Question

A quantity x is given by $\text{ }\\!\\!\varepsilon\\!\\!\text{ ,L}\dfrac{\vartriangle \text{V}}{\vartriangle \text{t}}$ where ${{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{q}}}$ is the permittivity of free space, L is a length, $\vartriangle \text{V}$ is a potential difference and $\vartriangle \text{t}$ is a time interval. The dimensional formula for X is the same as that of
(A) Resistance
(B) Charge
(C) Voltage
(D) Current

Answer 1

We know that the force acting on the two charges separated by distance ‘r’
$\begin{aligned} & \text{F=}\dfrac{{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}{\text{4 }\\!\\!\pi\\!\\!\text{ }{{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}{{\text{r}}^{\text{2}}}} \\\ & \text{=}{{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}\text{=}\dfrac{{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}{\text{4 }\\!\\!\pi\\!\\!\text{ F }{{\text{r}}^{\text{2}}}} \\\ \end{aligned}$
Dimensional formula for ${{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}\text{=}\left[ {{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{\text{ }\\!\\!\alpha\\!\\!\text{ }}}{{\text{A}}^{\text{2}}} \right]$
Electric potential is given by
$\text{V=}\dfrac{\text{work}}{\text{charge}}$
Dimensional formula for $\text{V=}\left[ {{\text{M}}^{\text{1}}}{{\text{L}}^{\text{1}}}{{\text{T}}^{-\text{3}}}{{\text{A}}^{-\text{1}}} \right]$.

Complete step by step solution
We know that ${{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}$ =frequently of free space and its dimensional formula is given by $\left[ {{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{4}}{{\text{I}}^{\text{2}}} \right]$
L= length having dimensional formula=[L]
$\vartriangle \text{V}$ =potential difference having dimensions formula $=\left[ \text{M}{{\text{L}}^{2}}{{\text{T}}^{-3}}{{\text{I}}^{-1}} \right]$
$\vartriangle \text{T}$ =time internal having dimensional formula $=\left[ {{\text{T}}^{1}} \right]$
Formula for X is $\text{=}{{\text{ }\\!\\!\varepsilon\\!\\!\text{ }}_{\text{0}}}\text{ L }\vartriangle \text{V/}\vartriangle \text{t}$
Therefore, dimensional formula for ‘X’ is
$=\dfrac{\left[ {{\text{M}}^{-\text{1}}}{{\text{L}}^{-\text{3}}}{{\text{T}}^{4}}{{\text{I}}^{\text{2}}} \right]\text{ }\left[ \text{L} \right]\left[ \text{M}{{\text{L}}^{2}}\text{ }{{\text{T}}^{-3}}\text{ }{{\text{I}}^{-1}} \right]}{\left[ \text{T} \right]}$
$\begin{aligned} & =\dfrac{\left[ {{\text{T}}^{4}}\text{ }{{\text{I}}^{2}} \right]\left[ \text{L} \right]\left[ \text{M}{{\text{L}}^{2}} \right]}{\left[ \text{M}{{\text{L}}^{3}}\text{ }{{\text{T}}^{3}}\text{ I T} \right]} \\\ & =\left[ \text{I} \right] \\\ \end{aligned}$
And, dimensional formula for $\text{Q}=\left[ \text{I} \right]$
$\therefore$ Quantities ‘X’ have the same dimensional formula as in charge.
$\therefore$ Option (D) is correct.

Note
While doing finding the dimensional formula for any quantity the formula for that quantity should be clear because from the formula we can make a dimensional formula. We can use dimensional analysis to convert the physical quantity from one system to another to check the correctness of a physical relation, and to obtain relationships among various physical quantities involved.