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Physics NEET MCQ
Quiz 8
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Q.1
If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is [ IIT 1983]
0%
a) ½ mg R
0%
b) 2 mg R
0%
c)mg R
0%
d)¼ mg R
Explanation
Formula for potential energy at earth surface=Object is raised to height R from surface thus final potential energy=Thus Change in P.E=U - U'=GMm/2R But g=GM/R2 Thus Change in P.E=mgR/2 Answer:(a)
Q.2
If the distance between the earth and the sun were half its present value, the number of days in a year would have been [ IIT 1996]
0%
a) 64.5
0%
b) 129
0%
c) 182.5
0%
d) 730
Explanation
According to Kepler's law Here T1=365 days , given R2=R1/2 Thus Answer: (b)
Q.3
An artificial satellite moving in a circular orbit around the Earth has a total ( kinetic + potential) Energy EIts potential energy is [ IIT 1997]
0%
a)- E0
0%
b) 1.5 E0
0%
c)2E0
0%
d)E0
Explanation
In case of circular motion of satellite around earth or electron moving around nucleus etc in circular orbit P.E=2×T.E and |K.E.|=|T.E.|Thus P.E=2×T.E.∴ P.E.=2E0Answer: (c)
Q.4
A go-stationary satellite orbits around the earth in a circular orbit of radius 36,000km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface ( Radius of earth=6,400 km) will approximately be [ IIT 2002]
0%
a) 0.5 hr
0%
b) 1 hr
0%
c)2 hr
0%
d)4 hr
Explanation
According to Kepler's law Here T1=24hr , R1=36,000 km and R2=6,400 kmOn substituting values in above equation we get T2=1.8 HrWe know that as height increases time period increases. Thus the time period of the spy satellite should be slightly greater than 1.8 hr Therefore 2hrAnswer: (c)
Q.5
A binary star system consists of two stars A and B which have time period TA and TB, radius RA And RB and mass MA and MB. Then [ IIT 2006]
0%
a) if TA > TB and RA > RB
0%
b) if TA > TB and MA > MB
0%
c)TA=TB
0%
d)
Explanation
The gravitational force of attraction between stars provides necessary centripetal forces. In this case angular velocity of both stars is the same. Therefore time period remains same Answer:(c)
Q.6
A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a is correctly shown in [ IIT 2002]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
The component of acceleration are tangential and radial as shown in figure a=ar + at The resultant of transverse and radial component of the acceleration is represented by aAnswer: (c)
Q.7
The kinetic energy needed to project body of mass m from the earth surface ( radius R) to infinity is [AIEEE 2002]
0%
a) mgR / 2
0%
b) 2 mg R
0%
c)mg R
0%
d)mg R/4
Explanation
Kinetic energy E=½ m v2 e Where ve is escape velocity=√(2gR) ∴ E=½ m × 2gR=mgRAnswer: (c)
Q.8
If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will [ AIEEE 2002]
0%
a)continue to move in its orbit with same velocity
0%
b) move tangentially to the original orbit in the same velocity
0%
c) become stationary in its orbit
0%
d)move towards the earth
Explanation
Due to inertia of motion it will move tangentially to the original orbit in the same velocity.Answer: (b)
Q.9
Energy required to move a body of mass m from an orbit of radius 2R to 3R is [ AIEEE 2002]
0%
a) GMm/ 12R2
0%
b) GMm / 3R2
0%
c)GMm / 8R
0%
d)GMm / 6R
Explanation
Energy Required=Change in potential energy from Orbit of 2R to 3RFrom the formula of potential energy Answer:(d)
Q.10
The escape velocity of a body depends upon mass as [ AIEEE 2002]
0%
a) m0
0%
b) m1
0%
c) m2
0%
d) m3
Explanation
Formula for escape velocity is ve=√(2gR) as escape velocity is independent of mass option (a) is correct Answer: (a)
Q.11
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become [ AIEEE 2003]
0%
a)10 hrs
0%
b) 80 hrs
0%
c)40 hrs
0%
d)20 hrs
Explanation
According to Keller's lawHere T1=5 hrs , R2=5R1 Substituting values in above equation we get T2=40 hrsAnswer: (c)
Q.12
Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centers equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is [ AIEEE 2003]
0%
a) 2.5R
0%
b) 4.5R
0%
c)7.5R
0%
d)1.5R
Explanation
Force acting on both sphere is same but mass of big sphere(M2) is 5 times the mass of small sphere( M1)Thus acceleration of small sphere a1 is five times the acceleration of big sphere (a2) Thus a1=5a2Now displacement of small sphere S1=½ a1t2 Displacement of big sphere S2=½ a2t2 By taking ratio of displacement we get S1 / S2=a1 /a2 But a1=5 a2 ∴ S1 / S2=5Before collision spheres should travel 9R distance Thus S1 + S2=9R But S2=(1/5)S1Therefore S1 + (1/5) S1=9RS1=(45/6) R=7.5RAnswer: (c)
Q.13
The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of 45° with the vertical, the escape velocity will be [ AIEEE 2003]
0%
a) 11√2 km/s
0%
b) 22 km/s
0%
c)11 km/s
0%
d)(11/√2) km/s
Explanation
The escape velocity is independent of angle of projection Ave=√(2gR) Answer:(c)
Q.14
A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is [ AIEEE 2004]
0%
a) gR2 /(R+x)
0%
b) gar / (R-x)
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c) g
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d) [gR2 /(R+x)]1/2
Explanation
Gravitational force provides necessary centripetal force. ∴ we know that g=GM/R2 GM=gR2 replacing value of GM in above equation Answer: (d)
Q.15
The time period of an earth satellite in circular orbit is independent of [ AIEEE 2004]
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a)both the mass and radius of the orbit
0%
b) radius of orbit
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c)the mass of the satellite
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d)neither the mass of the satellite nor the radius of its orbit
Explanation
Answer: (c)
Q.16
If 'g' is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an masses ' m' raised from the surface of the earth to height equal to the radius 'R' around the earth is [ AIEEE 2004]
0%
a) ¼ mgR
0%
b) ½ mgR
0%
c)2mg R
0%
d)mg R
Explanation
At earth surface Potential energy is=-GMn / R At a distance R from the earth's surface, P.E. of system is -GMm/2R Thus change in potential energy ΔU=-GMm/ 2R - GMm/ R=GMm / 2R Now GM/R2=g ∴ GM/R=gR ∴ ΔU=½ mgRAnswer: (b)
Q.17
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of planet in circular orbit of radius 'R' around the sun will be proportional to [ AIEEE 2004]
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a) Rn
