There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of the wire is
1
0.5
2
-0.5
3
0.25
4
-0.25
Official Solution
Correct Option:
(1)
Let the material of length 1 and side s If a material maintains constant volume during stretching Differentiate wrt dl dl.s 2l.ds
02
PYQ 2008
easy
physicsID: mht-cet-
If is the acceleration due to gravity on earth?s surface, the gain of the potential energy of an object of mass raised from the surface of the earth to a height equal to the radius of the earth is
1
2
3
4
Official Solution
Correct Option:
(3)
The potential energy of an object at the surface of the earth .....(i) The potential energy of the object at a height from the surface of the earth
.....(ii) Hence, the gain in potential energy of the object
But we know that Hence, or or
03
PYQ 2009
medium
physicsID: mht-cet-
In a satellite, if the time of revolution is , then is proportional to
1
2
3
4
Official Solution
Correct Option:
(4)
Velocity of satellite
and
04
PYQ 2019
medium
physicsID: mht-cet-
If , and represent the work done in moving a particle from to along three different paths , and (as shown in fig) in the gravitational field of the point mass . Find the correct relation between , and
1
2
3
4
Official Solution
Correct Option:
(3)
Gravitational force is a conservative froce. As we know that the work done by a conservative force is independent of the path followed and depends only on the end points of the motion. Since in the given pattern and have the same end point and as shown in the figure below,
05
PYQ 2020
medium
physicsID: mht-cet-
A satellite is revolving in a circular orbit around the earth with total energy . Its potential energy in that orbit is
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Total energy in a circular orbit. The total energy of a satellite in a circular orbit around the Earth is the sum of its kinetic energy and potential energy . For a satellite in orbit, the total energy is given by:
The kinetic energy is , and the potential energy is , where is the mass of the satellite, is its velocity, is the gravitational constant, and is the radius of the orbit. Step 2: Relationship between kinetic and potential energy. The kinetic energy of a satellite in orbit is related to its potential energy by:
Thus, the total energy is:
Therefore, the potential energy is:
Step 3: Conclusion. Thus, the potential energy of the satellite in its orbit is , which corresponds to option (B).
06
PYQ 2020
medium
physicsID: mht-cet-
A satellite of mass is revolving around earth of mass in an orbit of radius with constant angular velocity . The angular momentum of the satellite is (G = gravitational constant)
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Expression for orbital velocity. For a satellite revolving in a circular orbit around the earth, the gravitational force provides the centripetal force. Hence, Solving for velocity, Step 2: Formula for angular momentum. Angular momentum of a particle moving in a circular orbit is given by Step 3: Substituting the value of velocity.
Step 4: Conclusion. Thus, the angular momentum of the satellite is .
07
PYQ 2020
medium
physicsID: mht-cet-
Two satellites of masses and are revolving in a circular orbit of radius around the earth. The ratio of their frequencies of revolution will be
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Expression for orbital frequency. For a satellite in circular orbit:
Step 2: Dependence on mass of satellite. The orbital frequency depends only on the mass of the earth and radius of orbit, not on the mass of the satellite.
Step 3: Conclusion. Both satellites have the same frequency of revolution.
08
PYQ 2020
medium
physicsID: mht-cet-
A body situated on earth’s surface at its equator becomes weightless when the rotational kinetic energy of the earth reaches a critical value which is given by
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Understanding the concept. The body becomes weightless when the gravitational force is completely balanced by the centrifugal force due to the rotation of the Earth. The centrifugal force is given by: where is the angular velocity of Earth. The condition for weightlessness is when the centrifugal force equals the gravitational force: Step 2: Calculating the angular velocity. The rotational kinetic energy of the Earth is: where is the moment of inertia of the Earth. Setting this equal to the required condition, we solve for , and find that the critical value of the rotational kinetic energy is . Step 3: Conclusion. Thus, the critical rotational kinetic energy is , corresponding to option (B).
09
PYQ 2020
medium
physicsID: mht-cet-
The escape velocity from the surface of earth of mass and radius is . The escape velocity from the surface of a planet whose mass and radius are 3 times that of the earth will be
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Write formula for escape velocity.
