### Chapter 3: Electricity

Q
##### Electricity

Question:

A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Given:

Diameter, $d = 0.5\ mm$

Radius,$r = \frac{d}{2} = 0.25\ mm$

$r = 0.25 \times 10^{-3}\ m$

Resistivity, $\rho = 1.6 \times 10^{-8}\ \Omega m$

Resistance, $R = 10\ \Omega$

Length of wire,$L =\ ?$

We know,

$R = \frac{\rho L}{A} ----------(1)$

Here, the area of cross section

$A = \pi{r}^{2}$

$A = 3.14 \times 0.25 \times (10^{-3})^{2}$

$A = 3.14 \times 0.0625 \times 10^{-6}$

$A = 0.19625 \times 10^{-6}$

Putting this value in (1)

$10 = \frac{1.6 \times 10^{-8} L}{0.19625 \times 10^{-6}}$

$L = \frac{10 \times 0.19625}{1.6 \times 10^{-2}}$

$L = 10 \times 0.12265625 \times 10^2$

$L = 0.12265625 \times 10^3$

$L = 122.65\ m$

Again using,

$R = \frac{\rho L}{A}$

$R = \frac{\rho L}{\pi r^2}$

$R = \frac{\rho L}{\pi (\frac{d}{2})^{2}}$

$R = \frac{ \rho L}{\pi \frac{d^2}{4}}$

$R \propto \frac{1}{d^2}----------(2)$

The new resistance$R'$ when the diameter is double is given by,

$R' \propto \frac{1}{(2d)^2}$

$R' \propto \frac{1}{4(d)^2}---------(3)$

Dividing (3) by (2)

$\frac{R'}{R} = \frac{\frac {1}{4d^2}}{\frac{1}{d^2}}$

$\frac{{R}'}{R} = \frac{1}{4}$

$R' = \frac{R}{4}$

$R' = \frac{10}{4}$

$R' = 2.5\ \Omega$

So, the decrease in the resistance $= R - R'$

Decrease in the resistance $= 10 - 2.5 = 7.5\ \Omega$

Related Questions for Study

CBSE Class 10 Study Material