Question

The Young's modulus and Poisson's ratio of a material are respectively Y and $\sigma$. The force required to decrease the area of cross-section of a wire made of this material by $\Delta A$ is

• A. $\frac{Y\Delta A}{4\sigma}$
• B. $\frac{2Y\Delta A}{\sigma}$
• C. $\frac{Y\Delta A}{2\sigma}$
• D. $\frac{Y\Delta A}{\sigma}$

Hint:

Remember the definitions of the elastic moduli. Young's Modulus ($Y$) relates tensile stress and strain. Poisson's Ratio ($\sigma$) relates the lateral (sideways) strain to the longitudinal (lengthwise) strain. For small changes, the fractional change in area is twice the fractional change in radius: $\frac{\Delta A}{A} = 2\frac{\Delta r}{r}$.

Answer 1

The Correct Option is C

Let the wire be subjected to a tensile force $F$.
Longitudinal stress = $\frac{F}{A}$, where A is the original cross-sectional area.
Longitudinal strain = $\frac{\Delta L}{L}$, where L is the original length.
Young's modulus is defined as $Y = \frac{\text{Longitudinal stress}}{\text{Longitudinal strain}} = \frac{F/A}{\Delta L/L}$. (Eq. 1)
Poisson's ratio ($\sigma$) is the ratio of lateral strain to longitudinal strain.
$\sigma = -\frac{\text{Lateral strain}}{\text{Longitudinal strain}}$. The negative sign indicates that if length increases, the radius decreases.
Lateral strain is the change in radius divided by the original radius, $\frac{\Delta r}{r}$.
So, $\sigma = - \frac{\Delta r/r}{\Delta L/L}$. (Eq. 2)
The area of cross-section is $A = \pi r^2$. The change in area is $\Delta A$. For small changes, we can use differentials: $dA = 2\pi r dr$. The fractional change in area is $\frac{dA}{A} = \frac{2\pi r dr}{\pi r^2} = 2\frac{dr}{r}$. So, $\frac{\Delta A}{A} \approx 2\frac{\Delta r}{r}$. The decrease in area means $\Delta A$ is negative, so let's use magnitudes. Let the decrease be $\Delta A$. Then the change in radius is $\Delta r = -\frac{\Delta A}{2A} \frac{1}{\pi r}$. No, this is getting complicated. Let's use magnitudes: $\frac{\Delta A}{A} = 2\frac{|\Delta r|}{r}$. So, lateral strain $|\frac{\Delta r}{r}| = \frac{1}{2}\frac{\Delta A}{A}$.
From Eq. 2, in magnitude: Longitudinal strain $\frac{\Delta L}{L} = \frac{\text{Lateral strain}}{\sigma} = \frac{\frac{1}{2}\frac{\Delta A}{A}}{\sigma} = \frac{\Delta A}{2A\sigma}$.
From Eq. 1, the required force is $F = Y \cdot A \cdot (\frac{\Delta L}{L})$.
Substitute the expression for longitudinal strain:
$F = Y \cdot A \cdot \left(\frac{\Delta A}{2A\sigma}\right)$.
The 'A' terms cancel out:
$F = \frac{Y\Delta A}{2\sigma}$.