A) 3.5m
B) 7.0m
C) 10.5m
D) 14.5m
Correct Answer: B
Solution :
\[mgh=\frac{1}{2}mv_{CM}^{2}\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right)\] \[\Rightarrow \] \[\frac{7}{10}=mv_{Cm}^{2}=mgh\] \[\Rightarrow \] \[h=\frac{7}{10}\left( \frac{v_{CM}^{2}}{g} \right)\] Given, \[v=10\text{ }m{{s}^{-1}},\text{ }g=10\text{ }m{{s}^{-2}}\] \[\therefore \] \[h=\frac{7}{10}\times \frac{10\times 10}{10}\] \[\Rightarrow \] \[h=7.0\,m\]You need to login to perform this action.
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