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Quiz 5
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Q.1
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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a) v decreases as r increases
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b) both v and r decreases
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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.2
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.3
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.4
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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c)The total energy of S varies periodically with time
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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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d)The linear momentum of S remains constant in magnitude
Explanation
Answer: (a)
Q.5
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.6
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)
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c)
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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.7
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.8
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)
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c)
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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.9
in the 126 question, the gravitational potential energy of the particle at P is ..
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a)
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b)
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c)
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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.10
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)
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b)
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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.11
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.12
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.13
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)
Q.14
The period of moon's rotation around the Earth is nearly 29 days. If moon's mass were 2 fold of its present value, all other things remain unchanged, the period of moon's rotation would be nearly [ kerala CET 2002]
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a) 29√2
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b) 29 / √2
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c) 29×2
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d) 20 days
Explanation
If angular momentum changes with mass then period would be 29 days only, But we consider here Angular momentum as constant when mass of moon becomes double Angular momentum L=mrω2=constant Answer: (a)
Q.15
A planet has a mass 225 times and a radius 9 times that of the earth. The escape velocity for earth is known to be 11.2 km /s. The escape velocity for the planet is then [ Hariyana CET 2002]
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a)2.24 km/s
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b) 5 km/s
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c)56 km/s
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d) 280 km/s
Explanation
Formula for escape velocity Answer: (c)
Q.16
A point P lies on the axis of a ring of mass M and radius a, at a distance a from its center C. A small particle starts from P and reaches C under gravitation only. Its speed at C will be
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a)
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b)
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c)
0%
d)zero
Explanation
Gravitational potential at P Gravitational potential at c According to law of conservation of energy Answer: (b)
Q.17
Two small satellites move in a circular orbits around the earth, at distance r and (r + dr) from the center of the earth. Their time periods of rotation are T and T + dT ( Δr << r , ΔT << T) Then
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a)
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b)
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c)
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d)
Explanation
As T2 ∝ r3 T2=K r3 --eq(1) Differentiating it, we get2TΔT=3Kr2Δr --eq(2) Dividing eq(2) by eq(1) we have Answer:(a)
Q.18
A satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution in.. [ J and K CET 2005]
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a) (1/2) lunar month
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b) (2/3) lunar month
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c) 2-3/2 lunar month
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d) 23/2 lunar month
Explanation
Let r be the distance of moon from earth. For moon According to Kepler law T=K (r)3/2 [ Lunar month] For satellite T'=K (r/2)3/2 Answer: (c)
Q.19
If a rocket is to be launched from the surface of the earth so as to escape from the earth altogether, then
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a)it can have any velocity at start
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b) it must have a minimum velocity of 11.2 km/s
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c)the minimum velocity at start will be 11.2 km/s it it shoots up ony vertically
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d)none of above
Explanation
Answer: (b)
Q.20
The ratio of the radii of the planet P1 and P2 is kThe ratio of the acceleration due to the gravitation on them is kThe ratio of the escape velocities from them will be
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a) k1k2
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b) √(k1k2)
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c) √(k1 / k2)
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d) √(k2/ k1)
Explanation
Escape velocity Answer: (b)
Q.21
The orbital speed of Jupiter is
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a) greater than the orbital speed of earth
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b) less than the orbital speed of the earth
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c)equal to the orbital speed of the earth
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d)zero
Explanation
Answer:(b)
Q.22
The period of a satellite in a circular orbit around a planet is independent of
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a) the mass of the planet
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b) the radius of the planet
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c) the mass of the satellite
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d) all of there parameters a, b, c
Explanation
Formula for periodic time is Answer: (c)
Q.23
In planetary motion
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a)the angular speed remains constant
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b) the total angular momentum remains constant
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c)the linear speed remains constant
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d)neither the angular momentum nor angular speed remains constant
Explanation
Answer: (b)
Q.24
Inside the non uniform shell
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a) the gravitational field is zero
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b) the gravitational field is different everywhere
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c)gravitational potential is zero
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d)gravitational potential is not same everywhere
Explanation
Answer: (a)
Q.25
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite very small compared to the mass of the earth. Then
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a)the acceleration of S is don't change direction
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b) the angular momentum of S about the centre of the earth changes in direction, but magnitude remains constant
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c)the total mechanical energy energy of S varies periodically with time
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d)the linear momentum of S remains constant in magnitude
Explanation
Answer:(b)
Q.26
If the value of universal constant of gravitation is to decrease uniformly with time, then the satellite in orbit would still maintain its
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a) tangential speed
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b) radius of orbit
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c) period of revolution
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d) angular momentum
Explanation
Since no external force acts on satellite its angular momentum will not change Answer: (d)
Q.27
The speed of earth's rotation about its axis is ω. Its speed increases to x times to make the effective acceleration due to gravity equal to zero at the equator. Then x is ..
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a)1.7
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b) 8.5
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c)17
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d)3.4
Explanation
ω=2π / 86400 rad/s=7.27 × 10-5 Let ω be angular speed such that g - ω'2 R=0 ∴ ω=√(g/R)=√( 10 / 6.4 × 106)=1.25 × 10-3 rad/sω' / ω=1.25 × 10-3 / 7.27 × 10-5=17Answer: (c)
Q.28
A spaceship approaches the moon, whose mass and radius are M and R, along a line which is almost tangential o the moon's surface. At the moment of maximum approach the brake rocket is fired for a short while and the spaceship is transformed into a satellite of the moon. The change in velocity of the spaceship is
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a)
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b)
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c)
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d)
Explanation
The velocity is changed from escape velocity to orbital velocity ΔV=Ve - VoAnswer: (a)
Q.29
A person will get more kg-wt at [ J and K CET 2004]
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a) poles
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b) at latitude of 60°
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c)equator
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d)satellite
Explanation
Answer:(d)
Q.30
A body is suspended from a spring balance kept in a satellite. the reading of the balance is W1 when the satellite goes in an orbit of radius R and reading of the balance is W2 when it goes in an orbit of radius 2R. Then
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a) W1=W2
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b) W1 < W2
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c) W1 > W2
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d) W1 ≠ W2
Explanation
Spring balance measures mass. Not gravitational force Answer: (a)
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