A) 16 s
B) 4 s
C) \[\sqrt{2}\,s\]
D) 0.5 s
Correct Answer: C
Solution :
Here, time taken by the body to reach the bottom \[{{t}_{1}}=4\,s\] Initial distance \[{{s}_{1}}=s\] The final distance \[{{s}_{2}}=\frac{s}{8}\] The distance s is given by \[s=\frac{1}{2}a{{t}^{2}}\propto {{t}^{3}}\] (where \[{{t}^{2}}\]is the time change one eight of distance) \[\frac{{{t}_{1}}}{{{t}_{2}}}-\sqrt{\frac{{{s}_{1}}}{{{s}_{2}}}}=\frac{\sqrt{\frac{s}{s}}}{8}=\sqrt{8}=2\sqrt{2}\] \[{{t}_{2}}=\frac{{{t}_{1}}}{2\sqrt{2}}=\frac{4}{2\sqrt{2}}=\sqrt{2}\,s\]
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