QJEEM22Ph480d: JEE Main 2022 Physics - Class 11 JEEM22Ph480d Solution: The length of a seconds pendulum at a height h = 2R from earth…

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

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

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

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