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
132 views
in Physics by (86.7k points)
closed by
The force of attraction between two bodies of masses 100 kg and 1000 Kg separated by a distance of 10 m is
A. `6.67xx10^(-7)N`
B. `6.67xx10^(-8)N`
C. `6.67xx10^(-9)N`
D. `6.67xx10^(-10)`

1 Answer

0 votes
by (84.9k points)
selected by
 
Best answer
Correct Answer - b
`F=(GM_(1)M_(2))/(r^(2))`
`=(6.67xx10^(-11)xx100xx1000)/(10^(2))`
`=6.67xx10^(-8)N.`

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

...