The position of a particle at time t is given by the relation $x\left( t \right) = \left( {\begin{array}{*{20}{c}} {\frac{{{v_0}}}{\alpha }}\\ \; \end{array}} \right)\left( {1 - {e^{\alpha t}}} \right),\;$ where ${v_0}$ is constant and $\alpha > 0$ . The dimensions of ${v_0}$ and $\alpha $ are respectively

${M^0}{L^1}{T^0}\,and\,{T^{ - 1}}$

${M^0}{L^1}{T^{ - 1}}\,and\,L{T^{ - 2}}$

NEET UG PHYSICS MCQ: UNITS & MEASUREMENTS

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Application +4 | -1

The position of a particle at time t is given by the relation

x\left( t \right) = \left( {\begin{array}{cccccccccccccccccccc} {\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

View answer and explanation

Correct option: A

{M^0}{L^1}{T^{ - 1}}\,and\,{T^{ - 1}}

Option A (

{M^0}{L^1}{T^{ - 1}}\,and\,{T^{ - 1}}

) is correct because the dimensions of x(t) must be length (L), which implies

{v_0}

has dimensions of

LT^{-1}

and

\alpha

has dimensions of

T^{-1}

to make the exponent dimensionless.