question_answer

A block at rest slides down a smooth inclined plane which makes an angle 60° with the vertical and it reaches the ground in \[{{t}_{1}}\] seconds. Another block is dropped vertically from the same point and reaches the ground in \[{{t}_{2}}\] seconds. Then the ratio of \[{{t}_{1}}:{{t}_{2}}\] is

A)  1 : 2                      

B)         2 : 1                      

C)  1 : 3                      

D)         1: \[\sqrt{2}\]

E)  3 : 1

Correct Answer: B

Solution :

\[L=\frac{1}{2}g\cos {{60}^{o}}t_{1}^{2}\]                                    ...(i) \[L\cos \theta =\frac{1}{2}gt_{2}^{2}\]                       ...(ii)                 \[\frac{t_{1}^{2}}{t_{2}^{2}}=\frac{1}{{{\cos }^{2}}{{60}^{o}}}=\frac{4}{1}\]                 \[{{t}_{1}}:{{t}_{2}}::2:1\]