A particle falls on earth
(i) from infinity,
(ii)from a height 10 times the radius of earth.
The ratio of the velocities gained on reaching at the earth’s surface is:
A. $\sqrt {11} :\sqrt {10} $
B. $\sqrt {10} :\sqrt {11} $
C. 10:11
D. 11:10

Hint: In this question, we will use the energy conservation principle. First find the kinetic energy and potential energy at infinity and also at height of 10 times the radius of the earth and then find the same at the surface of the earth and apply energy conservation principle to find the final velocities in the two cases.

Complete answer:
We know that the kinetic energy of a body having mass ‘m’ and velocity ‘v’ is given by:
K.E. = $\dfrac{1}{2}m{v^2}$
The potential energy stored between two masses ‘M’ and ‘m’ and separation between their centres is ‘r’ is given by:
P,E, = $\dfrac{{GMm}}{r}$
Where ‘G’ is universal gravitational constant whose value = $6.673 \times {10^{ - 11}}kg{m^2}/{s^2}$
Also we know according to energy conservation principle, total energy of a body is always constant. Mathematically, we can write:
T.E = K.E.+P.E. = constant.

(I)Let the velocity of the particle ${v_1}$ at the earth's surface fall from infinity.

And ${v_2}$ be the velocity at the earth's surface when falling from a height of 10R where ‘R’ is the radius of the earth.
Now, total energy at infinity = 0.
And at surface of earth total energy = $\dfrac{1}{2}m{v_1}^2 + \left( { - \dfrac{{GMm}}{R}} \right)$ .
By using energy conservation principle, we have:
0 = $\dfrac{1}{2}m{v_1}^2 + \left( { - \dfrac{{GMm}}{R}} \right)$
$ \Rightarrow {v_1} = \sqrt {\dfrac{{2GM}}{R}} $ (1)

(II) T.E. at height, h= 10R = $\left( { - \dfrac{{GMm}}{{11R}}} \right)$

T.E. at the surface of earth = $\dfrac{1}{2}m{v_2}^2 + \left( { - \dfrac{{GMm}}{{11R}}} \right)$
By using energy conservation principle, we have:
$\left( { - \dfrac{{GMm}}{{11R}}} \right)$= $\dfrac{1}{2}m{v_2}^2 + \left( { - \dfrac{{GMm}}{R}} \right)$
$ \Rightarrow {v_2} = \sqrt {\dfrac{{20GM}}{{11R}}} $ (2)
Taking the ratio of the two velocities, we have:

${v_1}:{v_2} = \sqrt {11} :\sqrt {10} $

So, the correct answer is “Option A”.

Note:
In this question, you know the formula to find the kinetic energy of a body and also you should know to find the gravitational potential energy of a body on the earth surface. The acceleration due to gravity on the earth surface is found by using the energy conservation principle that we used here as well.