Two metallic wires P and Q have the same volume and are made up of the same material. If their area of cross-sections are in the ratio 4:1 and force is applied to P, an extension of is produced. The force required to produce the same extension in Q is ?

Question

Two metallic wires P and Q have the same volume and are made up of the same material. If their area of cross-sections are in the ratio 4:1 and force $F_{1}$ is applied to P, an extension of $∆ λ$ is produced. The force required to produce the same extension in Q is $F_{2}$ . The value of $\frac{F_{1}}{F_{2}}$ is _____. [2024]

Smart Achievers · Accepted Answer

(16)

$Y = \frac{Stress}{Strain} = \frac{F / A}{Δ ℓ / ℓ} = \frac{F ℓ}{A Δ ℓ}$

$Δ ℓ = \frac{F ℓ}{A Y}$

$V = A ℓ \Rightarrow ℓ = \frac{V}{A}$

$Δ ℓ = \frac{F V}{A^{2} Y}$

Y and V is same for both the wires

$Δ ℓ \propto \frac{F}{A^{2}}$

$\frac{Δ ℓ_{1}}{Δ ℓ_{2}} = \frac{F_{1}}{A_{1}^{2}} \times \frac{A_{2}^{2}}{F_{2}}$

$Δ ℓ_{1} = Δ ℓ_{2}$

$\frac{F_{1}}{F_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} = {(\frac{4}{1})}^{2} = 16$

Solution

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Solution

Ans. (16) Y=StressStrain=F/AΔℓ/ℓ=FℓAΔℓ Δℓ=FℓAY V=Aℓ⇒ℓ=VA Δℓ=FVA2Y Y and V is same for both the wires Δℓ∝FA2 Δℓ1Δℓ2=F1A12×A22F2 Δℓ1=Δℓ2 F1F2=A12A22=(41)2=16

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