A steel wire of length \(3.2 \mathrm{~m}\) (\(\mathrm{Y}_{\mathrm{S}}=2.0 \times 10^{11} \mathrm{Nm}^{-2}\)) and a copper wire of length 4.4 M ( \(\mathrm{Y}_{\mathrm{C}}=1.1 \times 10^{11} \mathrm{Nm}^{-2}\) ), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm . The load applied, in Newton, will be: (Given \(\pi=\frac{22}{7}\) )

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Elasticity & SHM

A steel wire of length

3.2 m

(

Y_{S} = 2.0 \times 1 0^{11} Nm^{- 2}

) and a copper wire of length 4.4 M (

Y_{C} = 1.1 \times 1 0^{11} Nm^{- 2}

), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm . The load applied, in Newton, will be: (Given

π = \frac{22}{7}

)

(A)360
(B)180
(C)1080
(D)154✓ Correct

Answer: (D)

Solution

Total elongation

Δ ℓ = Δ ℓ_{1} + Δ ℓ_{2} = \frac{F ℓ _{1}}{A Y _{1}} + \frac{F ℓ _{2}}{A Y _{2}}

. Since both wires have the same radius,

A = π r^{2}

is common. Substituting values:

1.4 \times 1 0^{- 3} = F (\frac{3.2}{π ( 1.4 \times 1 0 ^{- 3} ) ^{2} \times 2.0 \times 1 0 ^{11}} + \frac{4.4}{π ( 1.4 \times 1 0 ^{- 3} ) ^{2} \times 1.1 \times 1 0 ^{11}})

. Solving gives

F \approx 154

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Elasticity & SHM

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