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Gravitation Mcq
Quiz 4
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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)
Q.12
In a satellite if the time of revolution is T, then K.E. is proportional to [ BHU 1995]
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a) 1 /T
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b) 1/ T2
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c) 1 / T3
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d) T-2/3
Explanation
We know that K.E. ∝ v2 and v=ωR But ω=2π / T Thus K.E ∝=1/ t2 Answer: (b)
Q.13
A rubber ball is dropped from height of 5m on a planet where the acceleration due to gravity is not known. On bouncing it rises to 1.8m. The ball loses its velocity on bouncing by a factor of [ CBSE 1998]
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a)2/5
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b) 9/25
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c)3/5
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d)16/25
Explanation
Initial potential energy=mg(5) Now P.E = K.E mg(5) = ½m v12 v12 = (2×g×5) = (10g) v1 = √(10g) Final potential energy=mg(1.8) As above v2 = √(3.6g) Now fraction of loss of veloicty = 1 - ( final velocity/ initial veloicty) Fraction of loss of veloicty = 1 - [ √(3.6g) / √(10g)] Fraction of loss of veloicty = 1 - √ (3.6/10) Fraction of loss of veloicty = 1 -√(9/25) Fraction of loss of veloicty = 1 - (3/5) Fraction of loss of veloicty = 2/5 Answer: (a)
Q.14
The depth d, at which the value of acceleration due to gravity becomes (1/n) times the value at the surface is ( R=Radius of earth) [ MPPMT 1999]
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a) R/n
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b)
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c)
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d)
Explanation
We know that gravitational acceleration ∝ R when we go inside the earthLet when radius is R' acceleration is g' Taking the ratio of g/g'=R/R'Given g'/g=1/n 1/n=R'/R R'=R/n Depth h=R-R' thus R-R'=R - R/n h=R [ ( n-1) / n] Answer: (b)
Q.15
Spot the wrong statement: The acceleration due to gravity, g, decreases if : [ MPPMT 1994]
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a) we go down from the surface of the earth towards its center
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b) we go up from the surface of the earth
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c)we go from the equator towards the poles on the surface of the earth
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d)the rotational velocity of the earth is increased
Explanation
Answer:(b)
Q.16
A particle hanging from the spring stretches it by 1cm at the earth's surface. How much the same particle stretches the spring at a place 1600 km above the surface of earth ( R=6400 km)
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a) 16/50 cm
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b) 16/25 cm
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c) 25/16 cm
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d) 50/16 cm
Explanation
We know that when spring is in equilibrium mg=kx , here x is stretching Thus x ∝ g Gravitational acceleration g ∝ 1/R2 Thus x ∝ ( 1/R2) OR xR2=constant Thus x1R12=x2R22 1 × (6400)2=x2 ( 8000)2 x2=(6400 / 8000)2 x2=16/25 Answer: (b)
Q.17
A person brings a mass of 1 kg from infinity to a point P. Initially the mass was at rest but moves at a speed of 2m/s as it reaches at p. The work done by the person on the mass is -3J. The potential at P is
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a) - 2 K/kg
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b) -3 J/kg
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c) - 5J/kg
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d) -7 J/kg
Explanation
Work done = P.E. + K.E -3 = PE + ½ × 1 × 22 PE = -3 - 2 = -5J Gravitational potential = PE / mass = -5/1 = -5 J/kg Answer: (c)
Q.18
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.19
The time period of a second's pendulum in a satellite is
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a) zero
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b) 2
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c) infinity
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d) depends on mass of body
Explanation
Answer:(c)
Q.20
Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [ CBSE 1995]
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a)
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b)
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c)
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d)
Explanation
Centripetal force is due to gravitational force Answer: (c)
Q.21
Out of the following, the only correct statement about satellites is [ Hariyana C.E.E. 1996]
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a)A satellite can not move in a stable orbit in a plane passing through the earth's center
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b) Geostationary satellites are launched in equatorial plane
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c)we can use just one geostationary satellite for global communication around the globe
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d)The speed of satellite increases with an increase in the radius of orbit.
