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 $\mathrm{F}_{1}$ is applied to P, an extension of $\Delta l$ is produced. The force which is required to produce same extension in Q is $\mathrm{F}_{2}$. The value of $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}$ is ______.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Elasticity & SHM

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

Δ 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 ______.

Answer: 16

Both wires have the same volume

V

and are made of the same material (same Young's modulus

Y

). Since

V = A L

, the length is

L = V / A

, and from

Δ ℓ = \frac{F L}{A Y}

Δ ℓ = \frac{F ( V / A )}{A Y} = \frac{F V}{A ^{2} Y}

For the same extension

Δ ℓ

, the same

V

and the same

Y

F \propto A^{2} ⟹ \frac{F _{1}}{F _{2}} = (\frac{A _{1}}{A _{2}})^{2} = (\frac{4}{1})^{2} = 16

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer