Q. If force is proportional to square of velocity, then the dimensions of proportionality constant are
A) $\left\lbrack {ML}^{- 1}T \right\rbrack$
B) $\left\lbrack {ML}^{- 1}T^{0} \right\rbrack$
C) $\left\lbrack {MLT}^{0} \right\rbrack$
D) $\left\lbrack M^{0}{LT}^{- 1} \right\rbrack$
Explanation
Given, $F \propto v^{2}$ Or $F = kv^{2}$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k = \frac{F}{v^{2}}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack k\rbrack = \frac{\lbrack F\rbrack}{\left\lbrack v^{2} \right\rbrack} = \frac{\left\lbrack MLT^{- 2} \right\rbrack}{\left\lbrack L^{2}T^{- 2} \right\rbrack}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack ML^{- 1}T^{0} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.