Question

The Ohm-metre is the unit of
(A) Resistance
(B) Conductance
(C) Resistivity
(D) Voltmeter

Answer 1

Hint : The unit of any quantity can be derived from any of their expressions. The Ohm-metre is a composite unit which gives that the quantity whose unit is Ohm is multiplying the quantity (or set of quantities) whose end unit is meter in dimension.

Formula used: In this solution we will be using the following formula;
$\Rightarrow G = \dfrac{1}{R}$ where $G$ is conductance, and $R$ is the resistance.
$\Rightarrow \rho = \dfrac{{RA}}{L}$ where $\rho$ is the resistivity of a conductor, $R$ , again, is the resistance of a particular conductor, $A$ is the cross sectional area, and $L$ is the length of the conductor.
$\Rightarrow R = \dfrac{V}{I}$ where $V$ is voltage and $I$ is current.

Complete step by step answer
To answer the question, let's investigate the unit of the quantities in the options by using their expressions or formula as regard to other known units.
From the question, by observing the given options, we can eliminate voltmeter right away since, the voltmeter is not a quantity but a device used in measurement of a quantity Voltage, potential difference, and EMFs. All these all have units of Volts. Hence, option D is eliminated.
Investigating option A, from popular experience that the fundamental unit of the resistance of a body is the Ohms (not Ohms meter), given from Ohms law as $R = \dfrac{V}{I} = \dfrac{{\text{V}}}{{\text{A}}} = \Omega$ . Hence, option A can be eliminated.
The option B is given as conductance. Conductance is explained in simple terms as the opposite of the resistance and is given as its inverse, hence
$\Rightarrow G = \dfrac{1}{R}$ . Hence has a unit of Ohms inverse, ${\Omega ^{ - 1}}$ .
Since all other options are eliminated, the resistivity must hence if the correct option
Thus, the correct option is C.

Note
Alternatively, for certainty, you could derive the unit of resistivity from its expression given as
$\Rightarrow \rho = \dfrac{{RA}}{L}$ where $R$ is the resistance, $A$ is cross sectional area, and $L$ is length.
Hence the unit of resistivity is given as
$\Rightarrow {\rho _u} = \dfrac{{\Omega {m^2}}}{m} = \Omega m$ which is written as Ohm-meter.