JEE Main 2022 — Current Electricity Question with Solution

Initially, copper wire radius (r₁) = 9 mm

Resistance (R) = 14 $$\Omega$$

We know, $$R=\frac{\rho L}{A}=\frac{\rho L}{\pi r_1^2}=14$$

Now this copper wire is replaced by 7 parallel copper wire of resistance R₁.

$$\therefore$$ Equivalent resistance of 7 parallel copper wire,

$$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_1}$$

$$\Rightarrow \frac{1}{R_{eq}}=\frac{7}{R_1}$$

$$\Rightarrow R_{eq}=\frac{R_1}{7}$$

$$=\frac{1}{7}\times \frac{\rho L}{\pi r_2^2}$$

$$=\frac{1}{7}\times \frac{\rho L}{\pi {\left(\frac{r_1}{3}\right)}^2}$$ [as $$r_2=\frac{r_1}{3}$$ ; $$r_1=9$$ m and $$r_2=3$$]

$$=\frac{1}{7}\times \frac{\rho L}{\pi r_1^2}\times 9$$

$$=\frac{1}{7}\times 14\times 9$$

$$=18\Omega$$