Step 2: Substitute new mass and radius. For the planet: Step 3: Calculate new escape velocity.
Step 4: Conclusion. The escape velocity remains unchanged.
10
PYQ 2020
medium
physicsID: mht-cet-
The length and diameter of a metal wire used in a sonometer is doubled. The fundamental frequency will change from to
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Formula for fundamental frequency of a stretched string. where is length and is mass per unit length. Step 2: Effect of doubling the length. If is doubled, frequency becomes half:
Step 3: Effect of doubling the diameter. Mass per unit length . If diameter is doubled,
Step 4: Combined effect on frequency.
Step 5: Conclusion. The fundamental frequency becomes .
11
PYQ 2020
medium
physicsID: mht-cet-
Two monoatomic ideal gases and of molecular masses and respectively, are enclosed in separate containers kept at the same temperature. The ratio of speed of sound in gas to that in gas is given by
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Formula for speed of sound in a gas. Speed of sound in an ideal gas is given by: where is the molecular mass. Step 2: Same temperature and same type of gas. Since both gases are monoatomic and at the same temperature, , , and are the same for both gases. Step 3: Ratio of speeds of sound.
Step 4: Substituting given molecular masses.
Step 5: Conclusion. The required ratio is .
12
PYQ 2020
medium
physicsID: mht-cet-
For athermanous substances, coefficient of transmission is
1
less than one but greater than zero
2
zero
3
equal to one
4
greater than one
Official Solution
Correct Option:
(2)
Step 1: Meaning of athermanous substances. Athermanous substances do not transmit thermal radiation. Step 2: Transmission coefficient. Since no radiation passes through, the transmitted energy is zero. Step 3: Conclusion. Hence, the coefficient of transmission is zero.
13
PYQ 2020
medium
physicsID: mht-cet-
A microscope will have maximum resolving power if, to illuminate the specimen, it uses light of
1
red colour.
2
green colour.
3
yellow colour.
4
blue colour.
Official Solution
Correct Option:
(4)
Step 1: Resolving power of a microscope. Resolving power is inversely proportional to the wavelength of light used:
Step 2: Compare wavelengths of colours. Among visible colours, blue light has the shortest wavelength.
Step 3: Effect on resolving power. Shorter wavelength results in smaller diffraction limit and better resolution.
Step 4: Conclusion. Therefore, a microscope has maximum resolving power when blue light is used.
14
PYQ 2020
medium
physicsID: mht-cet-
A smooth sphere of mass moving with velocity directly collides elastically with another sphere of mass at rest. After collision, their final velocities are and respectively. The value of is given by
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Use elastic collision formula. For a one-dimensional elastic collision, the velocity of the second mass after collision is:
Step 2: Rewrite expression.
Step 3: Conclusion. The correct expression for velocity is .
15
PYQ 2020
medium
physicsID: mht-cet-
The length of the seconds pendulum is 1 m on the earth. If the mass and diameter of the planet is half that of the earth then the length of the seconds pendulum on the planet will be:
1
0.5 m
2
1 m
3
1.5 m
4
2 m
Official Solution
Correct Option:
(4)
Step 1: Relation between Length and Acceleration due to Gravity. The length of a seconds pendulum is related to the acceleration due to gravity by:
where is the time period of the pendulum. The time period of the pendulum depends on the gravitational acceleration, and for a planet with half the mass and diameter of the Earth, the value of would be proportional to the inverse of the square of the radius. Thus, the length of the seconds pendulum on the planet will be twice the length on Earth, as is halved. Step 2: Final Answer. Thus, the length of the seconds pendulum on the planet is 2 m.
16
PYQ 2020
medium
physicsID: mht-cet-
The gravitational potential energy of a rocket of mass 200 kg at a distance from the centre of the earth is . The weight of the rocket at a distance from the centre of the earth is:
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Gravitational Potential Energy Formula. The gravitational potential energy is related to the distance from the centre of the earth by:
where is the gravitational constant, is the mass of the earth, is the mass of the rocket, and is the distance from the centre of the earth. The weight of the rocket is given by:
At a distance , the weight can be calculated based on the inverse square law:
Step 2: Final Answer. Thus, the weight of the rocket is .
