JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Elasticity & SHM

Under the same load, wire A having length 5.0 m and cross section

2.5 \times 1 0^{- 5} m^{2}

stretches uniformly by the same amount as another wire B of length 6.0 m and a cross section of

3.0 \times 1 0^{- 5} m^{2}

stretches. The ratio of the Young's modulus of wire A to that of wire B will be:

Answer: (B)

Solution

Young's modulus is

Y = \frac{F L}{A Δ L}

. Both wires carry the same load

F

and are made to stretch by the same amount

Δ L

, so for each the modulus is proportional to

\frac{L}{A}

. The ratio is

\frac{Y _{A}}{Y _{B}} = \frac{L _{A} / A _{A}}{L _{B} / A _{B}} = \frac{L _{A} A _{B}}{L _{B} A _{A}} = \frac{5.0 \times ( 3.0 \times 1 0 ^{- 5} )}{6.0 \times ( 2.5 \times 1 0 ^{- 5} )} = \frac{15}{15}

. The two moduli stand in the ratio

1 : 1

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option