MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Gravitation Mcq
Quiz 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of 45° with the vertical, the escape velocity will be [ AIEEE 2003]
0%
a) 11√2 km/s
0%
b) 22 km/s
0%
c)11 km/s
0%
d)(11/√2) km/s
Explanation
The escape velocity is independent of angle of projection Ave=√(2gR) Answer:(c)
Q.2
A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is [ AIEEE 2004]
0%
a) gR2 /(R+x)
0%
b) gar / (R-x)
0%
c) g
0%
d) [gR2 /(R+x)]1/2
Explanation
Gravitational force provides necessary centripetal force. ∴ we know that g=GM/R2 GM=gR2 replacing value of GM in above equation Answer: (d)
Q.3
The time period of an earth satellite in circular orbit is independent of [ AIEEE 2004]
0%
a)both the mass and radius of the orbit
0%
b) radius of orbit
0%
c)the mass of the satellite
0%
d)neither the mass of the satellite nor the radius of its orbit
Explanation
Answer: (c)
Q.4
If 'g' is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an masses ' m' raised from the surface of the earth to height equal to the radius 'R' around the earth is [ AIEEE 2004]
0%
a) ¼ mgR
0%
b) ½ mgR
0%
c)2mg R
0%
d)mg R
Explanation
At earth surface Potential energy is=-GMn / R At a distance R from the earth's surface, P.E. of system is -GMm/2R Thus change in potential energy ΔU=-GMm/ 2R - GMm/ R=GMm / 2R Now GM/R2=g ∴ GM/R=gR ∴ ΔU=½ mgRAnswer: (b)
Q.5
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of planet in circular orbit of radius 'R' around the sun will be proportional to [ AIEEE 2004]
0%
a) Rn
0%
b)
0%
c)
0%
d)
Explanation
We know that centripetal force is provided by gravitational force Answer:(c)
Q.6
The change in the value of 'g' at height 'h' above the surface of the earth is same as at a depth 'd' below the surface of earth. When both 'd' and 'h' are much smaller than the radius of earth, then which one of the following is correct [ AIEEE 2005]
0%
a) d=3h/2
0%
b) d=h/2
0%
c) d=h
0%
d) d=2h
Explanation
Variation of 'g' with altitude is Variation of g with depth is Equating we get d=2hAnswer: (d)
Q.7
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere ( G=6.67 × 10-11 Nm2 / kg2) [ AIEEE 2005]
0%
a) 3.33 × 10-10 J
0%
b) 13.34 × 10-10 J
0%
c)6.67× 10-10 J
0%
d)6.67× 10-9 J
Explanation
Work done=Change in potential energy W=-GMm/R By substituting values we get W=6.67× 10-10 J Answer: (c)
Q.8
Average density of the earth [ AIEEE 2005]
0%
a) is a complex function of g
0%
b) does not depend on g
0%
c)is inversely proportional to g
0%
d)is directly proportional to g
Explanation
g=GM/R2 Let ρ is average density Thus density ρ is directly proportional to gAnswer: (d)
Q.9
A planet in a distant solar system is 10 times more massive than earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km/s, the escape velocity from the surface of the planet would be [ AIEEE 2008]
0%
a) 1.1 km/s
0%
b) 11 km/s
0%
c)110km/s
0%
d)0.11 km/s
Explanation
Answer:(c)
Q.10
The height at which the acceleration due to gravity becomes g/9 ( where g=the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is [ AIEEE 2009]
0%
a) R / √2
0%
b) R/ 2
0%
c) √2 R
0%
d) 2R
Explanation
We know that Answer: (d)
Q.11
If the earth is at one fourth of its present distance from the sun, the duration of the year will be [ EAMCET 1987]
0%
a)half the present year
0%
b) one-eight the present year
0%
c)one-fourth the present year
0%
d)one sixth the present year
Explanation
According to Kepler's law T2 ∝ R3 R2=(1/4) R1 Answer: (b)
Q.12
A satellite is moving round the earth with velocity v. To make the satellite escape, the minimum percentage increase in its velocity is nearly [ MPPET 1994]
0%
a) 41.4%
0%
b) 82.8%
0%
c)100%
0%
d)none of the above
Explanation
