The force of attraction between two masses each of $1 k g$ kept at a separation of $1 m$ is:

9.8

N

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6.7

N

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980

N

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6.7 \times 10^{- 11}

N

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Solution

The correct option is D $6.7 \times 10^{- 11}$ $N$
The force of attraction between two masses $m_{1}$ and $m_{2}$ , kept apart from each other at a distance of $r$ is, $F = G \frac{m_{1} \times m_{2}}{r^{2}}$
Here, $m_{1} = m_{2} = 1 k g$ and $r = 1 m$
Hence, $F = G \frac{1 \times 1}{1^{2}} = G = 6.7 \times 10^{- 11} N$

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The force of attraction between two masses each of 1kg kept at a separation of 1m is:

The correct option is D 6.7×10−11 NThe force of attraction between two masses m1 and m2, kept apart from each other at a distance of r is, F=Gm1×m2r2Here, m1=m2=1kg and r=1mHence, F=G1×112=G=6.7×10−11N

The force of attraction between two masses each of $1 k g$ kept at a separation of $1 m$ is: