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Toy cart tied to the end of an unstretched string of length a, when revolved moves in a horizontal circle
of radius 2a with a time period T. Now, the toy cart is speeded up until it moves in a horizontal circle of
radius 3a with a period T'. If Hooke's law (F = kx) holds, then

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$T^{'} = (\frac{\sqrt{3}}{2}) T$

$T^{'} = \sqrt{\frac{3}{2}} T$

$T^{'} = T$

$T^{'} = (\frac{3}{2}) T$

answer is B.

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Detailed Solution

In the given problem, centripetal force will be equal to

$F = k x$

$\Rightarrow k x = \frac{m v^{2}}{r} \Rightarrow v = \sqrt{\frac{k x r}{m}}$

$\Rightarrow T = \frac{2 π r}{v} = 2 π \sqrt{\frac{m r}{k x}}$

$\Rightarrow T \propto \sqrt{\frac{r}{x}}$

$∴ T^{'} = T \sqrt{\frac{r^{'} x}{r x^{'}}}$

$= T \sqrt{\frac{3 a \times a}{2 a \times 2 a}} = (\frac{\sqrt{3}}{2}) T$

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