MCQ
The position of a particle at time $t$ is given by the relation $x(t) = \left( {\frac{{{v_0}}}{\alpha }} \right)\,\,(1 - {e^{ - \alpha t}})$, where ${v_0}$ is a constant and $\alpha > 0$. The dimensions of ${v_0}$ and $\alpha $ are respectively
  • ${M^0}{L^1}{T^{ - 1}}$ and ${T^{ - 1}}$
  • B
    ${M^0}{L^1}{T^0}$ and ${T^{ - 1}}$
  • C
    ${M^0}{L^1}{T^{ - 1}}$ and $L{T^{ - 2}}$
  • D
    ${M^0}{L^1}{T^{ - 1}}$ and $T$

Answer

Correct option: A.
${M^0}{L^1}{T^{ - 1}}$ and ${T^{ - 1}}$
a
(a) Dimension of $\alpha t$$=$ $[{M^0}{L^0}{T^0}]$ $\therefore [\alpha] = [{T^{ - 1}}]$

Again $\left[ {\frac{{{v_0}}}{\alpha }} \right] = [L]$so $[{v_0}] = [L{T^{ - 1}}]$

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