Time period of a simple pendulum inside a lift that is accelerating up at $2 {m s}^{2}$ is $T_{1}$ . If lift is retarding down at $6 m s$ then time period of same pendulum is $T,$ Then $T, / T_{2}$ is

\sqrt{\frac{1}{3}}

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\sqrt{\frac{4}{3}}

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\sqrt{\frac{3}{4}}

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\sqrt{3}

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Solution

The correct option is B $\sqrt{\frac{4}{3}}$
$\begin{matrix} T_{1} = 2 π \sqrt{\frac{l}{g + 2}} T_{2} = 2 π \sqrt{\frac{l}{g + 6}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16}{12}} = \sqrt{\frac{4}{3}} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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Time period of a simple pendulum inside a lift that is accelerating up at 2ms2 is T1 . If lift is retarding down at 6ms then time period of same pendulum is T, Then T,/T2 is

The correct option is B √43T1=2π√lg+2T2=2π√lg+6T1T2=√1612=√43Hence,optionBiscorrectanswer.

Time period of a simple pendulum inside a lift that is accelerating up at $2 {m s}^{2}$ is $T_{1}$ . If lift is retarding down at $6 m s$ then time period of same pendulum is $T,$ Then $T, / T_{2}$ is

The correct option is B $\sqrt{\frac{4}{3}}$
$\begin{matrix} T_{1} = 2 π \sqrt{\frac{l}{g + 2}} T_{2} = 2 π \sqrt{\frac{l}{g + 6}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{16}{12}} = \sqrt{\frac{4}{3}} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$