Question:
5. Infinite number of masses, each 1 kg, are placed along the x-axis at x=±1m, ±2m, ±4m, ±8m, ±16m ….. The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0) is:
- a) G/2
- b) G
- c) 2G
- d) 4G
Question:
6. In the above problem, the ratio of the time duration of his jump on the moon to that of his jump on the earth is:
- a) 1 : 6
- b) 6 : 1
- c) √6 : 1
- d) 1 : √6
Answer:
6. (b)
ge / gm = (Re ρe) / (Rm ρm) = (2 / 3) × (4 / 1) = 6 or gm = ge / 6
For motion on earth, using the relation,
s = ut + (1 / 2) at2
We have, (1 / 2) = 0 + (1 / 2) × 9.8r2 or t = 1 / √9.8 s
For motion on the moon, 3 = 0 + (1 / 2) (9.8 / 6) t12
or t1 = 6√9.8 s ∴ t1 / t = 6 or t1 = 6t
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Kepler discovered:
Question:
9. Kepler discovered:
- a) Laws of motion
- b) Laws of rotational motion
- c) Laws of planetary motion
- d) Laws of curvilinear motion
Answer:
9. Kepler discovered:
a) Laws of motion
b) Laws of rotational motion
c) Laws of planetary motion
d) Laws of curvilinear motion
Answer: c) Laws of planetary motion
Explanation: Johannes Kepler is best known for his three laws of planetary motion, which describe the orbits of planets around the Sun. These laws were revolutionary in the field of astronomy, and they helped provide a clearer understanding of the motion of celestial bodies.
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What is the escape velocity for a planet with 6 times the mass and 2 times the radius of Earth?
Question:
8. The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is:
- a) √3 Ve
- b) 3 Ve
- c) √2 Ve
- d) 2 Ve
Answer:
8. (a) vp / ve = √(Mp / Me × Re / Rp) = √(6 × 1 / 2) = √3 ∴ vp = √3 ve
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Infinite number of masses, each 1 kg, are placed along the x-axis at x=±1m, ±2m, ±4m, ±8m, ±16m ….. The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0) is:
Question:
5. Infinite number of masses, each 1 kg, are placed along the x-axis at x=±1m, ±2m, ±4m, ±8m, ±16m ….. The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x=0) is:
- a) G/2
- b) G
- c) 2G
- d) 4G
Answer:
5. (c) Gravitational potential:
V = GM(1 / r1 + 1 / r2 + 1 / r3 + …)
V = G × 1(1 / 1 + 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + …)
V = G(1 / (1 – 1 / 2)) (∴ sum of GP = a / (1 – r))
V = 2G
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The escape velocity from the earth is 11 km/s. The escape velocity from a planet having twice the radius and the same mean density as that of earth is:
Question:
7. The escape velocity from the earth is 11 km/s. The escape velocity from a planet having twice the radius and the same mean density as that of earth is:
- a) 5.5 km/s
- b) 11 km/s
- c) 22 km/s
- d) None of these
Answer:
7. (c) Escape velocity,
vescape = √((2 GM) / R)
vescape = R√(8 / 3 πGρ)
∴ ve ∝ R if ρ = constant.
Since the planet has double the radius compared to Earth, the escape velocity becomes twice, i.e., 22 km/s.
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A body is taken to a height of nR from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is:
Question:
4. A body is taken to a height of nR from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is:
- a) (n+1)2
- b) (n+1)-2
- c) (n+1)-1
- d) (n+1)
Answer:
4. (a) Acceleration due to gravity at a height above the Earth’s surface:
g’ = g (R / (R + h))2
g / g’ = ((R + h) / R)2
g / g’ = ((R + nR) / R)2
g / g’ = (1 + n)2
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In the solar system, which is conserved:
Question:
10. In the solar system, which is conserved:
- a) Total Energy
- b) K.E.
- c) Angular Velocity
- d) Linear Momentum
Answer:
10. In the solar system, which is conserved:
a) Total Energy
b) K.E.
c) Angular Velocity
d) Linear Momentum
Answer: a) Total Energy
Explanation: In the solar system, total energy (the sum of kinetic and potential energy) is conserved in the absence of external forces. According to the law of conservation of energy, the total energy of a closed system remains constant over time. While individual components (kinetic and potential energies) can change, the total energy remains constant unless affected by an external force.
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Let “g” be the acceleration due to gravity at Earth’s surface, and “K” be the Earth’s rotational kinetic energy. If Earth’s radius decreases by 2%, keeping all other factors the same, what happens?
Question:
3. Let g be the acceleration due to gravity at earth’s surface and K be the rotational kinetic energy of the earth. Suppose the earth’s radius decreases by 2% keeping all other quantities the same, then:
- a) g decreases by 2% and K decreases by 4%
- b) g decreases by 4% and K increases by 2%
- c) g increases by 4% and K increases by 4%
- d) g decreases by 4% and K increases by 4%
Answer:
3. (c) g = GM / R2 and K = L2 / 2I
If mass of the earth and its angular momentum remains constant, then:
g ∝ 1 / R2 and K ∝ 1 / R2
i.e., if the radius of earth decreases by 2%, then g and K both increase by 4%
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Average density of the earth:
Question:
2. Average density of the Earth:
- a) does not depend on g
- b) is a complex function of g
- c) is directly proportional to g
- d) is inversely proportional to g
Answer:
2. (b) Acceleration due to gravity g = GM / R2, M = (4/3)πR3ρ
∴ g = (4G / 3) × (πR3 / R2) ρ
⟹ g = (4GπR / 3) ρ (ρ = average density)
⟹ g ∝ ρ or ρ ∝ g
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Halley’s comet takes 76 years to complete an orbit, and its closest approach to the sun is 8.9 × 10¹⁰ meters. Given the mass of the sun is 2 × 10³⁰ kg and the gravitational constant G = 6.67 × 10⁻¹¹ in MKS units, what is the comet’s farthest distance from the sun?
Question:
1. Halley’s comet has a period of 76 years, and its distance of closest approach to the sun is equal to 8.9×1010 m. The comet’s farthest distance from the sun, given the mass of the sun is 2×1030 kg and G = 6.67×10-11 in MKS units, is:
- a) 2×1012 m
- b) 2.7×1013 m
- c) 5.3×1012 m
- d) 5.3×1013 m
Answer:
1. (c) It is self-evident that the orbit of the comet is elliptic with the Sun being at one of the foci. Now, as for elliptic orbits, according to Kepler’s third law:
T2 = (4π2 a3) / GM
⇒ a = ((T2 GM) / (4π2 ))1/3
a = [((76 × 3.14 × 107) × 6.67 × 10-11 × 2 × 1010) / (4π2)]1/3
But in the case of an ellipse, 2a = rmin + rmax
∴ rmax = 2a – rmin = 2 × 2.7 × 1012 – 8.9 × 1010
≅ 5.3 × 1012 m
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