A) \[32\,h\]
B) \[\left( \frac{1}{8\sqrt[3]{2}} \right)h\]
C) \[8\sqrt[3]{2}\]
D) \[16\,h\]
Correct Answer: A
Solution :
We know,\[{{T}^{2}}\propto {{R}^{3}}\] \[\therefore \] \[{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3}}\Rightarrow \frac{{{T}_{2}}}{4}={{\left( \frac{4R}{R} \right)}^{3/2}}\] \[\Rightarrow \] \[{{T}_{2}}=4\times 8=32h\].
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