AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Force between two masses of $1 kg$ each, are separated by a distance $1 meter$ will be:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

6.67 \times 10^{- 11} N

9.8 N

6.67 \times 10^{- 8} N

6.67 \times 10^{- 12} N

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Force between two masses of

1 kg

each, are separated by a distance

1 meter

will be

6.67 \times 10^{- 11} N

.
Given,
Mass of two bodies is equal.
So,

m_{1} = m_{2} = 1 kg

Distance between them,

r = 1 meter

And gravitational constant,

G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

“Universal law of gravitation states that the force of attraction between two bodies of mass

m_{1}

and

m_{2}

which are kept at a distance

d

from each other is directly proportional to the product of masses and it is inversely proportional to the square of the distance in which they are kept.”
i.e.,

F α m_{1} m_{2}

and

F α \frac{1}{d^{2}}

or, combining these, we may write,

F = G \frac{m_{1} m_{2}}{d^{2}}

Putting the values in the above equation, we get
So,

F = 6.67 \times 10^{- 11} \frac{1 \times 1}{1^{2}}

\Rightarrow

F = 6.67 \times 10^{- 11} N

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!