PhysicsHard159×since 2002Q7169If R\mathrm{R}R is the radius of the earth and the acceleration due to gravity on the surface of earth is g=π2 m/s2g=\pi^2 \mathrm{~m} / \mathrm{s}^2g=π2 m/s2, then the length of the second's pendulum at a height h=2R\mathrm{h}=2 Rh=2R from the surface of earth will be, :A19 m\frac{1}{9} \mathrm{~m}91 mB89 m\frac{8}{9} \mathrm{~m}98 mC29 m\frac{2}{9} \mathrm{~m}92 mD49 m\frac{4}{9} \mathrm{~m}94 mCheck answerSkip