Problem 81 - A Complete Resource Book in Physics for JEE Main

Question

The same mass of copper is drawn into two wires \(1 \mathrm{~mm}\) and \(2 \mathrm{~mm}\) thick. Two wires are connected in series and current is passed through them. Heat produced in the wire is in the ratio (A) \(2: 1\) (B) \(1: 16\) (C) \(4: 1\) (D) \(16: 1\)

Step-by-Step Solution

Answer

The ratio of heat produced in the two wires is 4:1 (option C). This is calculated using the formula for heat produced, \(H=I^2*R*t\), along with the formula for resistance, \(R=\frac{\rho*L}{A}\). After finding the resistance of both wires, we can compute the heat produced in each and determine the ratio.

Step 1: Calculate the resistance of each wire

We will first find the resistance of each wire using the formula: $$ R=\frac{\rho*L}{A} $$, where R is the resistance, ρ (rho) is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. We are given that the thicknesses of the wires are 1 mm and 2 mm, so the cross-sectional areas will be: $$ A_1 = \pi (0.5 \times 10^{-3})^2 $$ $$ A_2 = \pi (1 \times 10^{-3})^2 $$ Let's assume that both wires have the same mass m. We know that mass = volume × density, and volume = area × length, so: $$ m = A_1*L_1*\rho_c = A_2*L_2*\rho_c $$ Here, ρ_c is the density of copper (which is a constant), so it cancels out: $$ \frac{A_1}{A_2} = \frac{L_2}{L_1} $$ Since $$ \frac{A_2}{A_1} = 4 $$ We conclude that $$ \frac{L_1}{L_2} = 4 $$ Now, we can calculate the resistance of both wires, using the given resistivity of copper ρ, and substituting the cross-sectional areas and lengths into the resistance formula: $$ R_1 = \frac{\rho*L_1}{A_1} = 4 *\rho*\frac{L_1}{\pi*(0.5\times10^{-3})^2} $$ $$ R_2 = \frac{\rho*L_2}{A_2}= \rho*\frac{L_1}{\pi*(1 \times 10^{-3})^2} $$

Step 2: Calculate the heat produced in each wire

Now, we will use the heat produced formula \( H=I^2*R*t \) for each wire: $$ H_1 = I^2*R_1*t = 4*I^2*\rho*L_1*\frac{t}{\pi*(0.5 \times 10^{-3})^2} $$ $$ H_2 = I^2*R_2*t = I^2*\rho*L_1*\frac{t}{\pi*(1 \times 10^{-3})^2} $$

Step 3: Find the ratio of heat produced in both wires

Finally, we will find the ratio of heat produced in both wires: $$ \frac{H_1}{H_2} = \frac{4*I^2*\rho*L_1*\frac{t}{\pi*(0.5 \times 10^{-3})^2}}{I^2*\rho*L_1*\frac{t}{\pi*(1 \times 10^{-3})^2}} $$ Some terms get canceled out, and we have: $$ \frac{H_1}{H_2} = \frac{4}{1} $$ Thus, the heat produced in the wires is in the ratio 4:1, which corresponds to option (C).