Explanation
Answer: (b)
Q.22
The angular speed of earth in rad/s, so that the object on equator may appear weightless ( g=10 m/s2 and radius of earth=6400 km) [ AMU PMT 1997]
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a) 1.56
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b) 1.25 × 10-1
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c) 1.56 × 10-3
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d)1.25 × 10-3
Explanation
The object at equator may appear weightless if mg=mRω2 Or ω=√ (g/R) Answer: (d)
Q.23
The value of g at a particular point is 9.8 m/sSuppose the earth suddenly shrinks to half its present size, without losing any mass. The value of g at the same point ( assume that distance of point from the center of earth does not shrink) will be [ Pb. CET 1998]
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a) 4.9 m/s2
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b) 3.1 m/s2
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c)9.8 m/s2
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d)19.6 m/s2
Explanation
Answer:(c)
Q.24
The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth [ SCRA 1998]
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a) is the same
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b) is smaller
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c) is greater
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d) varies with its phase
Explanation
Answer: (a)
Q.25
The escape velocity of rocket on the earth is 11.2 km/s. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of the earth will be in km/s [ M.N.R. 1999]
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a)11.2
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b) 5.6
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c)22.4
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d)33.6
Explanation
ve=√( 2gR)=11.2 km/v'e=√( 2g'R')=√( 2 × 2g × 2R) v'e=2 × 11.2=22.4 km/sAnswer: (c)
Q.26
If the radius of earth were to decrease by 1%, its mass remains same, the acceleration due to gravity on the surface of earth will [ IIT 1981]
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a) increase by 1%
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b) increase by 2%
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c) decrease by 1%
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d)decrease by 2%
Explanation
Formula for gravitational acceleration g=GM R-2 By talking first order derivative dg=GM(-2)R-3 dR thus dg/g=-2 dR/R Negative sign indicates increase in g when R is decreased Given dR/R=1%dg/g=2%,Answer: (b)
Q.27
How much energy will be necessary for making a body of 500 kg escape from the earth? ( g=9.8 km/s2, radius of the earth=6.4 × 106m) [ M.P. CEE 1999]
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a) About 9.8 × 106 J
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b) About 6.4 × 108 J
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c) About 3.1 × 1010 J
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d) About 27.4 × 1012 J
Explanation
Object velocity should be equal to escape velocity ve=√(2gR)Kinetic energy=½ (mve2 )Kinetic energy=mgR Kinetic Energy=500 × 9.8 × 6.4 × 106=3.1 × 1010 J Answer:(c)
Q.28
What should be the velocity of earth due to rotation about its own axis so that the weight of person become 3/5 of the present weight at equator. Radius of earth on equator is 6400km [ MNR 1999]
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a) 7.4 × 10-3 rad/s
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b) 6.7 ×10-4 rad/s
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c) 7.8 × 10-4 rad/s
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d) 8.7 × 10-4 rad/s
Explanation
Centrifugal force is acting away from center opposite to gravitational force mg - mRω2=(3/5) mg Answer: (c)
Q.29
If both the mass and the radius of the earth decreases by 1% [ karnataka CET 2001]
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a) the escape velocity would increase
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b) the acceleration due to gravity would increase
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c)the escape velocity would decrease
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d)the acceleration due to gravity would decrease
Explanation
Escape velocity Escape velocity don not change Formula for gravitational accelerationgravitational acceleration increasedAnswer: (b)
Q.30
The escape velocity on the surface of earth is 11.2km/s. What would be the escape velocity on the surface of another planet of same mass but 1/4 times the radius of the earth? [JIPMER 2002]
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a) 22.4 km/s
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b) 67.2 km/s
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c)44.8 km/s
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d) 89.6 km/s
Explanation
Escape velocity given ve=11.2 m/s escape velocity on the surface of another planet=2 × 11.2=22.4m/sAnswer: (a)
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