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JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Assertion: Linear mass density has the dimensions of

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

.
Reason: This is so because density is mass per unit volume.

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Correct option: B

If both assertion and reason are true but reason is not the correct explanation of assertion

Option B (If both assertion and reason are true but reason is not the correct explanation of assertion) is correct because both the assertion about linear mass density's dimensions and the reason about density are true, but the reason doesn't explain the assertion.

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

\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}},\;

{\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