A) 4 times
B) 8 times
C) 1/4 times
D) 1/8 times
Correct Answer: B
Solution :
\[{{T}^{2}}\propto {{R}^{3}}\] \[\frac{T{{'}^{2}}}{{{T}^{2}}}={{\left( \frac{R'}{R} \right)}^{3}}\] \[\frac{T'}{T}={{\left( \frac{4R}{R} \right)}^{3/2}}={{(64)}^{1/2}}\] \[T'=8T\]
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