Question

Two cylindrical conductors A and B, made of the same material, but of length $l$ and $2l$, and diameter $d$ and $\frac{d}{2}$, are connected in parallel across a battery. The drift speed of charge carriers inside the conductors is $v_A$ and $v_B$ respectively, and the rate of heat dissipation in the conductors is $H_A$ and $H_B$ respectively. Which of these options is/are correct?

• A. $\frac{v_A}{v_B}=\frac{1}{2}$
• B. $\frac{v_A}{v_B}=2$
• C. $\frac{H_A}{H_B}=4$
• D. $\frac{H_A}{H_B}=8$

Answer 1

Answer: (B), (D)

$\frac{v_A}{v_B}=2, \frac{H_A}{H_B}=8$

Answer 2

The problem involves two cylindrical conductors connected in parallel, meaning the potential difference across them is the same. We need to find the ratio of their drift speeds and the ratio of their rates of heat dissipation.

1. Calculate the Resistance of Each Conductor:

The resistance $R$ of a conductor is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area. The cross-sectional area of a cylinder is $A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$, where $D$ is the diameter.

For conductor A: Length $L_A = l$ Diameter $D_A = d$ Area $A_A = \frac{\pi d^2}{4}$ Resistance $R_A = \rho \frac{l}{\frac{\pi d^2}{4}} = \frac{4\rho l}{\pi d^2}$

For conductor B: Length $L_B = 2l$ Diameter $D_B = \frac{d}{2}$ Area $A_B = \frac{\pi (\frac{d}{2})^2}{4} = \frac{\pi \frac{d^2}{4}}{4} = \frac{\pi d^2}{16}$ Resistance $R_B = \rho \frac{2l}{\frac{\pi d^2}{16}} = \frac{32\rho l}{\pi d^2}$

Now, let's find the ratio of their resistances:

$$\frac{R_A}{R_B} = \frac{\frac{4\rho l}{\pi d^2}}{\frac{32\rho l}{\pi d^2}} = \frac{4}{32} = \frac{1}{8}$$

So, $R_B = 8R_A$.

2. Calculate the Ratio of Drift Speeds ($v_A/v_B$):

The current $I$ flowing through a conductor is related to the drift speed $v_d$ by the formula $I = n A v_d e$, where $n$ is the number density of charge carriers and $e$ is the elementary charge. From Ohm's Law, $I = \frac{V}{R}$, where $V$ is the potential difference across the conductor. Equating these two expressions for current:

$$\frac{V}{R} = n A v_d e$$

Thus, the drift speed $v_d = \frac{V}{R n A e}$.

Since the conductors are connected in parallel, the potential difference $V$ across them is the same. They are made of the same material, so the number density of charge carriers $n$ is the same for both. The elementary charge $e$ is a constant. Therefore, the drift speed is inversely proportional to the product of resistance and cross-sectional area ($v_d \propto \frac{1}{RA}$).

Let's calculate the product $RA$ for each conductor: For conductor A:

$$R_A A_A = \left(\frac{4\rho l}{\pi d^2}\right) \left(\frac{\pi d^2}{4}\right) = \rho l$$

For conductor B:

$$R_B A_B = \left(\frac{32\rho l}{\pi d^2}\right) \left(\frac{\pi d^2}{16}\right) = 2\rho l$$

Now, find the ratio of drift speeds:

$$\frac{v_A}{v_B} = \frac{\frac{1}{R_A A_A}}{\frac{1}{R_B A_B}} = \frac{R_B A_B}{R_A A_A} = \frac{2\rho l}{\rho l} = 2$$

So, $\frac{v_A}{v_B} = 2$.

3. Calculate the Ratio of Rate of Heat Dissipation ($H_A/H_B$):

The rate of heat dissipation, also known as power ($H$), in a resistor is given by $H = P = \frac{V^2}{R}$. Since the conductors are connected in parallel, the potential difference $V$ across them is the same. Therefore, the ratio of heat dissipation rates is:

$$\frac{H_A}{H_B} = \frac{\frac{V^2}{R_A}}{\frac{V^2}{R_B}} = \frac{R_B}{R_A}$$

From Step 1, we found $R_B = 8R_A$. Substitute this into the ratio:

$$\frac{H_A}{H_B} = \frac{8R_A}{R_A} = 8$$

So, $\frac{H_A}{H_B} = 8$.

Conclusion:

Based on our calculations:

• $\frac{v_A}{v_B} = 2$
• $\frac{H_A}{H_B} = 8$

Comparing these results with the given options, both $\frac{v_A}{v_B}=2$ and $\frac{H_A}{H_B}=8$ are correct.