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Q.1
What will be the formula of the mass in terms of g, R and G (R=radius of earth) [ CBSE-PMT 1996]
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a)g2(R/G)
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b) G (R2/g)
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c)G (R/g)
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d) g(R2/G)
Explanation
We know that g=GM / R2M=gR2 / G Answer: (d)
Q.2
The escape velocity from the surface of the earth is ve.The escape velocity from the surface of a planet whose mass and radius are three times those of the earth, will be [ CBSE-PMT 1995]
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a) ve
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b) 3ve
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c)9ve
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d)(1/3)ve
Explanation
Formula for escape velocity It is clear from formula if mass and radius is three times ,value of escape velocity from the surface of planet will be same as earth Answer: (a)
Q.3
The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps 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)(2/3)m
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b) (2/9) m
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c)18m
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d)6m
Explanation
On both the planet kinetic energy is same but height will be different According to law of conservation of energy ½ (m v2)=mgAhA=mgB hB ∴ hB=( gA / gB) hA hB=9 × 2=18m Answer:(c)
Q.4
A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest away at a distance of rIf v1 and v2 are the linear velocities at these points respectively, then the ratio is [2011]
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a) (r1/r2)2
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b) r2/r1
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c) (r2/r1)2
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d) r1/r2
Explanation
Angular momentum is conserved ∴ L1=L2 ∴ mr1v1=m r2v2 r1 v1=r2v2 ∴ v1 / v2=r2 / r1 Answer: (b)
Q.5
Two satellites of earth, S1 and S2 , are moving in the same orbit. The mass of S1 is four times the mass of S2 Which one of the following statements is true? [2007]
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a)The potential energies of earth satellites in the two cases are equal.
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b) S1 and S2, are moving with the same speed.
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c)The kinetic energies of the two satellites are equal.
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d)The time period of S1 is four times that of S2
Explanation
Since orbital velocity is independent of mass both will move with same speed.Answer: (b)
Q.6
The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is - [ CBSE-PMT 2006]
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a) 1/√2
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b) 1/3
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c)1/2
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d)√2
Explanation
Potential energy at height R=-GMm / 2R If ve is the velocity of mass 'm', so that it goes out of gravitational field from the distance 2R from center of earth½ (m ve2)=m × g ×R∴ ve=(gR) 1/2Now escape velocity if object in on surface of earth is v=(2gh) 1/2Thus v=ve × √2or ve=v / √2 ∴ f=1 / √ 2Answer: (a)
Q.7
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R. the radius of the planet would be [ CBSE-PMT 2004]
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a) R / 2
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b) 2R
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c)4R
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d)R/4
Explanation
g=GM / R2 Also Mass=D × (4/3)πR3∴ g=G(4/3) d π R At surface of planet gp=(4/3) G (2d) πR'At the surface of earth ge=(4/3) G (d) πR Now gp=ge∴ (4/3) G (2d) πR'=(4/3) G (d) πR∴ R'=R/2 Answer:(a)
Q.8
Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then [ CBSE-PMT 2005]
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a) g'=g/9
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b) g'=27g
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c) g'=9g
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d) g'=3g
Explanation
Density is same but size is three times bigger, hence mass of the planet will be three times the mass of earth g=GM / R 2 and M=(4/3) π R3 ρ ∴ g' / g=R'/R g'/g=3 g'=3g Answer: (d)
Q.9
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is [ CBSE-PMT 2005]
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a) 1/2
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b) 1/√2
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c)2
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d)√2
Explanation
Kinetic Energy of Satellite=½ (m v2 ) Now v=( GM/R)1/2 by substituting the value of v in equation of K.E. K.E=GMm/ 2R Potential energy=GMm / R ∴ K/U=1/2Answer: (a)
Q.10
The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun? [1997]
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a) 2
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b) 3
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c)4
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d)5
Explanation
Let TA and TB be time period of A and B about sunTA=8TBTA / TB=8According to Kepler's law T2 ∝ r3Answer: (c)
Q.11
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 ?[ CBSE-PMT 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
Periodic time T=πr / v0 Also v0=(gR2 / ω2)1/3 By substituting value of v0 in above equation we get Now Periodic time=2π/ω Answer:(a)
Q.12
The potential energy of a satellite, having mass m and rotating at a height of 6.4 χ 106 m from the earth surface, is [ CBSE-PMT 2001]
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a) -mgRe
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b) -0.67 mgRe
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c) -0.5mgRe
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d) -0.33 mgRe
Explanation
height of satellite is 6.4 χ 106 which is equal to radius of earth Re Now gravitational potential energy of the satellite at height Re U=-GMem / (Re + Re) But GMe=gRe2 ∴ U=gRe2m / 2R U=-gRem / 2=-0.5mgRe Answer: (c)
Q.13
A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively. M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is [2011M]
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a)
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b)
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c)
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d)
Explanation
Velocity 'u' should be equal to the escape velocity. That is u=(2gR)1/2But g=GM/R2∴ u=(2GM/R)1/2Answer: (a)
Q.14
Assuming earth to be a sphere of uniform density, what is the value of 'g' in a mine 100 km below the earth's surface? (Given, R=6400 km) [ CBSE-PMT 2001]
