A) 10 h
B) 80 h
C) 40 h
D) 20 h
Correct Answer: C
Solution :
According to Kepler's law \[{{T}^{2}}\propto {{r}^{3}}\] \[{{5}^{2}}\propto {{r}^{3}}\] ...(i) \[{{(T')}^{2}}\propto {{(4)}^{3}}\] ...(ii) From Eqs. (i) and (ii), we have \[\frac{25}{{{(T')}^{2}}}=\frac{{{r}^{3}}}{64{{r}^{3}}}\] \[T'=\sqrt{1600}\] \[T'=40\,h\]You need to login to perform this action.
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