17
PYQ 2020
medium
physicsID: mht-cet-
The period of revolution of a communication satellite is 24 hours. The period of a satellite in an orbit at a distance three times that of the earth’s radius above its surface will be
1
4 days
2
16 days
3
24 hours
4
8 days
Official Solution
Correct Option:
(4)
Step 1: Understanding the given data. The period of revolution of a communication satellite is given as 24 hours. A communication satellite revolves very close to the Earth, so its orbital radius can be taken approximately equal to the radius of the Earth, i.e., and The new satellite is at a height equal to three times the Earth's radius above the surface. Therefore, its orbital radius becomes: Step 2: Applying Kepler’s third law. According to Kepler’s third law of planetary motion: So, Substituting the values: Step 3: Calculating the time period. Step 4: Final conclusion. The period of revolution of the satellite at the given height is 8 days.
18
PYQ 2020
medium
physicsID: mht-cet-
For the weight of a body of mass to be zero on equator of the earth, angular velocity of the earth must be
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Condition for zero apparent weight at equator. For apparent weight to be zero, centrifugal force must balance gravitational force: Step 2: Cancel mass and substitute values. Step 3: Solve for angular velocity . Step 4: Substitute numerical values. Step 5: Simplify.
19
PYQ 2020
medium
physicsID: mht-cet-
The period of revolution of a satellite is
1
independent of mass of a satellite.
2
independent of radius of planet.
3
dependent on the mass of a satellite.
4
independent of height of the satellite from the planet.
Official Solution
Correct Option:
(1)
Step 1: Expression for time period of satellite. For a satellite in circular orbit: where is mass of planet.
Step 2: Analysis of formula. The mass of satellite does not appear in the expression of .
Step 3: Conclusion. Hence, the time period of revolution of a satellite is independent of its mass.
20
PYQ 2020
easy
physicsID: mht-cet-
Two satellites and are revolving with critical velocities and around the earth, in circular orbits of radii and , respectively. The ratio is
Official Solution
Correct Option:
(1)
21
PYQ 2020
medium
physicsID: mht-cet-
For a gas, , where is universal gas constant and is the molar specific heat at constant volume. The gas is made up of molecules which are
1
polyatomic.
2
rigid diatomic.
3
non-rigid diatomic.
4
monoatomic.
Official Solution
Correct Option:
(2)
Step 1: Understanding the relation. The relation suggests that the gas behaves like a diatomic gas with rigid molecules. For a monoatomic gas, the ratio would be different, and polyatomic or non-rigid diatomic gases would also show different values for this ratio. Step 2: Conclusion. Thus, the correct answer is (B) rigid diatomic.
22
PYQ 2020
medium
physicsID: mht-cet-
A charged particle is moving in a uniform magnetic field in a circular path of radius . When the energy of the particle becomes three times the original, the new radius will be
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Energy and radius relationship. The kinetic energy of the charged particle in a magnetic field is given by:
The radius of the circular path is related to the velocity by:
Since the energy becomes three times the original, the velocity must increase by a factor of . Therefore, the new radius is:
Step 2: Conclusion. Thus, the correct answer is (C) .
23
PYQ 2020
medium
physicsID: mht-cet-
Three particles each of mass are placed at the corners of an equilateral triangle of side . A particle of mass is placed at the midpoint of any one side of the triangle. Due to the system of particles the force acting on is (G = universal constant of gravitation)
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Force due to nearest masses. The midpoint mass is equidistant from the two nearest masses. The forces due to these two masses cancel each other due to symmetry.
Step 2: Force due to the third mass. The third mass is at a distance from .
Step 3: Net force direction and magnitude. Considering vector addition of forces from all three masses, the resultant force becomes:
Step 4: Conclusion. The net gravitational force acting on is .
