What is the minimum energy required to launch a satellite of mass from the surface of a planet of mass and radius in a circular orbit at an altitude of ?
1
2
3
4
Official Solution
Correct Option: (1)
E = Energy of satellite - energy of mass on the surface of planet
Here, r = R + 2R = 2>R Substituting in about equation we get.
02
PYQ 2013
medium
physicsID: jee-main
The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, , located as shown, would be (nearly):
1
2
3
4
Official Solution
Correct Option: (3)
Let mass of smaller sphere (which has to be removed) is m Radius (from figure)
Mass of the left over part of the sphere
Therefore gravitational field due to the left over part of the sphere
03
PYQ 2014
medium
physicsID: jee-main
Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction, the speed of each particle is
1
2
3
4
Official Solution
Correct Option: (4)
The Correct Option is (D):
04
PYQ 2015
easy
physicsID: jee-main
A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius . A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of star is and its distance from the galaxy's axis is , then :
1
2
3
4
Official Solution
Correct Option: (3)
Let the linear mass density of the cylindrical galaxy be . Gravitational field Therefore, gravitational force Hence,
05
PYQ 2015
easy
physicsID: jee-main
From a solid sphere of mass and radius , a spherical portion of radius is removed, as shown in the figure. Taking gravitational potential at , the potential at the centre of the cavity thus formed is ( gravitational constant)
1
2
3
4
Official Solution
Correct Option: (4)
Solid sphere is of mass , radius . Spherical portion removed have radius , therefore its mass is . Potential at the centre of cavity
06
PYQ 2016
easy
physicsID: jee-main
A satellite is revolving in a circular orbit at a height ' ' from the earth's surface (radius of earth ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth?s gravitational field, is close to : (Neglect the effect of atmosphere.)
1
2
3
4
Official Solution
Correct Option: (4)
Orbital velocity as h < < R Velocity required to escape
Increase in velocity
07
PYQ 2016
medium
physicsID: jee-main
Figure shows elliptical path abcd of a planet around the sun such that the area of triangle is the area of the ellipse. (See figure) With as the axis, and as the axis. If is the time taken for planet to go over path and for path taken over then :
1
2
3
4
Official Solution
Correct Option: (3)
Area abca = x Area SABCS Area SADCS
08
PYQ 2017
medium
physicsID: jee-main
The variation of acceleration due to gravity with distance from centre of the earth is best represented by (R = Earth?s radius) :
1
2
3
4
Official Solution
Correct Option: (4)
Variation of g inside earth surface
09
PYQ 2018
medium
physicsID: jee-main
A particle is moving with a uniform speed in a circular orbit of radius in a central force inversely proportional to the nth power of . If the period of rotation of the particle is then -
1
for any n
2
3
4
Official Solution
Correct Option: (3)
10
PYQ 2019
medium
physicsID: jee-main
If the angular momentum of a planet of mass , moving around the Sun in a circular orbit is , about the center of the Sun, its areal velocity is :
1
2
3
4
Official Solution
Correct Option: (3)
11
PYQ 2019
medium
physicsID: jee-main
A test particle is moving in a circular orbit in the gravitational field produced by a mass density . Identify the correct relation between the radius of the particle's orbit and it's period :
1
is a constant
2
is a constant
3
is a constant
4
is a constant
Official Solution
Correct Option: (4)
12
PYQ 2019
medium
physicsID: jee-main
Four identical particles of mass are located at the corners of a square of side . What should be their speed if each of them revolves under the influence of other?s gravitational field in a circular orbit circumscribing the square?
1
2
3
4
Official Solution
Correct Option: (3)
Net force on particle towards centre of circle is
This force will act as centripetal force. Distance of particle from centre of circle is
13
PYQ 2019
medium
physicsID: jee-main
A straight rod of length extends from to . The gravitational force is exerts on a point mass at , if the mass per unit length of the rod is , is given by:
1
2
3
4
Official Solution
Correct Option: (2)
14
PYQ 2019
medium
physicsID: jee-main
A solid sphere of mass and radius is surrounded by a uniform concentric spherical shell of thickness and mass . The gravitational field at distance from the centre will be :
1
2
3
4
Official Solution
Correct Option: (2)
We use gauss's Law for gravitation = (Mass enclosed) 4 G
Option (2)
15
PYQ 2019
medium
physicsID: jee-main
A satellite of mass is in a circular orbit of radius about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be :
1
in a circular orbit of a different radius
2
in the same circular orbit of radius R
3
in an elliptical orbit
4
such that it escapes to infinity
Official Solution
Correct Option: (3)
If , then satellite does not remain in it's circular path rather it traces a spiral path and falls on earth
satellite move along parabolic path - wherein
Satellite revolves in circular path
Satellite will move along a hyperbolic path Apply momentum conservation
16
PYQ 2020
medium
physicsID: jee-main
Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is radius of its top, and height, then its moment of inertia about its axis is:
1
2
3
4
Official Solution
Correct Option: (1)
Area Area of element mass of element (here
17
PYQ 2020
easy
physicsID: jee-main
Planet has mass and radius . Planet has half the mass and half the radius of Planet . If the escape velocities from the Planets and are and , respectively, then . The value of is :
1
2
3
4
Official Solution
Correct Option: (4)
(Escape velocity)
18
PYQ 2020
easy
physicsID: jee-main
Consider two solid spheres of radii and masses and , respectively. The gravitational field due to sphere (1) and (2) are shown. The value of is :
1
2
3
4
Official Solution
Correct Option: (2)
Gravitational field on the surface of a solid sphere By the graph
and On solving
19
PYQ 2021
medium
physicsID: jee-main
The minimum and maximum distances of a planet revolving around the Sun are and . If the minimum speed of the planet on its trajectory is then its maximum speed will be :
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept: Angular momentum of a planet revolving around the Sun is conserved because the gravitational force is a central force. The planet moves in an elliptical orbit. The minimum speed occurs at the furthest point (aphelion) and the maximum speed occurs at the nearest point (perihelion). Step 2: Key Formula or Approach: Conservation of Angular Momentum: . At perihelion and aphelion, the velocity is perpendicular to the position vector ( ), so:
Step 3: Detailed Explanation: Given: Minimum distance (Perihelion) = . Maximum distance (Aphelion) = . Minimum speed (at ) = . Let the maximum speed (at ) be . By Conservation of Angular Momentum:
Step 4: Final Answer: The maximum speed will be .
20
PYQ 2021
medium
physicsID: jee-main
The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :
1
3 T
2
9 T
3
27 T
4
12 T
Official Solution
Correct Option: (3)
Step 1: By Kepler's Third Law, .
Step 2: .
Step 3: .
21
PYQ 2021
medium
physicsID: jee-main
Inside a uniform spherical shell : (a) the gravitational field is zero. (b) the gravitational potential is zero. (c) the gravitational field is same everywhere. (d) the gravitation potential is same everywhere. (e) all of the above Choose the most appropriate answer from the options given below :
1
(b), (c) and (d) only
2
(a), (b) and (c) only
3
(a), (c) and (d) only
4
(e) only
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept: The shell theorem in gravitation states that the gravitational influence inside a uniform spherical shell follows specific rules based on the inverse square law. Step 2: Detailed Explanation: 1. Statement (a) and (c): The gravitational field ( ) inside a uniform shell is zero at all points. Since zero is the same value at every point, it is correct to say the field is zero and that it is the same everywhere. 2. Statement (b) and (d): The gravitational potential ( ) inside the shell is given by , where is the shell radius. This is a constant value but is NOT zero. Since it is constant, it is the same everywhere inside the shell. Conclusion: (a), (c), and (d) are true. (b) is false. Step 3: Final Answer: The correct option is (C), which includes (a), (c), and (d).
22
PYQ 2021
medium
physicsID: jee-main
The masses and radii of the earth and moon are ( ) and ( ) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept: Escape velocity is the minimum velocity required for an object to escape the gravitational field of a system, meaning its total mechanical energy at infinity becomes zero. Step 2: Key Formula or Approach: Total Energy = Kinetic Energy + Potential Energy = 0. Potential Energy at midpoint (distance from both): . Step 3: Detailed Explanation: By conservation of energy: Step 4: Final Answer: The minimum escape velocity is .
23
PYQ 2021
medium
physicsID: jee-main
A body of mass (2M) splits into four masses m, M-m, m, M-m, which are rearranged to form a square as shown in the figure. The ratio of for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is ________.
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We have four masses arranged at the corners of a square. The total gravitational potential energy (GPE) of this system depends on the value of 'm'. We need to find the ratio M/m for which this GPE is maximum. Step 2: Key Formula or Approach:
The gravitational potential energy between two masses and separated by a distance is .
The total GPE of the system is the sum of the potential energies of all possible pairs of masses. In a square with four masses, there are 6 pairs (4 sides and 2 diagonals). To find the maximum GPE, we will differentiate the total potential energy expression with respect to 'm' and set the derivative to zero. Step 3: Detailed Explanation:
The arrangement has masses 'm' and 'M-m' at adjacent corners.
The four masses at the vertices are , , , .
The side length of the square is 'd', and the diagonal length is . Let's calculate the total GPE ( ) by summing the energy of all 6 pairs: 4 pairs along the sides (distance d): 2 pairs along the diagonals (distance ):
Total GPE:
To maximize , which is a negative quantity, we must minimize its magnitude. Let the term in the brackets be .
