A) \[g({{l}^{2}}-1)\]
B) \[g({{l}^{2}}-1)\]
C) \[\frac{g}{\sqrt{{{l}^{2}}-1}}\]
D) \[\frac{g}{{{l}^{2}}-l}\]
Correct Answer: C
Solution :
[c]| The situation is as shown in the in the figure |
 |
| \[\tan \theta =\frac{1}{\sqrt{{{t}^{2}}-1}}\] |
| The free body diagram of block is shown in the below figure. |
 |
| For equilibrium of block w.r.t. incline,\[mg\,\sin \theta =ma\,\cos \theta \] |
| \[\Rightarrow \] \[a=g\,\tan \theta =\frac{g}{\sqrt{{{l}^{2}}-1}}\] |
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