Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4: 1 and force F₁ is applied to P, an extension of Δ l is produced. The force which is required to produce same extension in Q is F₂. The value of (F₁)/(F₂) is _________. Numerical answer type. Work it out, then open the answer.

Answer: 16. Same volume and material: F = (YA²Δ l)/(V) ∝ A² for the same Δ l. (F₁)/(F₂) = ((AP)/(AQ))² = (4/1)² (F₁)/(F₂) = 16

Two metallic wires P and Q have same volume and are made up of same… | JEE Main 2024 (29 Jan, Shift 2)

Q9JEE Main 2024 · 29 JAN · SHIFT 2Medium

Young's Modulus & Elongation

Two metallic wires $P$ and $Q$ have same volume and are made up of 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 $\Delta l$ is produced. The force which is required to produce same extension in $Q$ is $F_2$ .

The value of $\frac{F_1}{F_2}$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 16

Same volume and material: $F = \frac{YA^2\Delta l}{V} \propto A^2$ for the same $\Delta l$ .
$\frac{F_1}{F_2} = \left(\frac{A_P}{A_Q}\right)^2 = \left(\frac{4}{1}\right)^2$
$\frac{F_1}{F_2} = 16$

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