The pressure caused by gravitational pull inside the earth at ‘a’ distance, a measure from the center, when mass and radius of earth are ‘m’ and R’ respectively, is:
A: $\dfrac{{8G{m^2}}}{{3\pi {R^4}}}\left( {1 + \dfrac{{{a^2}}}{{{R^2}}}} \right)$
B: $\dfrac{{8G{m^2}}}{{3\pi {R^2}}}\left( {1 - \dfrac{{{a^2}}}{{{R^2}}}} \right)$
C: $\dfrac{{3G{m^2}}}{{8\pi {R^4}}}\left( {1 - \dfrac{{{a^2}}}{{{R^2}}}} \right)$
D: $\dfrac{{8G{m^2}}}{{3\pi {R^4}}}\left( {1 - \dfrac{a}{R}} \right)$
Answer
Verified
116.4k+ views
Hint: The gravitational pull below the surface of the earth also varies with the distance(it varies linearly) from the center. The pressure will vary too from the centre to the surface of the earth.
Complete step by step answer:
For all bodies below the surface of the earth the pressure opposes the weight of the body.
Since the body is now below the surface the value of acceleration due to gravity also changes.
And pressure will be the sum of all small pressures opposing the small masses.
So since we know that pressure opposes the weight of the body we have;
$dp = \rho {g_r}dr$
Here dp is the small pressure opposing a small mass of length dr.
Also $g_r$ is the acceleration due to gravity as a function of r.
Now we know that for a boy under the surface of the earth the acceleration due to gravity is given by;
${g_r} = \dfrac{{Gmr}}{{{R^3}}}$ here, G is the universal gravitational constant, r is the distance from the center, m is the mass of the body and R is the radius of earth.
Also th density of the body will be;
$\rho = \dfrac{m}{{\dfrac{4}{3}\pi {R^3}}}$
Hence, $dp = \dfrac{{3G{m^2}r}}{{4\pi {R^6}}}dr$
So, $P = \int\limits_a^R {\dfrac{{3G{m^2}r}}{{4\pi {R^6}}}dr} $
Hence $P = \left[ {\dfrac{{3G{m^2}{r^2}}}{{8\pi {R^6}}}} \right]_a^R$
$P = \dfrac{{3G{m^2}}}{{8\pi {R^4}}}\left( {1 - \dfrac{{{a^2}}}{{{R^2}}}} \right)$
Hence option C is correct.
Note: The concepts of gravitation and electrostatics are quite similar, so while finding the electric field as we use Gauss Law, for gravitational field problems too we can use Gauss law equivalent to find for example the field inside the earth.
Complete step by step answer:
For all bodies below the surface of the earth the pressure opposes the weight of the body.
Since the body is now below the surface the value of acceleration due to gravity also changes.
And pressure will be the sum of all small pressures opposing the small masses.
So since we know that pressure opposes the weight of the body we have;
$dp = \rho {g_r}dr$
Here dp is the small pressure opposing a small mass of length dr.
Also $g_r$ is the acceleration due to gravity as a function of r.
Now we know that for a boy under the surface of the earth the acceleration due to gravity is given by;
${g_r} = \dfrac{{Gmr}}{{{R^3}}}$ here, G is the universal gravitational constant, r is the distance from the center, m is the mass of the body and R is the radius of earth.
Also th density of the body will be;
$\rho = \dfrac{m}{{\dfrac{4}{3}\pi {R^3}}}$
Hence, $dp = \dfrac{{3G{m^2}r}}{{4\pi {R^6}}}dr$
So, $P = \int\limits_a^R {\dfrac{{3G{m^2}r}}{{4\pi {R^6}}}dr} $
Hence $P = \left[ {\dfrac{{3G{m^2}{r^2}}}{{8\pi {R^6}}}} \right]_a^R$
$P = \dfrac{{3G{m^2}}}{{8\pi {R^4}}}\left( {1 - \dfrac{{{a^2}}}{{{R^2}}}} \right)$
Hence option C is correct.
Note: The concepts of gravitation and electrostatics are quite similar, so while finding the electric field as we use Gauss Law, for gravitational field problems too we can use Gauss law equivalent to find for example the field inside the earth.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Charging and Discharging of Capacitor
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids