Question

Consider two cylindrical conductors A and B, made of the same metal connected in series to a battery. The length and the radius of B are twice that of A. If \( \mu_A \) and \( \mu_B \) are the mobility of electrons in A and B respectively, then \( \frac{\mu_A}{\mu_B} \) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. 2
• D. 1
• A. \( 2.5 \times 10^4 \)
• B. \( 3.0 \times 10^5 \)
• C. \( 3.75 \times 10^6 \)
• D. \( 5.0 \times 10^7 \)
• A. \( 250^\circ\text{C} \)
• B. \( 500^\circ\text{C} \)
• C. \( 850^\circ\text{C} \)
• D. \( 1000^\circ\text{C} \)
• A. I decreases and II increases.
• B. I increases and II is almost constant.
• C. Both I and II increase.
• D. I decreases and II is almost constant.
• A. II and III change, but I is constant.
• B. I and II change, but III is constant.
• C. I and III change, but II is constant.
• D. All I, II, and III change.

Answer 1

The Correct Option is D

Step 1: Define the given quantities.
Let the length of A be \( l_A = l \), radius of A be \( r_A = r \). Then, for B: \( l_B = 2l \), \( r_B = 2r \). Since A and B are in series, the current \( I \) is the same. Step 2: Calculate the resistance.
Resistance \( R = \frac{\rho l}{A} \), where \( A = \pi r^2 \).
For A: \( R_A = \frac{\rho l}{\pi r^2} \).
For B: \( R_B = \frac{\rho (2l)}{\pi (2r)^2} = \frac{\rho l}{2\pi r^2} \). Step 3: Find the electric field.
Potential difference: \( V_A = I R_A = I \frac{\rho l}{\pi r^2} \), \( V_B = I R_B = I \frac{\rho l}{2\pi r^2} \).
Electric field: \( E_A = \frac{V_A}{l_A} = \frac{I \rho}{\pi r^2} \), \( E_B = \frac{V_B}{l_B} = \frac{I \rho}{4\pi r^2} \). Step 4: Relate current to mobility.
Current density \( J = n e \mu E \). Since A and B are the same metal, \( n \) and \( e \) are the same.
For A: \( \frac{I}{\pi r^2} = (n e \mu_A) \frac{I \rho}{\pi r^2} \quad \Rightarrow \quad \rho = \frac{1}{n e \mu_A} \).
For B: \( \frac{I}{4\pi r^2} = (n e \mu_B) \frac{I \rho}{4\pi r^2} \quad \Rightarrow \quad \rho = \frac{1}{n e \mu_B} \). Step 5: Find the ratio.
Equate the expressions for \( \rho \): \( \frac{1}{n e \mu_A} = \frac{1}{n e \mu_B} \quad \Rightarrow \quad \frac{\mu_A}{\mu_B} = 1 \).
Thus, the answer is option (D).

Answer 2

Step 1: Define the given quantities.
Length \( l = 0.5 \, \text{m} \), cross-sectional area \( A = 1.0 \times 10^{-7} \, \text{m}^2 \), voltage \( V = 2 \, \text{V} \), current \( I = 1.5 \, \text{A} \). Step 2: Calculate the resistance.
Using Ohm’s law \( V = I R \): \[ R = \frac{V}{I} = \frac{2}{1.5} = \frac{4}{3} \, \Omega. \] Step 3: Relate resistance to resistivity.
Resistance \( R = \frac{\rho l}{A} \). Solve for resistivity \( \rho \): \[ \rho = \frac{R A}{l} = \frac{\frac{4}{3} \times 1.0 \times 10^{-7}}{0.5} = \frac{\frac{4}{3} \times 10^{-7}}{0.5} = \frac{8}{3} \times 10^{-7} \, \Omega \cdot \text{m}. \] Step 4: Calculate conductivity.
Conductivity \( \sigma = \frac{1}{\rho} \): \[ \sigma = \frac{1}{\frac{8}{3} \times 10^{-7}}} = \frac{3}{8} \times 10^7 = 3.75 \times 10^6 \, \Omega^{-1} \cdot \text{m}^{-1}. \] Thus, the answer is option (C).

Answer 3

Step 1: Define the given quantities.
Temperature coefficient \( \alpha = 1.70 \times 10^{-4} \, ^\circ\text{C}^{-1} \), percentage increase in resistance \( \frac{\Delta R}{R} = 8.5\% = 0.085 \). Step 2: Use the temperature coefficient formula.
The change in resistance is: \[ \frac{\Delta R}{R} = \alpha \Delta T. \] Rearrange to find \( \Delta T \): \[ \Delta T = \frac{\frac{\Delta R}{R}}{\alpha}. \] Step 3: Substitute the values.
\[ \Delta T = \frac{0.085}{1.70 \times 10^{-4}}} = \frac{0.085}{1.70} \times 10^4 = 0.05 \times 10^4 = 500 \, ^\circ\text{C}. \] Thus, the temperature should be increased by 500°C, which matches option (B).

Answer 4

Step 1: Understand resistivity in a metal.
The resistivity of a metal is given by: \[ \rho = \frac{m}{n e^2 \tau}, \] where \( m \) is the electron mass, \( n \) is the number of electrons per unit volume, \( e \) is the electron charge, and \( \tau \) is the relaxation time. Step 2: Analyze the effect of temperature.
- Factor I (Relaxation time \( \tau \)): As temperature increases, atomic vibrations increase, causing more frequent electron collisions, reducing \( \tau \). So, \( \tau \) decreases.
- Factor II (Number of electrons \( n \)): In a metal, \( n \) is nearly constant with temperature, as thermal energy does not significantly change the number of free electrons. Step 3: Relate to resistivity.
Since \( \rho \propto \frac{1}{\tau} \) and \( n \) is constant, a decrease in \( \tau \) increases \( \rho \). Thus, the resistivity increases because I decreases and II is almost constant, which matches option (D).

Answer 5

Step 1: Define the quantities.
Steady current \( I \) is constant. The cross-sectional area \( A \) varies along the wire.
- I: Electric field \( E \).
- II: Current density \( J \).
- III: Drift speed \( v_d \). Step 2: Analyze each quantity.
- Current density: \( J = \frac{I}{A} \). Since \( I \) is constant and \( A \) varies, \( J \) changes.
- Drift speed: \( I = n e A v_d \), so \( v_d = \frac{I}{n e A} \). Since \( n, e, I \) are constant and \( A \) varies, \( v_d \) changes.
- Electric field: \( J = \sigma E \), so \( E = \frac{J}{\sigma} = \frac{I}{\sigma A} \). Since \( \sigma, I \) are constant and \( A \) varies, \( E \) changes. Step 3: Conclusion.
All three quantities—I (electric field), II (current density), and III (drift speed)—change along the wire due to the varying cross-section. Thus, the answer is option (D).