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If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:
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The correct answer is:
$\sqrt{2}$
$\begin{aligned} & T^{\prime}=2 \pi \sqrt{\frac{\ell^{\prime}}{g}} \text { where } \ell^{\prime}=\frac{\ell}{2} \\ & T=2 \pi \sqrt{\frac{\ell}{g}} \\ & T^{\prime}=\frac{x}{2} T \\ & 2 \pi \sqrt{\frac{\ell}{2 g}}=\frac{x}{2} 2 \pi \sqrt{\frac{\ell}{g}} \\ & \frac{1}{\sqrt{2}}=\frac{x}{2} \Rightarrow x=\sqrt{2}\end{aligned}$
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