24
PYQ 2020
medium
physicsID: mht-cet-
A transistor has a voltage gain . If the amount of its output is applied to the input of the transistor, then the transistor becomes oscillator when
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Oscillator condition.
For a transistor to become an oscillator, the feedback must be such that the total gain around the loop is equal to one. This occurs when the product of the feedback factor and the voltage gain is equal to one. Step 2: Conclusion.
The condition for the transistor to oscillate is , so the correct answer is (B).
25
PYQ 2020
medium
physicsID: mht-cet-
Earth is assumed to be a sphere of radius and uniform density. The variation of acceleration due to gravity (g) according to the depth and the height (h) from the Earth's surface is shown correctly by the graph
1
(C)
2
(B)
3
(D)
4
(A)
Official Solution
Correct Option:
(1)
Step 1: Gravity at depth and height. The acceleration due to gravity at a depth below the Earth's surface decreases linearly with depth, and it is given by the equation:
where is the Earth's radius, and is the gravitational acceleration at the Earth's surface. At a height above the Earth's surface, the acceleration due to gravity decreases as:
Step 2: Graph analysis. Option (C) correctly shows the graph where gravity increases linearly with depth and decreases with height. Other options do not match this physical behavior. Step 3: Conclusion. Thus, the correct answer is (C).
26
PYQ 2020
medium
physicsID: mht-cet-
The acceleration due to gravity on moon is th times the acceleration due to gravity on earth. If the ratio of the density of earth to the density of moon is , then the radius of moon in terms of the radius of earth is
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Using the relation between acceleration and radius. The acceleration due to gravity is related to the radius and density of the planet by the formula: Where is the mass of the planet, is the radius, and is the gravitational constant. The mass of a planet is related to its density by , where . Step 2: Comparing gravity on Earth and Moon. The gravity on the moon is given by: Thus, comparing the gravity on Earth and the moon, we get: Thus, the radius of the moon is , corresponding to option (C).
27
PYQ 2020
medium
physicsID: mht-cet-
A body is projected vertically upwards from earth’s surface with velocity , where is escape velocity from earth’s surface. The velocity when body escapes the gravitational pull is
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Escape velocity and energy considerations. The escape velocity is the minimum velocity required for a body to escape the gravitational pull of the Earth. The total mechanical energy at the Earth's surface is: where is the gravitational constant, is the radius of the Earth, and is the mass of the body. For escape, the total energy at infinity is zero, so we set the total energy equal to zero.
Step 2: Energy after projection. After projecting with velocity , the total energy at the Earth's surface is: Since the body escapes the Earth, the total energy at infinity is zero, so we equate this energy to zero: Solving for the final velocity , we get: Step 3: Conclusion. The velocity of the body when it escapes the Earth's gravitational pull is , which corresponds to option (B).
28
PYQ 2020
medium
physicsID: mht-cet-
A particle of mass is performing U.C.M. along a circle of radius . The relation between centripetal acceleration and kinetic energy is given by
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Centripetal acceleration and kinetic energy.
The centripetal acceleration is given by:
where is the speed of the particle. The kinetic energy of the particle is:
Step 2: Relating acceleration and kinetic energy.
From the kinetic energy equation, solve for :
Substitute this into the equation for centripetal acceleration:
Step 3: Conclusion.
The correct relation is , so the correct answer is (A).
29
PYQ 2020
medium
physicsID: mht-cet-
In the case of earth, mean radius is , acceleration due to gravity on the surface is , angular speed about its own axis is . What will be the radius of the orbit of a geostationary satellite?
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Write condition for geostationary orbit. For a geostationary satellite, gravitational force provides the required centripetal force: Step 2: Express in terms of and . We know that: Step 3: Substitute value of . Step 4: Solve for orbital radius .
30
PYQ 2020
medium
physicsID: mht-cet-
The radius of the orbit of a geostationary satellite is (mean radius of the earth is , angular velocity about an axis is , and acceleration due to gravity on earth's surface is )
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Understanding the motion of a geostationary satellite. The radius of a geostationary satellite is derived from the balance between the gravitational force and the centrifugal force due to its orbital motion. The force of gravity is given by:
where is the gravitational constant, is the mass of the Earth, and is the radius of the orbit. The centrifugal force is:
Setting these two forces equal gives:
Simplifying and solving for yields:
where is related to by . Substituting this gives the final result:
Step 2: Conclusion. Thus, the correct answer is (A) .
