JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Elasticity & SHM

A uniform metallic wire is elongated by 0.04 m when subjected to a linear force

F

. The elongation, if its length and diameter is doubled and subjected to the same force will be ____ cm.

Answer: 2

Solution

Under a given load the elongation is

Δ L = \frac{F L}{A Y}

. Doubling the length doubles the numerator, while doubling the diameter makes the area

A = \frac{π d ^{2}}{4}

four times as large. For the same force and the same material,

Δ L^{'} = \frac{F ( 2 L )}{( 4 A ) Y} = \frac{1}{2} \cdot \frac{F L}{A Y} = \frac{Δ L}{2}

so the new elongation is

\frac{0.04}{2} = 0.02

m, that is

2

cm.

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer