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Physics NEET MCQ
Quiz 6
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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.24
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)
Q.25
A particle of mass M is situated at the center of a spherical shell of same mass and radius 'a'. The gravitational potential at a point situated at a/2 distance from the center, will be: [ CBSE-PMT 2010]
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a)
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b)
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c)
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d)
Explanation
Potential at the point=Potential at the point due to the shell + Potential due to the particle Note potential inside the shell is uniform=-GM / aPotential at point at a/2 due to mass at center=-2GM/aThus Potential at the given point=-3GM / a Answer:(a)
Q.26
The mean radius of earth is R, its angular speed on its own axis is ω and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite ?[1992]
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a) (R2g/ω2)1/3
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b) (Rg/ω2)1/3
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c) (R2 ω2/g)1/3
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d) (R2 g/ ω)1/3
Explanation
For a geostationary satellite it's angular velocity should be same as earth's angular velocity ω Centripetal acceleration of geostationary satellite here 'r' is the radius of satellite v is the linear velocity of satellite Now v=ωr since angular velocity of geostationary satellite must be equal to angular velocity of earth. substituting the value of v in above equation we get Now g=GM/R2 GM=gR2 substituting value in above equation we get Answer: (a)
Q.27
satellite A of mass m is at a distance of r from the earth's center. Another satellite B of mass 2m is at a distance of 2r from the earth's center. Their time periods are in the ratio of
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a) 1:2
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b) 1:16
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c) 1 : 32
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d) 1 : 2√2
Explanation
According to Kepler's law T2 ∝R3 note periodic time is independent of massTaking the ratio of periodic time we getAnswer: (d)
Q.28
The escape velocity from earth is 11.2 km/s. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is [1993]
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a) 11.2×2 km/s
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b) 11.2km/s
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c) 11.2 / √2 km/s
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d) 11.2 √2 km/s
Explanation
Escape velocity is independent of angle of projection Answer:(b)
Q.29
The distance of Neptune and Saturn from the sun is nearly 1013 and 1012 meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio [1994]
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a) 10
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b) 100
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c) 10√10
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d) 1000
Explanation
For solution refer Q2 Answer: (c)
Q.30
A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's center. Their time periods are in the ratio of..[ CBSE-PMT 1993]
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a)1:2
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b) 1:16
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c)1 : 32
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d)1 : 2√2
Explanation
Time period of satellite is independent of massT2 ∝ r3Answer: (d)
0 h : 0 m : 1 s
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