Consider the two bobs as shown in the figure. The bobs are connected to the hinges through massless rods. If $t_{A}$ be the time taken by the bob A to reach the lowest position and $t_{B}$ be the time taken by the bob B to reach the lowest position. (Both bobs being released from rest from a horizontal position) then the ratio $\frac{t_{A}}{t_{B}}$ is

\sqrt{3}

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\sqrt{5}

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\sqrt{2}

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\frac{1}{\sqrt{2}}

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Solution

The correct option is C $\sqrt{2}$

Consider a situation when the bob A has fallen through an angle 0.
Loss in PE = Gain in KE
$\Rightarrow \frac{1}{2} l ω^{2} = m g l s i n θ$
$\Rightarrow ω_{A} = \sqrt{\frac{2 m g l s i n θ}{I}} = \sqrt{\frac{2 m g l s i n θ}{m l^{2}}}$
$ω_{A} = \sqrt{\frac{2 g s i n θ}{l}}$
In the similar position, for B
$ω_{B} = \sqrt{\frac{4 g s i n θ}{l}}$
$\Rightarrow \frac{ω_{A}}{ω_{B}} = \frac{t_{A}}{B} = \sqrt{\frac{1}{2}}$
$\Rightarrow \frac{t_{B}}{t_{A}} = \sqrt{2}$

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