JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Application +4 | -1

The velocity

\mathrm{v}

of a particle at time

\mathrm{t}

is given by

\mathrm{v}=\mathrm{at}+\frac{\mathrm{b}}{\mathrm{t}+\mathrm{c}}

, where

\mathrm{a}, \mathrm{b}

and

\mathrm{c}

are constants. The dimensions of

\mathrm{a}

\mathrm{b}

and

\mathrm{c}

are respectively:

View answer and explanation

Correct option: A

\mathrm{LT^{-2}, L \,and\, T}

Option A (

\mathrm{LT^{-2}, L \,and\, T}

) is correct because it correctly identifies the dimensions of a, b, and c based on the principle of dimensional homogeneity.

More Questions from UNITS & MEASUREMENTS

Practice All Chapter MCQs →

\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}

\mathrm{ms}^{-1} \text {. }

\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t} \text {. }

\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{A}^2}{T^3}} \right]

{\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}},\;

{\varepsilon _0}

{\Delta V}

{\Delta t}