question_answer

A smooth inclined plane is inclined at an angle with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is:

A) \[\sqrt{\frac{2h}{g}}\]

B)                           \[\sqrt{\frac{2\ell }{g}}\]

C) \[\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}\]                      

D) \[\sin \theta \sqrt{\frac{2h}{g}}\]

Correct Answer: C

Solution :

For TIR at B \[\sin \theta  > sin C,~~~~~~~~~~sin\left( 90 - r \right) > sin C.\] \[\cos  r > sin C\]                            .....(1) By snell law \[1\times sini=n\,sin\,r\] \[\sin \,r=\frac{sin\,i}{n}\] \[\cos \,\,r=\sqrt{1-\frac{si{{n}^{2}}\operatorname{i}}{{{\operatorname{n}}^{2}}}}>sin\,C\] \[\sqrt{1-\frac{si{{n}^{2}}\operatorname{i}}{{{\operatorname{n}}^{2}}}}>\frac{1}{\operatorname{n}}\] \[\operatorname{On} solving {{n}^{2}} > si{{n}^{2}} i +1\] \[\operatorname{n} > \sqrt{2} \left( for max value of sin i = 1 \right)\]