A body is released from a height equal to the radius (R) of the earth. The velocity of the body when it strikes the surface of the earth will be

Since the initial velocity of the body is zero, its total energy is $v^2=2GM\left(\frac{1}{R}-\frac{1}{r}\right)$

$E=-\frac{GmM}{r} \left(i\right)$

where m is the mass of the body, M the mass of the earth and r its distance from the centre of the earth. When the body reaches the earth, let its velocity be v and its distance from the Centre of the earth is the earth's radius R. Therefore, the energy now is

$E=\frac{1}{2}m{v}^2-\frac{GmM}{R} \left(ii\right)$

Equating (i) and (ii) we get

$v^2=2GM \left(\frac{1}{R}-\frac{1}{r}\right)$

Also $g=\frac{GM}{R^2}$.

Therefore $GM=g{R}^2.$

Using this in above equation we get

$v=R{\left[2g\left(\frac{1}{R}-\frac{1}{r}\right)\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Now $r=2R \left(given\right).$

Therefore $v=R{\left[2g\left(\frac{1}{R}-\frac{1}{2R}\right)\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\sqrt{gR}$

Hence the correct choice is (1)