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A sphere of mass of $40 kg$ is attracted by another sphere of mass of $15 kg$ when their centers are apart $320 cm$ with a force of $0.1 mg$ weight. What will be the value of the gravitational constant

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{1.3 \times 10}^{- 8} N m^{2} / {kg}^{2}

1.4 \times 10^{- 8} N m^{2} / {kg}^{2}

1.6 \times 10^{- 8} N m^{2} / {kg}^{2}

1.7

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

answer is D.

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Detailed Solution

The value of gravitational constant,

G = 1.7 \times 1 0^{- 9} N m^{2} / k g^{2}

.
Given,

m_{1} = 40 kg

m_{2} = 15 kg

Distance between them,

r = 320 cm

= 3.2 m

F = 0.1 mg

= 0.1 \times 1 0^{- 2} \times 98 dyne

[as

1 mg = 98 \times {10^{- 2}}^{} dyne

]

=

98 \times 1 0^{- 3} dyne

=

98 \times 1 0^{- 8} N

[as

1 dyne = 10^{- 5} Newton

]

=

9.8 \times 1 0^{- 9} N

Putting values in the universal gravitational equation

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

⇒

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

⇒

G = \frac{9.8 х 1 0^{- 7} \times 3.2 \times 3.2}{40 х 15}

⇒

G = 1.7 \times 1 0^{- 9} N m^{2} / k g^{2}

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