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Quiz 7
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Q.1
A particle of mass M is situated at the center of a spherical shell of same mass and radius 'a'. The gravitational potential at a point situated at a/2 distance from the center, will be: [ CBSE-PMT 2010]
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a)
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b)
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c)
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d)
Explanation
Potential at the point=Potential at the point due to the shell + Potential due to the particle Note potential inside the shell is uniform=-GM / aPotential at point at a/2 due to mass at center=-2GM/aThus Potential at the given point=-3GM / a Answer:(a)
Q.2
The mean radius of earth is R, its angular speed on its own axis is ω and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite ?[1992]
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a) (R2g/ω2)1/3
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b) (Rg/ω2)1/3
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c) (R2 ω2/g)1/3
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d) (R2 g/ ω)1/3
Explanation
For a geostationary satellite it's angular velocity should be same as earth's angular velocity ω Centripetal acceleration of geostationary satellite here 'r' is the radius of satellite v is the linear velocity of satellite Now v=ωr since angular velocity of geostationary satellite must be equal to angular velocity of earth. substituting the value of v in above equation we get Now g=GM/R2 GM=gR2 substituting value in above equation we get Answer: (a)
Q.3
satellite A of mass m is at a distance of r from the earth's center. Another satellite B of mass 2m is at a distance of 2r from the earth's center. Their time periods are in the ratio of
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a) 1:2
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b) 1:16
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c) 1 : 32
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d) 1 : 2√2
Explanation
According to Kepler's law T2 ∝R3 note periodic time is independent of massTaking the ratio of periodic time we getAnswer: (d)
Q.4
The escape velocity from earth is 11.2 km/s. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is [1993]
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a) 11.2×2 km/s
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b) 11.2km/s
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c) 11.2 / √2 km/s
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d) 11.2 √2 km/s
Explanation
Escape velocity is independent of angle of projection Answer:(b)
Q.5
The distance of Neptune and Saturn from the sun is nearly 1013 and 1012 meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio [1994]
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a) 10
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b) 100
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c) 10√10
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d) 1000
Explanation
For solution refer Q2 Answer: (c)
Q.6
A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's center. Their time periods are in the ratio of..[ CBSE-PMT 1993]
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a)1:2
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b) 1:16
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c)1 : 32
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d)1 : 2√2
Explanation
Time period of satellite is independent of massT2 ∝ r3Answer: (d)
Q.7
A ball is dropped from a satellite revolving around the earth at a height of 120 km: The ball will -, [ CBSE-PMT 1996]
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a) continue to move with same speed along a straight line tangentially to the satellite at that time
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b) continue to move with the same speed along the original orbit of satellite
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c)fall down to earth gradually
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d)go far away in space
Explanation
The orbital speed of satellite is independent of mass of satellite, so the ball will behave as a satellite and will continue to move with the same speed in the original orbitAnswer: (b)
Q.8
The escape velocity on the surface of earth is 11.2 km/s. What would be the escape velocity on the surface of another planet of the same mass but 1/4 times the radius of the earth? [ CBSE-PMT 2000]
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a)22.4 km/s
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b) 44.8 km/s
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c) 5.6 km/s
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d)11.2 km/s
Explanation
Escape velocity of EarthEscape velocity of planetgiven Me=Mp Radius Re /4=RPOn substitution we getTaking the ratio of vp to ve we get∴ vp=2 × ve vp=2× 11.2=22.4 km/s Answer:(a)
Q.9
Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m. when it is taken from the earth's surface to a height 3R above its surface, is [ CBSE-PMT 2002]
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a) 3mgR
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b) (3/4) mgR
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c) mgR
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d) (3/2)mgR
Explanation
Gravitational potential energy (GPE) on the surface of earth E1=-GMm / R gravitational potential at 3R, E2=-GMm / 4R ∴ change in Gravitational potential Answer: (b)
Q.10
A roller coaster is designed such that riders experience "weight lessens" as they go round the lop of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between: [ CBSE-PMT 2008]
