Force ${F}$ depends on distance $(x)$ and time $(t)$ as $F=a x^2+b t^{\frac{1}{2}}$, find dimension of $\frac{b^2}{a}$

JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Application +4 | -1

Force

{F}

depends on distance

(x)

and time

(t)

F=a x^2+b t^{\frac{1}{2}}

, find dimension of

\frac{b^2}{a}

View answer and explanation

Correct option: C

M^1 L^3 T^{-3}

Option C (

M^1 L^3 T^{-3}

) is correct because it is derived from dimensional analysis of the given force equation.

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}