A simple pendulum has time period $T$ on the surface of earth. Its time period at a distance $R / 2$ from centre of earth will be $(R = r a d i u s o f e a r t h)$

T \sqrt{2}

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\frac{T}{\sqrt{2}}

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T

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2 T

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Solution

The correct option is A $T \sqrt{2}$
$\begin{matrix} T = 2 π \sqrt{\frac{l}{g_{effective}}} T_{surface} = T = 2 π \sqrt{\frac{l}{g}} T_{at R / 2 depth} = T_{2} = 2 x \sqrt{\frac{l}{g_{eff}}} gest at R / 2 from centre of earth = g_{e f f} at R / 2 depth \end{matrix}$

$\begin{matrix} = g (1 - \frac{d}{R}) = g (1 - \frac{\frac{R}{2}}{R}) = g / 2 \to T_{2} = 2 = \sqrt{\frac{l}{g / 2}} T_{2} = \sqrt{2} T \end{matrix}$

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A simple pendulum has time period T on the surface of earth. Its time period at a distance R/2 from centre of earth will be (R=radius of earth)

The correct option is A T√2T=2π√lg effectiveTsurface =T=2π√lgTat R/2 depth =T2=2x√lgeff gest at R/2 from centre of earth = geff at R/2 depth =g(1−dR)=g(1−R2R)=g/2→T2=2=√lg/2T2=√2T

A simple pendulum has time period $T$ on the surface of earth. Its time period at a distance $R / 2$ from centre of earth will be $(R = r a d i u s o f e a r t h)$