# JEE Main Physics Gravitation MCQ, JEE Physics Gravitation Mock Test

## JEE Main Physics Gravitation Online Test Series

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JEE Main Physics Gravitation MCQ, JEE Physics Gravitation Mock Test. JEE Main Online Test for Physics Gravitation. JEE Main Full Online Quiz **for Physics Gravitation**. **JEE Main Free Mock Test Paper 2018.** JEE Main 2018 Free Online Practice Test, Take JEE MCQ for All Subjects. JEE Main Question and Answers for Physics Gravitation. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n **Take JEE Main Physics Gravitation…**

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- Question 1 of 20
##### 1. Question

Figure shows the variation of energy with the orbit radius

*r*of a satellite in a circular motion. Mark the correct statement.CorrectKinetic energy KE =

*GMm*/*2r*Potential energy PE = –

*GMm*/*r*And the total energy E = –

*GMm*/*2r*Kinetic energy is always positive and KE ∝ 1/

*r*Potential energy is negative and |PE| ∝ 1/

*r*Similarly total energy is also negative and |E| ∝ 1/

*r*Also |E| < |PE|

∴ A is kinetic energy, B is potential energy and C is total energy of the satellite.

IncorrectKinetic energy KE =

*GMm*/*2r*Potential energy PE = –

*GMm*/*r*And the total energy E = –

*GMm*/*2r*Kinetic energy is always positive and KE ∝ 1/

*r*Potential energy is negative and |PE| ∝ 1/

*r*Similarly total energy is also negative and |E| ∝ 1/

*r*Also |E| < |PE|

∴ A is kinetic energy, B is potential energy and C is total energy of the satellite.

- Question 2 of 20
##### 2. Question

A planet revolving around sun in an elliptical orbit has a constant

CorrectIn elliptical orbit un is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet do not remain constant

IncorrectIn elliptical orbit un is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet do not remain constant

- Question 3 of 20
##### 3. Question

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon

CorrectMoon is revolving around earth in almost circular orbit. Sun exerts gravitational pull on both, earth and moon. When observed from sun, the orbit of the moon will not be strictly elliptical because they total gravitational force (i.e., force due to earth on moon and force due to sun on moon) is not central.

IncorrectMoon is revolving around earth in almost circular orbit. Sun exerts gravitational pull on both, earth and moon. When observed from sun, the orbit of the moon will not be strictly elliptical because they total gravitational force (i.e., force due to earth on moon and force due to sun on moon) is not central.

- Question 4 of 20
##### 4. Question

Three particles each having a mass of 100 g are placed on the vertices of an equilateral triangle of side 20 cm. The work done in increasing the side of this triangle to 40 cm is

CorrectW = U

_{f}– U_{i}m = 0.1 kg, r

_{f}= 0.4 m and r_{i}= 0.2 mSubstituting the value, we get

W = 5.0 × 10

^{-12}JIncorrectW = U

_{f}– U_{i}m = 0.1 kg, r

_{f}= 0.4 m and r_{i}= 0.2 mSubstituting the value, we get

W = 5.0 × 10

^{-12}J - Question 5 of 20
##### 5. Question

Two satellites S

_{1}& S_{2}of equal masses revolve in the same sense around a heavy planet in coplanar circular orbit of radii R & 4RCorrectIncorrect - Question 6 of 20
##### 6. Question

Two identical spherical masses are kept at some distance as shown. Potential energy when a mass

*m*is taken from surface of one sphere to the otherCorrectCentre point is the unstable equilibrium position where potential energy is maximum.

IncorrectCentre point is the unstable equilibrium position where potential energy is maximum.

- Question 7 of 20
##### 7. Question

If W

_{1}, W_{2}and W_{3}represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in a gravitational field of point mass*m*thenCorrectSince the gravitational field is conservative field hence, the work done in taking a particle from one point to another in a gravitational field is path independent.

IncorrectSince the gravitational field is conservative field hence, the work done in taking a particle from one point to another in a gravitational field is path independent.

- Question 8 of 20
##### 8. Question

Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth’s surface is:

CorrectApplying energy conservation

⇒ R + h = 4h ⇒ h = R/3

IncorrectApplying energy conservation

⇒ R + h = 4h ⇒ h = R/3

- Question 9 of 20
##### 9. Question

A thin spherical shell of mass M and radius R has a small hole. A particle of mass m is released at the mouth of the hole. Then

CorrectNet gravitational field inside a shell is zero. Hence, net force on the particle will be zero i.e., the particle stays at rest in its original position.

IncorrectNet gravitational field inside a shell is zero. Hence, net force on the particle will be zero i.e., the particle stays at rest in its original position.

- Question 10 of 20
##### 10. Question

A satellite in a circular orbit of radius r has time period of 4 hrs. A satellite with orbital radius of 4r around the same planet will have a time period of:

CorrectRadius of orbit of the first satellite

R

_{1}= RTime period of the first satellite T = 4 hrs. and radius of orbit of second satellite = 4r The time period of satellite is given by

T

_{2}= 8 × 4 = 32 hours.IncorrectRadius of orbit of the first satellite

R

_{1}= RTime period of the first satellite T = 4 hrs. and radius of orbit of second satellite = 4r The time period of satellite is given by

T

_{2}= 8 × 4 = 32 hours. - Question 11 of 20
##### 11. Question

A particle of mass M is at a distance a from the surface of a thin spherical shell of equal mass and having radius

*a*.CorrectAt centre, field is zero due to shell but non zero due to particle

∴ E

_{centre}= due to particlePotential at centre is non zero due to both shell and particle.

