Question

If, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, the quantity$\left( \frac{El^{2}}{m^{5}G^{2}} \right)$has the dimensions of

• A. Mass
• B. Length
• C. Time
• D. Angle

Answer 1

Answer: (D)

Angle

Answer 2

$\lbrack E\rbrack = \lbrack ML^{2}T^{- 2}\rbrack,\lbrack M\rbrack = \lbrack M\rbrack$

$\lbrack l\rbrack = \lbrack ML^{2}T^{- 1}\rbrack,\lbrack G\rbrack = \lbrack M^{- 1}L^{3}T^{- 2}\rbrack\therefore\left\lbrack \frac{Et^{2}}{m^{5}G^{2}} \right\rbrack = \frac{\lbrack ML^{2}T^{- 2}\rbrack\lbrack M^{2}L^{4}T^{- 2}\rbrack}{\lbrack M^{5}\rbrack\lbrack M^{- 2}L^{6}T^{- 4}\rbrack}$

$= \lbrack M^{0}L^{0}T^{0}\rbrack$

As angle has no dimensions, therefore $\frac{El^{2}}{m^{5}G^{2}}$has the same dimensions as that of angle.