Q. The position of a particle at time $t$ is given by the equation $x(t) = \frac{v_{0}}{A}\ \left( 1 - e^{At} \right),\ v_{0} =$ constant and $A > 0.$ Dimensions of $v_{0}\ and\ A$ respectively are
A) $\left\lbrack M^{0}{LT}^{0} \right\rbrack\ and\ \left\lbrack M^{0}L^{0}T^{- 1} \right\rbrack$
B) $\left\lbrack M^{0}{LT}^{- 1} \right\rbrack\ and\ \left\lbrack M^{0}{LT}^{- 2} \right\rbrack$
C) $\left\lbrack M^{0}{LT}^{- 1} \right\rbrack\ and\ \left\lbrack M^{0}L^{0}T \right\rbrack$
D) $\left\lbrack M^{0}{LT}^{- 1} \right\rbrack\ and\ \left\lbrack M^{0}L^{0}T^{- 1} \right\rbrack$
Explanation
The dimensions of $x =$ dimensions of $\frac{v_{0}}{A}$ Therefore, out of the given options $v_{0}$ has dimensions equal to $\left\lbrack M^{0}LT^{- 1} \right\rbrack$ and $A$ has dimensions equal to $\left\lbrack M^{0}L^{0}T^{- 1} \right\rbrack$ So, that $\frac{\left\lbrack v_{0} \right\rbrack}{\lbrack A\rbrack} = \frac{\left\lbrack M^{0}LT^{- 1} \right\rbrack}{\left\lbrack M^{0}L^{0}T^{- 1} \right\rbrack} = \lbrack L\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =$ dimension of $x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.