JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Analyzing +4 | -1

\left(P-\frac{a}{V^2}\right)(V-b)=n R T

, where

P, V, R

and

T

are pressure, volume, universal gas constant and temperature, then

\frac{a}{b^2}

has same dimensional formula as that of

View answer and explanation

Correct option: D

P

Option D is correct because dimensional analysis shows that a/b² has the same dimensions as pressure.

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}