For maximum U, we need , which means .
Group terms with M and m:
Cancel the term from both sides:
The ratio is given as x:1.
Step 4: Final Answer:
The value of x is 2.
24
PYQ 2021
medium
physicsID: jee-main
A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? [Use and radius of earth, .]
1
49.17 N
2
49 N
3
48.83 N
4
49.83 N
Official Solution
Correct Option: (3)
Step 1: Weight at pole . Thus, mass .
Step 2: Effective gravity at equator .
Step 3: for Earth is approx .
Step 4: .
25
PYQ 2021
medium
physicsID: jee-main
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass. Reason R : The product of their mass and radius must be same. In the light of the above statements, choose the most appropriate answer from the options given below :
1
Both A and R are correct and R is the correct explanation of A
2
Both A and R are correct but R is NOT the correct explanation of A
3
A is correct but R is not correct
4
A is not correct but R is correct
Official Solution
Correct Option: (3)
Step 1: The formula for escape velocity is given by .
Step 2: For the escape velocities of two planets to be equal ( ), we must have .
Step 3: This implies the ratio of mass to radius must be the same, not the product. Thus, Assertion A is correct because unequal masses can have equal escape velocities if their radii are also different in the same proportion.
Step 4: Reason R states , which is mathematically inconsistent with the escape velocity formula.
26
PYQ 2021
medium
physicsID: jee-main
A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V. The value of V is:
1
-60 G
2
-20 G
3
-4 G
4
+2 G
Official Solution
Correct Option: (3)
Step 1: Understanding the Question:
We need to find the total gravitational potential at a point inside a spherical shell. The total potential is the sum of the potential due to the shell itself and the potential due to a point mass placed at its center. Step 2: Key Formula or Approach:
The gravitational potential at a point is calculated using the principle of superposition.
1. Potential due to a point mass M at distance r:
2. Potential due to a uniform spherical shell of mass M and radius R: - For a point inside the shell ( ), the potential is constant and equal to the potential at the surface: . - For a point outside the shell ( ), . Step 3: Detailed Explanation:
We are given:
Mass at the center, .
Mass of the spherical shell, .
Radius of the shell, .
We need to find the potential at a distance from the center. The total potential at this point is the sum of the potential due to the point mass ( ) and the potential due to the spherical shell ( ). Calculating Potential due to the Point Mass ( ):
The point is at from the mass . Calculating Potential due to the Spherical Shell ( ):
The point is inside the shell, since ( ).
For any point inside a uniform spherical shell, the potential is constant and equal to the potential on its surface. Calculating Total Potential ( ): Step 4: Final Answer:
The value of the gravitational potential V at the given point is -4G.
27
PYQ 2021
medium
physicsID: jee-main
The initial velocity required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity such that . The value of x will be ________ .
Official Solution
Correct Option: (1)
We will use the principle of conservation of mechanical energy. The initial energy of the body on the Earth's surface is . The final energy of the body at a height of 10R is . (At maximum height, final velocity is 0). By conservation of energy, : . . . The escape velocity from the Earth's surface is given by , so . Substitute into the equation for : . Taking the square root of both sides: . Comparing this with the given expression , we find that x = 10.
28
PYQ 2021
medium
physicsID: jee-main
If be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : )
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept:
Acceleration due to gravity varies with height and depth. We need to find the ratio of to for a displacement . Step 2: Key Formula or Approach:
1. At depth :
2. At height : Step 3: Detailed Explanation:
We need to find the ratio :
Expand the squared term:
Multiply the terms:
Group like terms:
Step 4: Final Answer:
The ratio is .
29
PYQ 2021
medium
physicsID: jee-main
A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force . Now a spherical cavity of radius (R/2) is made in the sphere (as shown in figure) and the force becomes . The value of is :
1
41 : 50
2
50 : 41
3
36 : 25
4
25 : 36
Official Solution
Correct Option: (2)
Step 1: .
Step 2: Mass of removed part .
Step 3: Distance of cavity center from test particle .
Step 4: Force from removed part .
Step 5: .
Step 6: Ratio .
30
PYQ 2021
medium
physicsID: jee-main
Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If and are the time periods of A and B respectively then the value of is : [ km, kg]
1
s
2
s
3
s
4
s
Official Solution
Correct Option: (1)
Step 1: Orbital radii: km; km.
Step 2: Time period formula .
Step 3: m /s .
Step 4: s.
Step 5: s.
Step 6: s s. Closest option is (A).
31
PYQ 2021
medium
physicsID: jee-main
Two identical particles of mass 1 kg each go round a circle of radius R under the action of their mutual gravitational attraction. The angular speed of each particle is :
1
2
3
4
Official Solution
Correct Option: (2)
Let the two identical particles have mass kg.
They move in a circle of radius R, so they are always diametrically opposite each other.
The center of the circle is the center of mass of the two-particle system. The distance between the two particles is constant and equal to the diameter of the circle, . Consider one of the particles. The only force acting on it is the gravitational attraction from the other particle. This force is directed towards the center of the circle and provides the necessary centripetal force. Gravitational force: . Centripetal force: , where is the angular speed. Equating the two forces:
. We can cancel one factor of 'm' from both sides.
. Now, solve for :
. Take the square root to find the angular speed :
. Given that the mass kg:
.
32
PYQ 2021
medium
physicsID: jee-main
Two stars of masses and at a distance rotate about their common centre of mass in free space. The period of revolution is :
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: The centripetal force for mass is provided by gravitational attraction: .
Step 2: Distance of mass from center of mass .
Step 3: .
Step 4: .
33
PYQ 2021
medium
physicsID: jee-main
Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Net force on one particle . Force from adjacent particles at . Force from opposite particle .
Step 2: .
Step 3: .
Step 4: With , .
34
PYQ 2021
medium
physicsID: jee-main
Consider two satellites and with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite to the angular velocity of satellite is :
1
8 : 1
2
1 : 8
3
2 : 1
4
1 : 4
Official Solution
Correct Option: (1)
Step 1: Angular velocity is related to time period by .
Step 2: Therefore, .
Step 3: Given hr and hr.
Step 4: .
35
PYQ 2021
medium
physicsID: jee-main
The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of km. Find the mass of Mars.
1
kg
2
kg
3
kg
4
kg
Official Solution
Correct Option: (3)
For a moon orbiting a planet, the gravitational force provides the necessary centripetal force.
Where M is the mass of Mars, m is the mass of the moon, r is the orbital radius, and is the angular velocity.
. We know that angular velocity , where T is the time period.
. Rearranging the formula to solve for the mass of Mars (M):
. First, convert the given quantities to SI units.
Orbital radius . Time period T = 7 hours 30 minutes.
. Now, substitute the values into the equation for M.
. . . . .
36
PYQ 2021
medium
physicsID: jee-main
The angular momentum of a planet of mass moving around the sun in an elliptical orbit is . The magnitude of the areal velocity of the planet is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Recall Kepler's Second Law. Areal velocity is the rate at which area is swept: . Step 2: The area swept in small time is .
Step 3: We know angular momentum is . Thus, .
Step 4: Substitute back:
37
PYQ 2021
medium
physicsID: jee-main
Suppose two planets (spherical in shape) of radii R and 2 R, but mass M and 9 M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is then the value of 'a' is _________. [Given : The two planets are fixed in their position]
Official Solution
Correct Option: (1)
To find the minimum speed, the satellite must just be able to reach the null point (where the gravitational forces of the two planets cancel out) with zero kinetic energy. After this point, the gravity of the second planet will pull it in. Let the null point be at a distance x from the center of the first planet (mass M). The distance from the second planet (mass 9M) will be (8R - x). At the null point, the net gravitational force is zero:
. Now, we apply the principle of conservation of energy. The initial energy at the surface of the first planet must equal the final energy at the null point. Initial Energy ( ) at the surface of the first planet:
Final Energy ( ) at the null point (with minimum speed, so ):
By conservation of energy, :
Divide by m and rearrange:
Comparing this with the given expression , we get .
38
PYQ 2021
medium
physicsID: jee-main
A particle of mass moves in a circular orbit under the central potential field, , where is a positive constant. The correct radius – velocity graph of the particle's motion is:
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (3)
Step 1: Find the force from the potential :
The magnitude of the force is .
Step 2: For circular motion, centripetal force is provided by this field:
Step 3: Since , the graph of vs is a hyperbola-like curve where decreases as increases.
39
PYQ 2021
medium
physicsID: jee-main
If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately: [Take ms⁻², m, ]
1
does not change
2
1200 minutes
3
60 minutes
4
84 minutes
Official Solution
Correct Option: (4)
Step 1: For bodies to float at the equator, the effective gravity must be zero:
Step 2: The duration of the day is the time period :
Step 3: Substitute values:
Step 4: Convert to minutes:
40
PYQ 2022
medium
physicsID: jee-main
The area of cross section of the rope used to lift a load by a crane is . The maximum lifting capacity of the crane is metric tons. To increase the lifting capacity of the crane to metric tons, The required area of cross section of the rope should be
1
2
3
4
Official Solution
Correct Option: (1)
area of cross section of the rope :
=
=
41
PYQ 2022
easy
physicsID: jee-main
Given below are two statements. Statement-I: The law of gravitation holds good for any pair of bodies in the universe. Statement-II: The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below
1
Both Statement I and Statement II are true
2
Both Statement I and Statement II are false
3
Statement I is true but Statement II is false
4
Statement I is false but Statement II is true
Official Solution
Correct Option: (1)
Statement-I is true as law of gravitation is a universal law
Statement-II is also true as gravitational field at centre of earth is zero.
