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

Think about what happens if you swap the two ends of the tapering bar, calling the narrow end \( d_2 \) and the wide end \( d_1 \) instead. It is still physically the exact same bar, so its extension under the same load cannot change. This means the equivalent uniform diameter \( d \) has to come out the same whether it is written in terms of \( (d_1,d_2) \) or of \( (d_2,d_1) \), in other words the formula for \( d \) must treat \( d_1 \) and \( d_2 \) symmetrically. Checking this against the known extension relations, a uniform bar's extension goes as \( 1/d^2 \) while a tapering bar's extension goes as \( 1/(d_1 d_2) \), so equal extension forces \( d^2 = d_1 d_2 \), giving \( d = \sqrt{d_1 d_2} \). That is exactly the geometric mean of \( d_1 \) and \( d_2 \), option (A).