27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Mechanics
11. Gravitation
1. The escape velocity of a planet, double the size of the earth is (is p is same).
[BP 2014]
1 1.2 km/s
22. 4 km/s
44. 8 km/s
5.6 km/s
(c) Escape velocity ve = √(2GM/R) = R√(8πGρ/3). If radius is doubled and density is same, escape velocity doubles. Earth's escape velocity is 11.2 km/s. So, 11.2 * 2 = 22.4 km/s. Option b.
2. A satellite is orbiting a planet of radius 'R' and mass 'M' with a velocity 'V. If the radius of the planet is double keeping the density unchanged the final velocity of satellite, he would be?
[BP 2014]
V/2
V
2V
ISV
(b) Orbital velocity V = √(GM/r). Mass M = (4/3)πR³ρ. V = √(G(4/3)πR³ρ / r). If radius of planet R' = 2R, density remains same, then mass M' = (4/3)π(2R)³ρ = 8M. If orbital radius remains same r, then V' = √(G(8M)/r) = √8 V.
3. When earth doesn't rotate about its axis then what effect can be seen in its gravity? (Expect at poles) [BP 2014, 20131
[BP 2014, BP 2013]
Will decrease
Will increase
Remain same
None.
(b) The effective acceleration due to gravity at a point on the equator of the earth is given by g' = g - ω²R, where g is the acceleration due to gravity if the earth were not rotating, ω is the angular velocity of rotation, and R is the radius of the earth. If the earth stops rotating (ω = 0), then g' = g. Thus, the effective gravity will increase at all places except the poles.
4. Orbital velocity of satellite depends upon [BP 2013]
[BP 2013]
Mass of earth only.
Radius of earth only.
Mass of earth and radius of orbit.
Neither mass of nor radius of orbit.
(c) Orbital velocity vo = √(GM/r), where M is the mass of the Earth and r is the radius of the orbit.
5. What is true for a satellite orbiting around the earth [BP 2013]
[BP 2013]
Speed in maximum when it is near the earth because of constant angular momentum.
Sweeps out equal mass in equal interval as energy is constant.
Sweeps out maximum when it is near because angular momentum is variable.
Constant linear momentum.
(a) Angular momentum L = mvr = constant. So, v ∝ 1/r. Speed is maximum when it is near the earth (r is minimum).
6. Which is true? [BP 2013]
[BP 2013]
The velocity of earth is greater near the sun is on account of momentum.
The velocity of earth is greater near the sun is on account of repulsion.
The velocity of earth is less the sun.
The velocity of earth is same everywhere around the sun.
(a) According to Kepler's second law (related to conservation of angular momentum), a planet moves faster when it is closer to the star (Sun) and slower when it is farther away.
7. A plane is moving with velocity 200 m/s and an observer is just below it. If a shell is fired with 400 m/s by observer. What is minimum height for plane so would escape? [BP 2012]
[BP 2012]
6 km
8 km
4 km
3 km
() The question is incomplete or unclear about what needs to escape and how the shell is fired relative to the plane. Assuming it refers to the minimum height of the plane so that the shell fired vertically downwards by the observer below does not hit it.
8. The earth satellite is 4 times higher than the communication satellite from surface of the earth. Then time period of earth [BP 2012] satellite is:
[BP 2012]
192 hrs
372 hrs
384 hrs
(d) Time period T ∝ r3/2. Let radius of communication satellite orbit be R + h, and earth satellite be R + 4h. Ratio of radii ≈ 5h/h = 5 (approximately). Ratio of time periods = (5)3/2 = 11.18. Time period of communication satellite = 24 hrs. Time period of earth satellite = 24 * 11.18 ≈ 268 hrs. None of the options are close. There might be error in question. If radius of orbit is 4 times, T ∝ r3/2 => (4)3/2 = 8 times. 24 * 8 = 192 hrs. Option a.
9. The cause of day and night is due to [BP 2011]
[BP 2011]
Rotation of earth around its axis.
Revolution of earth around the sun.
Effect of gravity of moon on earth.
Effect of gravity of moon on earth.
(a) Day and night are caused by the Earth's rotation on its axis.
10. Work done in raising a body of mass 'm' from surface of earth height equal to radius of the earth is given by [BP 2011]
[BP 2011]
2 mgR
mgR/2
mgR
mgR
(b) Work done = change in potential energy = Uf - Ui = -GMm/(R+h) - (-GMm/R). Here h = R. Work done = -GMm/2R + GMm/R = GMm/2R = (gR²)m/2R = mgR/2.
