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

Hint: We know that the force acting on the two charges separated by distance ‘r’
$ \begin{align}
  & \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{align} $
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{align}
  & =\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{align} $
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.