Two block of mass $10 k g$ and $20 k g$ is separated by a distance $100 k m$ . What is the gravitational field (F) if $G = 6.674 08 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ ?

13.34 \times 10^{- 16} N

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1.334 \times 10^{- 11} N

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1334 \times 10^{- 16} N

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1.334 \times 10^{- 18} N

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Solution

The correct option is C $1.334 \times 10^{- 18} N$
Given : $m_{1} = 10 k g$ $m_{2} = 20 k g$ $r = 10^{5} m$
Gravitational force between the blocks $F = \frac{G m_{1} m_{2}}{r^{2}}$
$∴$ $F = \frac{6.67408 \times 10^{- 11} \times 10 \times 20}{(10^{5})^{2}}$
$⟹$ $1.334 \times 10^{- 18} N$

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Two block of mass 10kg and 20kg is separated by a distance 100km. What is the gravitational field (F) if G=6.67408×10−11m3kg−1s−2?

The correct option is C 1.334×10−18NGiven : m1=10kg m2=20kg r=105mGravitational force between the blocks F=Gm1m2r2∴ F=6.67408×10−11×10×20(105)2⟹ 1.334×10−18N

Two block of mass $10 k g$ and $20 k g$ is separated by a distance $100 k m$ . What is the gravitational field (F) if $G = 6.674 08 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ ?

The correct option is C $1.334 \times 10^{- 18} N$
Given : $m_{1} = 10 k g$ $m_{2} = 20 k g$ $r = 10^{5} m$
Gravitational force between the blocks $F = \frac{G m_{1} m_{2}}{r^{2}}$
$∴$ $F = \frac{6.67408 \times 10^{- 11} \times 10 \times 20}{(10^{5})^{2}}$
$⟹$ $1.334 \times 10^{- 18} N$