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. $4\times10^{11}\text{Nm}^{-2}$
• B. $8\times10^{11}\text{Nm}^{-2}$
• C. $4\times10^{8}\text{Nm}^{-2}$
• D. $400 \text{ Nm}^{-2}$

Hint:

Tie elasticity to mechanics: whenever a wire spins a mass, its tension force is $m\omega^2 r$.

Answer 1

The Correct Option is A

Step 1: Definition of Young's modulus.
It is stress divided by strain, which works out to $Y = \dfrac{F\,L}{A\,\Delta L}$, where $F$ stretches the wire of length $L$ and area $A$.

Step 2: What provides the stretching force?
The ball whirls in a horizontal circle, so the wire tension supplies the centre-pulling force. That tension is $F = m\omega^{2}L$.

Step 3: Find the force.
With $m=1$ kg, $\omega=20$ rad s$^{-1}$, $L=1$ m: \[ F = 1\times(20)^{2}\times 1 = 400 \text{ N} \]
Step 4: List the other values.
Area $A = 10^{-6}$ m$^{2}$ and stretch $\Delta L = 10^{-3}$ m.

Step 5: Put everything into the formula.
\[ Y = \frac{400\times 1}{10^{-6}\times 10^{-3}} = \frac{400}{10^{-9}} \]
Step 6: Simplify.
\[ Y = 400\times 10^{9} = 4\times 10^{11} \text{ N m}^{-2} \]This is option 1.
\[ \boxed{Y = 4\times 10^{11} \text{ N m}^{-2}} \]