Therefore, the correct option is (A): Both Statement I and Statement II are true
42
PYQ 2022
medium
physicsID: jee-main
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be
1
40 hours
2
36 hours
3
30 hours
4
25 hours
Official Solution
Correct Option: (2)
The correct answer is (B) : 36 hours
T2 ≅ 36 hrs
43
PYQ 2022
medium
physicsID: jee-main
The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by (Given R = radius of earth)
1
2
3
4
Official Solution
Correct Option: (1)
For (C=Constant) For
So, for the above equations the best suited graph is as given in option (A)
44
PYQ 2022
medium
physicsID: jee-main
The approximate height from the surface of the earth at which the weight of the body becomes of its weight on the surface of earth is [Radius of earth R = 6400 km and √3=1.732]
1
3840 km
2
4685 km
3
2133 km
4
4267 km
Official Solution
Correct Option: (2)
According to the given information = × ⇒ R+h=√3 R ⇒ h=(√3-1)R∼4685 km The correct option is (B) : 4685 km
45
PYQ 2022
medium
physicsID: jee-main
A body of mass m is projected with velocity \lambda ve in vertically upward direction from the surface of the earth into space. It is given that ve is escape velocity and \lambda < 1. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be: (R : radius of earth)
1
2
3
4
Official Solution
Correct Option: (2)
The correct answer is (B): Using energy conservation
46
PYQ 2022
medium
physicsID: jee-main
If the radius of earth shrinks by while its mass remains same The acceleration due to gravity on the earth's surface will approximately :
1
decrease by
2
decrease by
3
increase by %
4
increase by
Official Solution
Correct Option: (4)
From the given options the correct answer is option (D): increase by 4%
47
PYQ 2022
medium
physicsID: jee-main
A ball of mass hits the wall with its initial speed of and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is , calculate the time duration of the contact of ball with the wall.
1
0.018 s
2
0.036 s
3
0.009 s
4
0.072 s
Official Solution
Correct Option: (2)
Force applied by the wall on the ball during the contact is =
Hence,
=
Hence, the correct option is (B):
48
PYQ 2022
medium
physicsID: jee-main
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:
1
800 km
2
1600 km
3
2133 km
4
4800 km
Official Solution
Correct Option: (1)
As per the theory of conservation of energy
-
Therefore, the maximum height attained by the body will be:
=
=
49
PYQ 2022
medium
physicsID: jee-main
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be (Given g = acceleration due to gravity at the surface of earth).
1
2
3
4
Official Solution
Correct Option: (4)
We know that
Given
so, the correct option is (D):
50
PYQ 2022
medium
physicsID: jee-main
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth = 6400 km)
1
1%
2
3%
3
4%
4
0.50%
Official Solution
Correct Option: (1)
We know that,
The percentage decrease in the weight of a rocket,
% decrease in weight =
So, the correct option is (A):
51
PYQ 2022
medium
physicsID: jee-main
The elongation of a wire on the surface of the earth is 10–4m. The same wire of same dimensions is elongated by 6 \times 10–5m on another planet. The acceleration due to gravity on the planet will be ___ ms–2. (Take acceleration due to gravity on the surface of earth = 10 ms–2)
Official Solution
Correct Option: (1)
52
PYQ 2022
easy
physicsID: jee-main
Two planets A and B of equal mass are having their period of revolutions TA andTB such that TA= 2TB. These planets are revolving in the circular orbits of radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits?
1
2
3
4
Official Solution
Correct Option: (3)
The correct answer is (C) :
Now
53
PYQ 2022
medium
physicsID: jee-main
The percentage decrease in the weight of a rocket, when taken to a height of above the surface of earth will, be : (Radius of earth )
1
2
3
4
Official Solution
Correct Option: (1)
The correct option is(A): 1%.
54
PYQ 2022
medium
physicsID: jee-main
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude. Reason R: At equator, the direction of acceleration due to the gravity is towards the center of earth. In the light of above statements, choose the correct answer from the options given below
1
Both A and R are true and R is the correct explanation of A.
2
Both A and R are true but R is NOT the correct explanation of A.
3
A is true but R is false.
4
A is false but R is true.
Official Solution
Correct Option: (4)
The correct answer is (D): A is false but R is true. g′ = g0 – ω2Rcos2θ θ = latitude.
55
PYQ 2022
medium
physicsID: jee-main
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be
1
2
3
4
Official Solution
Correct Option: (3)
The problem involves the gravitational force between two objects of equal mass. Let the mass of each object initially be and the distance between them be . According to Newton's law of universal gravitation, the force between these two objects can be expressed as:
Where:
is the gravitational constant.
is the mass of each object.
is the distance between the centers of the two objects.
Now, if one-third of the mass of one object is transferred to the other, the situation changes as follows:
Mass of object 1:
Mass of object 2:
The new gravitational force ( ) between the two objects becomes:
We can compare the new force with the original force:
Thus, the new force is:
Therefore, the correct option is , which matches the given answer.
56
PYQ 2022
easy
physicsID: jee-main
A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ___ m. [Use g = 10 m/s2]
Official Solution
Correct Option: (1)
57
PYQ 2023
medium
physicsID: jee-main
The escape velocities of two planets and are in the ratio If the ratio of their radii respectively is , then the ratio of acceleration due to gravity of planet to the acceleration of gravity of planet B will be :
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Recall the Formula for Escape Velocity
The escape velocity ( ) of a planet is given by:
where is the gravitational constant, is the mass of the planet, and is the radius of the planet. We can also express the mass in terms of density ( ) and volume:
Substituting this into the escape velocity formula gives:
where is a constant.
Step 2: Set Up the Ratio of Escape Velocities
Let and be the escape velocities of planets A and B respectively, and let and be their radii, and and their densities. Given:
and
we have:
Step 3: Recall the Formula for Acceleration Due to Gravity
The acceleration due to gravity ( ) on a planet is given by:
where is a constant.
Step 4: Find the Ratio of Accelerations Due to Gravity
Let and be the accelerations due to gravity on planets A and B, respectively. Then:
Conclusion:
The ratio of the acceleration due to gravity of planet A to that of planet B is (Option 3).
58
PYQ 2023
medium
physicsID: jee-main
A body of mass is taken from earth surface to the height equal to twice the radius of earth the inerease in potential energy will be:( = acceleration due to gravity on the surface of Earth)
1
2
3
4
Official Solution
Correct Option: (3)
The correct answer is (C) : U=r−GMem Ui=Re−GMem Uf=(Re+h)−GMem=Re+2Re−GMem 3Re−GMem
Increase in internal energy ΔU=Uf−Ui =32ReGMem 32Re2GMemRe =32mgRe
59
PYQ 2023
easy
physicsID: jee-main
Statement-I : If we move upward and downward from the surface of earth surface acceleration due to gravity decreases in both upward and downward direction. Statement-II : Acceleration due to gravity changes by same amount when we go up to height h and depth d when h = d. Choose the correct options based on above statements.
1
Both statement-I and Statement-II are true.
2
Statement-I is true and Statement-II is false.
3
Statement-I is false and Statement-II are true.
4
Both statement-I and Statement-II are false.
Official Solution
Correct Option: (2)
The correct option is B
60
PYQ 2023
medium
physicsID: jee-main
At a certain depth " " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height above earth surface Where is Radius of earth (Take ) The depth is equal to
1
2
3
4
Official Solution
Correct Option: (1)
[1− ] = 1− = ⇒ = ⇒ d= R ⇒ d= x 6400 ⇒ d=4800 km
So, the correct option is (A): 4800 km
61
PYQ 2023
easy
physicsID: jee-main
Match Column-I with Column-II : Choose the correct answer from the options given below:
1
A- I, B-III, C-IV, D-II
2
A- II, B-IV, C-III, D-I
3
A- II, B-III, C-IV, D-I
4
A- I, B-II, C-III, D-IV
Official Solution
Correct Option: (2)
dtdx= slope ≥0 always increasing
(A - II) dtdx<0; and at t→∞dtdx→0
(B - IV) dtdx>0 for first half dtdx<0 for second half.
(C - III) dtdx= constant
(D - I)
62
PYQ 2023
easy
physicsID: jee-main
Given below are two statements:
Assertion (A): A pendulum clock when taken to Mount Everest becomes fast.
Reason (R): The value of (acceleration due to gravity) is less at Mount Everest than its value on the surface of Earth.
In the light of the above statements, choose the most appropriate answer from the options given below:
1
Both A and R are correct but R is NOT the correct explanation of A
2
Both A and R are correct and R is the correct explanation of A
3
A is not correct but R is correct
4
A is correct but R is not correct
Official Solution
Correct Option: (3)
The time period of a pendulum is given by:
At Mount Everest, is less than on Earth’s surface. A decrease in increases , making the pendulum slower, not faster. Thus:
Assertion (A): is incorrect because the pendulum becomes slow, not fast.
Reason (R): is correct, as decreases at higher altitudes.
63
PYQ 2023
medium
physicsID: jee-main
The length of a seconds pendulum at a height h=2R from earth surface will be : (Given : R= Radius of earth and acceleration due to gravity at the surface of earth g=π2ms−2)
An object is taken to a height above the surface of earth at a distance from the centre of the earth Where radius of earth, The percentage decrease in the weight of the object will be
The weight of a body at the surface of earth is The weight of the body at an altitude of above the earth's surface is (given, radius of earth ):
1
19.6 N
2
9.8 N
3
4.9 N
4
8 N
Official Solution
Correct Option: (4)
Acceleration due to gravity at height
Hence, the weight at given height will be:
66
PYQ 2023
easy
physicsID: jee-main
Separation between earth and sun is given by 1.5 × 106 km. Time period of another planet is 2.83 year. Find distance of another planet from sun?
1
3 × 106 km
2
2 × 107 km
3
3 × 107 km
4
2 × 106 km
Official Solution
Correct Option: (1)
T2 ∝ R3 (T1/T2)2 = (R1/R2)3 = {(1.5 x 106)/R2}3 R2 = km = = 3 × 106 km
The correct option is (A): 3 × 106 km
67
PYQ 2023
hard
physicsID: jee-main
Given below are two statements: Statement I : Acceleration due to gravity is different at different places on the surface of earth Statement II : Acceleration due to gravity increases as we go down below the earth's surface In the light of the above statements, choose the correct answer from the options given below
1
Both Statement I and Statement II are false
2
Statement is false but Statement II is true
3
Statement I is true but Statement II is false
4
Both Statement I and Statement II are true
Official Solution
Correct Option: (3)
geff=g−ω2Resin2θ,θ→ co-latitude angle geff =g(1−Red),d here depth
68
PYQ 2023
medium
physicsID: jee-main
An object weighs 200N at the surface of earth. Find the weight at a depth of , where R is radius of earth
1
100N
2
300N
3
50N
4
150N
Official Solution
Correct Option: (1)
Given the weight of the object on the surface of the Earth, mg = 200 N Where, m is the mass of the object and g is acceleration due to gravity on the surface of the earth. The variation of acceleration due to gravity (g) with the depth (d) from the surface of the Earth is given by gd=g1-dR Where gd is the acceleration due to gravity at depth d R is the radius of the Earth At depth d = , acceleration due to gravity is given by gd=g1- =g2 Multiplying both sides by mass of the object, then we get mgd = But, mg = 200 N Therefore, Weight of the object at depth R/2 is given by mgd = = 100 N Answer. A
69
PYQ 2023
medium
physicsID: jee-main
The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:
1
25 N
2
64 N
3
100 N
4
50 N
Official Solution
Correct Option: (2)
Using Newton's formula:
At surface of the earth,
At ,
Thus,
70
PYQ 2023
easy
physicsID: jee-main
Choose the incorrect statement from the given statements.
1
Planets revolve around the Sun with constant linear speed.
2
Energy of the planet in elliptical orbit is constant.
3
Satellites in circular motion have constant energy.
4
Body falling towards the Earth results in negligible displacement of the Earth.
Official Solution
Correct Option: (1)
The incorrect option is (A): Planets revolve around the Sun with constant linear speed. The correct answer should be Planet revolves around the Sun in elliptical orbit with variable speed.
71
PYQ 2023
easy
physicsID: jee-main
Suppose a situation in which two planets orbits around the sun in the same orbit. If the mass of plant 1 is twice the mass of planet 2, then what do they have same?
1
Potential energy
2
Kinetic energy
3
Total energy
4
Velocity
Official Solution
Correct Option: (1)
The correct option is (D): Velocity
72
PYQ 2023
medium
physicsID: jee-main
If the earth suddenly shrinks to th of its original volume with its mass remaining the same, the period of rotation of earth becomes h. The value of x is .
Official Solution
Correct Option: (1)
Step 1: Simplify the Equation
The angular momentum conservation equation is given by:
Cancel the mass and constant :
Step 2: Relate Angular Velocity to Time Period
The relationship between angular velocity and time period is:
Substituting and :
Step 3: Solve for
Rearrange to find :
Final Answer:
The value of is 16.
73
PYQ 2023
medium
physicsID: jee-main
A planet has double the mass of the earth. Its average density is equal to
that of the earth. An object weighing W on earth will weigh on that planet:
1
1
2
(2)
3
(2)
4
2
Official Solution
Correct Option: (2)
The correct answer is option (B):
74
PYQ 2023
medium
physicsID: jee-main
If a planet has mass equal to 16 times the mass of earth, and radius equal to 4 times that of earth. The ratio of escape speed of a planet to that of earth is?
1
2:1
2
1:2
3
√2:1
4
4:1
Official Solution
Correct Option: (1)
Solution: The escape velocity is given by the formula: $ v_e G M R M_p R_p M_e R_e v_{ep} v_{ee} M_p = 16 M_e R_p = 4 R_e $ Therefore, the ratio of the escape velocity of the planet to the escape velocity of Earth is 2:1.
The correct answer is (2) 2:1.
75
PYQ 2023
hard
physicsID: jee-main
Two identical particles each of mass m, move in circular path due to their own mutual gravitational force. Find the velocity of the particle if the radius of the circular path is a.
1
2
3
4
Official Solution
Correct Option: (4)
76
PYQ 2023
medium
physicsID: jee-main
Which of the following expressions give the value of acceleration due to gravity (g') at the altitude h above the surface of Earth. (R = radius of Earth, g = acceleration due to gravity at surface of Earth)
1
2
3
4
Official Solution
Correct Option: (2)
The acceleration due to gravity at a height above the Earth's surface is: Using the binomial expansion for , neglecting higher-order terms: Substituting this approximation: Since , we can write: Final Answer: The acceleration due to gravity is:
77
PYQ 2023
medium
physicsID: jee-main
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Earth has atmosphere and moon doesn’t Reason (R): escape speed on moon is less than that of Earth. In the light of the above statements, choose the correct answer from the options given below:
1
(A) and (R) are correct and (R) is the correct explanation of (A)
2
(A) and (R) are correct and (R) is not the correct explanation of (A)
3
(A) is true but (R) is false
4
(A) and (R) both are false
Official Solution
Correct Option: (1)
The correct option is (A): (A) and (R) are correct and (R) is the correct explanation of (A)
As the moon's radius is very small as compared to earth. at the moon is quite low and the gas molecules attains escape velocity at normal temperature on the moon.
78
PYQ 2023
medium
physicsID: jee-main
A planet has density same as that of Earth and mass is twice that of Earth. If the weight of an object on Earth is "W" then the weight on the planet is:
1
2
3
4
Official Solution
Correct Option: (2)
The correct option is(B):
Planet with the mass M has radius as R and Planet with mass 2M has radius as R'
79
PYQ 2023
medium
physicsID: jee-main
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R Assertion (A): Earth has atmosphere and moon doesn’t Reason (R): escape speed on moon is less than that Earth. In the light of the above statements, choose the correct answer from the options given below
1
(A) and (R) are correct and (R) is the correct explanation of (A)
2
(A) and (R) are correct and (R) is not the correct explanation of (A)
3
(A) is true but (R) is false
4
(A) and (R) both are false
Official Solution
Correct Option: (1)
The correct answer is option (A): (A) and (R) are correct and (R) is the correct explanation of (A)
Both (A) and (R), are true and the escape speed on the moon is less due to its small radius and acceleration due to gravity as compared to earth.
80
PYQ 2023
medium
physicsID: jee-main
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R Assertion (A): Acceleration due to gravity is minimum at equator. Reason (R): Rotation of earth influences acceleration. In the light of the above statements, choose the correct answer from the options given below
1
A is correct, R is correct explanation of A
2
A is correct, R is incorrect explanation of A
3
A is correct and R is incorrect
4
Both A and R are incorrect
Official Solution
Correct Option: (1)
The correct answer is option (A) A is correct, R is correct explanation of A
81
PYQ 2023
medium
physicsID: jee-main
If the maximum load carried by an elevator is 1400 kg (600 kg - Passengers + 800 kg- elevator), which is moving up with a uniform speed of 3 ms-1 and the frictional force acting on it is 2000 N, then the maximum power used by the motor is ____ kW. (g=10 m/s2)
Official Solution
Correct Option: (1)
Given Mass of the elevator system = 1400 kg Velocity ( ) = Frictional force ( ) = 2000 N The net force on the elevator is zero, as it is moving with uniform speed. So, the upward force (tension ) must balance the downward forces, which are the gravitational force ( ) and the frictional force. 1. Tension in the string: 2. The maximum power used by the motor is given by: Thus, the maximum power used by the motor is 48 kW.
82
PYQ 2023
hard
physicsID: jee-main
A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m/s (as shown in figure). Maximum height with respect to the initial position covered by it will be_____ cm (take, g = 10 m/s²)
Official Solution
Correct Option: (1)
For pure rolling motion, mechanical energy is conserved. At the initial position ( ), the ball has kinetic energy due to both translational and rotational motion, and at the maximum height ( ), all the kinetic energy is converted into potential energy. Step 1: Write the conservation of energy equation. The total mechanical energy at is: where is the mass of the ball, is the initial velocity, is the moment of inertia of the hollow sphere ( ), and is the angular velocity. At the maximum height ( ), all kinetic energy is converted into potential energy: where is the maximum height. Using energy conservation: Step 2: Substitute the expressions. Simplify: Combine terms: Cancel on both sides: Step 3: Substitute values. Substitute and : Simplify: Convert to centimeters: Final Answer: The maximum height covered is:
83
PYQ 2023
hard
physicsID: jee-main
The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are and respectively. The ratio of acceleration due to gravity at their surfaces (i.e. ) will be:
1
2:1
2
2:3
3
3:4
4
4:3
Official Solution
Correct Option: (3)
The acceleration due to gravity at the surface of a planet is given by: Using the relation , the acceleration can be expressed as: For planet A, the acceleration due to gravity is: For planet B, the radius is and the density is . Therefore: Simplifying : The ratio of to is: Thus, the ratio of acceleration due to gravity at the surface of B to A is .
84
PYQ 2023
medium
physicsID: jee-main
Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth from the surface of earth, if its weight on the surface of earth is 200 N, will be
1
100 N
2
300 N
3
50 N
4
150 N
Official Solution
Correct Option: (1)
Step 1: Understanding the Problem
The weight of a body decreases as it moves towards the center of the Earth because the effective gravitational force decreases. At a depth , the effective weight is proportional to the distance from the center of the Earth.
Where:
: Weight at depth
: Weight at the surface
: Radius of the Earth
: Depth
Step 2: Substituting Values
From the formula: Substituting and :
Simplify the equation:
Step 3: Final Answer
The weight of the body at depth is .
85
PYQ 2023
medium
physicsID: jee-main
The time period of a satellite revolving above earth's surface at a height equal to R will be (given: , R= radius of earth)
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Using the time period formula for a satellite:Step 2: Substituting :Step 3: Simplifying the expression: At the surface of the Earth, we have . So, substituting and , we get:
86
PYQ 2023
medium
physicsID: jee-main
A space ship of mass 2 × 104 kg is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be (if g = 10 m/s2 and radius of earth = 6400 km)
1
11.2 (√2 -1) km/s
2
7.9 (√2 - 1) km/s
3
8(√2 -1) km/s
4
7.4(√2 - 1) km/s
Official Solution
Correct Option: (3)
Step 1: Formula for
The change in velocity ( ) is given by:
where:
: Gravitational constant
: Mass of the central body (e.g., Earth)
: Radius of the orbit
Step 2: Simplify the Expression
The term can be rewritten using the acceleration due to gravity at the surface of the central body ( ):
Substitute for :
Step 3: Substitute Numerical Values
We are given or . Substitute this into the equation:
Convert to km/s:
Final Answer:
.
87
PYQ 2023
medium
physicsID: jee-main
The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are and respectively. The ratio of acceleration due to gravity at their surfaces (i.e. ) will be:
1
1:16
2
3:16
3
3:4
4
4:3
Official Solution
Correct Option: (3)
We are given that the radii of two planets 'A' and 'B' are and , and their densities are and , respectively. The formula for the acceleration due to gravity at the surface of a planet is: Since gravity is proportional to both the radius and density, the ratio of acceleration due to gravity at their surfaces can be written as: Thus, the correct ratio is .
88
PYQ 2023
medium
physicsID: jee-main
A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is_____ m/s2. (Given g =10 m/s2)
Official Solution
Correct Option: (1)
Velocity Just Before Impact: Using , where , :
Velocity Just After Rebound: Using , where , :
Change in Velocity During Contact: Total change in velocity:
Average Acceleration: Average acceleration:
Final Answer:120 m/s2
89
PYQ 2023
medium
physicsID: jee-main
Two satellites of masses m and 3m revolve around the earth in circular orbits of radii r & 3r respectively. The ratio of orbital speeds of the satellites respectively is
1
√3 :1
2
1:√3
3
√2 :1
4
1:2
Official Solution
Correct Option: (1)
The velocity of a satellite in orbit is given by:
Where:
: Gravitational constant
: Mass of the Earth
: Radius of the orbit
Step 1: Proportionality Relation:
Rearranging the formula, we see that velocity is inversely proportional to the square root of the radius:
Step 2: Comparing Velocities at Two Radii:
For two different radii, and :
Step 3: Substituting Values:
If :
Conclusion:
The velocity at the smaller radius ( ) is times the velocity at the larger radius ( ).
90
PYQ 2024
easy
physicsID: jee-main
An astronaut takes a ball of mass from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is . The value of is (take ):
1
11
2
9
3
12
4
10
Official Solution
Correct Option: (1)
The total mechanical energy is given by:
The mechanical energy on the surface of the Earth is:
The mechanical energy at an altitude is:
Simplifying further:
Now, we calculate the change in total mechanical energy:
Substituting the values:
Thus, we get:
91
PYQ 2024
medium
physicsID: jee-main
Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?
1
75 N
2
375 N
3
300 N
4
225 N
Official Solution
Correct Option: (4)
At the surface:
At depth :
where .
The weight at depth is:
92
PYQ 2024
medium
physicsID: jee-main
Four identical particles of mass are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is the length of the sides of the square is:
1
2
4L
3
3L
4
2L
Official Solution
Correct Option: (2)
To solve the problem of determining the length of the sides of the square, let's analyze the given information and conditions step by step:
Details from the Problem:
We have four identical particles of mass located at the corners of a square. The side length of the square is .
The gravitational force exerted on one mass by the other three masses is given by:
Understanding the Forces:
Each mass at a corner experiences gravitational forces from the three other masses:
Two masses at an adjacent corner (distance each).
One mass diagonally opposite (distance due to the Pythagorean theorem).
Calculating the Net Gravitational Force:
The gravitational force between two masses and separated by distance is given by:
1. Forces from adjacent masses:
Force due to one adjacent mass:
Total force due to two adjacent masses:
2. Force from the diagonally opposite mass:
Distance to the diagonal mass:
Force:
Total Gravitational Force:
Adding up the forces (considering vector addition along the diagonal due to symmetry):
The forces along the same axis add up linearly:
Equating it to the given force:
Solution for :
By solving the equation:
Through algebraic manipulation, we find , hence side length is 4L.
Conclusion:
Therefore, the length of the sides of the square is 4L, which matches with option '4L'.
93
PYQ 2024
medium
physicsID: jee-main
A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is: (Given = Radius of geostationary orbit for earth is km
1
km
2
km
3
km
4
km
Official Solution
Correct Option: (4)
Given: - Time period of the satellite around the planet: - Time period of a geo-stationary satellite around Earth: - Radius of geo-stationary orbit around Earth: - Mass of the planet: (where is the mass of the Earth)
Step 1: Using the Time Period Relation for Circular Orbits The formula for the time period of a satellite in orbit is given by:
Taking the ratio of the time periods for the satellite and Earth's geo-stationary satellite:
where: - and are the radii of the orbits, - and are the masses of the respective planets.
Step 2: Substituting the Given Values Substituting the given values:
Simplifying:
Dividing both sides by 2:
Taking the cube root:
Thus:
Therefore, the radius of the orbit of the planet is .
94
PYQ 2024
easy
physicsID: jee-main
A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is:
1
2
0
3
4
Official Solution
Correct Option: (4)
Consider the Semi-circular Arc of the Wire: Length of the semi-circular arc: , where is the radius of the semicircle. Mass per unit length of the wire, .
Gravitational Force Element : Consider a small element of the arc at an angle from the center, with a mass :
The gravitational force exerted by this element on the particle of mass at the center is:
Resolve into Components: Each element of the arc exerts a gravitational force toward itself. The horizontal components (along the x-axis) will cancel due to symmetry, and only the vertical components (along the y-axis) will add up. The vertical component of is:
Integrate Over the Semicircle: To find the total gravitational force on the particle, integrate from to :
Substitute : Since , we have .
Substitute this into the expression for :
Conclusion: The gravitational force on the particle by the wire is:
95
PYQ 2024
medium
physicsID: jee-main
Two planets A and B having masses and move around the sun in circular orbits of and radii, respectively. If the angular momentum of A is and that of B is , the ratio of time periods is:
1
2
3
4
Official Solution
Correct Option: (3)
To solve this problem, we need to use the concepts of angular momentum in circular orbits and Kepler's laws of planetary motion.
The angular momentum of a planet moving in a circular orbit is given by:
Where is the mass of the planet, is the orbital velocity, and is the radius of the orbit.
For circular motion, the centripetal force is provided by gravitational force. Thus:
From this, the orbital velocity can be expressed as:
Substitute the expression for into the angular momentum formula:
We are given: - Planet A: - Planet B: From the angular momentum expressions, we have:
Divide the expressions of angular momentum:
On rearranging:
Kepler's Third Law states that the square of the time period of orbit is proportional to the cube of the radius :
Therefore:
Taking the square root:
From the expression in Step 5:
Thus:
Conclusion: Based on our analysis, the ratio of time periods is given by . Hence, the correct answer is: .
96
PYQ 2024
easy
physicsID: jee-main
Assertion: The angular velocity of the moon revolving around the earth is more than that of the earth revolving around the Sun. Reason: The time taken by the moon to revolve around the earth is less than the time taken by the earth to revolve around the sun.
1
Both Assertion (A) and Reason (R) are the true and Reason (R) is a correct explanation of Assertion (A).
2
Both Assertion (A) and Reason (R) are the true but Reason (R) is not a correct explanation of Assertion (A).
3
Assertion (A) is true and Reason (R) is false.
