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(B) The acceleration due to gravity at a height $h$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Given $h = 2R$,we have $g' = g \left( \frac{R}

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$\frac{1}{9} \ m$

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(B) The acceleration due to gravity at a height $h$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Given $h = 2R$,we have $g' = g \left( \frac{R}{R+2R} \right)^2 = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9}$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g'}}$.For a second's pendulum,$T = 2 \ s$.Substituting the values: $2 = 2\pi \sqrt{\frac{\ell}{g/9}} = 2\pi \sqrt{\frac{9\ell}{g}}$.Dividing by $2$: $1 = \pi \sqrt{\frac{9\ell}{g}}$.Squaring both sides: $1 = \pi^2 \left( \frac{9\ell}{g} \right)$.Given $g = \pi^2 \ m/s^2$,we substitute this into the equation: $1 = \pi^2 \left( \frac{9\ell}{\pi^2} \right) = 9\ell$.Therefore,$\ell = \frac{1}{9} \ m$.

If $R$ is the radius of the earth and the acceleration due to gravity on the surface of the earth is $g = \pi^2 \ m/s^2$,then the length of the second's pendulum at a height $h = 2R$ from the surface of the earth will be:

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