Skip to main content
Back to UNITS & MEASUREMENTS

JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium
1 of 1
Understanding +4 | -1
The dimensions of mutual inductance (M)(\mathrm{M}) and capacitance (C)(\mathrm{C}) are respectively.
View answer and explanation
Correct option: C
[M1L2T−2A−2],[M−1L−2T4A2]\left[\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2} \mathrm{A}^{-2}\right],\left[\mathrm{M}^{-1} \mathrm{L}^{-2} \mathrm{T}^4 \mathrm{A}^2\right]
Option C is correct because the dimensions of mutual inductance are [ML²T⁻²A⁻²] and the dimensions of capacitance are [M⁻¹L⁻²T⁴A²].

More Questions from UNITS & MEASUREMENTS

Practice All Chapter MCQs →
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
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →