Question

A metallic cylindrical wire 'A' has length 10 cm and radius 3 mm. Another hollow cylindrical wire 'B' of the same metal has length 10 cm, inner radius 3 mm and outer radius 4 mm. The ratio of the resistance of the wires A to B is: 

• A. \(\frac{7}{9}\)
• B. \(\frac{9}{7}\)
• C. \(\frac{9}{16}\)
• D. \(\frac{16}{9}\)
• E. \(\frac{3}{4}\)

Answer 1

The Correct Option is A

Given:

• Wire A: Solid cylinder with length \( L = 10 \, \text{cm} \), radius \( r_A = 3 \, \text{mm} \)
• Wire B: Hollow cylinder with same length \( L = 10 \, \text{cm} \), inner radius \( r_{\text{inner}} = 3 \, \text{mm} \), outer radius \( r_{\text{outer}} = 4 \, \text{mm} \)
• Both wires are made of the same metal (same resistivity \( \rho \))

Step 1: Resistance Formula

The resistance of a cylindrical wire is given by:

\[ R = \rho \frac{L}{A} \]

where \( \rho \) is resistivity, \( L \) is length, and \( A \) is cross-sectional area.

Step 2: Calculate Cross-Sectional Areas

For Wire A (solid cylinder):

\[ A_A = \pi r_A^2 = \pi (3 \, \text{mm})^2 = 9\pi \, \text{mm}^2 \]

For Wire B (hollow cylinder):

\[ A_B = \pi (r_{\text{outer}}^2 - r_{\text{inner}}^2) = \pi (4^2 - 3^2) = \pi (16 - 9) = 7\pi \, \text{mm}^2 \]

Step 3: Calculate Resistances

Resistance of Wire A:

\[ R_A = \rho \frac{L}{A_A} = \rho \frac{10 \, \text{cm}}{9\pi \, \text{mm}^2} \]

Resistance of Wire B:

\[ R_B = \rho \frac{L}{A_B} = \rho \frac{10 \, \text{cm}}{7\pi \, \text{mm}^2} \]

Step 4: Compute Resistance Ratio

The ratio \( \frac{R_A}{R_B} \) is:

\[ \frac{R_A}{R_B} = \frac{\rho \frac{L}{9\pi}}{\rho \frac{L}{7\pi}} = \frac{7\pi}{9\pi} = \frac{7}{9} \]

Conclusion:

The ratio of the resistance of wire A to wire B is \( \frac{7}{9} \).

Answer: \(\boxed{A}\)

Answer 2

Step 1: Recall the formula for the resistance of a cylindrical conductor.

The resistance \( R \) of a cylindrical conductor is given by:

\[ R = \rho \frac{L}{A}, \]

where:

• \( \rho \) is the resistivity of the material,
• \( L \) is the length of the conductor, and
• \( A \) is the cross-sectional area of the conductor.

 

Step 2: Calculate the resistance of wire 'A'.

Wire 'A' is a solid cylindrical wire with radius \( r_A = 3 \, \text{mm} = 0.3 \, \text{cm} \). Its cross-sectional area is:

\[ A_A = \pi r_A^2 = \pi (0.3)^2 = 0.09\pi \, \text{cm}^2. \]

The resistance of wire 'A' is:

\[ R_A = \rho \frac{L}{A_A} = \rho \frac{10}{0.09\pi}. \]

Step 3: Calculate the resistance of wire 'B'.

Wire 'B' is a hollow cylindrical wire with inner radius \( r_{\text{inner}} = 3 \, \text{mm} = 0.3 \, \text{cm} \) and outer radius \( r_{\text{outer}} = 4 \, \text{mm} = 0.4 \, \text{cm} \). Its cross-sectional area is the difference between the areas of the outer and inner circles:

\[ A_B = \pi r_{\text{outer}}^2 - \pi r_{\text{inner}}^2 = \pi (0.4)^2 - \pi (0.3)^2 = \pi (0.16 - 0.09) = 0.07\pi \, \text{cm}^2. \]

The resistance of wire 'B' is:

\[ R_B = \rho \frac{L}{A_B} = \rho \frac{10}{0.07\pi}. \]

Step 4: Find the ratio of resistances \( R_A / R_B \).

The ratio of resistances is:

\[ \frac{R_A}{R_B} = \frac{\rho \frac{10}{0.09\pi}}{\rho \frac{10}{0.07\pi}} = \frac{0.07}{0.09} = \frac{7}{9}. \]

Final Answer: The ratio of the resistances of wires A to B is \( \mathbf{\frac{7}{9}} \), which corresponds to option \( \mathbf{(A)} \).

Answer 3

The correct option is (A): \(\frac{7}{9}\)
\(B=  \frac{ρL}{A}​\)
where:

• ρ is the resistivity of the material,
• L is the length of the wire,
• A is the cross-sectional area of the wire.
  Given:
• Wire A is a solid cylinder with length L=10 cm and radius r=3 mm.
• Wire B is a hollow cylinder with length L=10 cm, inner radius ri​=3 mm, and outer radius r_o​=4 mm.
  By calculating the cross-sectional  and resistances area using the resistance formula \(R=\frac{ρL}{A}\)
  ​Finally, we find the ratio \(\frac{R_A}{R_B}\)​​:
  \(\frac{R_A}{R_B}\)=\(\frac{7}{9}\)