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A satellite is moving with a constant speed v in a circular orbit about the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

1) (1/2) mv2 2) mv2 3) (3/2) mv2 4) 2mv2 Answer: 2) mv2 Solution: Escape velocity ve = √2 x orbital speed... View Article

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and R in a circular orbit at an altitude of 2R?

1) 5GmM/6R 2) 2GmM/3R 3) GmM/2R 4) GmM/3R Answer: 1) 5GmM/6R Solution: From the conservation of energy Total energy at the planet =... View Article

The total energy of an artificial satellite of mass m revolving in a circular orbit around the earth with a speed v is

1) (1/2)mv2 2) (1/4) mv2 3) (-1/4) mv2 4) -mv2 5) (-1/2) mv2 Answer: 5) (-1/2) mv2 Solution: Total energy of the satellite is \(\begin{array}{l}E = -\frac{1}{2}\frac{GM_{e}m}{R_{e}}\end{array} \)... View Article

The value of escape velocity on a certain planet is 2 kms ^-1. Then the value of orbital speed for a satellite orbiting close to its surface is

1) 12 kms -1 2) 1 kms -1 3) √2 kms -1 4) 2√2 kms -1 Answer: 3) √2 kms -1 Solution: ... View Article

If an object of mass m is taken from the surface of earth (radius R) to a height 2R, then the work done is

1) 2mgR 2) mgR 3) (2/3)mgR 4) (3/2)mgR 5) (1/3)mgR Answer: 3) (2/3)mgR Solution: Work done, W = ΔU = \(\begin{array}{l} \frac{mgh}{1+\left (... View Article

The gravitational potential energy of a body of mass m at a distance r from the centre of the earth is U . What is the weight of the body at this distance?

1) U 2) Ur 3) U/r 4) U/2r Answer: 3) U/r Solution: The gravitational potential energy, U = GMm/r U/r= GMm/r2... View Article

The weight of a body on the surface of the earth is 12.6 N. When it is raised to a height half the radius of the earth, its weight will be

1) 2.8 N 2) 5.6 N 3) 12.5 N 4) 25.2 N Answer: 2) 5.6 N Solution: Given: W0 = 12.6... View Article

The density of the earth in terms of acceleration due to gravity (g), the radius of the earth (R) and universal gravitational constant (G) is

1) 4πRG/3g 2) 3πRG/3g 3) 4g/3πRG 4) 3g/4πRG Answer: g = GM/R2 Density, ρ = Mass (M)/Volume(V) Mass (M) =... View Article

A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, R being the radius of the earth. What will be the time period of another satellite at a height 2.5R from the surface of the earth? (1) 6√2 h (2) 6√2.5 h (3) 6√3 h (4) 12 h

Answer: (1) 6√2 h For geostationary satellite: r1 = R + 6R = 7R For the second satellite r2 =... View Article

A planet of mass m moves around the sun of mass M in an elliptical orbit. The maximum and minimum distances of the planet from the sun are r₁ and r₂ respectively. The time period of the planet is proportional to

1) ( r1 + r2 ) 2) ( r1 + r2 )1/2 3) ( r1 – r2 )3/2 4) (... View Article

A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them. to take the particle far away from the sphere. (take, G = 6.67 × 10^-11 Nm²-kg ^-2)

1) 13.34 × 10-10 J 2) 3.33 × 10-10 J 3) 6.67 × 10-9 J 4) 6.67 × 10-10 J... View Article

Infinite number of masses each 1 kg are placed along the X–axis at x = ± lm, ± 2m, ± 4m, ± 8m, ± 16 m…. The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin (x = 0) is

1) G/2 2) G 3) 2G 4) 4G 5) 8G Answer: 4) 4G Solution: We know that the gravitational potential of... View Article

If g is the acceleration due to gravity on the earth’s surface, the gain of 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

1) 2 mgR 2) mgR 3) (1/2) mgR 4) (1/4) mgR Answer: 3) (1/2) mgR Solution: The gain in potential energy ΔU =... View Article

A body is released from a point distance r from the centre of earth. If R is the radius of the earth and r > R, then the velocity of the body at the time of striking the earth will be

1) √gR 2) √2gR 3) \(\begin{array}{l}\sqrt{\frac{2gR}{r-1}}\end{array} \) 4) \(\begin{array}{l}\sqrt{\frac{2gR(r-R)}r}\end{array} \) Answer: 4) \(\begin{array}{l}\sqrt{\frac{2gR(r-R)}r}\end{array} \) Solution: Total initial energy = Total final energy \(\begin{array}{l}\frac{-GMm}{r}=\frac{1}{2}mv^{2}-\frac{GMm}{R}\end{array} \) \(\begin{array}{l}\frac{v^{2}}{2}=\frac{GM}{R}-\frac{GM}{r} =... View Article

If the density of the earth is doubled keeping radius constant, find the new acceleration due to gravity ? (take, g = 9.8 m/s² )

1) 9.8 m/s2 2) 19.6 m/s2 3) 4.9 m/s2 4) 39.2 m/s2 Answer: 2) 19.6 m/s2 Solution: \(\begin{array}{l}g = \frac{GM}{R^{2}}=\frac{G}{R^{2}}\frac{4}{3}\pi R^{3}\times... View Article

A solid sphere of mass M and radius R has a spherical cavity of radius R/2 such that the centre of the cavity is at a distance of R / 2 from the centre of the sphere. A point mass m is placed inside the cavity at a distance of R / 4 from the centre of the sphere. The gravitational pull between the sphere and the point mass m is

1)\(\begin{array}{l}\frac{11GMm}{R^{2}}\end{array} \) 2)\(\begin{array}{l}\frac{14 GMm}{R^{2}}\end{array} \) 3)\(\begin{array}{l}\frac{GMm}{2R^{2}}\end{array} \) 4)\(\begin{array}{l}\frac{GMm}{R^{2}}\end{array} \) Answer:\(\begin{array}{l}\frac{GMm}{2R^{2}}\end{array} \) Solution: Field strength is uniform inside the cavity. Let... View Article

Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R^-5/2

1) T2 is proportional to R3 2) T2 is proportional to R7/2 3) T2 is proportional to R3/2 4) T2 is proportional to... View Article

If M is the mass of the earth and R its radius, the ratio of the gravitational acceleration and the gravitational constant is

1) R2/M 2) M/R2 3) MR2 4) M/R Answer: 2) M/R2 Solution: Gravitational acceleration, g = GM/R2 ⇒g/G = M/R2

if-rolles-theorem-holds-for-the-function-f-x-x3-ax2-bx-4-x-lies-in-1-2-with-f-4-3-0-then-ordered-pair-a-b-is-equal-to

a. (–5, 8) b. (5, 8) c. (5, –8) d. (–5, –8) Answer: (b) f(1) = f(2) ⇒ 1 –... View Article

What was the capital of the Rashtrakutas? Get the Answer at BYJU’S IAS Preparation

The capital of the Rashtrakutas was Manyakheta, modern Malkhaid. The city was founded in the 9th century by the Rashtrakuta... View Article

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