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Physics NEET MCQ
Gravitation Mcq
Quiz 6
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Q.1
A planet is revolving around the sun in an elliptical orbit, its closest distance from the sun is r and the farthest distance is R. If the orbital velocity of the planet closest to the sun be v, then what is the velocity at the farthest point.
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a) vr/R
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b) vR/r
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c) v(r/R)1/2
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d) v(r/R)1/2
Explanation
Applying law of conservation of angular momentum mvr = mVR V = vr/R Answer: (a)
Q.2
A uniform spherical shell gradually shrinks maintaining its shape. the gravitational potential at the centre
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a) increases
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b) decreases
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c) remains constant
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d) oscillates
Explanation
Gravitational potential due to a uniform spherical shell is given by formula At the centre of spherical shell, potential V = - 3GM / 2R as R decreases, V decreases Answer: (b)
Q.3
P is a point at distance r from centre of solid sphere of radius a. The gravitational potential at P is V. If V is plotted as a function of r, which is the correct curve?
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a)
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b)
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c)
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d)
Explanation
Gravitational potential inside the sphere is Gravitational potential outside the sphere is V = -GM/r Answer:(c)
Q.4
The eccentricity of the earth's orbit is 0.The ratio of its maximum speed in its orbit to its minimum speed is
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a) 1.67
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b) 1.034
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c) 1
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d) 0.167
Explanation
According to law of conservation of angular momentum mv1r1 = mv2r2 Where e is eccentricity of the earth's orbit Answer: (b)
Q.5
A body of mass 'm' taken from the earth's surface tothe height equal to twice the radius (R) of the earth.The change in potential energy of body will be…[NEET 2013]
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a) mg2R
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b) 2mgR/3
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c) 3mgR
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d) mgR/3
Explanation
Final potential energy Initial potential energy Change in potential energy from (i) and (ii) change in potential energy is 2mgR/3 Answer:(b)
Q.6
Infinite number of bodies, each of mass 2 kg aresituated on x-axis at distance 1 m, 2 m, 4 m, 8 m, .....respectively, from the origin. The resultinggravitational potential due to this system at theorigin will be
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a) - G
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b) -8G/3
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c) -4G/3
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d) -4G
Explanation
Gravitational potential is -Gm/r Potential due to n number of object kept on x axis is [ Note :For geometric progression Sum of G.P. up to ∞ when r < 1 is given by Here a =1 and r = ½] Answer:(d)
Q.7
A black hole is an object whose gravitationalfield is so strong that even light cannot escapefromit.To what approximateradiuswouldearth(mass = 5.98 × 1024 kg) have to becompressed to be a black hole?
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a) 10–2 m
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b) 100m
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c) 10–9 m
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d) 10–6 m
Explanation
Escape velocity of earth should become more than velocity of light r = 8.86×10-3 m ≈ 10-2 m Answer:(a)
Q.8
Kepler’s third law states that square of period of revolution (T) of a planet around the sun, isproportional to third power of average distance r between sun and planeti.e. T2 = Kr3 here K is constant.If the masses of sun and planet are M and m respectively then as per Newton’s law of gravitation forceof attraction between them is=G Mm/r2 , here G is gravitational constant. The relation between G and K is described as :
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a) K=1/G
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b) GK = 4π2
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c) GMK = 4π2
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d) K = G
Explanation
Gravitational force provides centripetal force But v = ωr and ω =2π/T so Given T2 = Kr3 or K=T2/r3 ∴ GMK= 4π2 Answer:(c)
Q.9
A satellite S is moving in an elliptical orbit aroundthe earth. The mass of thesatellite is very smallcompared to the mass of the earth. Then,.. [RE AIPMT 2015]
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a) the acceleration of S is always directed towardsthe centre of the earth.
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b) the angular momentum of S about the centreof the earth changes in direction, but itsmagnitude remains constant.
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c) the total mechanical energy of S variesperiodically with time.
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d) the linear momentum of S remains constant inmagnitude.
