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Gravitation Mcq
Quiz 10
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Q.1
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.2
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.3
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)
0%
c)
0%
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.4
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.5
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.6
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.7
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)
0%
b)
0%
c)
0%
d)
Explanation
Centripetal force is due to gravitational force Answer: (c)
Q.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
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)
Q.18
A satellite orbits around the earth in a circular orbit with speed v and orbital radius r. If it loses some energy, then v and r changes as [ JIPMER 2009]
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b) both v and r decreases
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a) v decreases as r increases
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c)v increases and t decreases
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d)both v and r increases
Explanation
Total energy of=-GMm / 2r If the satellite loses some energy, its total energy will decrease. It will be so if r decreases. Then from the equation for orbital velocity{v=√(GM/r)}, velocity increases Answer:(c)
Q.19
The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man can jump 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) 2/9 m
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b) 18 m
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c) 6 m
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d) 2/3 m
Explanation
Energy spend is same in both the cases Work=mgh mg1h1=mg2h2g1h1=g2h2 given g1=9 × g2 9 × g2 ×2=g2h2 h2=18 mAnswer: (b)
Q.20
A satellite with kinetic energy E is revolving round the earth in a circular orbit. The minimum additional kinetic energy required for it to escape into outer space is [ kerala PMT 2003]
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a)√2 × E
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b) 2E
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c)E/ √2
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d)E
Explanation
We have to provide energy equal to binding energy to satellite to escape into outer spaceBinding energy=Kinetic energy + potential energy Binding energy=½ mVo2 + GM/rBut vo2=GM/r Thus Binding energy=½ GM/r - GM/r Binding energy=-½ (GM/r) Total energy of Satellite=-½GM/r Thus binding energy=-ENegative sign indicates energy (E) should be provided to free the satellite Answer: (d)
Q.21
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the masses of earth... [ IIT 1998]
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a) The acceleration of S is always directed towards the center of the earth
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b) The angular momentum of S about the center of the earth changes in direction but its magnitude remains constant.
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c)The total energy of S varies periodically with time
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d)The linear momentum of S remains constant in magnitude
Explanation
Answer: (a)
Q.22
A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1% its speed will be [ Roorkee 1995]
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a) increased by 1%
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b) decrease by 0.5%
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c)increased by 0.5%
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d)decreased by 1%
Explanation
Orbital speed of satellite is inversely proportional to radius . If r increases orbital speed decreases negative sign in equation indicates inverse proportion of velocity and radius Answer:(c)
Q.23
The masses and radii of the earth and moon are M1, R1 and M2, R2 respectively. Their centers are distance d apart. the minimum speed with which a particle of mass m should be projected from a point mid way between the two centers so as to escape to infinity is
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a)
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b)
0%
c)
0%
d)
Explanation
potential energy of body at mid point between earth and moon is To escape the mass kinetic energy should be provided which should be equal to potential energy UKinetic energy given here K=½ m ve2 is escape velocity, Answer: (b)
Q.24
A rocket is launched vertically from the surface of earth with an initial velocity u the height up to which the rocket can go from the surface of earth, before falling back is
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a)
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b)
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c)
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d)
Explanation
According to law of conservation of energy Kinetic energy at surface + Potential energy at surface=Kinetic energy at height + Potential energy at height Answer: (d)
Q.25
Two concentric shells of uniform density having mass M1 and M2 are situated as shown in the figure. The force on the particle of mass m when it is located at r=b is
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a)
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b)
0%
c)
0%
d)
Explanation
There will not be any gravitational force at point P due to outer shell of mass M2 since it is interior point Thus gravitational force at point P will be only due to sphere of mass M1, b is the distance of point P form centre , option "a" is correctAnswer: (a)
Q.26
in the 126 question, the gravitational potential energy of the particle at P is ..
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a)
0%
b)
0%
c)
0%
d)
Explanation
Potential energy at all the points inside the shell mass M2 is uniform is given by Potential energy at point P due to shell of mass M1 is Total potential energy=U1 + U2 Answer:(c)
Q.27
The metallic bob of a simple pendulum has the relative density ρ. The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by [ SCRA 1998]
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a)
0%
b)
0%
c)
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d)
Explanation
Let V be the volume of the bob. Weight of the bob in air=vρg Weight of bob in water=weight of bob in air - Buoyant force Weight of bob in water=Vρg - V × 1 × g=Vg( ρ - 1) If g' is the effective gravitational acceleration in water then Weight of boob in water=Vρg' Thus Vρg'=Vρg - V × 1 × g=Vg( ρ - 1) g'=g( ρ-1) / ρ Now periodic time T=2π √(l/g) and T'=2π √(l/g') Answer: (d)
Q.28
A satellite is moving in a circular orbit at a height 100km above the earth's surface. A person inside the satellite feels weightless because [ CSEP 1999]
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a)acceleration due to gravity is almost zero at such a height
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b) the earth does not exerts any force on the person
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c)the centripetal force makes the satellite move in circular orbit
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d)the force due to earth and moon are almost compensated at such height
Explanation
Answer: (c)
Q.29
Four particles, each of mass m, are moving along a circle of radius r under the influence of mutual gravitational attraction. The speed of each particle will be
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a)
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b)
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c)
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d)
Explanation
As shown in figure resultant force on particle 1 is Fr=√F + F'Centripetal force is provided by Resultant force Answer: (d)
Q.30
A body is projected vertically upwards from the surface of a planet of radius R with velocity equal to half the escape velocity for the planet. The maximum height attended by the body is [ karnatak CET 2002]
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a) R/2
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b) R/3
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c)R/5
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d)R/4
Explanation
According to law of conservation of energy Energy at surface of planet=Energy at maximum height h given u=ve / 2 Answer:(b)
0 h : 0 m : 1 s
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