The diameter of a brass rod is 4 mm and Young’s modulus of brass is 9 x10¹⁰ N /m² . The force required to stretch by 0.1% of its length is

To find the force required to stretch the brass rod, we can use the definition of Young's modulus ($Y$).

Young's modulus is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$
where:
- $F$ is the stretching force applied.
- $A$ is the cross-sectional area of the rod.
- $\Delta L$ is the change in length.
- $L$ is the original length of the rod.
- $\frac{\mathrm{��}L}{L}$ is the longitudinal strain.

Rearranging the formula to solve for the force ($F$):
$F=Y·A·\left(\frac{\Delta L}{L}\right)$

Step 1: Calculate the cross-sectional area ($A$) of the rod
The diameter of the rod is given as $d=4\text{mm}=4·{10}^{-3}\text{m}$.
The radius ($r$) of the rod is:
$r=\frac{d}{2}=2\text{mm}=2·{10}^{-3}\text{m}$
The cross-sectional area ($A$) is:
$A=\pi ·r^2=\pi ·{\left(2·{10}^{-3}\text{m}\right)}^2=4\pi ·{10}^{-6}{\text{m}}^2$

Step 2: Determine the longitudinal strain ($\frac{\Delta L}{L}$)
The rod is stretched by $0.1\%$ of its length:
$\frac{\Delta L}{L}=0.1\%=\frac{0.1}{100}={10}^{-3}$

Step 3: Calculate the force ($F$)
Substitute the given values into the rearranged equation:
- Young's modulus of brass, $Y=9·{10}^{10}{\text{N/m}}^2$
- Cross-sectional area, $A=4\pi ·{10}^{-6}{\text{m}}^2$
- Longitudinal strain, $\frac{\Delta L}{L}={10}^{-3}$

Putting these values in:
$F=\left(9·{10}^{10}\right)·\left(4\pi ·{10}^{-6}\right)·\left({10}^{-3}\right)$
$F=36\pi ·{10}^{10-6-3}$
$F=36\pi ·{10}^1$
$F=360\pi \text{N}$

Therefore, the force required to stretch the brass rod is 360 πN.