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a)14 m/s and 15 m/s
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b) 15 m/s and 16 m/s
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c)16 m/s and 17 m/s
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d)13 m/s and 14 m/s
Explanation
At the top mg is down ward force while centrifugal is up wardsmg≤ mv2 / 2v≥ √(gr)v ≥ √(9.8)20 v ≥ 14 m/sAnswer: (a)
Q.11
If the gravitational force between two objects were proportional to 1/R (and not as 1/R2) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to [ CBSE-PMT 1989]
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a) 1/R2
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b) R0
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c) R
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d)1/R
Explanation
Now F ∝1/R ∴ F=k/R Gravitational force is centripetal force for Circular motion thus k/F=Mv2 / R Hence v ∝ R0Answer: (b)
Q.12
A seconds pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket [ CBSE-PMT 1991]
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a) comes down with uniform acceleration
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b) moves round the earth in a geostationary orbit
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c)moves up with a uniform velocity
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d)moves up with uniform acceleration
Explanation
Time period of pendulum is given by equation When rocket accelerates upwards g increases to (g + a) hence period decreases Answer:(d)
Q.13
A satellite in force free space sweeps stationary interplanetary dust at a rate dM/dt=αv where M is the mass and v is the velocity of the satellite and α is a constant. What is the deceleration of the satellite? [ CBSE-PMT 1994]
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a) -αv2
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b) -αv2/2M
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c) -αv2/M
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d) -2αv2/M
Explanation
For variable mass problems force= now dM/dt=αv substituting in above we get F=αv2 ∴ Retardation=-F / M=- αv2 / M Answer: (c)
Q.14
A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface [CBSE-PMT 2000]
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a)32 N
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b) 28 N
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c)16N
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d)72N
Explanation
gravitational acceleration at surface of earth g=GM / R2 At height H=R/2, g'=Body weight at height R/2 is mg'Answer: (a)
Q.15
With what velocity should a particle be projected so that its height becomes equal to radius of earth? [ CBSE-PMT 2001]
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a)
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b)
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c)
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d)
Explanation
Let u be the velocity According to law of conservation of energyAnswer: (a)
Q.16
For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be [ CBSE-PMT 1989]
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a)11 km/s
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b) 11 √3 km/s
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c)11/ √3 km/s
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d)33 km/s
Explanation
Escape velocity is independent of angle of projection, so it will not changeformula for escape velocity is Answer:(a)
Q.17
A planet is moving in an elliptical orbit around the sun. If T, V, E and L stand respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the center of force, which of the following is correct ? [1990]
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a) T is conserved
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b) V is always positive
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c) E is always negative
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d) L is conserved but direction of vector L changes continuously
Explanation
In circular or elliptical orbital motion,torque is always acting parallel to displacement or velocity. So, angular momentum is conserved. In attractive field, potential energy is negative. Kinetic energy changes as velocity increases when distance is less. So, option 'c' is correct Answer: (c)
Q.18
Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravityThe gravitational force will now be [ CBSE-PMT 2003]
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a) F/9
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b) 3F
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c)F
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d) F/3
Explanation
Gravitational force is independent of the medium. Hence, it will remain independentAnswer: (c)
Q.19
A satellite of mass m is orbiting around the earth in a circular orbit with a velocity v. What will be its total energy? [ CBSE-PMT 1991]
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a) (3/4) mv2
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b) (3/2) mv2
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c) mv2
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d) -(1/2)mv2
Explanation
total energy of the satellite=K.E. + P.E Now K.E=½ ( m v2) P.E=-GMm / R but v=( Gm/R)1/2 ∴ P.E=-mv Thus Total energy=½ ( m v2) Total energy=-½ ( m v2) Answer: (d)
Q.20
What will be the formula of the mass in terms of g, R and G (R=radius of earth) [ CBSE-PMT 1996]
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a)g2(R/G)
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b) G (R2/g)
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c)G (R/g)