31
PYQ 2020
medium
physicsID: mht-cet-
A satellite of mass , revolving round the earth of radius , has kinetic energy . Its angular momentum is
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Write expression for kinetic energy. For a satellite in circular orbit, Step 2: Express velocity in terms of kinetic energy.
Step 3: Write expression for angular momentum.
Step 4: Substitute the value of velocity.
Step 5: Conclusion. The angular momentum of the satellite is
32
PYQ 2020
medium
physicsID: mht-cet-
The length of seconds pendulum is 1 m on the earth. If the mass and diameter of the planet is double that of the earth, then the length of the seconds pendulum on the planet will be
1
0.2 m
2
0.4 m
3
0.3 m
4
0.5 m
Official Solution
Correct Option:
(3)
Step 1: Using the formula for the length of a seconds pendulum. The period of a seconds pendulum is given by: Where is the length of the pendulum and is the acceleration due to gravity. For a planet with double the mass and diameter of the earth, the gravity will be different. Step 2: Finding the new length. Since is proportional to the mass of the planet and inversely proportional to the square of the radius, we calculate the new gravity on the planet. For double the mass and diameter, gravity becomes half of the earth's gravity. Thus, the correct answer is (C) 0.3 m.
33
PYQ 2020
medium
physicsID: mht-cet-
A body is projected vertically upwards from earth's surface. If its K.E. of projection is equal to half of its minimum value required to escape from the gravitational influence, then the height upto which it rises is (R = radius of the earth)
1
4R
2
R
3
2R
4
3R
Official Solution
Correct Option:
(2)
Step 1: Understanding the energy required to escape. The minimum kinetic energy required for a body to escape from the Earth's gravitational field is equal to the gravitational potential energy at the Earth's surface. The total energy at the surface is the sum of the kinetic energy and potential energy:
For escape, this total energy must be zero. If the kinetic energy is half of this value, then the body will rise to a height equal to the radius of the Earth, , before it comes to rest. Step 2: Conclusion. Thus, the height upto which the body rises is , which is option (B).
34
PYQ 2022
easy
physicsID: mht-cet-
Two satellites A and B rotate round a planet’s orbit having radius 4R and R respectively. If the speed of satellite A is 3 V then speed of satellite B is
1
2
6V
3
4
12 V
Official Solution
Correct Option:
(2)
The speed of an object in a circular orbit is given by the formula: v = Let's compare the speeds of satellites A and B: For satellite A: vA = For satellite B: vB = vA = vA = vA = vB Therefore, the speed of satellite B is twice the speed of satellite A, which means: vB = 2 vA vB = 2 x 3V vB = 6V So, the correct option is (B) 6V.
35
PYQ 2022
easy
physicsID: mht-cet-
A satellite of mass ‘m’ is revolving around the earth of mass ‘M’ in an orbit of radius ‘r’. The angular momentum of the satellite about the centre of orbit will be
1
2
3
4
Official Solution
Correct Option:
(2)
The angular momentum (L) of a satellite revolving around a central body can be calculated using the formula: L = mvr To find the linear velocity (v) of the satellite, we can use the gravitational force equation: = Rearranging the equation, we can find the value of v: v = Substituting this value of v back into the formula for angular momentum: L = L = Therefore, the correct answer is (B) for the angular momentum of the satellite about the center of the orbit.
36
PYQ 2023
medium
physicsID: mht-cet-
If at depth ‘ ’ the gravitational force acting on a particle is 300 N, then what is the force on a particle at depth ‘ ’ ?
Official Solution
Correct Option:
(1)
The correct answer is 150 N.
We know that the gravitational force acting on a particle is directly proportional to the mass of the particle and the acceleration due to gravity. , where F = Force, m = mass, g = acceleration due to gravity
Now, if we consider the situation where we have two particles, one at depth ‘d’ and another at depth ‘d/2’ we can use the formula above.
