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Compute the resistance in ohms of a lead block 15 cm long and [tex]$0.10 \, \text{cm}^2$[/tex] in cross-sectional area. ([tex]\rho = 2.2 \times 10^{-5} \, \text{ohm-cm}[/tex])

A. [tex]3.3 \times 10^{-3} \, \text{ohms}[/tex]
B. [tex]2.7 \times 10^{-3} \, \text{ohms}[/tex]
C. [tex]1.8 \times 10^{-3} \, \text{ohms}[/tex]
D. [tex]2.1 \times 10^{-3} \, \text{ohms}[/tex]

Sagot :

GinnyAnsweridn
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]
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