Question

Two rods A and B (ohmic conductions at same temperatures) of same length are joined in series and a current / passes through it. The ratio of free electrons density ($n$), resistivity ($\rho$) and cross-section area ($A$) of both rods are 2, 2 and 0.5 respectively. List-I gives physical quantity of situation and List-II gives corresponding result. Match the list.

• A. Ratio of drift velocity of free electron in rod A to that of rod B is
• B. The ratio of electric field inside rod A to that of rod B is
• C. Ratio of potential difference between points 1 and 2 to that of point 2 and 3 is
• D. Ratio of average time taken by free electron to move from point 1 to 2 to that of point 2 to 3 is
• E. 0.5
• F. 1
• G. 2
• H. 4
• I. 3

Answer 1

P-2, Q-4, R-4, S-2

Answer 2

The problem involves two rods A and B of the same length joined in series, meaning the current $I$ flowing through them is the same. We are given the ratios of free electron density ($n$), resistivity ($\rho$), and cross-section area ($A$) for rod A to rod B. We need to find the ratios of drift velocity, electric field, potential difference, and time taken for electrons.

Given data:

1. Rods are in series, so $I_A = I_B = I$.
2. Length $L$ is the same for both rods: $L_A = L_B = L$.
3. Ratio of free electron density: $\frac{n_A}{n_B} = 2$.
4. Ratio of resistivity: $\frac{\rho_A}{\rho_B} = 2$.
5. Ratio of cross-section area: $\frac{A_A}{A_B} = 0.5$.

Let's calculate each quantity:

(P) Ratio of drift velocity of free electron in rod A to that of rod B ($v_{dA}/v_{dB}$): The current $I$ is related to drift velocity ($v_d$) by the formula: $I = n e A v_d$ Since $I$ is constant for both rods: $I_A = I_B$ $n_A e A_A v_{dA} = n_B e A_B v_{dB}$ $\frac{v_{dA}}{v_{dB}} = \frac{n_B A_B}{n_A A_A} = \left(\frac{n_B}{n_A}\right) \times \left(\frac{A_B}{A_A}\right)$ $\frac{v_{dA}}{v_{dB}} = \left(\frac{1}{n_A/n_B}\right) \times \left(\frac{1}{A_A/A_B}\right)$ Substitute the given ratios: $\frac{v_{dA}}{v_{dB}} = \left(\frac{1}{2}\right) \times \left(\frac{1}{0.5}\right) = 0.5 \times 2 = 1$ So, (P) matches with (2).

(Q) The ratio of electric field inside rod A to that of rod B ($E_A/E_B$): The electric field $E$ is related to resistivity $\rho$ and current density $J$ by: $E = \rho J$ Current density $J = \frac{I}{A}$. So, $E = \frac{\rho I}{A}$. For rod A and rod B: $E_A = \frac{\rho_A I}{A_A}$ $E_B = \frac{\rho_B I}{A_B}$ $\frac{E_A}{E_B} = \frac{(\rho_A I / A_A)}{(\rho_B I / A_B)} = \left(\frac{\rho_A}{\rho_B}\right) \times \left(\frac{A_B}{A_A}\right)$ $\frac{E_A}{E_B} = \left(\frac{\rho_A}{\rho_B}\right) \times \left(\frac{1}{A_A/A_B}\right)$ Substitute the given ratios: $\frac{E_A}{E_B} = (2) \times \left(\frac{1}{0.5}\right) = 2 \times 2 = 4$ So, (Q) matches with (4).

(R) Ratio of potential difference between points 1 and 2 to that of point 2 and 3 ($V_A/V_B$): Points 1 and 2 define rod A, and points 2 and 3 define rod B. The potential difference $V$ across a length $L$ with electric field $E$ is: $V = E L$ Since $L_A = L_B = L$: $\frac{V_A}{V_B} = \frac{E_A L_A}{E_B L_B} = \frac{E_A}{E_B}$ From (Q), we found $\frac{E_A}{E_B} = 4$. So, $\frac{V_A}{V_B} = 4$ So, (R) matches with (4).

(S) Ratio of average time taken by free electron to move from point 1 to 2 to that of point 2 to 3 ($t_A/t_B$): The time $t$ taken to travel a length $L$ with drift velocity $v_d$ is: $t = \frac{L}{v_d}$ Since $L_A = L_B = L$: $\frac{t_A}{t_B} = \frac{L_A/v_{dA}}{L_B/v_{dB}} = \frac{v_{dB}}{v_{dA}}$ From (P), we found $\frac{v_{dA}}{v_{dB}} = 1$. So, $\frac{t_A}{t_B} = \frac{1}{v_{dA}/v_{dB}} = \frac{1}{1} = 1$ So, (S) matches with (2).

Matching the lists: (P) - (2) (Q) - (4) (R) - (4) (S) - (2)