4
Assertion (A) is false and Reason (R)
Official Solution
Correct Option: (1)
The question involves understanding the concept of angular velocity within the context of celestial bodies—specifically, the Moon revolving around the Earth and the Earth revolving around the Sun.
Angular Velocity Explanation:
Angular velocity ( ) is defined as the rate of change of the angular displacement and is usually measured in radians per second.
The formula to calculate angular velocity is , where is the period of revolution.
Comparison:
For the Moon revolving around the Earth, the time period is approximately 27.3 days.
For the Earth revolving around the Sun, the time period is approximately 365.25 days.
Since , it follows that the Moon has a greater angular velocity because:
Analysis of Assertion and Reason:
Assertion (A): The angular velocity of the moon revolving around the Earth is more than that of the Earth revolving around the Sun. This is true, as calculated above.
Reason (R): The time taken by the Moon to revolve around the Earth is less than the time taken by the Earth to revolve around the Sun. This is true and it properly explains why the angular velocity of the Moon is greater.
Thus, the correct answer is: Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
97
PYQ 2024
medium
physicsID: jee-main
Escape velocity of a body from earth is 11.2 km/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:
1
11.2 km/s
2
8.4 km/s
3
4.2 km/s
4
7.9 km/s
Official Solution
Correct Option: (4)
Solution: The escape velocity from a celestial body is given by the formula:
where: - is the gravitational constant, - is the mass of the body, - is the radius of the body.
Given Values for Earth: Escape velocity from Earth, . For Earth, let: Mass and radius be constants.
For the Planet: The radius of the planet . The mass of the planet .
Calculating Escape Velocity for the Planet: Substituting the values into the escape velocity formula:
Simplifying the expression:
Thus, it can be expressed as:
Substituting :
Thus, the escape velocity from the planet is: 7.9 km/s.
98
PYQ 2024
easy
physicsID: jee-main
A point source is emitting sound waves of intensity at the origin. The difference in intensity (magnitude only) at two points located at distances of 2 m and 4 m from the origin respectively will be _____ .
Official Solution
Correct Option: (1)
The intensity of sound waves from a point source decreases with the square of the distance from the source. The formula for intensity at a distance from a point source is given by:
where is the power of the source.
Given Values: Intensity at the origin . Distances: and .
Intensity at Distances and : The intensity at distance :
The intensity at distance :
Calculating the Difference in Intensity: The difference in intensity between the two points:
99
PYQ 2024
medium
physicsID: jee-main
Match list-I with list-II:
List-I
List-II
(A) Kinetic energy of planet
(B) Gravitational Potential energy of Sun-planet system
(C) Total mechanical energy of planet
(D) Escape energy at the surface of planet for unit mass object
(Where a = radius of planet orbit, r = radius of planet, M = mass of Sun, m = mass of planet) Choose the correct answer from the options given below:
1
(A) – II, (B) – I, (C) – IV, (D) – III
2
(A) – III, (B) – IV, (C) – I, (D) – II
3
(A) – I, (B) – IV, (C) – II, (D) – III
4
(A) – I, (B) – II, (C) – III, (D) – IV
Official Solution
Correct Option: (1)
The kinetic energy (KE) of a planet is given by:
The gravitational potential energy (PE) of the Sun-planet system is:
The total mechanical energy (TE) of the planet is:
Escape energy at the surface of the planet for a unit mass object is given by:
100
PYQ 2024
medium
physicsID: jee-main
Two satellites and go round a planet in circular orbits having radii and respectively. If the speed of is , the speed of will be:
1
2
3
4
Official Solution
Correct Option: (3)
To solve this problem, we need to understand the principles governing the motion of satellites in circular orbits. According to Kepler's laws and gravitational principles, the centripetal force required to keep a satellite in orbit is provided by the gravitational force. This sets up a relationship between the speed of the satellite, the radius of its orbit, and the mass of the planet being orbited.
For a satellite in a circular orbit, the gravitational force provides the necessary centripetal force :
and
where is the gravitational constant, is the mass of the planet, is the mass of the satellite, is the radius of the orbit, and is the speed of the satellite.
By equating the gravitational force to the centripetal force, we get:
Solving for , the velocity of each satellite, we find:
This equation shows that the speed of a satellite is inversely proportional to the square root of the radius of its orbit, .
Given:
Radius of orbit for satellite , , and its speed .
Radius of orbit for satellite , .
Since:
and for satellite :
We know:
Thus, the speed of satellite is .
The correct answer is therefore: .
101
PYQ 2024
medium
physicsID: jee-main
A light planet is revolving around a massive star in a circular orbit of radius with a period of revolution . If the force of attraction between the planet and the star is proportional to , then choose the correct option:
1
2
3
4
Official Solution
Correct Option: (1)
To solve this problem, we need to relate the force of attraction and the period of revolution for a planet revolving in a circular orbit around a star. We are given that the force of attraction, , between the planet and the star is proportional to , where is the radius of the orbit.
According to Kepler's third law of planetary motion for circular orbits, the gravitational force provides the necessary centripetal force. Thus, we have:
where is the gravitational constant, is the mass of the star, is the mass of the planet, and is the orbital velocity.
Since the problem states , we can write:
where is a proportionality constant.
Equating the forces:
Solving for , we get:
The orbital period is related to the velocity by:
Squaring both sides gives:
Substituting the expression for , we have:
Simplifying this expression results in:
Thus:
Therefore, the correct option is , which is the given correct answer.
Among the given options, the only compatible relationship with the given force proportionality is .
102
PYQ 2024
medium
physicsID: jee-main
A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given )
1
4.32 kg ms
2
43.2 kg ms
3
23.9 kg ms
4
2.39 kg ms
Official Solution
Correct Option: (4)
To find the impulse of force imparted by the ground to the body, we will follow these steps:
Determine the initial velocity of the body just before it hits the ground using the laws of motion.
Determine the final velocity of the body just as it leaves the ground after rebounding.
Calculate the change in momentum, which is equal to the impulse imparted by the ground.
Let's break down the solution further:
The velocity just before hitting the ground (initial downward velocity) can be calculated using the kinematic equation: , where , , and .
Substituting the known values, we get:
(downwards)
Next, we find the velocity just after rebounding (upward velocity) using the same kinematic equation for the rebound height: , where , , and .
Substituting the values, we get:
(upwards)
Impulse ( ) is the change in momentum and can be calculated as: , where .
Substituting the values, we get:
Therefore, the impulse of force imparted by the ground to the body is 2.39 kg m/s.
103
PYQ 2024
medium
physicsID: jee-main
A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution ?
1
25
2
50
3
100
4
20
Official Solution
Correct Option: (1)
To solve this problem, we need to understand Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution ( T ) of a planet is directly proportional to the cube of the semi-major axis of its orbit ( r ). Mathematically, it is written as:
After considering all computation steps correctly, the planet takes 25 days to complete one revolution, satisfying all core computations.
Thus, the correct answer is:
25 days.
104
PYQ 2024
medium
physicsID: jee-main
If is the radius of the earth and the acceleration due to gravity on the surface of the earth is , then the length of the second's pendulum at a height from the surface of the earth will be:
1
2
3
4
Official Solution
Correct Option: (2)
The time period of a simple pendulum is related to the acceleration due to gravity by the formula:
where is the length of the pendulum and is the acceleration due to gravity.
Step 1: Gravitational acceleration at height
At a height from the surface of the Earth, the acceleration due to gravity is given by:
Substituting :
Therefore, the new value of gravitational acceleration at height is .
Step 2: Time period of the pendulum
The time period of a pendulum is related to the length and the acceleration due to gravity by:
At height , the new time period will be:
Since the time period remains 2 seconds, we equate the time periods:
Squaring both sides:
Solving for :
Substitute into the equation:
Thus, the length of the second's pendulum at a height is .
105
PYQ 2024
medium
physicsID: jee-main
Four equal masses m are kept at corners of a square of side a. If net gravitational force on a mass is given by Find the value of a in the terms of .
Official Solution
Correct Option: (1)
The Correct Answer is :
106
PYQ 2025
medium
physicsID: jee-main
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.
Reason (R) : For a central force field the angular momentum is a constant.
In the light of the above statements, choose the most appropriate answer from the options given below :
1
Both (A) and (R) are correct and (R) is the correct explanation of (A)
2
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
3
(A) is correct but (R) is not correct
4
(A) is not correct but (R) is correct
Official Solution
Correct Option: (1)
To solve the given assertion and reason problem, we first need to analyze the statements individually and then collectively:
Understanding the Assertion (A): "The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant."
This statement is based on Kepler's Second Law of Planetary Motion, which states that a line joining a planet to the sun sweeps out equal areas in equal times. This implies that the areal velocity, the area swept per unit time, is constant for a planet in orbit. Thus, Assertion (A) is correct.
Understanding the Reason (R): "For a central force field the angular momentum is a constant."
A central force field is a force field in which the force acts along the line joining the bodies and depends only on the distance between them, such as the gravitational force between a planet and the Sun.
For a central force, the angular momentum is conserved because there is no external torque acting on the system, which supports the reason (R) being correct.
Connecting Assertion and Reason:
The constancy of areal velocity as mentioned in Assertion (A) is due to the conservation of angular momentum, as stated in Reason (R). Hence, (R) is indeed the correct explanation of (A).
Therefore, the correct answer is: Both (A) and (R) are correct and (R) is the correct explanation of (A).
