JEE MAIN - Physics (2022 - 25th July Evening Shift - No. 11)

The length of a seconds pendulum at a height h = 2R from earth surface will be:

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $$\pi$$² ms^$$-$$2)

$${2 \over 9}$$ m

$${4 \over 9}$$ m

$${8 \over 9}$$ m

$${1 \over 9}$$ m

$$g = {{GM} \over {{{(R + h)}^2}}} = {{GM} \over {9{R^2}}} = {{{g_0}} \over 9}$$

$$ \Rightarrow T = 2\pi \sqrt {{l \over g}} = 2\pi \sqrt {{l \over {{{{g_0}} \over 9}}}} $$

$$ \Rightarrow 2 = 2\pi \sqrt {{{9l} \over {{g_0}}}} $$

$$ \Rightarrow l = {{{g_0}} \over {9{\pi ^2}}} = {1 \over 9}\,m$$