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Physics NEET MCQ
Quiz 9
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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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c) - αv2 / 2M
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d) - αv2
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a) -2αv² / M
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b) - αv² / M
Explanation
Thrust=v (dM/dt)=- αv2 ∴ retardation=-αv2 / M Answer: (b)
Q.16
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.17
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.18
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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d)K/2x2
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c)K/x²
Explanation
dV=- EdxOn integrating we get V=k/2x2 Answer:(d)
Q.19
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.20
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.21
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.22
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.23
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.24
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.25
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.26
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.27
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.28
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.29
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.30
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)
0 h : 0 m : 1 s
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