A) \[1/(1-{{n}^{2}})\]
B) \[1-(1/{{n}^{2}})\]
C) Ö\[\left\{ 1-(1/{{n}^{2}}) \right\}\]
D) Ö\[\left\{ 1/(1-{{n}^{2}}) \right\}\]
Correct Answer: B
Solution :
| Acceleration when there is no friction is \[a=g\sin \theta \] and acceleration when friction is there is |
| \[a'=(g\sin \theta -\mu g\cos \theta )\] |
| We know that \[s=ut+\frac{1}{2}a{{t}^{2}},\] under condition of \[u=0,\] |
| gives,\[t\text{ }=\sqrt{\left( 2s/a \right)}\]. |
| Therefore \[t't\text{ }=\sqrt{\left( a/a' \right)}\]. |
| For our case we get |
| \[n=\sqrt{[g\sin \theta /(g\sin \theta -\mu g\cos \theta )]}\] |
| \[\Rightarrow \,{{n}^{2}}=g\sin \theta /(g\sin \theta -\mu g\cos \theta )=1/(1-\mu \cot \theta )\]but \[\theta ={{45}^{o}}\] |
| \[\Rightarrow \,\,\cot \theta =1\,\,\therefore \,\,{{n}^{2}}=1/(1-\mu )\] or \[\mu =1-(1/{{n}^{2}})\] |
You need to login to perform this action.
You will be redirected in
3 sec
