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JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
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Let [$${\varepsilon _0}$$] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:
A
$${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]$$
B
$${\varepsilon _0} = $$$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$
C
$${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]$$
D
$${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]$$
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