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JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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The dimensional formula for gravitational constant is
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Correct option: A
[M−1L3T−2]\left[{M}^{-1} {L}^3 {T}^{-2}\right]
Option A ([M−1L3T−2]\left[{M}^{-1} {L}^3 {T}^{-2}\right]) is correct because it is derived from the formula for gravitational force and isolating the gravitational constant G.

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Assertion :  Modulus of elasticity  Density \sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}} has the unit ms−1. \mathrm{ms}^{-1} \text {. }

Reason: Acceleration has the dimensions of 1(ε0μ0)t. \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 [M−1L−3A2T3]\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{A}^2}{T^3}} \right] is A quantity X is given by ε0LΔVΔt,  {\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}},\;, where ε0{\varepsilon _0} is the permittivity of free space, L is a length, ΔV{\Delta V} is a potential difference and Δt{\Delta t} is a time interval. The dimensional formula for X is the same as that of
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