The deflection of a uniform circular bar of diameter d and length l , which extends by an amount under a tensile pull W, when it carries the same load at its mid-span, is

Answer: Option C

We have a circular bar with diameter d and length l.
It extends by an amount e under a tensile pull W.
We want to find the deflection when the same load W is applied at the mid-span of the bar, making it act like a beam.

1. Relate Elongation (e) to Tensile Stress and Strain:
When the bar is under a tensile pull W, we know:
Stress $\sigma$ = Force (W) / Area (A), where $A=\frac{\pi d^2}{4}$.
Strain $\epsilon$ = Change in Length (e) / Original Length (l).
Also, Stress $\sigma$ = Young's Modulus $E$ * Strain $\epsilon$.
Therefore, $E=\frac{\sigma }{\epsilon }$.

2. Find Young's Modulus (E) in terms of e, l, d, and W:
$E=\frac{W}{\left(\frac{\pi d^2}{4}\right)}÷\frac{e}{l}$
$E=\frac{4Wl}{\pi d^{2}e}$.

3. Deflection Formula for a Simply Supported Beam with a Point Load at Mid-Span:
The standard formula for the deflection ($\delta$) of a simply supported beam with a point load (W) at the mid-span is:
$\delta =\frac{W{l}^3}{48EI}$, where $I$ is the area moment of inertia.

4. Area Moment of Inertia (I) for a Circular Cross-Section:
$I=\frac{\pi d^4}{64}$.

5. Substitute I and E into the Deflection Formula:
$\delta =\frac{W{l}^3}{48\times E\times \left(\frac{\pi d^4}{64}\right)}$
$\delta =\frac{W{l}^3}{48\times \left(\frac{4Wl}{\pi d^{2}e}\right)\times \left(\frac{\pi d^4}{64}\right)}$
$\delta =\frac{W{l}^3\times \pi d^{2}e\times 64}{48\times 4Wl\times \pi d^4}$
$\delta =\frac{64W{l}^3\pi d^{2}e}{192Wl\pi d^4}$
$\delta =\frac{e{l}^2}{3d^2}$.

Therefore, the deflection is: $\frac{e{l}^2}{3d^2}$.
Correct answer: Option C $\frac{e{l}^2}{3d^2}$.