A simple pendulum has a time period T₁ when on earth's surface and T₂ when taken to a height R above the earth's surface, where R is the radius of earth. The value of T₂ / T₁ is

The correct option is 2.

The time period of a simple pendulum is given by the formula:
$T=2\pi \sqrt{\frac{L}{g}}$
where L is the length of the pendulum and g is the acceleration due to gravity.

On the Earth's surface, the acceleration due to gravity $g_1$ is:
$g_1=\frac{GM}{R^2}$
where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth. The time period on the surface is:
$T_1=2\pi \sqrt{\frac{L}{g_1}}$

At a height R above the Earth's surface, the distance from the center of the Earth is $r=R+R=2R$. The acceleration due to gravity $g_2$ at this height is:
$g_2=\frac{GM}{{\left(2R\right)}^2}=\frac{GM}{4R^2}=\frac{g_1}{4}$

Substituting $g_2$ into the equation for the time period $T_2$ at height R, we get:
$T_2=2\pi \sqrt{\frac{L}{g_2}}=2\pi \sqrt{\frac{L}{\left(\frac{g_1}{4}\right)}}=2\pi \sqrt{\frac{4L}{g_1}}=2\left(2\pi \sqrt{\frac{L}{g_1}}\right)=2T_1$

Therefore, the ratio of the time periods is:
$\frac{T_2}{T_1}=2$