Two metallic wires P and Q have the same volume and are made up o… — Solution | JEE Physics

MCQMediumJEE 2024Young's Modulus, Bulk & Rigidity Modulus

Two metallic wires P and Q have the same volume and are made up of the same material. If their areas of cross-section are in the ratio $4:1$ and force $F_1$ is applied to P, an extension of $\Delta \ell$ is produced. The force required to produce the same extension in Q is $F_2$ . The value of $F_1/F_2$ is:

A
$16$
B
$3$
C
$2$
D
$4$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The two wires have the same volume and are made of the same material. Their cross-sectional areas satisfy

\frac{A_P}{A_Q} = 4

and both wires have the same extension $\Delta \ell$ .

Find: The ratio $F_1/F_2$ .

Since volume is the same for both wires,

V = A \ell

A_P \ell_P = A_Q \ell_Q

Using

A_P = 4A_Q

we get

4A_Q \ell_P = A_Q \ell_Q

Hence,

\ell_Q = 4\ell_P

Using Young's Modulus Relation

For the same material, Young's modulus is constant. Using

Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta \ell / \ell}

we obtain

F = \frac{YA\Delta \ell}{\ell}

Direct Proportionality Trick

For fixed material and same extension $\Delta \ell$ , the force is proportional to

F \propto \frac{A}{\ell}

Now,

\frac{A_P}{A_Q} = 4

and from equal volume,

\frac{\ell_Q}{\ell_P} = 4

Therefore,

\frac{F_1}{F_2} = \frac{A_P/\ell_P}{A_Q/\ell_Q} = \frac{A_P}{A_Q} \cdot \frac{\ell_Q}{\ell_P} = 4 \times 4 = 16

Therefore, the correct option is A.

Substitution and Ratio

For wire P,

F_1 = \frac{YA_P \Delta \ell}{\ell_P}

For wire Q,

F_2 = \frac{YA_Q \Delta \ell}{\ell_Q}

Substitute

A_Q = \frac{A_P}{4}, \qquad \ell_Q = 4\ell_P

Then

F_2 = \frac{Y(A_P/4)\Delta \ell}{4\ell_P} = \frac{YA_P \Delta \ell}{16\ell_P}

Hence,

\frac{F_1}{F_2} = \frac{\frac{YA_P \Delta \ell}{\ell_P}}{\frac{YA_P \Delta \ell}{16\ell_P}} = 16

Therefore, the value of $F_1/F_2$ is $16$ , so the correct option is A.

Common mistakes

Using $F \propto A\ell$ instead of $F \propto A/\ell$ is incorrect because extension in a wire depends inversely on length. Start from Young's modulus and derive
$F = \frac{YA\Delta \ell}{\ell}$
before comparing the two wires.
Assuming equal volume means equal length is wrong. Equal volume gives
$A_P \ell_P = A_Q \ell_Q$
so when area changes, length must adjust inversely. Here the thinner wire is longer.
Taking $A_P:A_Q = 4:1$ but then writing $\ell_P:\ell_Q = 4:1$ is incorrect. For equal volume, length ratio is the inverse of area ratio, so
$\ell_P:\ell_Q = 1:4 $$.$

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