11. The time period of communication satellite [BP 2011]
[BP 2011]
12 hours
24 hours
6 hours.
(b) The time period of a communication satellite orbiting the Earth is approximately 24 hours, which allows it to stay above the same point on the Earth's equator.
12. Two bodies of mass is 20 kg and 30 kg a 30 m apart. Then the gravitational for between the bodies is: IMOE 201
13. Earth's escape velocity for a satellite launched vertically is 11km/s. If the satellite is launched at 60 with verticle, th escape velocity would be.
1 1km/s
1 1√/3 km/s
- km/s
11.2km/s
(a) Escape velocity is independent of the angle of projection. It only depends on the mass and radius of the planet from which the object is escaping.
14. An earth satellite revolves round a circular orbit at a height 300 km above the earth surface. If the radius of earth is 6400 k the velocity of the satellite will be nearly MOE 2011
15. The mass of the earth is 80 times that o the moon and their diameters are 1280 km and 3200 km respectively. The value of ion due to gravity on earth is 9.8 m/s', the value on the moon will be: [MOE 2011
[MOE 2011]
16.6 m/s
9.8 m/s
1.66 m/s'
0
(c) g = GM/R². Let mass of moon be Mm and radius Rm. Mass of earth Me = 80Mm. Diameter of earth De = 12800 km, Re = 6400 km. Diameter of moon Dm = 3200 km, Rm = 1600 km = Re/4. ge = GMe/Re² = 9.8. gm = GMm/Rm² = G(Me/80) / (Re/4)² = GMe/80 / Re²/16 = 16/80 * GMe/Re² = 1/5 * ge = 9.8/5 = 1.96 m/s². Closest option is 1.66 m/s².
16. The escape velocity of moon of mass 7.2 10" kg and radius 1.7 x 10' m
1.2 km/s
2.4 km/s
10 km/
. 11.62 km/s .
(a) Escape velocity ve = √(2GM/R) = √(2 * 6.67 × 10⁻¹¹ * 7.2 × 10²² / 1.7 × 10⁶) = √(57.6 × 10⁵) ≈ 2.4 × 10³ m/s = 2.4 km/s. Option a is closest.
17. If 'A' is the areal velocity of the planet of mass 'M' then angular momentum is: [TE 2010]
[TE 2010]
M
AM
AM
2AM
(d) Areal velocity dA/dt = L / 2M. So Angular momentum L = 2MA.
18. Two solid spheres of radii 'r' and '2r made of same material are kept in contact mutual gravitational force of attraction between them is proportional to [TE 2010]
[TE 2010]
() Gravitational force F = G(m₁m₂)/r². Mass m = density * volume ∝ r³. So m₁ ∝ r³ and m₂ ∝ (2r)³ = 8r³. Distance between centers = r + 2r = 3r. F ∝ (r³ * 8r³) / (3r)² = 8r⁶ / 9r² = 8/9 r⁴. So F ∝ r⁴.
19. The escape velocity of the planet is V. It the mass of the planet is increased four times keeping the radius same. The escape velocity becomes: [1.E 2011)
[1.E 2011)]
4V.
2V.
2V/2 V.
V2V.
(b) Escape velocity Ve = √(2GM/R). If mass is increased four times (M' = 4M) and radius remains same, then new escape velocity Ve' = √(2G(4M)/R) = √4 √(2GM/R) = 2Ve = 2V.
20. The escape velocity from earth surface is 11 km/sec then what is the escape velocity from another planet of double radius having mean density same [TOM 2012)
[TOM 2012)]
5.5 km/sec
11 km/sec
22 km/sec
15.5 km/sec.
(c) Escape velocity ve ∝ R√ρ. If radius is doubled and density is same, escape velocity becomes 2 times. 2 * 11 = 22 km/sec.
21. X of 10 kg, Y of 20 kg and acceleration due to gravity on the surface of moon(gm) is 1.6 m/s then gravitation field intensity of: [KU 2014]
[KU 2014]
X and Y on the surface of earth is same.
X on the surface of moon is equal to that of Y on the surface of earth
X on the surface of moon is twice that of Y on the surface of moon.
All
(a) Gravitational field intensity g = GM/R², which is acceleration due to gravity and is independent of the mass of the object. So, gravitational field intensity of X and Y on the surface of earth is the same (approximately 9.8 m/s²).
22. The planet of mass and diameter is three times than the earth. Then find the acceleration due to gravity on planet is: [KU 2013, 2011]
25. What will be the time period of satellite moving around the earth if the radius of revolution is increases by 1 and half time
[MOE 2008]
b.
() Time period T ∝ r3/2. If radius increases by 1 and half time, new radius r' = r + 1.5r = 2.5r = 5/2 r. T' / T = (r'/r)3/2 = (5/2)3/2 = (5√5) / (2√2) = (5√10) / 4 ≈ 3.95.
26. A satellite is moving round the earth in a circular orbit of radius 8000Km with a velocity of 800m/s. The acceleration due to gravity on the satellite is nearly.[MOE 2065
[MOE 2065]
.0.04ms
0.06ms?
0.08ms
0.01ms
(c) Orbital velocity v = √(GM/r). Acceleration due to gravity a = GM/r². a = v²/r = (800)² / (8 × 10⁶) = 64 × 10⁴ / 8 × 10⁶ = 8 × 10⁻² = 0.08 m/s².
27. Two planets have radii r, and r, and densities d, and d, respectively. The ratio of th acceleration due to gravity on them will be [MOE 2010]
[MOE 2010]
ridz: radi
rid: ryd
rid,: 12 dz
rid, : rad,
(a) g = GM/R² = G(4/3)πR³ρ / R² = (4/3)πGRρ. g₁/g₂ = (R₁ρ₁) / (R₂ρ₂). Option a has d as z and 1.
28. The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v. For a satellite orbiting at an altitude of half the earth's radius, the orbital velocity is [MOE 2010]
[MOE 2010]
(3/2) v
V(3/2)
(2/3) V
V(2/3)
(d) Orbital velocity v = √(GM/r) ∝ 1/√r. v' / v = √(R / (R + R/2)) = √(R / (3R/2)) = √(2/3). v' = v√(2/3). Option d.
29. g is the acceleration due to gravity at the equator. Its value at the pole is [MOE 2000]
[MOE 2000]
less than that at equator
greater than
lesser than g
none of the above.
(b) The acceleration due to gravity is greater at the poles than at the equator due to the Earth's shape (oblate spheroid) and rotational effects.
30. Let G be the gravitational constant and R be the radius of the earth. If the radius of the earth becomes half, the new value of the gravitational constant will be [MOE]
[MOE]
G/2
2G
G
(c) The gravitational constant G is a universal constant and its value does not change with the change in the Earth's radius or any other physical parameters.
31. An astronaut feels weightlessness when [TE-02
[TE-02]
it is leaving earth
it is provided with escape velocity
in any orbit
in certain orbit
(c) An astronaut feels weightlessness in orbit because both the spacecraft and the astronaut are in continuous free fall around the Earth.
32. The satellite of mass M and 9M are orbiting a planet in a circular orbit or radius R. Their time period of revolution will be in the ratio of [TE-08]
[TE-08]
9:1
3:1
1:1
1:3
(c) The time period of a satellite orbiting a planet depends on the mass of the planet and the radius of the orbit, but not on the mass of the satellite itself. Therefore, both satellites will have the same time period, and the ratio will be 1:1.
33. The earth radius is 'R', acceleration due to gravity at its surface is 'g'. If the body of mass 'M' falls from height h = R/5 from earth surface, its potential energy decreases by [IE-08]
[IE-08]
Mgh
Mgh
Mgh
- Mgh
(a) Decrease in potential energy = mgh / (1 + h/R) = Mg(R/5) / (1 + R/(5R)) = MgR/5 / (6/5) = MgR/6. Option a is Mgh which should be MgR/5.
34. A man has weight 80 kg on earth's surface. The height above ground where he will have weight 40 kg is, (radius of earth 'R'= 6400 km) [IE-08]
36. A satellite is revolving around a planet of radius 'r' with speed 'V.'. If the satellite is made to revolve in another planet of twice the mass than original and constant radius then ratio of initial to final velocity is [BPKIHS-06]
[BPKIHS-06]
1:2
(c) Orbital velocity V = √(GM/r). V₁/V₂ = √(M₁/M₂) * √(r₂/r₁) = √(M / 2M) * √(r/r) = 1/√2.
37. At what height above the earth's surface will the weight of a body be half as that on earth's surface?
3200 km
3050 km
2624 km
1600 km
(c) Weight at height h, W' = W / (1 + h/R)². 1/2 W = W / (1 + h/R)² => (1 + h/R)² = 2 => 1 + h/R = √2 => h/R = √2 - 1 ≈ 0.414. h = 0.414 * 6400 km ≈ 2649 km. Closest option is 2624 km.
