If in a system the force of attraction between two point masses of `1kg` each situated `1km` apart is taken as a unit force and is called notwen (newton written in reverse order) If `G = 6.67 xx 10^(-11) N-m^(2)kg^(-2)` in `\SI` units, the relation of newton and nowton is

Options

A. `1 "notwen" = 6.67xx 10^(-11) "newton"`
B. `1 "newton" = 6.67xx 10^(-17) "notwen"`
C. `1 "notwen" = 6.67xx 10^(-17) "newton"`
D. `1 "newton" = 6.67xx 10^(-12) "notwen"`

Answer

c

Explanation

If two point masses each of masses `1kg` are seperated a distance `1m` they experience a force `6.67 xx 10^(-11)` newton
We know `,F = G(m_(1)m_(2))/(r^(2))`
`1 "notwen" = 6.67 xx 10^(-11) xx ((1 kg)^(2))/((1 km)^(2))`
`= 6.67 xx 10^(-11) xx 10^(-6) ((1 kg)^(2))/((1 m)^(2))`
`= 6.67 xx 10^(-11) xx 10^(-6)"newton"`
`1 "notwen" = 6.67 xx 10^(-17) "newton"`