IncorrectAt centre, field is zero due to shell but non zero due to particle

∴ E

_{centre}= due to particlePotential at centre is non zero due to both shell and particle.

- Question 12 of 20
##### 12. Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F

_{1}on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F_{2}on the same particle. The ratio F_{2}/F_{1}isCorrectFrom superposition principle, F

_{1}= F_{r}+ F_{c}Here, F

_{r}= force due to remaining part = F_{2}And F

_{c}= force due to cavityIncorrectFrom superposition principle, F

_{1}= F_{r}+ F_{c}Here, F

_{r}= force due to remaining part = F_{2}And F

_{c}= force due to cavity - Question 13 of 20
##### 13. Question

If the distance between the earth And the sun were half its present value the number of days in a year would have been

CorrectT

^{2}∝ R^{3}IncorrectT

^{2}∝ R^{3} - Question 14 of 20
##### 14. Question

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg, mass of the earth = 6.0 × 10

^{24}kg, radius of the earth = 6.4 × 10^{6}m, G = 6.67 × 10^{-11}N-m^{2}/kg^{2}.CorrectGiven, height of the satellite above the earth’s surface (h) = 400 km = 0.4 × 10

^{6}mMass of the satellite (m) = 200 kg

Radius of earth (R

_{e}) = 6.4 × 10^{6}mMass of earth (M

_{e}) = 6.0 × 10^{24}kgGravitational constant (G) = 6.67 × 10

^{-11}N-m^{2}/kg^{2}Energy required send a satellite out of earth’s gravitational influence is called its binding energy.

Binding energy of a satellite =

= 5.885 × 10

^{9}J= 5.9 × 10

^{9}JIncorrectGiven, height of the satellite above the earth’s surface (h) = 400 km = 0.4 × 10

^{6}mMass of the satellite (m) = 200 kg

Radius of earth (R

_{e}) = 6.4 × 10^{6}mMass of earth (M

_{e}) = 6.0 × 10^{24}kgGravitational constant (G) = 6.67 × 10

^{-11}N-m^{2}/kg^{2}Energy required send a satellite out of earth’s gravitational influence is called its binding energy.

Binding energy of a satellite =

= 5.885 × 10

^{9}J= 5.9 × 10

^{9}J - Question 15 of 20
##### 15. Question

A particle of mass

*m*is moved from A to B as shown in figure. Then potential energy of the particleCorrectOutside the shell V

_{0}=Therefore, from A to surface v and hence the potential energy will decrease.

Inside the shell potential is constant. Hence, the potential energy is constant.

IncorrectOutside the shell V

_{0}=Therefore, from A to surface v and hence the potential energy will decrease.

Inside the shell potential is constant. Hence, the potential energy is constant.

- Question 16 of 20
##### 16. Question

A satellite is in a circular orbit around the earth has kinetic energy E

_{k}. Minimum amount of energy that is added so that it escapes the earth’s gravitational field is:CorrectTotal Mechanical energy (kinetic energy)

∴ TME = – E

_{k}For escape, TME = 0

i.e., If, E

_{k}is provided then TME. Becomes ZeroHence. The minimum amount of energy that is added so that it escapes the earth’s gravitational field is E

_{k}.IncorrectTotal Mechanical energy (kinetic energy)

∴ TME = – E

_{k}For escape, TME = 0

i.e., If, E

_{k}is provided then TME. Becomes ZeroHence. The minimum amount of energy that is added so that it escapes the earth’s gravitational field is E

_{k}. - Question 17 of 20
##### 17. Question

**Assertion**From a solid sphere of radius R, a hole of radius R/2 is cut as shown in figure. To find the magnitude of gravitational potential and gravitational field strength at O, we can directly subtract Potential due to hole (before removing) from potential due to whole sphere (before removing). The same can be done to find field strength at O, although potential is a scalar quantity and field strength is a vector quantity.**Reason**In gravity, it is done like this. It makes no difference, whether the field strength is added/subtracted by vector method or by scalar method.CorrectIncorrect - Question 18 of 20
##### 18. Question

At a distance 320 km above the surface of earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly (radius of earth = 6400 km)

CorrectIncorrect - Question 19 of 20
##### 19. Question

**Assertion**A planet may orbit around a star either in orbit P or orbit Q. The speed of planet at O is same for both orbits.**Reason**Planets orbit around a star with uniform velocity.CorrectIn circular path, star should lie at centre.

IncorrectIn circular path, star should lie at centre.

- Question 20 of 20
##### 20. Question

If the distance between the earth and the sun were half its present value, the number of days in a year would have been

CorrectFrom Kepler’s third law

T

_{2}= 129 daysIncorrectFrom Kepler’s third law

T

_{2}= 129 days