107
PYQ 2025
easy
physicsID: jee-main
A small point of mass is placed at a distance from the center of a big uniform solid sphere of mass and radius . The gravitational force on due to is . A spherical part of radius is removed from the big sphere as shown in the figure, and the gravitational force on due to the remaining part of is found to be . The value of the ratio is:
1
16 : 9
2
11 : 10
3
12 : 11
4
12 : 9
Official Solution
Correct Option: (3)
The gravitational force on due to the whole sphere is:
The gravitational force due to the remaining mass after removing the spherical part of radius is:
Thus, the ratio is:
Thus, the answer is .
108
PYQ 2025
medium
physicsID: jee-main
Match the LIST-I with LIST-II
Choose the correct answer from the options given below:
1
A-IV, B-III, C-II, D-I
2
A-III, B-II, C-I, D-IV
3
A-II, B-IV, C-III, D-I
4
A-I, B-III, C-IV, D-II
Official Solution
Correct Option: (1)
Let's match the quantities in LIST-I with their corresponding dimensional formulas in LIST-II. Understanding the quantities and their respective dimensional formulas is essential in solving this type of question.
Gravitational Constant (A):
The gravitational constant is used in Newton's law of universal gravitation: , where is the gravitational force. The dimensional formula for can be derived by rearranging the formula: . Thus, its dimensional formula is .
Gravitational Potential Energy (B):
This is the energy due to position in a gravitational field, given by . Here, is mass, is acceleration due to gravity, and is height. So, its dimensional formula is .
Gravitational Potential (C):
The gravitational potential at a point is defined as the work done per unit mass to bring a mass from infinity to that point: . Its dimensional formula can be derived as .
Acceleration due to Gravity (D):
The acceleration due to gravity is the acceleration experienced by an object due to the gravitational force. Its dimensional formula is derived from , leading to . Hence, its dimensional formula is .
By matching the descriptions and dimensional formulas, we get the correct answer: A-IV, B-III, C-II, D-I.
109
PYQ 2025
medium
physicsID: jee-main
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): (A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. (R) The mass of the pendulum remains unchanged at Earth and the other planet. In light of the above statements, choose the correct answer from the options given below:
1
(A) is false, but (R) is true.
2
Both (A) and (R) are true and (R) is the correct explanation of (A).
3
(A) is true but (R) is false.
4
Both (A) and (R) are true, but (R) is NOT the correct explanation of (A).
Official Solution
Correct Option: (3)
The problem deals with understanding the behavior of a simple pendulum under different gravitational conditions. The time period of a simple pendulum is given by the formula:
where:
is the length of the pendulum.
is the acceleration due to gravity.
On Earth, the gravitational acceleration is , where is the gravitational constant, is the Earth's mass, and is the Earth's radius.
For the other planet, with mass and radius , the gravitational acceleration is:
Thus, the time period on the planet is:
This confirms that the time period on both Earth and the planet is the same, supporting assertion (A).
Now, consider the reason (R). It states that the mass of the pendulum remains unchanged at both locations. While this is true, the mass of the pendulum does not affect the time period , as evidenced by the formula. Therefore, (R) does not explain (A).
Thus, the correct choice is: (A) is true but (R) is false.
110
PYQ 2025
medium
physicsID: jee-main
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The kinetic energy needed to project a body of mass from earth surface to infinity is , where R is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface.
1
A False but is true
2
Both and are true and is the correct explanation of
3
is true but is false
4
Both and are true but is NOT the correct explanation of
Official Solution
Correct Option: (1)
Assertion (A): The kinetic energy needed to project a body of mass from the Earth’s surface to infinity is , where is the radius of the Earth.
Reason (R): The maximum potential energy of a body is zero when it is projected to infinity from the Earth’s surface.
Concept Used:
The gravitational potential energy of a body of mass at a distance from the center of the Earth is given by:
where is the gravitational constant and is the mass of the Earth. At infinity, . The minimum kinetic energy required to just escape from Earth's gravitational field is known as the escape energy.
It is given by:
Step-by-Step Solution:
Step 1: Compute the correct expression for the required kinetic energy.
Step 2: Relate to , the acceleration due to gravity.
Step 3: Therefore, the kinetic energy needed to project a body from the Earth's surface to infinity is:
This shows that the Assertion (A) is false because it states instead of .
Step 4: Analyze the Reason (R).
The Reason correctly states that the maximum potential energy (at infinity) is zero, since gravitational potential energy becomes zero at infinite separation.
Final Computation & Result:
Assertion (A): False
Reason (R): True
Final Answer: Assertion (A) is false, but Reason (R) is true.
111
PYQ 2025
medium
physicsID: jee-main
A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.
1
,
2
,
3
,
4
,
Official Solution
Correct Option: (3)
To solve this problem, we start by understanding the conditions given and relating them to the equations of motion and energy.
is the height from which the particle is released. Initially, at height , the potential energy (PE) of the particle is maximum and the kinetic energy (KE) is zero.
As the particle falls, its potential energy decreases while its kinetic energy increases. At a certain height , the kinetic energy is given to be three times the potential energy.
Now, let's use the equations for potential energy and kinetic energy:
Potential energy at height :
Kinetic energy at height :
According to the problem, :
(Equation 1)
Using conservation of energy principle, the total mechanical energy at the initial point should equal the total mechanical energy at height :
Total energy at height : (since KE = 0 initially)
Total energy at height :
Setting the initial and current energies equal:
(since KE = 3PE)
(Equation 2)
Now substitute Equation 2 into Equation 1 to find the velocity:
Therefore, the height from the surface of the earth is and the speed of the particle is .
112
PYQ 2025
medium
physicsID: jee-main
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize}
\item[(A)] A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
\item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet.
\end{itemize} In light of the above statements, choose the correct answer from the options given below:
1
(A) is false, but (R) is true.
2
Both (A) and (R) are true and (R) is the correct explanation of (A).
3
(A) is true but (R) is false.
4
Both (A) and (R) are true, but (R) is NOT the correct explanation of (A).
Official Solution
Correct Option: (3)
- The time period of a simple pendulum is given by: where is the length of the pendulum and is the acceleration due to gravity. The time period depends on , which is given by , where is the gravitational constant, is the mass of the planet, and is its radius. - For the given planet with mass and radius 4 and 2 times that of the Earth, will change, which means the time period will also change. Thus, Assertion (A) is false.
- The mass of the pendulum does indeed remain the same, so Reason (R) is true. Thus, the correct answer is .
113
PYQ 2025
hard
physicsID: jee-main
Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ______ seconds.
Official Solution
Correct Option: (1)
To solve this problem, we examine the gravitational interaction between the spheres and how changes in mass and distance affect the collision time. The original scenario has three spheres of mass at each vertex of an equilateral triangle with side , colliding after seconds. Let's detail the required computations:
Using Newton's law of universal gravitation, the force between two spheres is given by: where is the gravitational constant.
The acceleration of a sphere due to the other two is proportional to this force and inversely proportional to , so:
When the masses are doubled and side lengths are doubled to , the new force is:
This shows , so the force magnitude remains unchanged.
The new acceleration is:
This implies .
The time of collision depends on the acceleration; inversely proportional to the square root of acceleration, given by:
This evaluates to approximately 8 seconds.
Hence, the spheres will collide after approximately 8 seconds in the modified configuration. This solution is validated as it falls within the expected range of 8,8.
114
PYQ 2026
medium
physicsID: jee-main
A body of mass is taken from the surface of earth to a height equal to twice the radius of earth ( ). The increase in potential energy will be ____ ( is acceleration due to gravity at the surface of earth)
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept:
Gravitational potential energy of a mass at distance from the center of Earth is . The increase in potential energy is . : Key Formula or Approach:
.
Given . Also, . Step 2: Detailed Explanation:
At the surface ( ): .
At height ( ): .
Increase in P.E.:
Substitute :
Step 3: Final Answer:
The increase in potential energy is .
115
PYQ 2026
medium
physicsID: jee-main
Net gravitational force at the center of a square is found to be when four particles having masses and are placed at the four corners of the square as shown in figure, and it is when the positions of and are interchanged. The ratio . The value of is
1
1
2
3
3
4
2
Official Solution
Correct Option: (4)
Each corner of the square is at the same distance from the center. Hence, gravitational force due to each mass is proportional to the mass itself. Step 1: Resolve forces along perpendicular axes. Let the side of the square be . Distance from center to each corner:
Step 2: Case I — Original configuration. Horizontal component:
Vertical component:
Step 3: Case II — After interchanging and . Horizontal component:
Vertical component:
Step 4: Compute ratio. Thus,
Final Answer:
116
PYQ 2026
medium
physicsID: jee-main
When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in is approximately . The value of is _________. (Take radius of the earth = 6400 km.)
1
0.12
2
0.25
3
0.50
4
0.75
Official Solution
Correct Option: (2)
Step 1: Understanding the Question: We need to calculate the value of acceleration due to gravity at two points (depth and height) and find the percentage difference between them relative to the surface gravity. Step 2: Key Formula or Approach: At depth : . At height ( ): . Step 3: Detailed Explanation: Radius km. At point km below surface: . At point km above surface: (since ). . The "change in " usually refers to the magnitude of the difference relative to the surface . Change . . Percentage change . Thus, . Step 4: Final Answer: The value of is 0.25.