38. The value of g will 1% of its value at the surface of earth at a height (R. = 6400 km)
6400 km
2560 km
64000 km
$7600 km
(c) g' = g / (1 + h/R)² = 0.01 g => (1 + h/R)² = 100 => 1 + h/R = 10 => h/R = 9 => h = 9R = 9 * 6400 = 57600 km.
39. If the value of 'g' is same at depth 'd' inside earth and height 'h' above carth then
d= h
d= 2h
d = h/2
d = 3h
(b) gdepth = g(1 - d/R). gheight = g / (1 + h/R)² ≈ g(1 - 2h/R) for h< d = 2h.
40. A mass 'M' is broken into two parts one of which has mass 'm'. To have maximum gravitational force of attraction between the broken masses
m = 2M
m = M/2
m = M/4
m = M/6
(b) Force F = G m (M-m) / r². For maximum force, dF/dm = 0 => G(M - 2m) / r² = 0 => m = M/2.
41. A body weighs
very slightly greater at night
very slightly less at night
exactly equal at day and night
zero at night
(b) The Earth's rotation causes centrifugal force, which is maximum at the equator and decreases towards the poles. This centrifugal force opposes gravity, making the effective weight slightly less. At night, the observer is farther from the Sun, and the combined gravitational effect of the Sun and Earth might slightly change the weight, but the primary difference between day and night is due to the Earth's rotation relative to the stars.
42. If a body weighs xN on the surface of earth, its weight half way down to the centre of the earth, assuming earth to be of uniform density is
2xN
2x N
x N
(a) Weight at depth d = W(1 - d/R). Here d = R/2. Weight' = W(1 - R/2R) = W(1 - 1/2) = W/2 = x/2 N. Option c has x N.
43. The gravitational force between two identical steel balls each of radius R touching their surfaces is F. Then gravitational force between two identical steel balls each of radius 2R touching their surfaces is
16F
64F
F
F/4
(a) F = G(m*m) / (2R)². Mass m ∝ R³. So m' ∝ (2R)³ = 8m. F' = G(8m * 8m) / (4R)² = 64Gm² / 16R² = 4 * G(m*m) / (2R)² * 4 = 16F.
44. The gravitational force between ty identical copper spheres each of radius R touch their surface is F. Then gravitational force between two copper spheres of radius R and 2R touching their surface is now
b.
32
SE
(a) F = G(m*m) / (2R)². F' = G(m * 8m) / (3R)² = 8Gm² / 9R² = 8/9 * G(m*m) / (2R)² * 4 = 32/9 F.
45. The gravitational force on a small particle lying on earth's surface is F. If a concentric spherical cavity of radius R/2 is made inside earth, then force on the particle is
E
F
17F 18
18
(c) Force on surface F = GMm/R². Force inside cavity at distance r from center F' = GM'm/r². Mass of cavity = (1/8)M. Effective mass M' = M - M/8 = 7M/8. If particle is on surface, r=R. F' = G(7M/8)m / R² = 7/8 F.
46. In the above question, if the cavity of radius ? is made touching the particle then the force on the particle will
E 8
17F 8
7F 8
(d) Force = GMm/R². Remove sphere touching the surface. Force due to removed sphere F' = G(M/8)m / (R/2)² = GMm / 2R². Net force = GMm/R² - GMm/2R² = GMm/2R² = F/2. Option a has F/8.
47. A small particle is held inside an isolated hollow sphere at distance x from centre of the sphere. The gravitational force on that particle by the sphere is [IOM]
[IOM]
directly proportional to x
inversely proportional to x
. inversely proportional to x
zero
(d) The gravitational force on a particle inside a uniform hollow sphere is zero.
48. The gravitational potential at surface of an isolated soap bubble is Vo. Thus potential at its centre will be
() Gravitational potential inside a spherical shell is constant and equal to the potential at the surface. So potential at centre will be V₀.
49. The gravitational potential at the surface of earth (or mercury drop) is V.. The potential at its centre is
V.
infinite
zero
(a) Gravitational potential at the surface of a solid sphere is Vs = -GM/R. Potential at the center is Vc = -3GM/2R = 3/2 Vs.
