JEE Main 2019 (Online) 10th April Evening Slot | Gravitation Question 98 | Physics

JEE Main 2019 (Online) 10th April Evening Slot

MCQ (Single Correct Answer)

-1

A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?

[Given ; Mass of planet = 8 × 10²² kg, Radius of planet = 2 × 10⁶ m, Gravitational constant G = 6.67 × 10^–11 Nm² /kg²]

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

The value of acceleration due to gravity at Earth's surface is 9.8 ms^–2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms^–2, is close to : (Radius of earth = 6.4 × 10⁶ m)

1.6 × 10⁶ m

9.0 × 10⁶ m

6.4 × 10⁶ m

2.6 × 10⁶ m

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

A test particle is moving in a circular orbit in the gravitational field produced by a mass density $$\rho (r) = {K \over {{r^2}}}$$ . Identify the correct relation between the radius R of the particle's orbit and its period T

T²/R³ is a constant

TR is a constant

T/R² is a constant

T/R is a constant

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

A solid sphere of mass 'M' and radius 'a' is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance '3a' from the centre will be :

$${{GM} \over {3{a^2}}}$$

$${{2GM} \over {9{a^2}}}$$

$${{GM} \over {9{a^2}}}$$

$${{2GM} \over {3{a^2}}}$$

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