The cross-sectional area of the lead bar is [tex]9.188 \times 10^{-4} \ m^2[/tex].
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;
[tex]E = \frac{stress}{strain} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \\\\A = \frac{Fl}{Ee}[/tex]
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
[tex]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[/tex]
Thus, the cross-sectional area of the lead bar is [tex]9.188 \times 10^{-4} \ m^2[/tex].
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