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JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Assertion: Dimensions of coefficient of viscosity are

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

Reason: Coefficient of viscosity is force acting per unit area per unit velocity gradient

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Correct option: A

If both assertion and reason are true and reason is the correct explanation of assertion

Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because the dimensions of viscosity are indeed [ML⁻¹T⁻¹], and the coefficient of viscosity is defined as force per unit area per unit velocity gradient.

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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}

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{\Delta V}

is a potential difference and

{\Delta t}

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