Explanation
Answer:(a)
Q.10
A remote - sensing satellite of earth revolves in acircular orbit at a height of 0.25 × 106 m above thesurface of earth. If earth's radius is 6.38 × 106 m and g=9.8 ms-2, then the orbital speed of thesatellite is :
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a) 6.67 km s-1
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b) 7.76 km s-1
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c) 8.56 km s-1
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d) 9.13 km s-1
Explanation
On substituting values we get v0 = 7.76 × 103 m/s = 7.76 km/s Answer:(b)
Q.11
At what height from the surface of earth thegravitation potential and the value of g are –5.4 × 107 J kg–2 and 6.0 ms–2 respectively ?Take the radius of earth as 6400 km : …[ AIPMT 2016]
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a) 2600 km
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b) 1600 km
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c) 1400 km
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d) 2000 km
Explanation
r= 0.9×107 m =9000 Km Now height from surface (h) = height (r) - radius (R) Height = 8000 – 6400 =2600 km Answer:(a)
Q.12
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp)whose radius and mean density are twice as that of earth is :-
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a) 1 : 2
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b) 1 : 2√2
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c) 1 : 4
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d) 1 : √2
Explanation
Given: whose radius and mean density are twice as that of earth Answer:(b)
Q.13
Starting from the centre of the earth having radius R, the variation of g ( acceleration due to gravity) is shown by …[ NEET II – 2016]
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a)
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b)
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c)
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d)
Explanation
Answer:(d)
Q.14
A satellite of mass m is orbiting the earth(of radius R) at a height h from its surface. The totalenergy of the satellite in terms of g0, the value ofacceleration due to gravity at the earth's surface, is …[ NEET II – 2016]
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a)
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b)
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c)
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d)
Explanation
Answer:(c)
Q.15
The acceleration due to gravity at a height 1 kmabove the earth is the same as at a depth d belowthe surface of earth. Then …[NEET 2016]
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a) ½ m
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b) 1 km
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c) 3/2 km
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d) 2 km
Explanation
Acceleration due to gravity above earth Acceleration due to gravity below earth given g'=g" d =2h Given h = 1 km ∴ d = 2km Answer:(d)
Q.16
Which of the following statements are correct? …[NEET 2017] (i) Centre of mass of a body always coincides withthe centre of gravity of the body. (ii) Centre of mass of a body is the point at whichthe total gravitational torque on the body is zero (iii) A couple on a body produce both translationaland rotational motion in a body. (iv) Mechanical advantage greater than one meansthat small effort can be used to lift a large load.
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a) (ii) and (iv)
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b) (i) and (ii)
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c) (ii) and (iii)
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d) (iii) and (iv)
Explanation
Statement i is wrong Centre of mass may or may not coincide with centreof gravity. Statement iii is wrong: couple is equal and opposite force produce only rotational motion Answer:(a)
Q.17
Two astronauts are floating in gravitational free spaceafter having lost contact with their spaceship. Thetwo will … [ NEET 2017]
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a) Keep floating at the same distance between them
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b) Move towards each other
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c) Move away from each other
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d) Will become stationary
Explanation
Both the astronauts are in the condition ofweightless. Gravitational force between them pulls towards each other. Answer:(b)
Q.18
Consider a drop of rain water having mass 1 g fallingfrom a height of 1 km. It its the ground with a speedof 50 m/s. Take g constant with a value10 m/sThe work done by the (i) gravitational force and the (ii) resistive force of air is
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a) (i) – 10 J (ii) –8.25 J
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b) (i) 1.25 J (ii) –8.25 J
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c) (i) 100 J (ii) 8.75 J
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d) (i) 10 J (ii) –8.75 J
Explanation
Work done by gravitational force = potential energy of drop W =mgh = 10-3×10×103 = 10 J Work done by resistive force WR= P.E. – K.E of drop when it reach surface WR = 10 – 1.250 = 8.75. This work is done by resistive force so Work is negative = -8.75 J Answer:(d)
Q.19
What is the minimum energy required to launch a satellite of mass m from the surface of a planetof mass M and radius R in a circular orbit at an altitude of 2R ?
