JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Easy

1 of 1

Application +4 | -1

The displacement of an oscillating particle is given by

y=A \sin (B x+C t+D)

. The dimensional formula for

(A B C D)

View answer and explanation

Correct option: B

\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]

Option B ([M⁰ L⁰ T^-1]) is correct because the dimensional formula for (A B C D) is [T^-1] based on dimensional analysis.

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}