JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Elasticity & SHM

A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is

2.0 \times 1 0^{11} N / m^{2}

. The longitudinal strain produced in the wire is

____

\times 1 0^{- 5}

Answer: 25

Solution

Strain is related to stress by Young's modulus:

Strain = \frac{Stress}{Y} = \frac{F / A}{Y}

. The cross-sectional area is

A = π r^{2} = π (0.02)^{2} = 4 π \times 1 0^{- 4}

m². The stress is

\frac{62.8 \times 1 0 ^{3}}{4 π \times 1 0 ^{- 4}} = \frac{62.8 \times 1 0 ^{3}}{1.256 \times 1 0 ^{- 3}}

. Therefore, strain

= \frac{62.8 \times 1 0 ^{3}}{1.256 \times 1 0 ^{- 3} \times 2 \times 1 0 ^{11}} = 2.5 \times 1 0^{- 4} = 25 \times 1 0^{- 5}

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer