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Physics NEET MCQ
Quiz 12
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Q.1
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.2
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere ( G=6.67 × 10⁻¹¹ Nm2 / kg2) [ AIEEE 2005]
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a) 3.33 × 10⁻¹⁰ J
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b) 13.34 × 10⁻¹⁰ J
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c)6.67× 10⁻¹⁰ J
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d)6.67× 10⁻⁹ J
Explanation
Work done=Change in potential energy W=-GMm/R By substituting values we get W=6.67× 10⁻¹⁰ J Answer: (c)
Q.3
The gravitational potential energy of a rocket of mass 100kg at a distance 109 m from earth surface is 4 × 107 joule. The weight of the rocket in Newton's at distance 109 m from earth is ( Re=6400 km)
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a) 8 ×10⁻² N
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b) 8 × 10⁻³ N
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c)4 × 10⁻³ N
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d)4 × 10⁻²N
Explanation
Potential energy ∴ weight of the rocket at 109 m from earth=4 × 10⁻⁴ × 100=4 × 10⁻²N Answer:(d)
Q.4
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.5
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.6
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]
0%
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.7
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere ( G=6.67 × 10⁻¹¹ Nm2 / kg2) [ AIEEE 2005]
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a) 3.33 × 10⁻¹⁰ J
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b) 13.34 × 10⁻¹⁰ J
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c)6.67× 10⁻¹⁰ J
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d)6.67× 10⁻⁹ J
Explanation
Work done=Change in potential energy W=-GMm/R By substituting values we get W=6.67× 10⁻¹⁰ J Answer: (c)
Q.8
The gravitational potential energy of a rocket of mass 100kg at a distance 109 m from earth surface is 4 × 107 joule. The weight of the rocket in Newton's at distance 109 m from earth is ( Re=6400 km)
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a) 8 ×10⁻² N
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b) 8 × 10⁻³ N
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c)4 × 10⁻³ N
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d)4 × 10⁻²N
Explanation
Potential energy ∴ weight of the rocket at 109 m from earth=4 × 10⁻⁴ × 100=4 × 10⁻²N Answer:(d)
Q.9
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.10
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)
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