JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Analyzing +4 | -1

Dimensional formula for

\frac{h^2 v^2 / m c^2}{1+\frac{h v}{m c^2}}

is (where

h

is planck's constant,

v

is frequency of light,

m

is mass and

c

is speed of light).

View answer and explanation

Correct option: C

\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

Option C (

\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

) is correct because after simplifying the expression using dimensional analysis, the result matches the dimensions of energy.

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}