The dimensions of universal gravitational constant are
Answer & explanation
Correct answer: option 1
F = G$\frac{m_1m_2}{2^2}$
[G] = $\frac{Fr^2}{m_1m_2}$ = $\frac{[MLT^{-2}][L^2]}{[M^2]}$ = $[M^{-1}L^3T^{-2}]$
The dimensions of universal gravitational constant are
Correct answer: option 1
F = G$\frac{m_1m_2}{2^2}$
[G] = $\frac{Fr^2}{m_1m_2}$ = $\frac{[MLT^{-2}][L^2]}{[M^2]}$ = $[M^{-1}L^3T^{-2}]$