0%
b)
0%
c)
0%
d)
Explanation
We know that centripetal force is provided by gravitational force Answer:(c)
Q.18
The change in the value of 'g' at height 'h' above the surface of the earth is same as at a depth 'd' below the surface of earth. When both 'd' and 'h' are much smaller than the radius of earth, then which one of the following is correct [ AIEEE 2005]
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a) d=3h/2
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b) d=h/2
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c) d=h
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d) d=2h
Explanation
Variation of 'g' with altitude is Variation of g with depth is Equating we get d=2hAnswer: (d)
Q.19
Average density of the earth [ AIEEE 2005]
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a) is a complex function of g
0%
b) does not depend on g
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c)is inversely proportional to g
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d)is directly proportional to g
Explanation
g=GM/R2 Let ρ is average density Thus density ρ is directly proportional to gAnswer: (d)
Q.20
A planet in a distant solar system is 10 times more massive than earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km/s, the escape velocity from the surface of the planet would be [ AIEEE 2008]
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a) 1.1 km/s
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b) 11 km/s
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c)110km/s
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d)0.11 km/s
Explanation
Answer:(c)
Q.21
The height at which the acceleration due to gravity becomes g/9 ( where g=the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is [ AIEEE 2009]
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a) R / √2
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b) R/ 2
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c) √2 R
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d) 2R
Explanation
We know that Answer: (d)
Q.22
If the earth is at one fourth of its present distance from the sun, the duration of the year will be [ EAMCET 1987]
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a)half the present year
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b) one-eight the present year
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c)one-fourth the present year
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d)one sixth the present year
Explanation
According to Kepler's law T2 ∝ R3 R2=(1/4) R1 Answer: (b)
Q.23
A satellite is moving round the earth with velocity v. To make the satellite escape, the minimum percentage increase in its velocity is nearly [ MPPET 1994]
0%
a) 41.4%
0%
b) 82.8%
0%
c)100%
0%
d)none of the above
Explanation
We know that area velocity of satellite v=√(gr) and escape velocity ve=√(2gr) ∴ Increase in velocity=(√2 - 1)√(gr) Increase in velocity=0.414√(gr) ∴ Percentage increase in velocity=[0.414√(gr) / √(gr)] × 100=41.4%Answer: (a)
Q.24
A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energy is [ MNR 1986]
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a) positive
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b) negative
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c) zero
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d) may be positive or negative depending upon its initial velocity
Explanation
Negative energy shows that object is bounded Answer: (b)
Q.25
Consider a satellite going round the earth in a circular orbit. Which of the following statements is wrong?
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a)It is freely falling body
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b) It is acted upon by a force directed away from the center of the earth which counter-balances the gravitational pull
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c)It is moving with constant speed
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d)Its angular momentum remains constant
Explanation
Answer: (b)
Q.26
Imagine a light planet revolving around a very massive star in circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between planet and star is proportional t R-5/2, then T2 is proportional to ..[ IIT 1989]
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a) R3
0%
b) R7/2
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c)R5/2
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d)R3/2
Explanation
From given Answer: (b)
Q.27
A spherical planet far out in space has a mass Mo and diameter Do. A particle of mass m falling freely near the surface of this planet will experience acceleration due to gravity which is equal to [ MPPMT 1987]
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a) GMo / D20
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b) 4m GMo / D20
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c) 4GMo / D20
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d) GmMo / D20
Explanation
In gravitational formula radius is used while in problem Diameter is given R=D/2 substituting value of R in equation for acceleration we get option c. Answer:(c)
Q.28
Which of the following represents the Newton's law of universal gravitation correctly
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.29
The gravitational potential at any point distant r from a point mass m is work done in moving a unit mass from
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a)the point of interest to infinity
0%
b) from infinity to the point of interest
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c)from the position of point mass m to the point of interest
0%
d)from infinity to the position of point mass m
Explanation
Answer: (a)
Q.30
If earth ware to cease rotating about its own axis, the increase in the value of g is C.G.S system at equator is nearly
0%
a) 3.36 cm/s²
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b) 6.72 cm/s²
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c)1.68 cm/s²
0%
d)0.84 cm/s²
Explanation
Observed value of g g'=g - Rω2cos2λ , for equator cosλ=1 ∴ g'=g - Rω2If earth ceases rotation, then increase in the value of g at equator =Rω2 =6400 × 103 × ( 2π / 86400)2=3.36 cm/s2 Answer: (a)
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