JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 12)
Which of the following combinations has the dimension of electrical resistance ($$ \in $$0 is the permittivity of
vacuum and $$\mu $$0 is the permeability of vacuum)?
$$\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $$
$${{{{ \in _0}} \over {{\mu _0}}}}$$
$$\sqrt {{{{\mu _0}} \over {{ \in _0}}}} $$
$${{{{\mu _0}} \over {{ \in _0}}}}$$
Explanation
According to Coulomb's law
F = $${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$$
$$ \therefore $$ $${ \in _0} = {1 \over {4\pi }}{{{q^2}} \over {F{r^2}}}$$
$$\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}$$ = $$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$
Force between two parallel current carrying wires,
$${F \over L} = {{{\mu _0}} \over {2\pi }}{{{i^2}} \over r}$$
$$ \therefore $$ $${{\mu _0}}$$ = $${{2\pi rF} \over {{i^2}L}}$$
$$\left[ {{\mu _0}} \right] = {{\left[ L \right]\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ L \right]}}$$ = $$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$$
From Ohm's law,
V = IR
$$ \therefore $$ $$R = {V \over I}$$ $$ = {U \over {It}} \times {1 \over I}$$
[R] = $${{\left[ U \right]} \over {{{\left[ I \right]}^2}\left[ t \right]}}$$ = $${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ T \right]}}$$ = $$\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$$
Let, R $$ \propto $$ $${\left[ {{ \in _0}} \right]^a}{\left[ {{\mu _0}} \right]^b}$$
$$ \Rightarrow $$ $$\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$$ =
$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$$$a$$ $$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$$b
$$ \Rightarrow $$ $$\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$$ =
[ M-$$a$$ + b L-3$$a$$ + b T4$$a$$ - 2b A2$$a$$ - 2b ]
By comparing both sides we get,
- $$a$$ + b = 1
- 3$$a$$ + b = 2
By solving we get,
$$a$$ = $$ - {1 \over 2}$$ and b = $${1 \over 2}$$
$$ \therefore $$ R = $${\left[ {{ \in _0}} \right]^{ - {1 \over 2}}}{\left[ {{\mu _0}} \right]^{{1 \over 2}}}$$ = $$\sqrt {{{{\mu _0}} \over {{ \in _0}}}} $$
F = $${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$$
$$ \therefore $$ $${ \in _0} = {1 \over {4\pi }}{{{q^2}} \over {F{r^2}}}$$
$$\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}$$ = $$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$
Force between two parallel current carrying wires,
$${F \over L} = {{{\mu _0}} \over {2\pi }}{{{i^2}} \over r}$$
$$ \therefore $$ $${{\mu _0}}$$ = $${{2\pi rF} \over {{i^2}L}}$$
$$\left[ {{\mu _0}} \right] = {{\left[ L \right]\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ L \right]}}$$ = $$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$$
From Ohm's law,
V = IR
$$ \therefore $$ $$R = {V \over I}$$ $$ = {U \over {It}} \times {1 \over I}$$
[R] = $${{\left[ U \right]} \over {{{\left[ I \right]}^2}\left[ t \right]}}$$ = $${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ T \right]}}$$ = $$\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$$
Let, R $$ \propto $$ $${\left[ {{ \in _0}} \right]^a}{\left[ {{\mu _0}} \right]^b}$$
$$ \Rightarrow $$ $$\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$$ =
$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$$$a$$ $$\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$$b
$$ \Rightarrow $$ $$\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$$ =
[ M-$$a$$ + b L-3$$a$$ + b T4$$a$$ - 2b A2$$a$$ - 2b ]
By comparing both sides we get,
- $$a$$ + b = 1
- 3$$a$$ + b = 2
By solving we get,
$$a$$ = $$ - {1 \over 2}$$ and b = $${1 \over 2}$$
$$ \therefore $$ R = $${\left[ {{ \in _0}} \right]^{ - {1 \over 2}}}{\left[ {{\mu _0}} \right]^{{1 \over 2}}}$$ = $$\sqrt {{{{\mu _0}} \over {{ \in _0}}}} $$
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