Q. The position of a particle at time t is given by the relation $x(t) = \left( \begin{array}{r} \frac{v_{0}}{\alpha} \\ \ \end{array} \right)\left( 1 - e^{\alpha t} \right),\$where $v_{0}$ is constant and $\alpha > 0$. The dimensions of $v_{0}$ and $\alpha\$are respectively
A) $M^{0}L^{1}T^{- 1}$ and $T^{- 1}$
B) $M^{0}L^{1}T^{0}$ and $T^{- 1}$
C) $M^{0}L^{1}T^{- 1}$ and $LT^{- 2}$
D) $M^{0}L^{1}T^{- 1}$ and $T$
Explanation
Dimension of $\alpha t = \left\lbrack M^{0}L^{0}T^{0} \right\rbrack\therefore\lbrack\alpha\rbrack = \lbrack T^{- 1}\rbrack$ Again $\left\lbrack \frac{v_{0}}{\alpha} \right\rbrack = \lbrack L\rbrack$ so $\left\lbrack v_{0} \right\rbrack = \lbrack LT^{- 1}\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.