\[(a)\;2^{\frac{1}{3}}\;times \quad (b)\;2^{\frac{2}{3}}\;times \quad (c)\;2^{\frac{3}{2}}\;times \quad (d)\;4\;times \]

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$T^2 \; \alpha \;r^3$

$T \; \alpha \; r^{3/2}$

$ \large\frac{2 \pi}{w}$$ \; \alpha \; r^{3/2}$

$w \; \alpha \; r^{-3/2}$

If w is halved . distance will become

$w_2 =\large\frac{1}{2} w_1$

$\large\frac{w_2}{w_1}=\bigg(\large\frac{r_2}{r_1}\bigg)^{-3/2}$

$\large\frac{1}{2}=\bigg(\frac{r_2}{r_1}\bigg)^{-3/2}$

$r_2 =r_1 \bigg(\frac{1}{2}\bigg)^{-2/3}$

$r_2=r_1 \;2^{2/3}$

it become $2^{2/3}$ times

Hence b is the correct answer.

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