We know that area velocity of satellite v=√(gr) and escape velocity ve=√(2gr) ∴ Increase in velocity=(√2 - 1)√(gr) Increase in velocity=0.414√(gr) ∴ Percentage increase in velocity=[0.414√(gr) / √(gr)] × 100=41.4%Answer: (a)
Q.13
The gravitational potential energy of a rocket of mass 100kg at a distance 109 m from earth surface is 4 × 107 joule. The weight of the rocket in Newton's at distance 109 m from earth is ( Re=6400 km)
0%
a) 8 ×10-2 N
0%
b) 8 × 10-3 N
0%
c)4 × 10-3 N
0%
d)4 × 10-2N
Explanation
Potential energy ∴ weight of the rocket at 109 m from earth=4 × 10-4 × 100=4 × 10-2N Answer:(d)
Q.14
A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is [ MNR 1986]
0%
a) positive
0%
b) negative
0%
c) zero
0%
d) may be positive or negative depending upon its initial velocity
Explanation
Negative energy shows that object is bounded Answer: (b)
Q.15
Consider a satellite going round the earth in a circular orbit. Which of the following statements is wrong?
0%
a)It is freely falling body
0%
b) It is acted upon by a force directed away from the center of the earth which counter-balances the gravitational pull
0%
c)It is moving with constant speed
0%
d)Its angular momentum remains constant
Explanation
Answer: (b)
Q.16
Imagine a light planet revolving around a very massive star in circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between planet and star is proportional t R-5/2, then T2 is proportional to ..[ IIT 1989]
0%
a) R3
0%
b) R7/2
0%
c)R5/2
0%
d)R3/2
Explanation
From given Answer: (b)
Q.17
A spherical planet far out in space has a mass Mo and diameter Do. A particle of mass m falling freely near the surface of this planet will experience acceleration due to gravity which is equal to [ MPPMT 1987]
0%
a) GMo / D20
0%
b) 4m GMo / D20
0%
c) 4GMo / D20
0%
d) GmMo / D20
Explanation
In gravitational formula radius is used while in problem Diameter is given R=D/2 substituting value of R in equation for acceleration we get option c. Answer:(c)
Q.18
Which of the following represents the Newton's law of universal gravitation correctly
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.19
The gravitational potential at any point distant r from a point mass m is work done in moving a unit mass from
0%
a)the point of interest to infinity
0%
b) from infinity to the point of interest
0%
c)from the position of point mass m to the point of interest
0%
d)from infinity to the position of point mass m
Explanation
Answer: (a)
Q.20
If earth ware to cease rotating about its own axis, the increase in the value of g is C.G.S system at equator is nearly
0%
a) 3.36 cm/s2
0%
b) 6.72 cm/s2
0%
c)1.68 cm/s2
0%
d)0.84 cm/s2
Explanation
Observed value of g g'=g - Rω2cos2λ , for equator cosλ=1 ∴ g'=g - Rω2If earth ceases rotation, then increase in the value of g at equator =Rω2 =6400 × 103 × ( 2π / 86400)2=3.36 cm/s2 Answer: (a)
Q.21
An astronaut orbiting the earth in a circular orbit 120km above the surface of earth, gently drops a spoon from the spaceship. The spoon will [Raj-PMT 1996]
0%
a) fall vertically down to the earth
0%
b) move towards the moon
0%
c)will move along with the spaceship
0%
d)will move in an irregular way and then fall down to earth
Explanation
Answer:(c)
Q.22
The earth revolves round the sun in an elliptical orbit. Its speed is
0%
a) goes on decreasing continuously
0%
b) greatest when it is closest to the sun
0%
c) greatest when it is farthest from the sun
0%
d) constant at all the points on the orbit
Explanation
Answer: (b)
Q.23
Two satellites A and B go around a planet P in circular orbit having radii 4R and R respectively if the speed of the satellite a is 3V, the speed of the satellite B will be [ MNR 1991]
0%
a)12V
0%
b) 6V
0%
c)(4/3)V
0%
d)(3/2)V
Explanation
VA=(GM/4R)1/2 VB=(GM/R)1/2 By taking the ratio of above two equations we get BB=6VA Answer: (b)
