Question

A steel rod with a circular cross-section of diameter 1cm and another steel rod with a square cross-section of side 1cm have equal mass. If the two rods are subjected to same tension, the ratio of the elongations of the two rods is

• A. 1
• B. 2/$\pi$
• C. 4/$\pi^2$
• D. 16/$\pi^2$

Hint:

In problems comparing properties of objects with equal mass and material, first use the mass equality ($M = \rho L A$) to find the ratio of their lengths or areas. Then use this ratio in the formula for the property you are comparing (like elongation, resistance, etc.).

Answer 1

The Correct Option is D

Step 1: Define variables and formulas.
Let the circular rod be rod 1 and the square rod be rod 2. Elongation formula: $\Delta L = \frac{FL}{AY}$, where $F$ is tension, $L$ is original length, $A$ is cross-sectional area, and $Y$ is Young's modulus. Mass formula: $M = \rho L A$, where $\rho$ is density.

Step 2: Determine the properties of the rods.
Rod 1 (circular): diameter $d_1=1$ cm, radius $r_1=0.5$ cm. Area $A_1 = \pi r_1^2 = \pi(0.5)^2 = 0.25\pi$ cm$^2$. Rod 2 (square): side $s_2=1$ cm. Area $A_2 = s_2^2 = 1^2 = 1$ cm$^2$. Both rods are steel (same $Y$, same $\rho$) and have equal mass ($M_1=M_2$) and are under the same tension ($F_1=F_2$).

Step 3: Relate the lengths of the rods.
Since the masses are equal: $M_1 = M_2 \implies \rho L_1 A_1 = \rho L_2 A_2$. \[ \frac{L_1}{L_2} = \frac{A_2}{A_1} = \frac{1}{0.25\pi} = \frac{4}{\pi}. \]

Step 4: Calculate the ratio of elongations.
We need to find the ratio $\frac{\Delta L_1}{\Delta L_2}$. \[ \frac{\Delta L_1}{\Delta L_2} = \frac{F_1 L_1 / (A_1 Y_1)}{F_2 L_2 / (A_2 Y_2)}. \] Since $F_1=F_2$ and $Y_1=Y_2$, this simplifies to: \[ \frac{\Delta L_1}{\Delta L_2} = \frac{L_1/A_1}{L_2/A_2} = \left(\frac{L_1}{L_2}\right) \left(\frac{A_2}{A_1}\right). \] From Step 3, we know both these ratios are equal to $4/\pi$. \[ \frac{\Delta L_1}{\Delta L_2} = \left(\frac{4}{\pi}\right) \left(\frac{4}{\pi}\right) = \frac{16}{\pi^2}. \]