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Physics NEET MCQ
Gravitation Mcq
Quiz 9
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Q.1
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)
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a) 8 ×10-2 N
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b) 8 × 10-3 N
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c)4 × 10-3 N
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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.2
A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is [ MNR 1986]
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a) positive
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b) negative
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c) zero
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d) may be positive or negative depending upon its initial velocity
Explanation
Negative energy shows that object is bounded Answer: (b)
Q.3
Consider a satellite going round the earth in a circular orbit. Which of the following statements is wrong?
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a)It is freely falling body
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b) It is acted upon by a force directed away from the center of the earth which counter-balances the gravitational pull
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c)It is moving with constant speed
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d)Its angular momentum remains constant
Explanation
Answer: (b)
Q.4
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]
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a) R3
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b) R7/2
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c)R5/2
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d)R3/2
Explanation
From given Answer: (b)
Q.5
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]
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a) GMo / D20
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b) 4m GMo / D20
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c) 4GMo / D20
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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.6
Which of the following represents the Newton's law of universal gravitation correctly
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a)
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b)
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c)
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d)
Explanation
Answer: (c)
Q.7
The gravitational potential at any point distant r from a point mass m is work done in moving a unit mass from
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a)the point of interest to infinity
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b) from infinity to the point of interest
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c)from the position of point mass m to the point of interest
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d)from infinity to the position of point mass m
Explanation
Answer: (a)
Q.8
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
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a) 3.36 cm/s2
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b) 6.72 cm/s2
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c)1.68 cm/s2
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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.9
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]
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a) fall vertically down to the earth
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b) move towards the moon
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c)will move along with the spaceship
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d)will move in an irregular way and then fall down to earth
Explanation
Answer:(c)
Q.10
The earth revolves round the sun in an elliptical orbit. Its speed is
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a) goes on decreasing continuously
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b) greatest when it is closest to the sun
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c) greatest when it is farthest from the sun
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d) constant at all the points on the orbit
Explanation
Answer: (b)
Q.11
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]
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a)12V
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b) 6V
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c)(4/3)V
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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.12
In a gravitational field, at a point where the gravitational potential is zero
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a) the gravitational field is necessarily zero
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b) the gravitational field is not necessarily zero
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c)nothing can be said definitely about gravitational field
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d)none of the above
Explanation
Answer: (b)
Q.13
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=
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a)
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b)
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c)
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d)
Explanation
Relative angular speed of the satellite with respect to earth =ωs - ωethus option a is correct Answer:(a)
Q.14
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
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a) (31/1024) GM
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b) GM/ 1024
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c) 31 GM
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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.15
The tail of the Comet Halley is directed away from sun due to the fact that..[ CPMT 1988]
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a)As the comet rotates around the sub the lighter mass of the comet is pushed away due to centrifugal force only
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b) as the comet rotates, the lighter mass of the comet is attracted by some star situated in direction of the tail
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c)the radiation emitted by the sun exerts a radiant pressure on the comet throwing its tail away from sun
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d)the tail of the comet always exists in the same orientation
Explanation
Answer: (c)
Q.16
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
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a) total acceleration is zero
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b) transverse acceleration is zero
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c)tangential acceleration is zero
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d)radial acceleration is zero
Explanation
Answer: (b)
Q.17
Suppose the gravitational force varies inversely as the nth power of the distance. Then the time period of the planet in circular orbit of radius R around the sun will b proportional to
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a) R-n
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b) Rn
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c) R(n-1)/2
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d) R(n+1)/2
Explanation
Gravitational force provides centripetal acceleration Answer: (d)
Q.18
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 ;
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a)
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b)
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c)
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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.19
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]
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a) d12v1 / d22
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b) d2v1 / d1
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c) d1v1 / d2
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d) d22v1 / d12
Explanation
By law of conservation of angular momentum mv1d1=mv2d2 v2=v1d1 / d2 Answer: (c)
Q.20
A body weight with a spring balance in a train at rest, shows a weight WWhen the train begins to move with a velocity v around the equator from west to east and if the angular velocity of earth is ω
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a)W0
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b) W0 ( 1 + 2vω/g)
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c)W0( 1 - 2vω/g)
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d)W0( 1 - v2 / R)
Explanation
linear velocity of train=v linear velocity of train due to motion of earth=ωR Since train both moves from west to east, linear velocity of train with respect to earth=v + ωR Thus relative acceleration of train with respect to earth a=(v + ωR)2 / R, and is towards the center Therefore apparent weight W=m ( g - a)v2 / Rg and Rω2 / g are negligibly small W=W0 ( 1 - 2vω/g)Answer: (c)
Q.21
A hydrogen balloon released on the moon would
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a) climb with an acceleration of 9.8×6 m/s2
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b) climb with an acceleration of 9.8 / 6 m/s2
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c)neither climb nor fall
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d)fall with an acceleration of 9.8 / 6 m/s2
Explanation
Answer: (d)
Q.22
A body is dropped from a communication satellite around the earth. Which of the following statements is not correct?
