JEE Advance - Physics (1998 - No. 1)
Let [$${\mathrm\varepsilon}_\mathrm o$$] denote the dimentional formula of the permittivity of the vacuum, and [$$\mu_o$$] that of the permeability of the vacuum.If M = mass, L = length, T = time and I = electric current,
$$\left[{\mathrm\varepsilon}_\mathrm o\right]\;=\;\mathrm M^{-1}\mathrm L^{-3}\mathrm T^2\;\mathrm I$$
$$\left[{\mathrm\varepsilon}_\mathrm o\right]\;=\;\mathrm M^{-1}\mathrm L^{-3}\mathrm T^4\mathrm I^2$$
$$\left[{\mathrm\mu}_\mathrm o\right]\;=\;\mathrm{MLT}^{-2}\mathrm I^{-2}$$
$$\left[{\mathrm\mu}_\mathrm o\right]\;=\;\mathrm{ML}^2\mathrm T^{-1}\;\mathrm I$$
Explanation
From coulombs law, we know that
$$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} \Rightarrow {\varepsilon _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {{r^2}F}}$$
Therefore, $${\varepsilon _0} = {{[{I^2}{T^2}]} \over {[{L^2}][ML{T^{ - 2}}]}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$$
Now, $$c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} \Rightarrow {c^2} = {1 \over {{\varepsilon _0}{\mu _0}}}$$
Therefore, $${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}} = {1 \over {{{[L{T^{ - 1}}]}^2}[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}} = [ML{T^{ - 2}}{I^{ - 2}}]$$
Answer (B), (C)
$$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} \Rightarrow {\varepsilon _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {{r^2}F}}$$
Therefore, $${\varepsilon _0} = {{[{I^2}{T^2}]} \over {[{L^2}][ML{T^{ - 2}}]}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$$
Now, $$c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} \Rightarrow {c^2} = {1 \over {{\varepsilon _0}{\mu _0}}}$$
Therefore, $${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}} = {1 \over {{{[L{T^{ - 1}}]}^2}[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}} = [ML{T^{ - 2}}{I^{ - 2}}]$$
Answer (B), (C)
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