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A sphere of mass $40 kg$ is attracted by another sphere of mass $15 kg$ , when their centers are $20 cm$ apart, with a force of $0.1 milligram weight$ . Find the value of $G$ .

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6.53 \times 10^{- 11} N m^{2} {kg}^{- 2}

6.53 \times 10^{- 9} N m^{2} {kg}^{- 2}

6.53 \times 10^{- 12} N m^{2} {kg}^{- 2}

6.53 \times 10^{- 12} N m^{2} {kg}^{- 2}

answer is A.

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Detailed Solution

The value of G is

6.53 \times 10^{- 11} N m^{2} {kg}^{- 2}

.
By the universal law of gravitation the attractive force between two bodies of masses

M

and

m

separated by a distance

r

is:

F = \frac{GMm}{r^{2}} . . . . . . . . . . . . . . \dots (1)

\Rightarrow G = \frac{F r^{2}}{Mm} . . . . . . . . . . . . . . . . . \dots (2)

Where

G

is the gravitational constant and

F

is the force of attraction due to the effect of gravitation.
Given:

Mass of the first object,

M = 40 kg

Mass of the second object,

m = 15 kg

Distance separating them,

r = 0.2 m

Acceleration due to gravity,

g = 9.8 m / s^{2}

F = 0.1 milligram weight

\Rightarrow F = 0.1 \times 10^{- 6} kg wt

Therefore,

\Rightarrow G = \frac{10^{- 7} \times 9.8 \times (0.2 \times 0.2)}{40 \times 15}

\Rightarrow G = 6.53 \times 10^{- 11} N m^{2} {kg}^{- 2}

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