JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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In the equation

y=a \sin (\omega t+k x)

, the dimensional formula of

\omega

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Correct option: A

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

Option A ([M⁰ L⁰ T⁻¹]) is correct because the argument of the sine function must be dimensionless, so ωt has dimensions of 1, making ω have dimensions of T⁻¹.

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Assertion :

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

has the unit

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

Reason: Acceleration has the dimensions of

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

The method of dimensional analysis can be used to derive which of the following relations? Statement-I: Dimensions of constant of proportionality of constants can be derived from dimensional method
Statement-II: Numerical value of constant of proportionality can be found from experiments only The dimensions of torque are The physical quantity having the dimensions

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

is A quantity X is given by

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

, where

{\varepsilon _0}

is the permittivity of free space, L is a length,

{\Delta V}

is a potential difference and

{\Delta t}

is a time interval. The dimensional formula for X is the same as that of