution :
| [b] For a body moving with constant acceleration, the kinematics equation is |
| \[s=ut+\frac{1}{2}a{{t}^{2}}\] |
| If the initial speed is zero, then the time taken to reach a distance \[s\] is \[t=\sqrt{\frac{2s}{a}}\] |
| i.e.,\[t\propto {{a}^{-0.5}}\] |
| In the case of a smooth inclined plane,\[{{a}_{1}}=g\sin \theta \] |
| In the case of rough inclined plane,\[{{a}_{2}}=g(\sin \theta -\mu \cos \theta )\] |
| Time taken to travel down the smooth inclined plane is \[{{t}_{1}}=t\] |
| Time taken to travel down the smooth inclined plane is \[{{t}_{2}}=nt\] |
| \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{a}_{2}}}{{{a}_{1}}}}\Rightarrow \frac{t_{1}^{2}}{t_{2}^{2}}=\frac{{{a}_{2}}}{{{a}_{1}}}\Rightarrow \frac{1}{{{n}^{2}}}=\frac{\sin \theta -\mu cos\theta }{\sin \theta }\Rightarrow \mu =\tan \theta (1-\frac{1}{{{n}^{2}}})\] |
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