Q. If the velocity $v\left( is\ {cms}^{- 1} \right)$ of a particle is given in terms of $t$ (in second) by the relation $\ \ v = at + \frac{b}{t + c}$ then, the dimensions of $a,\ b\ and\ c$ are $a$ $b$ $c$
A) $\lbrack L\rbrack\ \ \ \ \ \ \ \ \lbrack LT\rbrack\ \ \ \ \ \ \ \ \left\lbrack T^{2} \right\rbrack$
B) $\left\lbrack L^{2} \right\rbrack\ \ \ \ \ \ \lbrack T\rbrack\ \ \ \ \ \ \ \ \left\lbrack {LT}^{- 2} \right\rbrack$
C) $\left\lbrack {LT}^{2} \right\rbrack\ \ \ \ \ \lbrack LT\rbrack\ \ \ \ \ \ \ \lbrack L\rbrack$
D) $\left\lbrack {LT}^{- 2} \right\rbrack\ \ \ \ \lbrack L\rbrack\ \ \ \ \ \ \ \ \ \lbrack T\rbrack$
Explanation
Given, $v = at + \frac{b}{t + c}$ Since, LHS is equal to velocity, so $at$ and $\frac{b}{t + c}$ must have the dimensions of velocity. $\therefore\ \ \ at = v$ Or $\ \ a = \frac{v}{t} = \frac{\left\lbrack LT^{- 1} \right\rbrack}{\lbrack T\rbrack} = \left\lbrack LT^{- 2} \right\rbrack$ Now, $\ \ c = time\ \ \ \ \ (\because like\ quantities\ are\ added)\$ $\therefore\ \ \ \ c = t = \lbrack T\rbrack$ Now, $\ \ \frac{b}{t + c} = v$ $\therefore\ \ \ b = v \times time = \left\lbrack {LT}^{- 1} \right\rbrack\lbrack T\rbrack = \lbrack L\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.