50. A hollow sphere shrinks maintaining its shape. Then gravitational potential at its centre
decreases
increases
remains unchanged
-increases and decrease alternatively
(b) Gravitational potential at the center of a hollow sphere V = -GM/R. As the sphere shrinks, R decreases, so the magnitude of potential increases, becoming more negative, thus potential decreases. Option Error in my thought. As sphere shrinks, mass remains same, R decreases. V = -GM/R. So V becomes more negative, so decreases. Magnitude increases. If potential increases, it becomes less negative. So option b.
51. $1. Two planets of equal density having radii R and 2R. Then ratio of acceleration due to gravity on their surface will be
1:2
2:1
(a) g = (4/3)πGρR. g₁/g₂ = R₁/R₂ = R / 2R = 1/2.
52. 2, If earth shrinks so that its radius decreases by 50%, then the value of acceleration due to gravity at its surface
increases by 300%
decreases by 300%
increases by 100%
decreases by 100%
(a) g = GM/R². If R' = R/2, then g' = G M / (R/2)² = 4GM/R² = 4g. Increase = g' - g = 3g = 300% increase.
53. 53. At what depth below earth surface, value of acceleration due to gravity becomes (If R = radius of the earth) [MOE]
56. 56. If g be the acceleration due to gravity at earth's surface and R be radius of earth then escape velocity of a particle at height R will be:
V2gR
VER
VER
VER
(a) Escape velocity ve = √(2GM/r) = √(2gR²/r). At height R, r = R + R = 2R. ve = √(2gR²/2R) = √(gR). Option b has VER.
57. 57. The ratio of escape velocity of a particle at height R to orbital velocity of a satellite in orbit close to earth is
V2:1
1:1
1:12
(b) Escape velocity at height R, ve = √(gR). Orbital velocity close to earth, vo = √(gR). Ratio = √2:1.
58. 58. Two planets of equal densities have radii R and 2R. Then ratio of escape velocities of particles from their surface is
1 : V2
a 1:1
V2:1
1:2
(b) Escape velocity ve ∝ R√ρ. Ratio v₁/v₂ = R₁/R₂ = R / 2R = 1/2. Option d.
59. 59. In a missile launched with velocity less than escape velocity, the sum of its KE and PE is always [IOM]
[IOM]
positive
zero
negative
arbitary
(c) If the initial velocity is less than the escape velocity, the total mechanical energy (KE + PE) is negative, indicating that the missile is bound to the gravitational field.
60. 60. Two satellites of masses m and 9m are orbiting a planet in a circular orbit of radius R. Their time periods of revolution will be in the ratio of
9:1
3:1
1:1
(c) The time period of a satellite orbiting a planet depends only on the mass of the planet and the radius of the orbit, not on the mass of the satellite. Therefore, the time periods of both satellites will be the same, and the ratio will be 1:1.
61. 61. A satellite with kinetic energy E is revolving round the earth in a circular orbit. The minimum additional K.E. required for it to escape into outer space is
2. V2 E
2E
E/2
(a) Orbital velocity vo = √(GM/r). KEo = 1/2 mvo² = GMm/2r = E. Escape velocity ve = √(2GM/r). KEe = 1/2 mve² = GMm/r = 2E. Additional KE required = KEe - KEo = 2E - E = E.
62. 62. By what percentage is escape speed greater than the speed of satellite close to earth surface
50%
41.4%
100%
75%
(a) Escape velocity ve = √(2gR). Orbital velocity vo = √(gR). ve = √2 vo = 1.414 vo. Percentage increase = (ve - vo) / vo * 100 = (1.414 vo - vo) / vo * 100 = 0.414 * 100 = 41.4%. Option b.
63. 63. The maximum height reached by a rocket fired with 90% of escape velocity from surface of earth is
90 R 19
ST
81 F 10
100
(d) KE + PE = 0 for escape. Initial KE = 1/2 m(0.9ve)² = 0.81 * 1/2 mve² = 0.81 * GMm/R. Initial PE = -GMm/R. Total energy = 0.81 GMm/R - GMm/R = -0.19 GMm/R. At max height h, velocity is zero. PE = -GMm/(R+h). -0.19 GMm/R = -GMm/(R+h) => 0.19 = R/(R+h) => 0.19R + 0.19h = R => 0.19h = 0.81R => h = 0.81/0.19 R = 81/19 R. Option a has 90/19 R.