Q.24
In a gravitational field, at a point where the gravitational potential is zero
0%
a) the gravitational field is necessarily zero
0%
b) the gravitational field is not necessarily zero
0%
c)nothing can be said definitely about gravitational field
0%
d)none of the above
Explanation
Answer: (b)
Q.25
A satellite is orbiting around the earth in the equatorial plane rotating from west to east as the earth does. If ωe is angular velocity of earth and ωs is angular velocity of satellite, then the satellite will appear at the same location after a time t=
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Relative angular speed of the satellite with respect to earth =ωs - ωethus option a is correct Answer:(a)
Q.26
A solid sphere of uniform density and mass M has a radius of 4m. Its center is at the origin of coordinate system. Two sphere of radii 1 m are taken out so that their centers are at P(0, -2, 0) and Q(0, 2, 0) respectively. This leaves two spherical cavities. What is the gravitational field at the center of each cavity
0%
a) (31/1024) GM
0%
b) GM/ 1024
0%
c) 31 GM
0%
d) GM
Explanation
Density of sphere Radius=4 m and Mass=M A) If we consider cavities are not made, then intensity at the point P or Q Radius of sphere=2 m having mass M' M'=V × ρ Gravitational field intensity at Point P or Q if we consider cavities are not made I= B) Gravitational field intensity at point P due to small sphere at Q Mass of small sphere m= Gravitational intensity at point P due to small sphere at Q= C) Gravitational Intensity at point P due to sphere at P=0 D) Intensity at P or Q I'=I - IP - IQ Answer: (a)
Q.27
The tail of the Comet Halley is directed away from sun due to the fact that..[ CPMT 1988]
0%
a)As the comet rotates around the sub the lighter mass of the comet is pushed away due to centrifugal force only
0%
b) as the comet rotates, the lighter mass of the comet is attracted by some star situated in direction of the tail
0%
c)the radiation emitted by the sun exerts a radiant pressure on the comet throwing its tail away from sun
0%
d)the tail of the comet always exists in the same orientation
Explanation
Answer: (c)
Q.28
Kepler's second law states that the straight line joining the planet to the sun sweeps out equal areas in equal times. This statement is equivalent to saying that
0%
a) total acceleration is zero
0%
b) transverse acceleration is zero
0%
c)tangential acceleration is zero
0%
d)radial acceleration is zero
Explanation
Answer: (b)
Q.29
The point masses m1 and m2 are initially at rest and at infinite distance apart. They start moving towards one another under their mutual gravitation field. There relative speed when they are at distance d apart is ;
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Initial potential energy at infinity=0 Potential energy at distance d=-(Gm1m2) /d Change in P.E=-(Gm1m2) /d let V1 be the velocity of mass m1 and V2 be the velocity of mass m2 Then kinetic energy at distance d= ½ m1 V12 + ½ m2 V22 Now according to law of conservation of energy ΔK=- ΔU thus ½ m1 V12 + ½ m2 V22=(Gm1m2) /d -- eq(1) Since no external force acts according to law of conservation of momentum Final momentum=Initial momentum m1 V1 + m2 V2=0 Thus V1 / V2=- ( m2 / m1) Putting it in equation 1 we get Similarly Relative speed=V2 - (-V1) Relative speed=V2 + V1 Answer:(b)
Q.30
A planet moves around the sun. At a point P it is closest from the sun at a distance d1 and has a speed vAt another point Q, when it is farthest fron the sun at a distance d2, its speed will be [ MPPMT 1987]
0%
a) d12v1 / d22
0%
b) d2v1 / d1
0%
c) d1v1 / d2
0%
d) d22v1 / d12
Explanation
By law of conservation of angular momentum mv1d1=mv2d2 v2=v1d1 / d2 Answer: (c)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page