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d) g(R2/G)
Explanation
We know that g=GM / R2M=gR2 / G Answer: (d)
Q.21
The escape velocity from the surface of the earth is ve.The escape velocity from the surface of a planet whose mass and radius are three times those of the earth, will be [ CBSE-PMT 1995]
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a) ve
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b) 3ve
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c)9ve
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d)(1/3)ve
Explanation
Formula for escape velocity It is clear from formula if mass and radius is three times ,value of escape velocity from the surface of planet will be same as earth Answer: (a)
Q.22
The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B? [ CBSE-PMT 2003]
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a)(2/3)m
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b) (2/9) m
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c)18m
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d)6m
Explanation
On both the planet kinetic energy is same but height will be different According to law of conservation of energy ½ (m v2)=mgAhA=mgB hB ∴ hB=( gA / gB) hA hB=9 × 2=18m Answer:(c)
Q.23
A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest away at a distance of rIf v1 and v2 are the linear velocities at these points respectively, then the ratio is [2011]
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a) (r1/r2)2
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b) r2/r1
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c) (r2/r1)2
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d) r1/r2
Explanation
Angular momentum is conserved ∴ L1=L2 ∴ mr1v1=m r2v2 r1 v1=r2v2 ∴ v1 / v2=r2 / r1 Answer: (b)
Q.24
Two satellites of earth, S1 and S2 , are moving in the same orbit. The mass of S1 is four times the mass of S2 Which one of the following statements is true? [2007]
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a)The potential energies of earth satellites in the two cases are equal.
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b) S1 and S2, are moving with the same speed.
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c)The kinetic energies of the two satellites are equal.
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d)The time period of S1 is four times that of S2
Explanation
Since orbital velocity is independent of mass both will move with same speed.Answer: (b)
Q.25
The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is - [ CBSE-PMT 2006]
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a) 1/√2
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b) 1/3
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c)1/2
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d)√2
Explanation
Potential energy at height R=-GMm / 2R If ve is the velocity of mass 'm', so that it goes out of gravitational field from the distance 2R from center of earth½ (m ve2)=m × g ×R∴ ve=(gR) 1/2Now escape velocity if object in on surface of earth is v=(2gh) 1/2Thus v=ve × √2or ve=v / √2 ∴ f=1 / √ 2Answer: (a)
Q.26
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R. the radius of the planet would be [ CBSE-PMT 2004]
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a) R / 2
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b) 2R
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c)4R
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d)R/4
Explanation
g=GM / R2 Also Mass=D × (4/3)πR3∴ g=G(4/3) d π R At surface of planet gp=(4/3) G (2d) πR'At the surface of earth ge=(4/3) G (d) πR Now gp=ge∴ (4/3) G (2d) πR'=(4/3) G (d) πR∴ R'=R/2 Answer:(a)
Q.27
Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then [ CBSE-PMT 2005]
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a) g'=g/9
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b) g'=27g
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c) g'=9g
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d) g'=3g
Explanation
Density is same but size is three times bigger, hence mass of the planet will be three times the mass of earth g=GM / R 2 and M=(4/3) π R3 ρ ∴ g' / g=R'/R g'/g=3 g'=3g Answer: (d)
Q.28
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is [ CBSE-PMT 2005]
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a) 1/2
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b) 1/√2
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c)2
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d)√2
Explanation
Kinetic Energy of Satellite=½ (m v2 ) Now v=( GM/R)1/2 by substituting the value of v in equation of K.E. K.E=GMm/ 2R Potential energy=GMm / R ∴ K/U=1/2Answer: (a)
Q.29
The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun? [1997]
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a) 2
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b) 3
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c)4
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d)5
Explanation
Let TA and TB be time period of A and B about sunTA=8TBTA / TB=8According to Kepler's law T2 ∝ r3Answer: (c)
Q.30
The mean radius of earth is R, its angular speed on its own axis is ω and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite ?[ CBSE-PMT 1992]
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a)(R2g/ω2)1/3
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b)(Rg/ω2)1/3
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c)(R2 ω2/g)1/3
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d)(R2 g/ ω)1/3
Explanation
Periodic time T=πr / v0 Also v0=(gR2 / ω2)1/3 By substituting value of v0 in above equation we get Now Periodic time=2π/ω Answer:(a)
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