Question

A sphere of mass 40kg is attracted by another sphere of mass 80kg. by a force of 2.5×10 raise to the power -6 N, when there centres are 30mm apart. Find value of G.

Answer 1

» GIVEN :-

$= > M _{1} = 40 \: Kg \\ \\ = > M _{2} = 80 \: Kg \\ \\ = > F = 2.5 \times {10}^{ - 6} \: N \\ \\ = > R = 30 \: \: mm = 3 \: cm$

___________

» WE KNOW THAT :-

$= > G = \frac{F \times {R}^{2} }{M _{1}.M_{2} } \\ \\ = > G = \frac{2.5 \times {10}^{ - 6} \times {3}^{2} }{40 \times 80} \\ \\ = > G = \frac{25 \times 9 \times {10}^{ - 6} }{32 \times 1000} \\ \\ = > G = \frac{225}{32} \times {10}^{ - 9} \\ \\ = > G = 7.03 \times {10}^{ - 9} \: \frac{ {N}^{} . {m}^{2} }{ {kg}^{2} }$
________________________[ANSWER]☺

====================================

_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

Answer:

$G=7.03\times 10^{-13}\ Nm^2/kg^2$

Explanation:

Mass of sphere 1, $m_1=40\ kg$

Mass of sphere 2, $m_2=80\ kg$

Force acting between two spheres, $F=2.5\times 10^{-6}\ N$

Distance between spheres, d = 30 mm = 0.03 m

The gravitational force acting between two spheres is given by :

$F=G\dfrac{m_1m_2}{d^2}$

$G=\dfrac{F.d^2}{m_1m_2}$

$G=\dfrac{2.5\times 10^{-6}\times (0.03)^2}{40\times 80}$

$G=7.03\times 10^{-13}\ Nm^2/kg^2$

Hence, this is the required solution.