Question

A tapering bar varying from the diameters from ‘d\(_1\)' to ‘d\(_2\)' and a bar of uniform cross section ‘d' have the same length and subjected to same axial pull attains the same extension. Then the ‘d' is equal to

• A. Geometric mean of d\(_1\) and d\(_2\)
• B. Arithmetic mean of d\(_1\) and d\(_2\)
• C. Harmonic mean of d\(_1\) and d\(_2\)
• D. Meridian of d\(_1\) and d\(_2\)

Hint:

To remember this result easily, notice that the product of the two end diameters \( d_1 d_2 \) in the tapering bar's elongation formula simply replaces the squared diameter term \( d^2 \) from the uniform bar's formula. Setting them equal gives \( d^2 = d_1 d_2 \) instantly!

Answer 1

The Correct Option is A

Concept: According to Hooke's Law in axial deformation, the total elongation \( \delta \) of a bar under an axial tensile load \( P \) depends on its length \( L \), material Young's Modulus \( E \), and its cross-sectional area \( A \).
• For a bar with a uniform circular cross-section of diameter \( d \), the area is \( A = \frac{\pi}{4}d^2 \). Its elongation is: \[ \delta_{\text{uniform}} = \frac{PL}{AE} = \frac{PL}{\left(\frac{\pi}{4}d^2\right)E} = \frac{4PL}{\pi E d^2} \]
• For a circularly tapering bar whose diameter changes linearly from \( d_1 \) to \( d_2 \) over its length, integration of the varying cross-section yields the elongation formula: \[ \delta_{\text{tapering}} = \frac{4PL}{\pi E d_1 d_2} \]

Step 1: Setting up the equal elongation condition.
The problem states that both bars have the same length \( L \), are made of the same material (same \( E \)), are subjected to the same axial tensile pull \( P \), and undergo the exact same total extension: \[ \delta_{\text{uniform}} = \delta_{\text{tapering}} \]

Step 2: Substituting the elongation formulas into the equation.
Substitute our two elongation formulas into the equality: \[ \frac{4PL}{\pi E d^2} = \frac{4PL}{\pi E d_1 d_2} \]

Step 3: Canceling common algebraic factors.
We can cancel out the common multiplier terms \( \frac{4PL}{\pi E} \) from both sides of the equation: \[ \frac{1}{d^2} = \frac{1}{d_1 d_2} \]

Step 4: Solving for the uniform diameter variable \( d \).
Taking the reciprocal of both sides of the equation: \[ d^2 = d_1 d_2 \] Taking the square root of both sides isolates the uniform diameter \( d \): \[ d = \sqrt{d_1 d_2} \]

Step 5: Identifying the mathematical mean type.
The mathematical term \( \sqrt{a \cdot b} \) defines the Geometric Mean of two numbers. Therefore, the uniform diameter \( d \) is equal to the geometric mean of the two end diameters \( d_1 \) and \( d_2 \). This matches option (A).