64. 64. A body is projected vertically from th earth's surface with the K.E. equal to ha the minimum value needed for it to escape. The height through which it rises above the earth is
R/2
2R
R /4
(a) Escape KE = GMm/R. Given KE = GMm/2R. Total energy = KE + PE = GMm/2R - GMm/R = -GMm/2R. At height h, PE = -GMm/(R+h), KE = 0. -GMm/(R+h) = -GMm/2R => R+h = 2R => h = R.
65. 66. Mass of moon is 1/81 times that of earth and its radius is 1/4 times the earth's radius. If escape velocity on earth's surfac is 11.2 km/s, its value at the surface of the moon is
66. 67. Time taken by a radiowave to go and come back from a communication satellite to earth is nearly [BPKIHS]
[BPKIHS]
Is
(b) Distance to geostationary satellite ≈ 36000 km. Total distance = 72000 km. Speed of radiowave = speed of light = 3 × 10⁸ m/s = 3 × 10⁵ km/s. Time = Distance / Speed = 72000 / 3 × 10⁵ = 0.24 s.
67. A satellite A of mass 'm' is at a distance 'r' from earth's surface. Another satellite of mass 2m is at a distance 2r from earth's surface. Their time periods are in the ratio of
1:2
: 212
(d) Time period T ∝ (radius of orbit)3/2. For A, radius = R+r. For B, radius = R+2r. Ratio TA/TB = ((R+r) / (R+2r))3/2.
68. Earth's escape velocity for a satellite launched vertically is 11 km/s. If the satellite is launched at 60" with vertical, the escape velocity would be
11 km/s
1 1\/3 km/s
-km/s V3
11.2 km/s
(a) Escape velocity is independent of the angle of projection.
69. If the earth were to suddenly contract to half the present radius (without any external torque acting on it), what would be the time of rotation of earth?
6 hours
12 hour
18 hours
48 hours
(a) By conservation of angular momentum, Iω = constant. (2/5)MR²(2π/T) = constant. If R becomes R/2, then (2/5)M(R/2)²(2π/T') = (2/5)MR²(2π/T) => T' = T/4 = 24/4 = 6 hours.
70. A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is
R
R 2
R 3
R 4
(c) v = 1/2 ve = 1/2 √(2GM/R). Using conservation of energy, 1/2 mv² - GMm/R = 0 - GMm/(R+h). 1/2 m (1/4 * 2GM/R) - GMm/R = -GMm/(R+h) => GMm/4R - GMm/R = -GMm/(R+h) => 1/4 - 1 = -1/(R+h) => -3/4 = -1/(R+h) => R+h = 4/3 R => h = R/3.
71. Two particles of equal masses (each of mass M) go round a circle of radius R under the action of their mutual attraction. The speed of each particle is:
(a) g is acceleration due to gravity which is approximately constant near the Earth's surface.
73. `The acceleration due to gravity in the planet A is 9 times the acceleration due to gravity on earth. A man jumps to a height of 2m on the surface of planet What is the height of jump by the same person on earth?
74. Which one of the following statement is correct?
[KU 2015]
value of g is same at all places
the value of g is more at the equator than 6. at the poles
the value of g is more at the poles that at the equator.
the value of g is maximum at the center of the earth.
(c) The value of g is greater at the poles than at the equator due to the Earth's shape and rotation.
75. Escape velocity of a body projected from earth:
[KU 2016]
2Gm R
Gm R
Gm V2R
(a) Escape velocity ve = √(2GM/R).
76. Geostationary satellite of earth is:
[KU 2016]
Rotate around the polar axis
Rest in positional
it stays stationary in space
Its time period is less than that of the earth
(c) A geostationary satellite appears stationary from the Earth.
77. If a body is projected with velocity V less. than the escape velocity of earth, then the total energy of the projected body is:
[KU 2016]
+ve
-ve
zero
may be+ve or -ve
(b) Total energy is negative.
78. A 5kg mass with a string Im length moves in a vertical circle with the velocity of 4m/s. The net tension in the string is 130N. Then, the position of body is.
[KU 2016]
, at the top position
at the bottom position
at mid way
can't be predicted
(b) Tension is maximum at the bottom position.
79. At particular point, the acceleration due to gravity is always constant for same or different mass due to :
[KU 2016]
gravitation force is constant
inertial mass & gravitational mass constant
gravitational force is directly proportional to inertial mass
All
(c) g = GM/R², independent of the object's mass.
80. Star that appear stationary form the earth
[KU 2017]
Sirius
Pole star
Orion
(b) The Pole Star (North Star) appears stationary because it is located close to the Earth's axis of rotation in the Northern Hemisphere.
