(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.At height $h$ above the surface, the acceleration due to gravity is

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.At height $h$ above the surface, the acceleration due to gravity is $g' = \frac{g}{(1 + \frac{h}{R})^2}$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g / (1 + \frac{h}{R})^2}} = T(1 + \frac{h}{R})$.Given $T' = 2T$, we have $2T = T(1 + \frac{h}{R})$.$2 = 1 + \frac{h}{R} \Rightarrow \frac{h}{R} = 1$.Therefore, $h = R = 6400 \text{ km}$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

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The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

AP EAMCET 2023 All AP EAMCET

A
$3200$
B
$6400$
C
$19200$
D
$800$

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Gravitation

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Acceleration Due to Gravity and its Variation

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The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

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