Given: Similarly for the particle at :
The acceleration due to gravity changes with depth because the gravitational field strength decreases as we move away from the center of the Earth. Therefore,
where, g’ = gravitational field strength, G = gravitational Constant, M = Mass of the Earth and r = distance from the center of the earth.
As the gravitational field strength at depth 'd/2' is half of the gravitational field strength at depth 'd', the force on the particle at depth ' ' is half of the force at depth 'd'.
Therefore, the force on the particle at depth
37
PYQ 2024
medium
physicsID: mht-cet-
Two identical particles each of mass go around a circle of radius under the action of their mutual gravitational attraction. The angular speed of each particle will be:
1
2
3
4
Official Solution
Correct Option:
(3)
The gravitational force acts as the centripetal force for circular motion: The gravitational force between two particles is expressed as: where: - is the gravitational constant, - , - , the distance between the two particles. Substitute and : For circular motion: where . Equating : Simplify the equation: Take the square root to find : Final Answer:
38
PYQ 2024
medium
physicsID: mht-cet-
"The height from Earth's surface at which acceleration due to gravity becomes , where is acceleration due to gravity on the surface of Earth and is the radius of Earth?"
1
.
2
.
3
.
4
.
Official Solution
Correct Option:
(2)
Step 1: Variation of gravity with height. The acceleration due to gravity at a height above the Earth's surface is given by: where , is the radius of Earth, and is the acceleration due to gravity on the surface.
Step 2: Substituting . Cancel from both sides: Taking the square root: Rearranging:
39
PYQ 2024
easy
physicsID: mht-cet-
The height from Earth's surface at which acceleration due to gravity becomes is ? (Where is the acceleration due to gravity on the surface of the Earth and is the radius of the Earth.)
1
.
2
.
3
.
4
.
Official Solution
Correct Option:
(2)
Step 1: Formula for acceleration due to gravity at a height from Earth's surface.
The acceleration due to gravity at a height is given by:
where: : Acceleration due to gravity at height , : Acceleration due to gravity on the surface of the Earth, : Radius of the Earth. Step 2: Setting .
Substitute into the equation:
Step 3: Simplifying the equation.
Cancel from both sides:
Take the square root of both sides:
Cross-multiply:
Step 4: Solve for . Step 5: Conclusion.
The height from Earth's surface at which acceleration due to gravity becomes is:
40
PYQ 2025
medium
physicsID: mht-cet-
Two long straight wires A and B carrying equal current 'I' were kept parallel to each other at distance ' d ' apart. Magnitude of magnetic force experienced by length of wire A is ' '. If the distance between the wires is made half and currents are doubled, force on length of wire A will be
1
2
3
4
Official Solution
Correct Option:
(3)
Concept:
Force between parallel wires:
Step 1: Initial condition. Step 2: New condition. Step 3: Compare. Step 4: Conclusion. Force becomes
41
PYQ 2025
easy
physicsID: mht-cet-
The gravitational potential energy of an object of mass 5 kg at a height of 10 m above the surface of the Earth is:
1
2
3
4
Official Solution
Correct Option:
(1)
The gravitational potential energy is given by the formula:
Where:
- is the mass of the object,
- is the acceleration due to gravity,
- is the height above the Earth's surface. Substitute the values:
Thus, the gravitational potential energy of the object is .
42
PYQ 2025
medium
physicsID: mht-cet-
A wire of length carries current along x - axis. A magnetic field acts on the wire. The magnitude of magnetic force acting on the wire is
1
2
3
4
Official Solution
Correct Option:
(4)
Step 1: Vector Formula
.
. Step 2: Cross Product
.
.
. Step 3: Magnitude
. Final Answer: (D)
43
PYQ 2025
medium
physicsID: mht-cet-
The gravitational potential energy of a 2 kg object at a height of 5 m above the surface of the Earth is?
