Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
297 views
in Physics by (48.9k points)
closed by

What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is (3.1)2 m/s2 and whose radius is 8100 km?

(a) 2790 km/s

(b) 27.9 km/s

(c) \(\frac{27.9}{\sqrt 5}\) km/s

(d) 27.9√5 km/s

1 Answer

+1 vote
by (49.0k points)
selected by
 
Best answer

Correct option is (b) 27.9 km/s

Escape velocity, v = \(\sqrt {2gR}\)

\(= \sqrt{2 \times (3.1)^2 \times 8100 \times(10)^3}\)

\(= 27. 9 km/s\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...