Question

A copper wire has a diameter 0.5 mm and resistivity 1.6× 10^-8

ohm m.

(i) What will be the length of this wire to make the resistance of 12 Ohm

(ii) How much will be the resistance of another copper wire of same length but half the

diameter?​

Answer 1

➡ Given:

✏ Diameter of cross-section = 0.5mm

✏ Resistivity = $\rm{1.6×10^{-8}\Omega{m}}$

➡ To Find:

1. Length of wire to make the resistance of 12$\Omega$
2. Resistance of another wire of same length but half the diameter.

➡ Formula:

✏ Formula of resistance is given by...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: \underline{ \boxed{ \bold{ \rm{ \pink{R = \frac{ \rho \times l}{A}}}}}} \: \star$

✏ R denotes resistance of wire.

✏ $\rho$ denotes resistivity of wire

✏ l denotes length of wire

✏ A denotes area of cross-section

➡ Calculation:

(1) For length of wire

$\implies \rm \: 12 = \frac{ 1.6 \times {10}^{ - 8} \times l}{\pi ({0.5 \times {10}^{ - 3}) }^{2} } \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \red{l = 588 \: m}}}}} \: \star$

(2) Resistance of other wire

$\implies \rm \: R \propto \frac{1}{ {r}^{2} } \\ \\ \therefore \rm \: \frac{R_1}{R_2} = \frac{ {r_2}^{2} }{ {r_1}^{2} } = 4 \\ \\ \therefore \rm \: \frac{12}{R_2} = 4 \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \purple{R_2 = 3 \Omega}}}}} \: \star$