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a) 9.65 m/s2
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b) 7.65 m/s2
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c)5.06 m/s2
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d) 3.10 m/s2
Explanation
We know that effective gravity g' at depth below earth surface is given by Here d=100km, R=6400 kmAnswer: (a)
Q.15
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 [ CBSE-PMT 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)-(1/2)mv2 km/s
Explanation
Escape velocity does not depend on the angle of projection Answer:(b)
Q.16
The largest and the shortest distance of the earth from the sun are r1 and rIts distance from the sun when it is at perpendicular to the major-axis of the orbit drawn from the sun is [ CBSE-PMT 1988]
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a) (r1 + r2)/2
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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
Applying properties of ellipse, we have Answer: (c)
Q.17
The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCO is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then : [ CBSE-PMT2009]
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a) t1=4t2
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b) t1=2t2
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c)t1=t2
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d)t1 > t2
Explanation
According to Kepler's law, the aerial velocity of a planet around the sun always remains constantArea of SCD: A1 ∝ t1 ( aerial velocity constant)Area of SAB: A2 ∝ t2A1 / t1=A2 / t2t1=t2 ( A1 / A2)∴ t1=2 t2Answer: (b)
Q.18
If radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth's surface would [ IIT 1981]
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a) decrease
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b) remain unchanged
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c)increase
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d)be zero
Explanation
Radius decreases without changing mass thus g ∝ 1/R2 Thus g increases Answer: (c)
Q.19
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]
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a) ½ mg R
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b) 2 mg R
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c)mg R
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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.20
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]
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a) 64.5
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b) 129
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c) 182.5
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d) 730
Explanation
According to Kepler's law Here T1=365 days , given R2=R1/2 Thus Answer: (b)
Q.21
An artificial satellite moving in a circular orbit around the Earth has a total ( kinetic + potential) Energy EIts potential energy is [ IIT 1997]
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a)- E0
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b) 1.5 E0
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c)2E0
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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.22
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]
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a) 0.5 hr
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b) 1 hr
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c)2 hr
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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.23
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]
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a) if TA > TB and RA > RB
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b) if TA > TB and MA > MB
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c)TA=TB
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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.24
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]
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a)
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b)
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c)
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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.25
The kinetic energy needed to project body of mass m from the earth surface ( radius R) to infinity is [AIEEE 2002]
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a) mgR / 2
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b) 2 mg R
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c)mg R
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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.26
If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will [ AIEEE 2002]
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a)continue to move in its orbit with same velocity
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b) move tangentially to the original orbit in the same velocity
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c) become stationary in its orbit
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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.27
Energy required to move a body of mass m from an orbit of radius 2R to 3R is [ AIEEE 2002]
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a) GMm/ 12R2
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b) GMm / 3R2
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c)GMm / 8R
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d)GMm / 6R
Explanation
Energy Required=Change in potential energy from Orbit of 2R to 3RFrom the formula of potential energy Answer:(d)
Q.28
The escape velocity of a body depends upon mass as [ AIEEE 2002]
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a) m0
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b) m1
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c) m2
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d) m3
Explanation
Formula for escape velocity is ve=√(2gR) as escape velocity is independent of mass option (a) is correct Answer: (a)
Q.29
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]
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a)10 hrs
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b) 80 hrs
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c)40 hrs
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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.30
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]
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a) 2.5R
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b) 4.5R
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c)7.5R
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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)
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