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Question

The time period of oscillations of a simple pendulum is $$1$$ minute. If its length is increased by $$44\%$$ then its new time period of oscillation will be 


A
96 s
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B
58 s
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C
12 s
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D
72 s
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Solution

The correct option is C $$12\ s$$
We know $$T= 2\pi \sqrt{\dfrac{l}{g}}$$
$$T \propto \sqrt{l_{1}}$$
Given data $$T_{1} = 1\, sec$$
$$T_{2}=2$$
$$l_{1}=l$$
$$l_{2} = 44\% = 100+44 = 144l$$
$$\therefore \dfrac{T_{1}}{T_{2}}=\sqrt{\dfrac{l_{1}}{l_{2}}}\Rightarrow \dfrac{1}{T_{2}}=\sqrt{\dfrac{1}{144}}$$
$$T_{2}= 12\, sec$$

Physics

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