A) 16 : 1
B) 1 : 16
C) 1 : 4
D) 4 : 1
Correct Answer: A
Solution :
We know that acceleration down the inclined plane is uniform and initial velocity being zero. Let \[{{t}_{1}}\] is the time to cover whole distance and \[{{t}_{2}}\] time the cover \[\frac{1}{4}\] of the distance from the relation \[s=\frac{1}{2}a{{t}^{2}}\,\,\,\,\Rightarrow \,\,\,\,s\propto {{t}^{2}}\] hence \[\frac{{{s}_{1}}}{{{s}_{2}}}={{\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}=\left( \frac{4}{1} \right)\] \[\frac{{{t}_{1}}}{{{t}_{2}}}=2\] So, \[{{t}_{1}}=4s,\] \[{{t}_{2}}=2\,s\]
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