Question

A 10^3 kg load is placed on top of a lead bar of length 3.0 m. If the bar is compressed by 2.0
mm, calculate the cross-sectional area of the bar.

Answer 1

The cross-sectional area of the lead bar is $9.188 \times 10^{-4} \ m^2$.

Young's modulus of lead bar

The Young's modulus (E)of lead bar is given as 1.6 x 10¹⁰ Pa.

Cross-sectional area of the bar

The cross-sectional area of the bar is calculated as follows;

$E = \frac{stress}{strain} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \\\\A = \frac{Fl}{Ee}$

where;

• A is cross-sectional area
• F is applied force due to its weight
• E is Young's modulus
• e is compression
• l is length

$A = \frac{(mg) l}{Ee} \\\\A = \frac{(10^3 \times 9.8) \times 3}{1.6 \times 10^{10} \times 2 \times 10^{-3}} \\\\A = 9.188 \times 10^{-4} \ m^2$

Thus, the cross-sectional area of the lead bar is $9.188 \times 10^{-4} \ m^2$.

Learn more about Young's modulus here: https://brainly.com/question/6864866