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a)
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b)
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c)
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d)
Explanation
Potential energy at surface Potential energy at altitude 2R Kinetic energy of orbiting satellite at altitude 2R Thus total energy at altitude 2R = U+K Kinetic energy to be supplied = Energy at 3R – Energy at surface Answer:(a)
Q.20
then one correct answer Q169) Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m isprojected from the midpoint of the line joining their centres, perpendicular to the line. Thegravitational constant is G. The correct statement(s) is (are) … [ IIT Advance 2013]
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a) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
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b) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
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c) The minimum initial velocity of the mass m to escape the gravitational field of the twobodies is
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d) The energy of the mass m remains constant
Explanation
Binding energy of mass m This much kinetic energy should be provided to free the body b correct d correct, since gravitational field is conservative. Answer:(b, d)
Q.21
A planet of radius R =X/10 , (X radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and of linearmass density 10−3 kgm−1 into it. If the wire is not touching anywhere, the force applied atthe top of the wire by a person holding it in place is (take the radius of Earth = 6 × 106 m and the acceleration due to gravity on Earth is 10 ms−2) [IIT Advance 2014]
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a) 96 N
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b) 108 N
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c) 120 N
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d) 150 N
Explanation
R= 6×105 Fore on surface is maximum and as we go down force is reduced. Also as wire goes down the well mass of wire also increases If λ is mass per unit length then mass of the small element lowered wire is λdr. Gravitational force depends up on mass of the sphere at depth ∴ df=λdr×g' F = 1.08×102=108N Answer:(b)
Q.22
A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the sun and the Earth. The Sun is 3 × 105 times heavier than the Earth and is at a distance 2.5 ×104 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is ve = 11.2 km s–The minimum initial velocity (vs) required for the rocket to be able to leave the Sun- Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet) [IIT Advance 2017]
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a) vs = 22 km s–1
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b) vs = 72 km s–1
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c) vs = 42 km s–1
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d) vs = 62 km s–1
Explanation
Binding energy Answer:(c)
Q.23
The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3 V, then the speed of satellite B will be: [ CBSE-PMT 2010]
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a) 3V/4
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b) 6V
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c)12V
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d)3V/2
Explanation
Orbital velocity of a satellite in a circular orbit of radius 'R' inversely proportional ro square root of radius Thus If R1 and R2 Answer: (b)
Q.24
The earth ( mass=6 × 1024kg) revolves around the sun with an angular velocity 2 × 10-7 rad/sec in a circular orbit of radius=1.5 × 108km. The force exerted by sun on earth in newton is ...[ AFMC 1997, 1999]
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a)36 × 1021
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b) 18 × 1025
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c)29 × 1039
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d)zero
Explanation
Force exerted by sun on the earth is centripetal force.Centripetal force=mω2rm=6 × 1024kgω=2 × 10-7 rad/secr=1.5 × 1011kmsubstituting values in above equation we getCentripetal force=6 × 1024 × (2 × 10-7)2 × 1.5 × 1011=36 × 1021Answer: (a)
Q.25
The distance of two planets from the sun are 1013 and 1012 meters respectively. The ratio of time periods of these two planets is [ CBSE PMT 1988]
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a) 1/√10
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b) 100
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c) 10√10
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d) √10
Explanation
According to Kepler's law T2 ∝R3Taking the ratio of periodic time we getAnswer: (c)
Q.26
The largest and the shortest distance of the earth from the sun are r1and rIts distance from the sun when it is at perpendicular to the major-axis of the orbit drawn from the sun is [1988]
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a) (r1 + r2)/4
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b) (r1 + r2)/ (r1 - r2)
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c) 2(r1 × r2)/ (r1 + r2)
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d) (r1 + r2)/ 3
Explanation
Planets move around sun in elliptical path, Applying the properties of ellipse we have F and S are focus point and sun is at one focus point Verticle line is semi-minor axis Horizontal line is semi-major axis Point E represents position of Earth FE is the distance of earth from other focus F denoted by "p" SE is perpendicular distance of earth from sun denoted by "q" Answer:(c)
Q.27
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 [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
Given that Gravitational force is proportional to 1/R Hence Gravitational force=GMm/R Formula for centripetal force=mv2/R From above equation we get mv2/R=GMm/R ∴ v2=GM v ∝R0Answer: (b)
Q.28
A planet is moving in an elliptical orbit around the sun. If T, V, E, and L stands for kinetic energy, gravitational potential, total energy, and magnitude of angular moment bout the center of force respectively
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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 a 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 increase when distance is less. So, option (c) is correct.Answer: (c)
Q.29
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 [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
Formula for escape velocity ve=√(2gRe) is independent of angle of projection, so it will not change it will be 11 km/s Answer: (a)
Q.30
A satellite of mass am is orbiting around the earth in a circular orbit with a velocity v. What will be its total energy? [1991]
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a) (3/4) mv2
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b) (1/2) mv2
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c) mv2
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d)-(1/2)mv2
Explanation
Total energy of the satellite=kinetic energy + Potential energy=½mV2 - GMm/ h --re(1) here M is the mass of earth and h is the height of satellite from the center of the earthNow Centripetal force=Gravitational force mv2 / h=GMm/ h2multiplying both sides by h we getmv2=GMm/ hsubstituting value of GMm/ h in eq(1) we getTotal Energy=½mV2 - mv2Total energy=-(1/2)mv2negative sign indicates binding energy Answer:(d)
0 h : 0 m : 1 s
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