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Question

Answers

$\text{A}. \quad ML^{3}T^{-1}Q^{-2}$

$\text{B}. \quad ML^{3}T^{-2}Q^{-1}$

$\text{C}. \quad ML^{2}T^{-1}Q^{-1}$

$\text{D}. \quad MLT^{-1}Q^{-1}$

Answer
Verified

$R = \dfrac{{\rho}L}A, \ V=iR, \ V= \dfrac {W_{done}}Q$

Where R is the resistance of a wire of length ‘L’ and cross-section area ‘A’ and the resistivity of the material is $\rho$. W is the work done on charge Q.

Ohm’s law states, V=iR.

Hence $R=\dfrac{V}{i}$

And $V=\dfrac {W_{done}}Q$

Hence dimensions of $V =\dfrac{ML^2T^{-2}}Q={ML^2T^{-2}}Q^{-1}$ [$W=\int { F.ds }$]

Now, current is the rate of flow of charge, hence dimensions of $R = \dfrac{{ML^2T^{-2}}Q^{-1}}{QT^{-1}}= {ML^2T^{-1}}Q^{-2}$

Finally, using $R = \dfrac{{\rho}L}A \ or \ \rho = \dfrac{{R}A}L$

Putting the dimensions of R, A and L, we get

$\rho = \dfrac{[{ML^2T^{-1}}Q^{-2}] \times [L^2]}{[L]}$

$\rho = {ML^3T^{-1}Q^{-2}}$

It is to be noted that many physical quantities have the same physical dimensions by they can be completely different in every aspect like work and torque, both have the same dimensions $[M^1 L^{2}T^{-2}]$ which are also the dimensions of heat given.

Students are advised to learn basic quantities’ dimensions only (for example displacement, angle, intensity, mass, etc.) and complex quantities’ dimensions (for example torque, pressure, force, etc.) can be easily derived by knowing the formulae and proceeding as shown in the above example. Also in the above example for calculating the dimensions of V we have many more formulae like $V=\int { E.dr }$. But it’s our duty to choose the formula so that it becomes the easiest to get evaluated.

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