Question

All the edges of a block with parallel faces are unequal. Its longest edge is twice the shortest edge. The ratio of the maximum to the minimum resistance between the parallel faces is
a) 2
b) 4
c) 8
d) intermediate until the length of the third edge is specified

Answer 1

It is given in the question the edges of a block with parallel faces are unequal and the length of the longest edge is equal to the length of the shortest edge. Let us consider the third edge to be of length b. The resistance of any material is proportional to its length and inversely proportional to the area of cross section. The resistance of the above block will be maximum when we measure its resistance along its maximum length and across the minimum area of cross section and minimum resistance across the smallest length and maximum area. Therefore we will take this into consideration and accordingly take the ratio.

Formula used:
$R=\dfrac{\rho l}{A}$

Complete answer:
The resistance of an object is given by $R=\dfrac{\rho l}{A}...(1)$ where $\rho$ determines the nature of the material or the type of material, l is the length of the object along which the resistance is to be calculated and A is the area of cross section.
Let the minimum edge of the block be $l$, hence its longest edge will be equal to $2l$. Let us consider the third edge as b. The resistance of the above block will be maximum when we measure its resistance along its maximum length i.e. $2l$ and across the minimum area of cross section i.e. equal to $A=lb$. Therefore the maximum resistance of the block from equation 1 we get,
$\begin{aligned} & R=\dfrac{\rho l}{A} \\\ & R(\max )=\dfrac{\rho 2l}{lb} \\\ \end{aligned}$
Similarly the resistance of the block will be minimum when measured across its shortest edge and across the maximum area of cross section i.e. equal to $A=\left( 2l \right)b$. Therefore the minimum resistance of the block from equation 1 we get,
$\begin{aligned} & R=\dfrac{\rho l}{A} \\\ & R(\min )=\dfrac{\rho l}{2lb} \\\ \end{aligned}$
Hence the ratio of the maximum to the minimum resistance is,
$\begin{aligned} & \dfrac{R(\max )}{R(\min )}=\dfrac{\dfrac{\rho 2l}{lb}}{\dfrac{\rho l}{2lb}} \\\ & \Rightarrow \dfrac{R(\max )}{R(\min )}=\dfrac{\rho 2l}{lb}\times \dfrac{2lb}{\rho l} \\\ & \Rightarrow \dfrac{R(\max )}{R(\min )}=2\times 2=4 \\\ \end{aligned}$

Hence the correct answer is option b.

Note:
The material whose resistance we measured should be in pure form. Until this condition is satisfied, all the above expressions hold true. The length of the third edge does not play any role i.e. b. This is because the edge b is common to both the minimum as well as the maximum area.