Question:
A copper wire of negligible mass, length 1 m and area of cross-section $10^{-6}\text{m}^{2}$ is kept on a smooth horizontal table. One end of the wire is fixed and other end of the wire attached with a ball of mass 1 kg. If the ball and wire are rotating with 20 rad $\text{s}^{-1}$, an elongation of $10^{-3}\text{m}$ is observed in the wire, find its Young's modulus
A copper wire of negligible mass, length 1 m and area of cross-section $10^{-6}\text{m}^{2}$ is kept on a smooth horizontal table. One end of the wire is fixed and other end of the wire attached with a ball of mass 1 kg. If the ball and wire are rotating with 20 rad $\text{s}^{-1}$, an elongation of $10^{-3}\text{m}$ is observed in the wire, find its Young's modulus
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Tie elasticity to mechanics: whenever a wire spins a mass, its tension force is $m\omega^2 r$.
Updated On: Jun 3, 2026
- $4\times10^{11}\text{Nm}^{-2}$
- $8\times10^{11}\text{Nm}^{-2}$
- $4\times10^{8}\text{Nm}^{-2}$
- $400 \text{ Nm}^{-2}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Young's Modulus ($Y$) of a material is defined as the ratio of tensile stress to tensile strain: $Y = \frac{F \cdot L}{A \cdot \Delta L}$, where $F$ is the stretching force, $L$ is the original length, $A$ is the cross-sectional area, and $\Delta L$ is the elongation.
Step 2: Meaning
As the ball rotates in a horizontal circle on a smooth table, the required centripetal force is provided entirely by the tension $F$ in the wire: $F = m\omega^2 L$.
Step 3: Analysis
Given data: $m = 1 \text{ kg}$, $\omega = 20 \text{ rad s}^{-1}$, $L = 1 \text{ m}$, $A = 10^{-6} \text{ m}^2$, and $\Delta L = 10^{-3} \text{ m}$. Calculating the tension force: $F = 1 \times (20)^2 \times 1 = 400 \text{ N}$. Now substituting into Young's modulus expression: $Y = \frac{400 \times 1}{10^{-6} \times 10^{-3}} = \frac{400}{10^{-9}} = 400 \times 10^9 = 4 \times 10^{11} \text{ Nm}^{-2}$.
Step 4: Conclusion
Hence, the calculated value of Young's modulus for the wire is $4 \times 10^{11} \text{ Nm}^{-2}$.
Final Answer: (A)
Young's Modulus ($Y$) of a material is defined as the ratio of tensile stress to tensile strain: $Y = \frac{F \cdot L}{A \cdot \Delta L}$, where $F$ is the stretching force, $L$ is the original length, $A$ is the cross-sectional area, and $\Delta L$ is the elongation.
Step 2: Meaning
As the ball rotates in a horizontal circle on a smooth table, the required centripetal force is provided entirely by the tension $F$ in the wire: $F = m\omega^2 L$.
Step 3: Analysis
Given data: $m = 1 \text{ kg}$, $\omega = 20 \text{ rad s}^{-1}$, $L = 1 \text{ m}$, $A = 10^{-6} \text{ m}^2$, and $\Delta L = 10^{-3} \text{ m}$. Calculating the tension force: $F = 1 \times (20)^2 \times 1 = 400 \text{ N}$. Now substituting into Young's modulus expression: $Y = \frac{400 \times 1}{10^{-6} \times 10^{-3}} = \frac{400}{10^{-9}} = 400 \times 10^9 = 4 \times 10^{11} \text{ Nm}^{-2}$.
Step 4: Conclusion
Hence, the calculated value of Young's modulus for the wire is $4 \times 10^{11} \text{ Nm}^{-2}$.
Final Answer: (A)
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