1
100 J
2
150 J
3
50 J
4
25 J
Official Solution
Correct Option:
(1)
The gravitational potential energy (GPE) is given by the formula:
Where:
- is the mass of the object,
- is the acceleration due to gravity,
- is the height above the surface. Now, calculate the potential energy:
Thus, the gravitational potential energy is .
44
PYQ 2025
medium
physicsID: mht-cet-
A radioactive element A decays into radioactive element C by the following processes in succession.
Then elements}
1
A and B are isobars.
2
A and C are isobars.
3
A and C are isotopes.
4
A and B are isotopes.
Official Solution
Correct Option:
(3)
Step 1: First Decay ( decay)
. Step 2: Second Decay ( decay)
.
So, element C is . Step 3: Compare A and C
Element A and C have the same atomic number but different mass numbers. Therefore, they are isotopes. Final Answer: (C)
45
PYQ 2025
easy
physicsID: mht-cet-
A body of mass is dropped from a height of . What is its speed just before hitting the ground? (Assume )
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Use the equation for final velocity in free fall For an object dropped from a height, the final velocity just before hitting the ground can be calculated using the equation: where:
- is the acceleration due to gravity,
- is the height from which the object is dropped. Step 2: Substitute the given values Given: - - Answer: Therefore, the speed of the body just before hitting the ground is . So, the correct answer is option (1).
46
PYQ 2025
hard
physicsID: mht-cet-
At a height above the Earth's surface, the acceleration due to gravity becomes . What is the value of in terms of the Earth's radius ?
1
2
3
4
Official Solution
Correct Option:
(1)
The acceleration due to gravity at a height above the Earth's surface is given by the formula: Where: - is the acceleration due to gravity at the Earth's surface, - is the acceleration due to gravity at a height , - is the radius of the Earth. We are given that the acceleration due to gravity at height is . Thus: Canceling on both sides: Taking the square root of both sides: Solving for : Thus, the height is . The correct answer is option (a).
47
PYQ 2025
medium
physicsID: mht-cet-
A satellite is orbiting the Earth at a height of above the Earth's surface. If the radius of the Earth is , calculate the orbital speed of the satellite. (Gravitational constant and Earth's mass )
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3
4
Official Solution
Correct Option:
(1)
Step 1: Use the formula for the orbital speed of a satellite The orbital speed of a satellite orbiting at a height above the Earth's surface is given by: where:
- is the gravitational constant,
- is the mass of the Earth,
- is the distance from the center of the Earth to the satellite, which is , where is the radius of the Earth. Step 2: Substitute the given values Given:
- ,
- ,
- Radius of the Earth ,
- Height of the satellite . The total distance from the center of the Earth to the satellite is: Now, substitute these values into the orbital speed formula: Answer: Therefore, the orbital speed of the satellite is . So, the correct answer is option (1).
48
PYQ 2025
hard
physicsID: mht-cet-
What is the gravitational force between two objects of masses and , separated by a distance of ?(Gravitational constant )
1
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3
4
Official Solution
Correct Option:
(1)
The gravitational force is given by the formula:
Substituting the values:
Correct Answer:
Option 1:
49
PYQ 2025
medium
physicsID: mht-cet-
The formula for the physical quantity is and the percentage error in the determination of physical quantities are and respectively. The percentage error in the measurement of is
The value of at height above Earth's surface is . Find in terms of the radius of the Earth.
1
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4
Official Solution
Correct Option:
(3)
We are given that the value of gravitational acceleration at a height above the Earth's surface is , where is the acceleration due to gravity at the Earth's surface. The formula for gravitational acceleration at a height above the Earth's surface is given by: where: - is the acceleration due to gravity on the surface of the Earth, - is the radius of the Earth, - is the height above the Earth's surface. We are given that: Substitute this into the equation: Cancel out from both sides: Now, square both sides to eliminate the square root: Cross-multiply: Expand the right-hand side: Simplify: Rearrange the equation: Solve for : Approximate for (assuming , so is much smaller than ): Thus, the height is approximately .
About Gravitation - MHT-CET
Gravitation is a vital chapter for MHT-CET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Gravitation PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Gravitation carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
