Q. Taking frequency $f$, velocity $v$ and density $\rho$ to be the fundamental quantities, then the dimensional formula for momentum will be
A) $\lbrack\rho v^{4}f^{- 3}\rbrack$
B) $\lbrack\rho v^{3}f^{- 1}\rbrack$
C) $\lbrack\rho vf^{2}\rbrack$
D) $\lbrack\rho^{2}v^{2}f^{2}\rbrack$
Explanation
Momentum $p \propto f^{a}v^{b}\rho^{c}$ [$MLT^{- 1}$]=$\left\lbrack T^{- 1} \right\rbrack^{a}\left\lbrack LT^{- 1} \right\rbrack^{b}\left\lbrack ML^{- 3} \right\rbrack^{c}$ $\left\lbrack MLT^{- 1} \right\rbrack = \lbrack M^{c}L^{b - 3c}T^{- a - b}\rbrack$ $\Longrightarrow \ c = 1$. $b - 3c = 1 \Longrightarrow b = 4$ $- a - b = - 1$ $a + b = 1,a = - 3$ $\therefore\ \lbrack p\rbrack = \lbrack f^{- 3}v^{4}\rho\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.