question_answer

A block released from rest from the top of a smooth inclined plane of angle \[{{\theta }_{1}}\] reaches the bottom in time t\[{{\operatorname{t}}_{1}}\]. The same block released from rest from the top of another smooth inclined plane of angle\[{{\theta }_{2}}\], reached the bottom in time\[{{\operatorname{t}}_{2}}\]. If the two inclined planes have the same height, the relation between\[{{\operatorname{t}}_{1}} and {{t}_{2}}\] is:

A) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}=1\]                    

B) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}={{\left[ \frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}} \right]}^{{}^{1}/{}_{2}}}\]

C) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\]                       

D) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}\]

Correct Answer: B

Solution :

In both cases black starts from rest u = 0 \[\operatorname{S}=ut+\frac{1}{2}{{a}_{1}}{{t}^{2}}s=ut+\frac{1}{2}{{a}_{2}}{{t}^{2}}\] \[{{\operatorname{S}}_{1}}=0+\frac{1}{2}gsin\theta {{t}^{2}}{{s}_{2}}=0+\frac{1}{2}gsin{{\theta }_{2}}{{t}^{2}}\]Two inclined plane have same height \[{{\operatorname{S}}_{1}}={{\operatorname{S}}_{2}}\] \[\frac{1}{2}gsin{{\theta }_{1}}{{t}_{1}}^{2}=\frac{1}{2}gsin{{\theta }_{2}}{{t}^{2}}_{2}\] \[\frac{{{t}^{2}}_{1}}{{{t}^{2}}_{2}}=\frac{sin{{\theta }_{2}}}{sin{{\theta }_{1}}}\] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{sin{{\theta }_{2}}}{sin{{\theta }_{1}}}}\] \[\frac{{{t}_{2}}}{{{t}_{1}}}={{\left[ \frac{sin{{\theta }_{2}}}{sin{{\theta }_{1}}} \right]}^{{}^{1}/{}_{2}}}\]