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# The resistivity of a material is $S$ ohm meter. The resistance between opposite faces of a solid cube of edge $10\,cm$ is (in ohm):A) $\dfrac{S}{2}$B) $\dfrac{S}{{10}}$C) $100S$D) $10S$

Last updated date: 13th Jun 2024
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Hint: To solve the asked question, you firstly need to notice that we’ve been given the resistivity of the material, hence you need to find the resistance using the relation between resistance, resistivity, length and the surface area of the wire, mathematically given as: $R = \rho \dfrac{l}{A}$

Complete step by step solution:
We will proceed with the solution exactly as explained in the hint section of the solution to the question. First, we need to define the resistance in terms of resistivity $\left( \rho \right)$ , length of the conducting part $\left( l \right)$ and the cross-sectional area of the conducting part $\left( A \right)$ . Once we do that, we can easily find the resistance between two opposite faces of the given cube with edges of $10\,cm$.
Let us first share with you the formula that gives us a relation between resistance $\left( R \right)$ , resistivity $\left( \rho \right)$ , length of the conducting part $\left( l \right)$ and the cross-sectional area of the conducting part $\left( A \right)$ :
$R = \rho \dfrac{l}{A}$
We can clearly see that in the equation mentioned above, $\rho$ is the resistivity of the material of the wire or the conducting part which carries the current.
$l$ is the length of the wire or the conducting part which carries the current
$A$ is the cross-sectional area of the wire or the conducting part which carries current and,
$R$ is the resistance provided by the wire to the current, or by the conducting part to the current flowing in it.
In the question, we have been given the value of the resistivity of the material as: $S$ ohm meter.
The value of edges of the cube are given to be $10\,cm$ or $0.1$ meters.
Using this, we can confidently say that the cross-sectional area is given as: $A = {\left( {0.1} \right)^2}{\kern 1pt} {m^2}$
$A = 0.01\,{m^2}$
Now, let us assume that at a distance of $x$ from one face of the cube, an integral element of the cube lies with a thickness of $dx$ .
For this small element of the cube, we can write its resistance as:
$\Rightarrow dR = \rho \dfrac{{dx}}{A}$
Now, let us integrate both the sides:
$\Rightarrow \int {dR} = \int {\rho \dfrac{{dx}}{A}}$
We know that the value of $x$ ranges from zero to $0.1\,m$ , similarly, the value of resistance varies from zero to $R$ , which is the net resistance between two opposite faces of the given cube.
We also have been given that resistivity and area are constant and thus, can be taken out of the integration.
Now, we can write:
$\Rightarrow \int\limits_0^R {dR} = \dfrac{\rho }{A}\int\limits_0^{0.1} {dx}$
After integrating, we get:
$\Rightarrow \left( {R - 0} \right) = \dfrac{\rho }{A}\left( {0.1 - 0} \right)$
Substituting in the values of $\rho$ and $A$ , we get:
$\Rightarrow R = \dfrac{S}{{0.01}} \times 0.1$
Upon solving, we get:
$\Rightarrow R = 10S$

We can see that this matches the value of option (D), hence, option (D) is the correct answer.

Note: One main mistake that many students make is that instead of noticing that the question has given us the resistivity of the material they assume that the resistance of the edges is given, since it is similar to the famous concept of resistances of outline of a cube.