JEE Exam  >  JEE Questions  >  Gravitational potential energy of a solid hem... Start Learning for Free
Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?

A) (-6GM^2)/5R
B) (-3GM^2)/10R
C) Less than (-6GM^2)/5R
D) Greater than (-6GM^2)/5R?
Most Upvoted Answer
Gravitational potential energy of a solid hemispherical object of mass...
The gravitational potential energy of an object is the work done in bringing the object from an infinite distance away to its current position. The gravitational potential energy of a solid hemispherical object of mass M and radius R can be found using the formula:
U = (-3GM^2)/(5R)

Where U is the gravitational potential energy, G is the universal gravitational constant, M is the mass of the object, and R is the radius of the object.

Therefore, the gravitational potential energy of a solid hemispherical object of mass M and radius R is (-3GM^2)/(5R). The correct answer is (B) (-3GM^2)/10R.

Community Answer
Gravitational potential energy of a solid hemispherical object of mass...
Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. To calculate the gravitational potential energy of a solid hemispherical object, we need to consider the gravitational potential energy of each infinitesimally small element of the object and integrate it over the entire object.

Gravitational Potential Energy of an Infinitesimally Small Element:
Consider an infinitesimally small element of the hemispherical object with mass dm and radius r. The gravitational potential energy of this element is given by:

dU = -G * (M * dm) / r

Here, G is the gravitational constant.

Integration to Find Total Gravitational Potential Energy:
To find the total gravitational potential energy of the hemispherical object, we need to integrate the above expression over the entire object. Since the mass distribution is symmetric, we can integrate over the limits from 0 to R for the radius and from 0 to π/2 for the angle.

U = ∫∫ (-G * (M * dm) / r) dA

Where dA is the infinitesimal area element on the surface of the hemisphere.

Calculating the Infinitesimal Area Element:
In spherical coordinates, the infinitesimal area element dA is given by:

dA = r^2 * sinθ * dθ * dφ

Here, θ is the polar angle and φ is the azimuthal angle.

Substituting the expression for dA into the integral, we get:

U = ∫∫ (-G * (M * dm) / r) * r^2 * sinθ * dθ * dφ

Simplifying the expression, we have:

U = -G * M * ∫∫ dm * sinθ * dθ * dφ / r

Since the mass distribution is uniform, dm = (M / V) * dV, where V is the volume of the hemisphere.

U = -G * M * ∫∫ (M / V) * dV * sinθ * dθ * dφ / r

U = -G * M^2 * ∫∫ (1 / V) * dV * sinθ * dθ * dφ / r

Finding the Total Volume:
The volume of the hemisphere can be calculated as:

V = (2/3) * π * R^3

Substituting this into the expression for U, we get:

U = -G * M^2 * ∫∫ (1 / [(2/3) * π * R^3]) * dV * sinθ * dθ * dφ / r

Simplifying further, we obtain:

U = (-3GM^2) / (2πR^3) * ∫∫ dV * sinθ * dθ * dφ / r

U = (-3GM^2) / (2πR^3) * ∫∫ r^2 * sinθ * dθ * dφ / r

U = (-3GM^2) / (2πR^3) * ∫∫ r * sinθ * dθ * dφ

Evaluating the Integral:
Integrating over the limits, we find:

U = (-3GM^2) / (2πR^3) * [(2/3) * R^3
Explore Courses for JEE exam
Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R?
Question Description
Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R?.
Solutions for Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? defined & explained in the simplest way possible. Besides giving the explanation of Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R?, a detailed solution for Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? has been provided alongside types of Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? theory, EduRev gives you an ample number of questions to practice Gravitational potential energy of a solid hemispherical object of mass M and radius R will be?A) (-6GM^2)/5RB) (-3GM^2)/10RC) Less than (-6GM^2)/5RD) Greater than (-6GM^2)/5R? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev