Questions & Answers

Question

Answers

\[\begin{align}

& A)\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right] \\

& B)\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right] \\

& C)\left[ M{{L}^{3}}{{T}^{-3}}{{A}^{-2}} \right] \\

& D)\left[ M{{L}^{-1}}{{L}^{3}}{{T}^{3}}{{A}^{2}} \right] \\

\end{align}\]

Answer
Verified

We must know that the electrical resistance is defined as the ratio of the electric voltage to electric current. i.e.

\[\text{Resistance=}\dfrac{\text{Voltage}}{\text{Current}}=\dfrac{V}{I}\]

But, the voltage here is defined as work done by an electric charge divided by the amount of charge which is expressed as product of current and time. Because current is the rate of change of charge in a unit area of the cross section.

\[\text{Voltage=}\dfrac{\text{Work}}{\text{Charge}}\text{=}\dfrac{\text{Work}}{\text{Current }\!\!\times\!\!\text{ Time}}=\dfrac{W}{I\times t}\]

Substituting this in the equation of resistance, we will get

\[R=\dfrac{W}{{{I}^{2}}\times t}\]

We know, the dimensional formula of work is \[\left[ M{{L}^{2}}{{T}^{-2}} \right]\], dimensional formula of current is \[\left[ A \right]\] and dimensional formula of time is\[\left[ T \right]\]. Let us substitute these dimensional formulas in the equation for resistance.

\[\Rightarrow R=\dfrac{W}{{{I}^{2}}\times t}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ {{A}^{2}} \right]\left[ T \right]}\]

\[R=\left[ M{{L}^{2}}{{A}^{-2}}{{T}^{-3}} \right]\]

Therefore, the dimensional formula of resistance is found to be \[\left[ M{{L}^{2}}{{A}^{-2}}{{T}^{-3}} \right]\].

In the solution we are using ohm’s law to determine the expression for resistance. Ohm’s law states that current is directly proportional to the voltage across the conductor. i.e. \[V=IR\]. Here, electrical resistance is a barrier for current having S.I. unit \[\Omega \]. The resistance of a material depends upon the resistivity of the particular material (\[\rho \]), length (\[l\]) and area of cross section (\[A\]). Here, resistance is also expressed as \[R=\dfrac{\rho l}{A}\].