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 \fracF_1F_2 is \underline\hspace2cm .

Properties of Matter PYQ — JEE Main 2024 · 29 Jan · Shift 2 · Q87 with Solution

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Correct answer: 16

For a wire under tension,

\Delta l = \frac{F L}{A Y}

where:

Since both wires are of the same material, $Y$ is same for both.

Volume of a wire is

V = A L

Since both wires have the same volume,

A_P L_P = A_Q L_Q

Given,

A_P : A_Q = 4:1

So let

A_P = 4A, \quad A_Q = A

Then from equal volumes,

4A \cdot L_P = A \cdot L_Q

L_Q = 4L_P

Thus,

L_P : L_Q = 1:4

For wire $P$ ,

\Delta l = \frac{F_1 L_P}{A_P Y}

For wire $Q$ ,

\Delta l = \frac{F_2 L_Q}{A_Q Y}

Since the extension is same,

\frac{F_1 L_P}{A_P Y} = \frac{F_2 L_Q}{A_Q Y}

Cancel $Y$ :

\frac{F_1 L_P}{A_P} = \frac{F_2 L_Q}{A_Q}

So,

\frac{F_1}{F_2} = \frac{L_Q}{L_P}\cdot\frac{A_P}{A_Q}

Substitute ratios:

\frac{F_1}{F_2} = 4 \times 4 = 16

\boxed{16}

Stored correct answer = $16$ .

Our derived answer matches the stored answer.