Question

A tunnel is dug across the diameter of the earth. A ball is released from the surface of Earth into the tunnel. The velocity of the ball when it is at a distance $\dfrac{R}{2}$ from the centre of the earth is (where R = radius of Earth and M = mass of Earth)
\eqalign{ & A.\,\sqrt {\dfrac{{3GM}}{{4R}}} \cr & B.\,\sqrt {\dfrac{{2GM}}{{3R}}} \cr & C.\,\sqrt {\dfrac{{GM}}{{2R}}} \cr & D.\,\sqrt {\dfrac{{2GM}}{R}} \cr}

Answer 1

Here we have to find the velocity of the ball, for it we will apply the law of conservation of energy. The total energy remains the constant in the universe. The sum of kinetic energy and the potential energy at the initial point is equal to the sum of the kinetic and the potential energy at the final point.

Complete step by step answer:
By the law of conservation of energy –
${K_i} + {U_i} = {K_f} + {U_f}\,{\text{ }}......{\text{(1)}}$
Where, ${K_i} =$ initial kinetic energy
${U_i} =$ Initial potential energy
${K_f} =$ Final kinetic energy
${U_f} =$ Final potential energy
The initial velocity is zero implies that the initial kinetic energy is zero, ${K_i} = 0\;\;\;\;{\text{ }}....{\text{(2)}}$
${U_i} = - \dfrac{{GMm}}{R}{\text{ }}.......{\text{(3)}}$
$m =$ mass of ball
${K_f} = \dfrac{1}{2}m{v^2}\,{\text{ }}......{\text{(4)}}$
Consider the Earth as the solid, potential energy is unknown.
${U_f} = m{v_e}\,{\text{ }}.....{\text{(5)}}$
${v_e} = \dfrac{{ - GM}}{{2{R^3}}}(3{R^2} - {r^2})$
Now, given that- $r = \dfrac{R}{2}$ , so place this value in the above equation
${v_e} = \dfrac{{ - GM}}{{2{R^3}}}(3{R^2} - {(\dfrac{R}{2})^2})$
Simplify –
\eqalign{ & {v_e} = \dfrac{{ - GM}}{{2{R^3}}}(3{R^2} - {\dfrac{R}{4}^2}) \cr & {v_e} = \dfrac{{ - GM}}{{2{R^3}}}\left( {\dfrac{{12{R^2} - {R^2}}}{4}} \right) \cr & {v_e} = \dfrac{{ - GM}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right)\,{\text{ }}.....{\text{(6)}} \cr}
Place value of the equation $(6){\text{ in equation (5)}}$
\eqalign{ & {U_f} = m\dfrac{{ - GM}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right) \cr & {U_f} = \dfrac{{ - GMm}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right)\,{\text{ }}.....{\text{(7)}} \cr}
Place values of equations $(2),{\text{ (3), (4) and (7) in equation (1)}}$
$0 - \dfrac{{GMm}}{R} = \dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right)$
Simplify and make velocity, “v” as the subject-
\eqalign{ & \dfrac{{GMm}}{R} + \dfrac{{GMm}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right) = \dfrac{1}{2}m{v^2} \cr & \dfrac{1}{2}m{v^2} = \dfrac{{GMm}}{R} + \dfrac{{GMm}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right) \cr}
Take mass “m” from both sides of the equation and so remove it.
\eqalign{ & \dfrac{1}{2}{v^2} = \dfrac{{ - GM}}{R} + \dfrac{{GM}}{{2{R^3}}}\left( {\dfrac{{11{R^2}}}{4}} \right) \cr & \dfrac{1}{2}{v^2} = \dfrac{{GM}}{R}\left( { - 1 + \dfrac{{11{R^2}}}{{8{R^2}}}} \right) \cr & \dfrac{1}{2}{v^2} = \dfrac{{GM}}{R}\left( { - 1 + \dfrac{{11}}{8}} \right) \cr & \dfrac{1}{2}{v^2} = \dfrac{{GM}}{R}\left( {\dfrac{3}{8}} \right) \cr}
Make velocity “v” the subject
\eqalign{ & {v^2} = 2 \times \dfrac{{GM}}{R}\left( {\dfrac{3}{8}} \right) \cr & {v^2} = \dfrac{{GM}}{R}\left( {\dfrac{3}{4}} \right) \cr}
Take square-root on both the sides of the equation
$\sqrt {{v^2}} = \sqrt {\dfrac{{GM}}{R}\left( {\dfrac{3}{4}} \right)}$
Square and square-root cancels each other on left hand side of the equation –
$v = \sqrt {\dfrac{{3GM}}{{4R}}}$
Therefore, the required answer- A tunnel is dug across the diameter of the earth. A ball is released from the surface of Earth into the tunnel. The velocity of the ball when it is at a distance $\dfrac{R}{2}$ from the centre of the earth is $\,\sqrt {\dfrac{{3GM}}{{4R}}}$
Hence, from the given multiple choices- option A is the correct answer.

Note: The law of the conservation states that the energy cannot be created nor destroyed but can be changed from one form to another or can be transferred from one object to another object. The total energy of an isolated system remains constant. Also, remember different parameters such as the initial velocity remains zero.