Question

A property R of a circular disc of thickness and area A is defined as $R=k\frac{dx}{A}$. Find the same property R for the cylinder shown in figure.

• A. $\frac{2kL}{\pi Dd}$
• B. $\frac{2kL}{\pi (D-d)d}$
• C. $\frac{kL}{\pi Dd}$
• D. $\frac{4kL}{\pi \left(Dd\right)}$

Answer 1

The correct option is D $\frac{4kL}{\pi \left(Dd\right)}$

We know that,

$\begin{array}{c}{d}^{\prime }=d+\frac{\left(D-d\right)}{L}.x\\ dR=k\frac{dx\times 4}{\pi {\left\{d+\frac{\left(D-d\right)}{L}.x\right\}}^2}\\ dR=\frac{4k}{\pi d^2}\frac{dx}{{\left(1++\frac{\left(D-d\right)}{L}.x\right)}^2}\\ R=\frac{4k}{\pi d^2}{\int }_0^L\frac{dx}{{\left(1++\frac{\left(D-d\right)}{L}.x\right)}^2}\\ =\frac{4k}{\pi d^2}.\frac{dL}{\left(D-d\right)}.\frac{\left(D-d\right)}{D}\\ R=\frac{4kL}{\pi \left(Dd\right)}\end{array}$

Hence,

Option $D$ is correct.