JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Elasticity & SHM

Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are

Y_{1}

and

Y_{2}

. The combination behaves as a single wire then its Young's modulus is:

Answer: (D)

Solution

For each wire, Young's modulus is

Y = \frac{F / A}{Δ ℓ / ℓ}

, so the elongation is

Δ ℓ = \frac{F ℓ}{A Y}

. For wire 1:

Δ ℓ_{1} = \frac{F ℓ}{A Y _{1}}

. For wire 2:

Δ ℓ_{2} = \frac{F ℓ}{A Y _{2}}

. The total elongation is

Δ ℓ_{t o t a l} = Δ ℓ_{1} + Δ ℓ_{2} = \frac{F ℓ}{A} (\frac{1}{Y _{1}} + \frac{1}{Y _{2}})

. The effective Young's modulus for the combination of total length

2 ℓ

Y_{e f f} = \frac{F / A}{Δ ℓ _{t o t a l} / ( 2 ℓ )} = \frac{2 F ℓ}{A Δ ℓ _{t o t a l}} = \frac{2 F ℓ}{A \cdot \frac{F ℓ}{A} ( \frac{1}{Y _{1}} + \frac{1}{Y _{2}} )} = \frac{2}{\frac{1}{Y _{1}} + \frac{1}{Y _{2}}} = \frac{2 Y _{1} Y _{2}}{Y _{1} + Y _{2}}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option