Question

The dimension of R in the equation \[{\text{Q = }}{{\text{Q}}_{\text{0}}}{\text{(1 - }}{{\text{e}}^{{\text{ - t/RC}}}}{\text{)}}\]
(A) \[{\text{[}}{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{ - 3}}{{\text{A}}^{ - 2}}{\text{]}}\]
(B) \[{\text{[}}{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{ - 2}}{{\text{A}}^{ - 3}}{\text{]}}\]
(C) \[{\text{[}}{{\text{M}}^2}{{\text{L}}^{\text{2}}}{{\text{T}}^{ - 3}}{{\text{A}}^{ - 2}}{\text{]}}\]
(D) \[{\text{[}}{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^1}{{\text{A}}^{ - 2}}{\text{]}}\]

Answer 1

Hint Manipulate the equation and put R on the left side of the equation and the other terms on the right side. Then substitute the values of each quantity in terms of the fundamental quantities to get the required answer.

Complete step-by-step solution
As given in the question,
As we know that the energy stored in the capacitor is:
\[{\text{E = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{V}}^{\text{2}}}\]
\[
  {\text{[}}{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{{\text{ - 2}}}}{{\text{A}}^{\text{0}}}{\text{] = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C[}}{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{{\text{ - 3}}}}{{\text{A}}^{{\text{ - 1}}}}{{\text{]}}^{\text{2}}} \\
  {\text{C = [}}{{\text{M}}^{{\text{ - 1}}}}{{\text{L}}^{{\text{ - 2}}}}{{\text{T}}^4}{{\text{A}}^{\text{2}}}{\text{]}} \\
 \]
So the value of capacitance in terms of fundamental quantities is \[{\text{[}}{{\text{M}}^{{\text{ - 1}}}}{{\text{L}}^{{\text{ - 2}}}}{{\text{T}}^{\text{2}}}{{\text{A}}^{\text{2}}}{\text{]}}\]. Substituting this in the above equation we get,
\[{\text{Q = }}{{\text{Q}}_{\text{0}}}{\text{(1 - }}{{\text{e}}^{{\text{ - t/RC}}}}{\text{)}}\]
\[{\text{[}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{1}}}{{\text{A}}^{\text{1}}}{\text{] = [}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{1}}}{{\text{A}}^{\text{1}}}{\text{](1 - }}{{\text{e}}^{{\text{ - t/RC}}}})\]
\[
  {\text{(1 - }}{{\text{e}}^{{\text{ - t/RC}}}}{\text{) = [}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}}{{\text{A}}^{\text{0}}}{\text{]}} \\
  {{\text{e}}^{{\text{ - t/RC}}}}{\text{ = [}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}}{{\text{A}}^{\text{0}}}{\text{]}} \\
  \dfrac{{{\text{ - t}}}}{{{\text{RC}}}}{\text{ = [}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}}{{\text{A}}^{\text{0}}}{\text{]}} \\
  {\text{[}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{1}}}{{\text{A}}^{\text{0}}}{\text{] = R[}}{{\text{M}}^{ - 1}}{{\text{L}}^{ - 2}}{{\text{T}}^4}{{\text{A}}^2}{\text{]}} \\
  {\text{R = [}}{{\text{M}}^1}{{\text{L}}^2}{{\text{T}}^{ - 3}}{{\text{A}}^{ - 2}}{\text{]}} \\
 \]

Therefore the correct answer is option A

Note You can also do it in the way that the power of e is always going to be a dimensionless quantity. You can simply substitute t/rc to 1 and then solve the equation. Also we do not consider the negative sign if any in the equations, as we are supposed to only compare its dimensions and they will stay the same regardless of the negative sign.