JEE Main 2017 (Offline) | Gravitation Question 119 | Physics

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

The variation of acceleration due to gravity $$g$$ with distance d from centre of the earth is best represented by (R = Earth’s radius):

JEE Main 2017 (Offline) Physics - Gravitation Question 119 English Option 1

JEE Main 2017 (Offline) Physics - Gravitation Question 119 English Option 2

JEE Main 2017 (Offline) Physics - Gravitation Question 119 English Option 3

JEE Main 2017 (Offline) Physics - Gravitation Question 119 English Option 4

JEE Main 2016 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth’s surface. The radius of the earth is R, while its mass is M. The gravitational pull F_G on the astronaut is :

Zero since astronaut feels weightless

0 < F_G < $${{GMm} \over {{R^2}}}$$

$${{GMm} \over {{{\left( {R + h} \right)}^2}}}$$ < F_G < $${{GMm} \over {{R^2}}}$$

F_G = $${{GMm} \over {{{\left( {R + h} \right)}^2}}}$$

JEE Main 2016 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is $${1 \over 4}$$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t₁ is the time taken for planet to go over path abc and t₂ for path taken over cda then :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Gravitation Question 113 English

JEE Main 2016 (Online) 9th April Morning Slot Physics - Gravitation Question 113 English

t₁ = t₂

t₁ = 2t₂

t₁ = 3t₂

t₁ = 4t₂

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)

-1

A satellite is revolving in a circular orbit at a height $$'h'$$ from the earth's surface (radius of earth $$R;h < < R$$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.)

$$\sqrt{2 g R}$$

$$\sqrt{g R}$$

$$\sqrt{g R / 2}$$

$$\sqrt{g R}(\sqrt{2}-1)$$

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