Question

A planet of radius $ R = \dfrac{1}{{10}} \times $ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $ \dfrac{R}{5} $ on it and lower a wire of the same length and of linear mass density $ {10^{ - 3}}kg{m^{ - 1}} $ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $ = 6 \times {10^6}m $ and the acceleration due to gravity of the Earth is $ 10m{s^{ - 2}} $ )
(A) $ 96N $
(B) $ 108N $
(C) $ 120N $
(D) $ 150N $

Answer 1

Hint From the formula for the acceleration due to gravity, we can find its value in the planet with respect to that of the Earth. Then using the formula for the decrease in the acceleration due to gravity with depth, we can find its value at $ \dfrac{R}{5} $ . Using this value we can find the force on the fire.

Formula Used: In this solution we will be using the following formula,
 $\Rightarrow g = \dfrac{{GM}}{{{R^2}}} $
where $ g $ is the acceleration due to gravity, $ G $ is the universal gravitational, $ M $ is the mass and $ R $ is the radius.
 $\Rightarrow {g_d} = g\left( {1 - \dfrac{d}{R}} \right) $
where $ {g_d} $ is the acceleration due to gravity at a depth $ d $ from the surface of the Earth.

Complete step by step answer
Let us consider the radius of the said planet as $ {R_p} $ , where the radius of the Earth is $ {R_e} $ . Now according to the question we have,
 $\Rightarrow {R_p} = \dfrac{{{R_e}}}{{10}} $ .
Now the acceleration due to gravity on the surface of the Earth is given by the formula,
 $\Rightarrow {g_e} = \dfrac{{G{M_e}}}{{{R_e}^2}} $
Now the mass of the earth can be broken down in the terms of the radius and the mean density as,
 $\Rightarrow {M_e} = \dfrac{4}{3}\pi {R_e}^3\rho $
So substituting we get,
 $\Rightarrow {g_e} = \dfrac{G}{{{R_e}^2}} \times \dfrac{4}{3}\pi {R_e}^3\rho $
On cancelling the like terms, we get $ {g_e} = \dfrac{4}{3}\pi G{R_e}\rho $
Now for the planet, the acceleration due to gravity will similarly be given by the formula,
 $\Rightarrow {g_p} = \dfrac{G}{{{R_p}^2}} \times \dfrac{4}{3}\pi {R_p}^3\rho $
Again on cancelling the common terms we get, $ {g_p} = \dfrac{4}{3}\pi G{R_p}\rho $
Now we can take the ratio of the two accelerations due to gravities. Therefore,
 $\Rightarrow \dfrac{{{g_p}}}{{{g_e}}} = \dfrac{{\dfrac{4}{3}\pi G{R_p}\rho }}{{\dfrac{4}{3}\pi G{R_e}\rho }} $
On cancelling all the common terms we get,
 $\Rightarrow \dfrac{{{g_p}}}{{{g_e}}} = \dfrac{{{R_p}}}{{{R_e}}} $
Therefore,
 $\Rightarrow {g_p} = \dfrac{{{R_p}}}{{{R_e}}} \times {g_e} $
Substituting the values $ {R_p} = \dfrac{{{R_e}}}{{10}} $ and $ {g_e} = 10m{s^{ - 2}} $ we get,
 $\Rightarrow {g_p} = \dfrac{{{R_e}}}{{10{R_e}}} \times 10 $
Hence, we have $ {g_p} = 1m{s^{ - 2}} $
Now the value of the acceleration due to gravity decreases with the increase in depth below the surface of the earth. This is given by the formula, $ {g_{pd}} = {g_p}\left( {1 - \dfrac{x}{{{R_p}}}} \right) $
Since, $ {g_p} = 1m{s^{ - 2}} $ , so we have $ {g_{pd}} = \left( {1 - \dfrac{x}{{{R_p}}}} \right) $
Now for the wire, first let us consider a small portion of the wire of length $ dx $ at a depth $ x $ . So the force on this elementary part will be given as,
 $\Rightarrow dF = \left( {\lambda dx} \right){g_{pd}} $
Substituting the value,
 $\Rightarrow dF = \left( {\lambda dx} \right)\left( {1 - \dfrac{x}{{{R_p}}}} \right) $
Therefore, the total force will be,
 $\Rightarrow \int {dF} = \int {\left( {\lambda dx} \right)\left( {1 - \dfrac{x}{{{R_p}}}} \right)} $
So we get the total force as,
 $\Rightarrow F = \lambda \int\limits_0^{\dfrac{{{R_p}}}{5}} {\left( {1 - \dfrac{x}{{{R_p}}}} \right)} dx $
On integrating we get,
 $\Rightarrow F = \lambda \left[ {x - \dfrac{{{x^2}}}{{2{R_p}}}} \right]_0^{\dfrac{{{R_p}}}{5}} $
Therefore on substituting the limits,
 $\Rightarrow F = \lambda \left[ {\dfrac{{{R_p}}}{5} - \dfrac{{{R_p}^2}}{{2 \times 25{R_p}}}} \right] $
Now substituting $ {R_p} = \dfrac{{{R_e}}}{{10}} $ and $ \lambda = {10^{ - 3}}kg{m^{ - 1}} $
 $\Rightarrow F = {10^{ - 3}} \times \left[ {\dfrac{{{R_e}}}{{50}} - \dfrac{{{R_e}}}{{500}}} \right] $
Hence on calculating we get,
 $\Rightarrow F = {10^{ - 3}} \times \left[ {\dfrac{{10{R_e} - {R_e}}}{{500}}} \right] $
In the question we are given $ {R_e} = 6 \times {10^6}m $
So we get,
 $\Rightarrow F = {10^{ - 3}} \times \dfrac{{9 \times 6 \times {{10}^6}m}}{{500}} $
On calculating we get the force as,
 $\Rightarrow F = 108N $
Therefore the correct answer is option B.

Note
The acceleration due to gravity for any object on the surface of any planet is the acceleration that is gained by that object due to the gravitational force on that planet. It is represented by $ g $ and has a unit of $ m/{s^2} $ .