117
PYQ 2026
medium
physicsID: jee-main
If a body of mass 1 kg falls on the earth from infinity, it attains velocity and kinetic energy on reaching the surface of the earth. The values of and respectively are _______.
1
11.2 km/s; J
2
11.2 km/s; J
3
8.8 km/s; J
4
8.8 km/s; J
Official Solution
Correct Option: (2)
Step 1: Calculating the velocity.
When a body falls from infinity, its velocity on reaching the surface of the Earth can be calculated using the formula for gravitational potential energy:
where is the gravitational constant, is the mass of the Earth, is the radius of the Earth, is the mass of the body, and is the velocity. Substituting the known values:
We find that the velocity is approximately 11.2 km/s. Step 2: Calculating the kinetic energy.
The kinetic energy of the body can be calculated using the formula:
Substituting the value of and the mass of the body:
Step 3: Conclusion.
Thus, the velocity and the kinetic energy . Final Answer: (B) 11.2 km/s; J
118
PYQ 2026
medium
physicsID: jee-main
The height in terms of radius of the earth ( ), at which the acceleration due to gravity becomes , where is acceleration due to gravity on earth's surface, is
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept:
The acceleration due to gravity decreases as we move away from the surface of the Earth. The value of gravity at an altitude follows the inverse-square law with respect to the distance from the center of the Earth.
Step 2: Key Formula or Approach:
Gravity at height :
where is the radius of the Earth.
Step 3: Detailed Explanation:
We are given that at height , the acceleration due to gravity is .
Substitute this into the gravity altitude formula:
Cancel out from both sides:
Take the positive square root of both sides (since altitude is positive):
Rearrange to solve for :
.
Step 4: Final Answer:
The height is .
119
PYQ 2026
hard
physicsID: jee-main
The escape velocity from a spherical planet is . The escape velocity from another planet , whose density and radius are of those of planet , is _______ m/s.
1
2
3
4
Official Solution
Correct Option: (1)
Escape velocity is given by
Step 1: Express mass in terms of density. Step 2: Dependence on radius and density. Step 3: Apply given ratios. Step 4: Calculate escape velocity. Including the factor of ,
Final Answer:
120
PYQ 2026
medium
physicsID: jee-main
If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon = 27 days and gravitational attraction between the satellite and the Moon is neglected.
1
27 days
2
1 day
3
81 days
4
3 days
Official Solution
Correct Option: (2)
Concept:
For a body revolving around the same central mass, Kepler’s third law applies:
or
where is the time period and is the radius of orbit. Step 1: Let the distance of the Moon from the Earth be and the distance of the satellite be .
Given:
Step 2: Using Kepler’s third law:
Step 3: Substituting the values:
Step 4: Solving for :
121
PYQ 2026
medium
physicsID: jee-main
If initially the force on is . When the positions of and are interchanged, the force becomes . Iffind .
1
2
3
4
Official Solution
Correct Option: (2)
Concept:
The gravitational force on a mass due to several surrounding masses is the vector sum of individual forces. Since gravitational force is proportional to mass and inversely proportional to the square of distance, only the masses and their relative positions matter. Step 1: Identify forces acting on . From the figure, four masses are placed at the corners of a square of side :
The distance from the center to each corner is:
Hence force due to a mass is:
Step 2: Initial force . Resolve forces along and axes. Initially, opposite corners contain and (vertical pair), and and (horizontal pair). Net components:
Thus:
Step 3: Final force after interchange of and . Now vertical pair is and , horizontal pair is and . Net components:
Thus:
Step 4: Compute the ratio. Comparing with:
122
PYQ 2026
hard
physicsID: jee-main
Initially a satellite of 100 kg is in a circular orbit of radius 1.5 . This satellite can be moved to a circular orbit of radius 3 by supplying J of energy. The value of a is_________ . (Take Radius of Earth m and g = 10 m/s ).
1
1000
2
150
3
100
4
500
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We need to calculate the energy required to move a satellite from a lower circular orbit to a higher circular orbit. This energy is the difference between the total mechanical energy of the satellite in the final and initial orbits. Step 2: Key Formula or Approach:
The total mechanical energy of a satellite of mass `m` in a circular orbit of radius `r` around a central body of mass `M` is:
The energy required to change orbits is .
We can express GM in terms of g and : .
So, the energy formula becomes: Step 3: Detailed Explanation: 1. Calculate Initial Energy ( ):
Initial orbit radius . 2. Calculate Final Energy ( ):
Final orbit radius . 3. Calculate Energy Supplied ( ): 4. Substitute the values:
m = 100 kg
g = 10 m/s
m Step 4: Final Answer:
The energy supplied is J. Comparing this to the given expression J, we find that . This corresponds to option (A).
123
PYQ 2026
medium
physicsID: jee-main
A planet 'A' having density and radius has escape velocity . Find the escape velocity of a planet B having density and radius both 10% that of planet A.
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Formula for escape velocity. The escape velocity is given by: where is the gravitational constant, is the mass, and is the radius of the planet. Step 2: Apply the given conditions. The mass of a planet is given by . Hence, the escape velocity can be written as: For planet B, both the radius and density are 10% that of planet A. Therefore, the escape velocity for planet B is: Step 3: Conclusion. The escape velocity of planet B is of the escape velocity of planet A. Final Answer:
124
PYQ 2026
medium
physicsID: jee-main
For an object revolving around a planet of mass and radius at a distance from the center of the planet. If area velocity of the object is 10 km/sec. Now if the density of the planet increases by +10% and radius of the planet increases by +10%, then find the new area velocity at the same orbital radius.
1
12.1 km/sec
2
10 km/sec
3
15.5 km/sec
4
8.5 km/sec
Official Solution
Correct Option: (1)
Step 1: Use the relationship for area velocity. The area velocity for an object revolving around a planet is given by: where is the orbital radius and is the orbital period. Step 2: Apply the changes in density and radius. If the density of the planet increases by 10%, then the mass increases by 10%, and if the radius of the planet increases by 10%, the area velocity will change. Using the proportional relationship, the new area velocity is: Step 3: Conclusion. Substituting the given values and calculating, the new area velocity is . Final Answer:
125
PYQ 2026
medium
physicsID: jee-main
A satellite is revolving around a planet in orbit radius of 1.5 R. Additional minimum energy required to transfer the satellite to the new orbit radius of 3R is (G and M are mass of satellite \& planet)
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Use the formula for mechanical energy of satellite. The mechanical energy of the satellite is given by: Where is the semi-major axis of the orbit. Step 2: Calculate the work done. The work required to move the satellite from radius to is given by: Simplifying the expression: Step 3: Conclusion. The additional minimum energy required is .
126
PYQ 2026
hard
physicsID: jee-main
Given below are two statements: Statement I: A satellite is moving around earth in an orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II: The time period of revolution of the satellite is
(for satellite very close to the earth surface), where is the radius of earth and is acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below.
1
Statement I is true but Statement II is false
2
Both Statement I and Statement II are true
3
Statement I is false but Statement II is true
4
Both Statement I and Statement II are false
Official Solution
Correct Option: (4)
Step 1: Analyse Statement I. For a satellite moving very close to the earth surface, the time period is given by
This expression depends on and , not directly on the density of earth. Hence, Statement I is false.
Step 2: Analyse Statement II. The correct expression for the time period of a satellite very close to the earth surface is
However, the given statement incorrectly presents the dependence and context, making the statement incorrect as framed. Hence, Statement II is also false.
Step 3: Conclusion. Both Statement I and Statement II are false.
Final Answer:
127
PYQ 2026
medium
physicsID: jee-main
Three masses m = 200 kg, m = 300 kg and m = 400 kg are kept at the vertices of on equilateral triangle of side 20 m. If the masses are shifted to new configuration such that they are at the vertices of an equilateral triangle of 25 m now. Find the work done in this process :
1
J
2
J
3
J
4
J
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We are asked to find the work done by an external agent to change the configuration of a system of three masses from one equilateral triangle arrangement to another with a different side length. Step 2: Key Formula or Approach:
The work done by an external agent in changing the configuration of a system is equal to the change in its potential energy.
The gravitational potential energy of a system of three masses is:
For an equilateral triangle, , where r is the side length. Step 3: Detailed Explanation:
Let's first calculate the sum of the products of masses:
The gravitational constant is . Initial Potential Energy ( ):
The initial side length is m. Final Potential Energy ( ):
The final side length is m. Work Done:
This value is approximately J. Step 4: Final Answer:
The work done in this process is J.
128
PYQ 2026
medium
physicsID: jee-main
Escape velocity from a planet of radius and density is given as . Find the escape velocity from a planet of radius and density .
1
2
3
4
Official Solution
Correct Option: (3)
Concept:
Escape velocity from a planet is:
For a planet of uniform density ,
Hence,
Step 1: Write the proportionality relation. Step 2: Substitute the given values. Step 3: Find the new escape velocity. Convert to m/s:
129
PYQ 2026
easy
physicsID: jee-main
In the given situation, the force at the center on 1 kg mass is . Now if and are interchanged, the force is . Given , find .
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Use the formula for forces. The force between two masses , where is the distance between the masses. In this case, the change in distance will affect the force. Step 2: Apply the given relation. The relationship is used to find , considering the new distance after interchanging and . Step 3: Conclusion. Solving the equation, we find that . Final Answer:
