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Test: Gravitation (10 August) - JEE MCQ


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15 Questions MCQ Test - Test: Gravitation (10 August)

Test: Gravitation (10 August) for JEE 2024 is part of JEE preparation. The Test: Gravitation (10 August) questions and answers have been prepared according to the JEE exam syllabus.The Test: Gravitation (10 August) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Gravitation (10 August) below.
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Test: Gravitation (10 August) - Question 1

A straight rod of length L extends from x = a to x = L + a. Find the gravitational force exerted by it on a point mass m at x = 0 if the linear density of rod μ = A+ Bx2

Detailed Solution for Test: Gravitation (10 August) - Question 1

Test: Gravitation (10 August) - Question 2

With what angular velocity the earth should spin in order that a body lying at 30° latitude may become weightless [R is radius of earth and g is acceleration due to gravity on the surface of earth]

Detailed Solution for Test: Gravitation (10 August) - Question 2

g' = acceleration due to gravity at latitude
g = acceleration due to gravity at poles
ω = angular velocity of earth
(θ) = latitude angle
Now,
g' = g - Rω^2 cos^2(θ)
As, g' = 0 (weightless):
g = Rω^2 cos^2(θ)
g = Rω^2 cos^2(30)
g = Rω^2 (3/4)
ω^2 = 4g/3R
ω = √(4g/3R)

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Test: Gravitation (10 August) - Question 3

The work done in shifting a particle of mass m from centre of earth to the surface of earth is (0m at both places are fixed)

Detailed Solution for Test: Gravitation (10 August) - Question 3

Here u i = gravitational potential energy at the centre of earth

u = gravitational potential energy at the surface of earth

Test: Gravitation (10 August) - Question 4

A man of mass m starts falling towards a planet of mass M and radius R. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass 2M/3 and a point mass M/3 at the centre. Change in the force of gravity experienced by the man is

Detailed Solution for Test: Gravitation (10 August) - Question 4
The gravitational force on the man when he was falling towards the planet was (GMm)/r^2. G here is the constant. M is mass of the whole planet ie M/3 + 2M/3. m is mass of the man. and r is the radius of the planet. once he has entered the hole in the planet. the force he was experiencing earlier due to the outer shell of the planet becomes 0. and now he experiences force only due to the point mass at the centre. this force is equal to (GMm)/3r^2. thus the change in force is initial - final force experienced which is equal to (2GMm)/3r^2. thus the answer.
Test: Gravitation (10 August) - Question 5

A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth’s surface. The height to which it would rise is

Detailed Solution for Test: Gravitation (10 August) - Question 5

  (orbital speed υ0 of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

Test: Gravitation (10 August) - Question 6

Two masses m1 & m2 are initially at rest and are separated by a very large distance. If the masses approach  each  other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separation distance of d is :

Detailed Solution for Test: Gravitation (10 August) - Question 6

We use the energy balance
Initial potential energy equals final kinetic energy
Gm1​m2/d​​=1mv12​/2+1​mv22​/2
also, from the conservation of momentum we have
m1​v1​=m2​v2
or
v1​= ​m1​v1/ m2
Substituting this we get 
v1​=√2Gm22​​​/d(m1​+m2​)
Similarly, we have 
v2​= √2Gm12​​​/ d(m1​+m2​)
Now as velocities are in opposite direction their relative velocity is v1​−(−v2​)=v1​+v2
or
[2G(m1​+m2​)​​/ d]1/2

Test: Gravitation (10 August) - Question 7

The ratio of the energy required to raise a satellite to a height h above the earth to that of the kinetic energy of satellite in the orbit there is (R = radius of earth)

Detailed Solution for Test: Gravitation (10 August) - Question 7

Energy required to raise a satellite upto a height h:

 

Test: Gravitation (10 August) - Question 8

Two concentric shells of uniform density of mass M1 and M2 are situated as shown in the figure. The forces experienced by a particle of mass m when placed at positions A, B and C respectively are (given OA = p, OB = q and OC = r)

Detailed Solution for Test: Gravitation (10 August) - Question 8

We know that attraction at an external point due to spherical shell of mass M is GM/r2 while at an internal point is zero. So, for particle at point C
OA = G (M1 + M2)m/p2
OB = G M1m/q2
OC = 0
 

Test: Gravitation (10 August) - Question 9

A spherical hole is made in a solid sphere of radius R. The mass of the original sphere was M.The gravitational field at the centre of the hole due to the remaining mass is

Detailed Solution for Test: Gravitation (10 August) - Question 9

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

Test: Gravitation (10 August) - Question 10

At what height above the earth’s surface does the accelerati on due to gravity fall to 1% of its value at the earth’s surface?

Detailed Solution for Test: Gravitation (10 August) - Question 10
1/(R+h)^2 =1/100(R)^2
Solve and you will get h=9R
Test: Gravitation (10 August) - Question 11

A ring has a total mass m but not uniformly distributed over its circumference. The radius of the ring is R. A point mass m is placed at the centre of the ring. Work done in taking away the point mass from centre to infinity is

Detailed Solution for Test: Gravitation (10 August) - Question 11

W = increase in potential energy of system

=Uf - Ui

=m (Vf - Vi)

(V = gravitational potential)

Note: Even if mass is nonuniformly distributed potential at centre would be -GM/R

Test: Gravitation (10 August) - Question 12

If the radius of the earth be increased by a factor of 5, by what factor its density  be changed to keep the value of g the same?

Detailed Solution for Test: Gravitation (10 August) - Question 12

Since density = M/V and M = gr^2/G and V = 4/3πr^3
Therefore density = (3/4) * (g/πrG) and new density = (3/4)*(g/πr'G)
As r’ = 5r then the new density = ¾ * (g/5rGπ)
d’ = 1/5*3/4(g/πrG)
d'=1/5 d

Test: Gravitation (10 August) - Question 13

Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is R−5/2 , then T2 is proportional to

Detailed Solution for Test: Gravitation (10 August) - Question 13

Test: Gravitation (10 August) - Question 14

Two point masses of mass 4m and m respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be :

Detailed Solution for Test: Gravitation (10 August) - Question 14

Let’s say the velocity of ‘m’ is ‘v’ and velocoty of ‘4m’ is ‘u’
Therefore, mv = 4mu ---- (Given)
So, v = 4u
Now, we know that kinetic energy = (1/2) mv2
Thus for the ratio, we need to calculate
(1/2 * m * (4u)2) / (1/2 * 4m * u2)
It turns out to be 4/1.
Which means the ratio of kinetic energies of both masses is 1:4

Test: Gravitation (10 August) - Question 15

If G is the universal gravitational constant and ρ is the uniform density of spherical planet.
Then shortest possible period of the planet can be

Detailed Solution for Test: Gravitation (10 August) - Question 15

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Let M be the mass of the planet R its radius and m the mass of a particle on its surface. Then

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