To compute the resistance in ohms of a lead block given the length, cross-sectional area, and resistivity, we can use the formula for calculating electrical resistance in a uniform material:
[tex]\[ R = \rho \frac{L}{A} \][/tex]
Where:
- [tex]\( R \)[/tex] is the resistance in ohms (Ω)
- [tex]\( \rho \)[/tex] is the resistivity in ohm-centimeters (Ω·cm)
- [tex]\( L \)[/tex] is the length of the conductor in centimeters (cm)
- [tex]\( A \)[/tex] is the cross-sectional area in square centimeters (cm²)
Given the values:
- Resistivity [tex]\( \rho = 2.2 \times 10^{-5} \)[/tex] ohm-cm
- Length [tex]\( L = 15 \)[/tex] cm
- Cross-sectional area [tex]\( A = 0.10 \)[/tex] cm²
We substitute these values into the formula:
[tex]\[ R = 2.2 \times 10^{-5} \frac{15}{0.10} \][/tex]
First, simplify the fraction:
[tex]\[ \frac{15}{0.10} = 150 \][/tex]
Now, we multiply the resistivity by this value:
[tex]\[ R = 2.2 \times 10^{-5} \times 150 \][/tex]
Perform the multiplication:
[tex]\[ R = 0.0033 \][/tex]
Therefore, the resistance of the lead block is [tex]\( 0.0033 \)[/tex] ohms.
Thus, the correct answer is:
[tex]\[ 3.3 \times 10^{-3} \, \text{ohms} \][/tex]
