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

The acceleration due to gravity at height $2R$ from the surface of the Earth will be,

$$\begin{gathered} g^{\text{'}}=\frac{g{R}^2}{{\left(R+h\right)}^2} \\ \Rightarrow g^{\text{'}}=\frac{g{R}^2}{{\left(R+2R\right)}^2} \\ \Rightarrow g^{\text{'}}=\frac{g}{9} \end{gathered}$$

Length of the second's pendulum at the surface of the Earth is $1 \mathrm{m}$.

Since, time period of a pendulum is given by,

$$\begin{gathered} T=2\pi \sqrt{\frac{l}{g}} \\ \Rightarrow \frac{T^{\text{'}}}{T}=\sqrt{\frac{l^{\text{'}}}{l}\times \frac{g}{g^{\text{'}}}} \\ \Rightarrow 1=\sqrt{\frac{l^{\text{'}}}{l}\times \frac{g}{g^{\text{'}}}} \\ \Rightarrow l^{\text{'}}=\frac{l}{9}=\frac{1}{9} \mathrm{m} \end{gathered}$$