
A tunnel is dug along a diameter of the earth. If \[{M_e}\] and ${R_e}$ are the mass and radius, respectively, of the earth, then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre is :
A. $\dfrac{{G{M_e}m}}{{{R_e}^3}}r$
B. $\dfrac{{G{M_e}m}}{{{R_e}^3r}}$
C. $\dfrac{{G{M_e}mR_e^3}}{r}$
D. $\dfrac{{G{M_e}m}}{{{R_e}^2}}r$
Answer
559.8k+ views
Hint: When a tunnel is dug, the value of acceleration due to gravity will vary at different points from the centre of the earth. Calculate mass of the inner sphere of radius $r$ and then calculate acceleration due to gravity, $g$ at a distance $r$ from centre of the earth.
Force on a particle of mass $m$ where acceleration due to gravity is $g$ is given by $F = mg$.
Complete solution:
As given in the question, a tunnel is dug along the diameter of the earth and we are asked to find the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre.
So, let us first calculate the mass of the inner solid sphere of radius $r$.
Total volume of the earth is $\dfrac{4}{3}\pi R_e^3$ where ${R_e}$ is the radius of the earth.
Volume of the inner sphere of radius $r$ is $\dfrac{4}{3}\pi {r^3}$
So, mass of the inner solid sphere of radius $r$ is given by
$M = \dfrac{{{M_e}}}{{\dfrac{4}{3}\pi R_e^3}} \times \dfrac{4}{3}\pi {r^3} = \dfrac{{{M_e}{r^3}}}{{R_e^3}}$
Now, we calculate acceleration due to gravity, $g$ at a distance $r$ from centre of the earth by using the formula $\dfrac{{GM}}{{{R^2}}}$ where $G$ is Gravitational constant whose value is $6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$ and $R = r$ .
So,$g=\dfrac{{G{M_e}{r^3}}}{{R_e^3}} \times \dfrac{1}{{{r^2}}} = \dfrac{{G{M_e}r}}{{R_e^3}}$
Now we know that the force on a particle of mass $m$ where acceleration due to gravity is $g$ is given by $F = mg$ .
Then, $F = m \times \dfrac{{G{M_e}r}}{{R_e^3}} = \dfrac{{G{M_e}mr}}{{R_e^3}}$
Hence, the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre is $\dfrac{{G{M_e}m}}{{R_e^3}}r$ .
So, option A is correct.
Note: When a body is falling freely, it acquires an acceleration equal to the acceleration due to gravity. Remember that at a different location from the earth the value of g is different and can be calculated by the formula $\dfrac{{GM}}{{{R^2}}}$ by carefully calculating the mass. This formula also clarifies that the acceleration due to gravity doesn’t depend upon the mass of the body. It means when a big rock and a small ball will drop from the same height, both will reach the ground at the same time.
Force on a particle of mass $m$ where acceleration due to gravity is $g$ is given by $F = mg$.
Complete solution:
As given in the question, a tunnel is dug along the diameter of the earth and we are asked to find the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre.
So, let us first calculate the mass of the inner solid sphere of radius $r$.
Total volume of the earth is $\dfrac{4}{3}\pi R_e^3$ where ${R_e}$ is the radius of the earth.
Volume of the inner sphere of radius $r$ is $\dfrac{4}{3}\pi {r^3}$
So, mass of the inner solid sphere of radius $r$ is given by
$M = \dfrac{{{M_e}}}{{\dfrac{4}{3}\pi R_e^3}} \times \dfrac{4}{3}\pi {r^3} = \dfrac{{{M_e}{r^3}}}{{R_e^3}}$
Now, we calculate acceleration due to gravity, $g$ at a distance $r$ from centre of the earth by using the formula $\dfrac{{GM}}{{{R^2}}}$ where $G$ is Gravitational constant whose value is $6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$ and $R = r$ .
So,$g=\dfrac{{G{M_e}{r^3}}}{{R_e^3}} \times \dfrac{1}{{{r^2}}} = \dfrac{{G{M_e}r}}{{R_e^3}}$
Now we know that the force on a particle of mass $m$ where acceleration due to gravity is $g$ is given by $F = mg$ .
Then, $F = m \times \dfrac{{G{M_e}r}}{{R_e^3}} = \dfrac{{G{M_e}mr}}{{R_e^3}}$
Hence, the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre is $\dfrac{{G{M_e}m}}{{R_e^3}}r$ .
So, option A is correct.
Note: When a body is falling freely, it acquires an acceleration equal to the acceleration due to gravity. Remember that at a different location from the earth the value of g is different and can be calculated by the formula $\dfrac{{GM}}{{{R^2}}}$ by carefully calculating the mass. This formula also clarifies that the acceleration due to gravity doesn’t depend upon the mass of the body. It means when a big rock and a small ball will drop from the same height, both will reach the ground at the same time.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