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a)the body is falling continuously towards the center of earth
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b) the body will hit the surface of the earth
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c)the body will move into circular orbit around the earth
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d)the body will appear to be stationary when observed from the earth
Explanation
Answer:(b)
Q.23
A satellite in force-free space sweeps stationary interplanetary dust at rate dM/dt=αv, where M is the mass and v is the velocity of the satellite and α is constant. What is the deceleration of the satellite: [ CBSE 1994]
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a) -2αv2 / M
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b) - αv2 / M
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c) - αv2 / 2M
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d) - αv2
Explanation
Thrust=v (dM/dt)=- αv2 ∴ retardation=-αv2 / M Answer: (b)
Q.24
Acceleration due to gravity, g and the mean density of the earth ρ are related by the relation [ BHU 1998]
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a)
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b)
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c)
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d)
Explanation
g=GM / Re2 Now M=V× ρ Answer: (b)
Q.25
Choose the correct statement from the following:The radius of the orbit of a geostationary satellite depends upon [ PMT 1995]
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a) mass of the satellite, its time period and gravitational constant
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b) mass of the satellite, mass of the earth and the gravitational constant
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c)mass of the earth, mass of the satellite, time period of the satellite and the gravitational constant
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d)mass of the earth, time period of the satellite and the gravitational constant
Explanation
From the formula for orbital velocityAnswer: (d)
Q.26
The gravitational field due to a mass distribution is E=K/x3 in the direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is ..[ MPPMT 1994]
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a) K/x
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b) K/2x
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c)K/x2
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d)K/2x2
Explanation
dV=- EdxOn integrating we get V=k/2x2 Answer:(d)
Q.27
To purchase a sugar will be beneficial at [ Raj PET 1996]
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a) Poles
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b) Equator
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c) At 45° altitudes
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d) At altitude 40°
Explanation
Answer: (b)
Q.28
A satellite moves east wards with a speed V0 in equatorial plane, very near to the surface of earth. Another satellite moves West wards in the same orbit with same speed. if R=radius of earth and ω, the angular speed of earth about its own axis, then the difference in time periods of satellites as observed from earth is
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a)
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b)
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c)
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d)
Explanation
Velocity of earth=v=RωSatellite moving westward periodic times Tw satellite moves in opposite direction of earth Satellite moving eastward periodic times Te satellite moves in same direction of earth Answer: (b)
Q.29
At what height over the earth's pole, the free fall acceleration decreases by one percent? ( Assume the radius of earth to be 6400 km) [ CET 1994]
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a) 32 km
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b) 64 km
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c)80 km
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d)1.253 km
Explanation
Decrease in the value of g as we go above the surface of the earth by amount Δg=2gh/R Thus (Δg/g)=2h/R thus h=[(Δg/g) × R] / 2h=(0.01) × 6400 /2Answer: (a)
Q.30
the diameters of two planets are in the ratio 4:1 and their mean densities in the ratio 1:The acceleration due to gravity on the planets will be in ratio....[ ISM Dhanbad 1994]
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a) 1:2
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b) 2:3
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c)2:1
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d)4:1
Explanation
Let R1 and ρ1 be the radius and density of first planet Let R2 and ρ2 be the radius and density of second planet From the formula for gravitational accelerationGiven 4R2=R1and 2ρ1=ρ2On substituting values and taking ratios we get g1 : g2=2:1 Answer:(c)
0 h : 0 m : 1 s
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