Question

Calculate the area of cross section of a wire of length 2m, its resistance is 25Ω and the resistivity of material of wire is 1.84*10-6 Ωm.
Try to answer by own

Answer 1

l = 1.0 m

R = 23 Ohm

$p = 1.84 × {10}^{ - 6}$

ohm - metre

We have ,

$R = p \times \frac{l}{a}$

$23 = 1.84 \times {10}^{ - 6} \times \frac{1}{A}$

$A = 1.84 \times {10}^{ - 6} \times \frac{1}{A}$

$= 0.08 \times 10 - 6 \: m {}^{2}$

$= 8 \times 10 - {10}^{ - 6}$

$= 8 \times 10 - 8 {m}^{2}$

Answer 2

★ Given :-

• Length of wire = l = 2m
• Resistance = R = 25 Ω
• Resistivity = ρ = 1.84 × 10⁻⁶ Ω m

★ To find :-

→ Area of cross section = A = ?

★ Solution :-

Use the below formula:

$\boxed{ \purple{ \bf{R = \frac{ \rho \: l}{A} }}}$

$\implies \sf{A = \dfrac{1.84 \times {10}^{ - 6} \times 2}{25} }$

$\implies \sf{A = \dfrac{3.68 \times 10 {}^{ - 6} }{25} }$

$\implies \sf{A = \dfrac{3.68}{25} \times {10}^{ - 6} }$

$\implies \sf{A = 14.72 \times {10}^{ - 6} }$

$\therefore \boxed{ \boxed{ \red{ \bf{Area = 1.472 \times {10}^{ - 7} \: {m}^{2} }}}}$

Know more:

♦ Electrical resistivity is the constant of proportionality which increases when resistance and area of cross section of wire increases and decreases when the length of the wire decreases.

♦ It's SI unit is Ω m