JEE Main 2020 — Gravitation Question with Solution
From: JEE Main 2020 (Online) 8th January Evening Slot
Question
An asteroid is moving directly towards the
centre of the earth. When at a distance of
10R (R is the radius of the earth) from the earths
centre, it has a speed of 12 km/s. Neglecting
the effect of earths atmosphere, what will be the
speed of the asteroid when it hits the surface
of the earth (escape velocity from the earth is
11.2 km/s) ? Give your answer to the nearest
integer in kilometer/s _____.
Enter your answer
Show full solutionCorrect answer: 16
Correct answer
16
Step-by-step explanation
U1 + K1 = U2 + K2
=
....(1)
Given escape velocity Ve = 11.2 km/s
= 11.2
So from (1)
= + 112.896
V2 = 16 km/s
=
....(1)
Given escape velocity Ve = 11.2 km/s
= 11.2
So from (1)
= + 112.896
V2 = 16 km/s
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This is a previous-year question from JEE Main 2020, covering the Gravitation chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.