Question

A sphere of mass 40 kg is attracted by another sphere of mass 80 kg, by a force of 2.5 x 10-6 N when their centres are 300 mm apart. Find the value of G​

Answer 1

Answer:

GIVEN :-

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

=>M

1

=40Kg

=>M

2

=80Kg

=>F=2.5×10

−6

N

=>R=30mm=3cm

___________

» WE KNOW THAT :-

\begin{lgathered}= > 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} }\end{lgathered}

=>G=

M

1

.M

2

F×R

2

=>G=

40×80

2.5×10

−6

×3

2

=>G=

32×1000

25×9×10

−6

=>G=

32

225

×10

−9

=>G=7.03×10

−9

kg

2

N

.m

2

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Answer 2

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