Question:
Two wires made of the same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If \(x\) is the increase in the length of the first wire, then the increase in the length of the second wire is
Two wires made of the same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If \(x\) is the increase in the length of the first wire, then the increase in the length of the second wire is
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For a stretched wire,
\[
\Delta L=\frac{FL}{AY}.
\]
Extension is directly proportional to length and inversely proportional to cross-sectional area. Since area depends on \(r^2\), even a small change in radius significantly affects the extension.
Updated On: Jun 26, 2026
- \(\dfrac{x}{3}\)
- \(3x\)
- \(9x\)
- \(\sqrt{3}x\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for extension of a wire.
The extension produced in a wire under a stretching force is given by \[ \Delta L=\frac{FL}{AY} \] where \[ F=\text{applied force}, \] \[ L=\text{length of wire}, \] \[ A=\text{cross-sectional area}, \] and \[ Y=\text{Young's modulus}. \] Since both wires are made of the same material, \[ Y=\text{constant}. \] Also, the same force \(F\) is applied to both wires.
Step 2: Express the dimensions of the two wires.
Let the length and radius of the second wire be \[ L \quad \text{and} \quad r. \] Then for the first wire, \[ L_1=3L, \] \[ r_1=3r. \] The cross-sectional areas are \[ A_1=\pi r_1^2 =\pi (3r)^2 =9\pi r^2, \] and \[ A_2=\pi r^2. \]
Step 3: Calculate the extension of the first wire.
Using \[ \Delta L=\frac{FL}{AY}, \] for the first wire, \[ \Delta L_1 = \frac{F(3L)} {(9\pi r^2)Y} = \frac{FL} {3\pi r^2Y}. \] Given that \[ \Delta L_1=x. \] Thus, \[ x=\frac{FL} {3\pi r^2Y}. \]
Step 4: Calculate the extension of the second wire.
For the second wire, \[ \Delta L_2 = \frac{FL} {\pi r^2Y}. \] Comparing with the expression for \(x\), \[ \Delta L_2 = 3\left( \frac{FL} {3\pi r^2Y} \right). \] Therefore, \[ \Delta L_2=3x. \]
Step 5: Final conclusion.
Hence, the increase in the length of the second wire is \[ \boxed{3x} \]
The extension produced in a wire under a stretching force is given by \[ \Delta L=\frac{FL}{AY} \] where \[ F=\text{applied force}, \] \[ L=\text{length of wire}, \] \[ A=\text{cross-sectional area}, \] and \[ Y=\text{Young's modulus}. \] Since both wires are made of the same material, \[ Y=\text{constant}. \] Also, the same force \(F\) is applied to both wires.
Step 2: Express the dimensions of the two wires.
Let the length and radius of the second wire be \[ L \quad \text{and} \quad r. \] Then for the first wire, \[ L_1=3L, \] \[ r_1=3r. \] The cross-sectional areas are \[ A_1=\pi r_1^2 =\pi (3r)^2 =9\pi r^2, \] and \[ A_2=\pi r^2. \]
Step 3: Calculate the extension of the first wire.
Using \[ \Delta L=\frac{FL}{AY}, \] for the first wire, \[ \Delta L_1 = \frac{F(3L)} {(9\pi r^2)Y} = \frac{FL} {3\pi r^2Y}. \] Given that \[ \Delta L_1=x. \] Thus, \[ x=\frac{FL} {3\pi r^2Y}. \]
Step 4: Calculate the extension of the second wire.
For the second wire, \[ \Delta L_2 = \frac{FL} {\pi r^2Y}. \] Comparing with the expression for \(x\), \[ \Delta L_2 = 3\left( \frac{FL} {3\pi r^2Y} \right). \] Therefore, \[ \Delta L_2=3x. \]
Step 5: Final conclusion.
Hence, the increase in the length of the second wire is \[ \boxed{3x} \]
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