question_answer

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

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

B)  \[\frac{{{t}_{2}}}{{{t}_{1}}}=1\]

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

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

Correct Answer: C

Solution :

                 From first inclined plane                 \[\sin {{\theta }_{1}}=\frac{h}{{{l}_{1}}}\] Or           \[{{l}_{1}}=\frac{h}{\sin {{\theta }_{1}}}\] From second inclined plane                 \[\sin {{\theta }_{2}}=\frac{h}{{{l}_{2}}}\]or \[{{l}_{2}}=\frac{h}{\sin {{\theta }_{2}}}\] Acceleration of the block down the two planes are                 \[{{a}_{1}}=g\sin {{\theta }_{1}}\] and        \[{{a}_{2}}=g\sin {{\theta }_{2}}\] For the first inclined plane, distance \[{{l}_{1}}=\frac{1}{2}{{a}_{1}}t_{1}^{2}\]                 \[(\because u=0)\] For the second inclined plane, distance \[{{l}_{2}}=\frac{1}{2}{{a}_{2}}t_{2}^{2}\] \[\therefore \]  \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{a}_{1}}t_{1}^{2}}{{{a}_{2}}t_{2}^{2}}\] \[\therefore \]  \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{g\sin {{\theta }_{1}}t_{1}^{2}}{g\sin {{\theta }_{2}}t_{2}^{2}}\] Or           \[\frac{t_{1}^{2}}{t_{2}^{2}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\times \frac{{{l}_{2}}}{{{l}_{1}}}\]                 \[=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\times \frac{h}{\sin {{\theta }_{2}}}\times \frac{\sin {{\theta }_{1}}}{h}\] \[\therefore \]  \[\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\]