Question

S.I. unit of gravitational constant is:
A. ${N^2}{m^2}k{g^2}$
B.$Nmk{g^2}$
C. ${N^2}mk{g^{ - 2}}$
D.$N{m^2}k{g^{ - 2}}$

Answer 1

Hint The gravitational constant G is the small quantity and its measurement needs very sensitive arrangement . The first important successful measurement of this quantity was made by Cavendish in 1736 about 71 years after the law was formulated.

Complete Step by step solution
As we all know that in the earth if two particles of masses ${m_1}$ and ${m_2}$are placed in a distance of $r$unit in the earth. There between these two particles a force of attraction works called gravitational force.
And which is equal to :
$gravitational{\text{ force (F) = }}\dfrac{{{\text{G}}{{\text{m}}_1}{{\text{m}}_2}}}{{{r^2}}}$, where each symbol have its usual meaning.
Now from above given formula we can write-
$G = \dfrac{{F{r^2}}}{{{m_1}{m_2}}}$
Unit of force is newton(N)
Unit of distance between masses will me meter(m)
Unit of mass will be kilogram (kg)
So from above formula
Unit of gravitational constant (G) will be $\dfrac{{N{m^2}}}{{k{g^2}}} = N{m^2}k{g^{ - 2}}$
Hence S.I. unit of gravitation constant is $N{m^2}k{g^{ - 2}}$

Therefore, answer number D will be correct option

Note The value of gravitation constant is very small.
For numerical purpose we use value of gravitational constant as $6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
Value of gravitation constant is very low in comparison with the electrical force constant
Therefore electric force is more dominant over the gravitational force.
Gravitational force is always attractive in nature whereas electric force is attractive or repulsive with respect to the charges.