The length of a seconds pendulum at a height | vedclass

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(D) seconds pendulum has a time period $T = 2 \ s$.The formula for the time period of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g'}}$.At a height

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$\frac{1}{9} \ m$

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(D) seconds pendulum has a time period $T = 2 \ s$.The formula for the time period of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g'}}$.At a height $h = 2R$ from the surface,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^{2} = g \left( \frac{R}{R+2R} \right)^{2} = g \left( \frac{1}{3} \right)^{2} = \frac{g}{9}$.Given $g = \pi^{2} \ m/s^{2}$,so $g' = \frac{\pi^{2}}{9} \ m/s^{2}$.Substituting the values into the time period formula: $2 = 2 \pi \sqrt{\frac{L}{\pi^{2}/9}}$.$1 = \pi \sqrt{\frac{9L}{\pi^{2}}} = \pi \cdot \frac{3\sqrt{L}}{\pi} = 3\sqrt{L}$.$\sqrt{L} = \frac{1}{3} \Rightarrow L = \frac{1}{9} \ m$.

The length of a seconds pendulum at a height $h=2R$ from the earth's surface will be. (Given: $R =$ radius of earth and acceleration due to gravity at the surface of earth $g = \pi^{2} \ m/s^{2}$)

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