Question

Two wires P and Q are made of the same material. Wire Q has twice the diameter and half the length of wire P. If the resistance of wire P is \( R \), the resistance of wire Q will be:

• A. \( R \)
• B. \( \frac{R}{2} \)
• C. \( \frac{R}{8} \)
• D. \( 2R \)

Hint:

Remember that resistance is inversely proportional to the cross-sectional area and directly proportional to the length of the wire.

Answer 1

The Correct Option is C

To determine the resistance of wire Q, we employ the resistance formula: \[ R = \frac{\rho L}{A} \], where \( \rho \) denotes resistivity, \( L \) represents length, and \( A \) signifies cross-sectional area.

For wire P, the resistance is expressed as \( R = \frac{\rho L}{A} \).

Given that wire Q has twice the diameter of wire P, its cross-sectional area, \( A_Q \), is four times that of wire P, as area \( A \) is proportional to the square of the diameter (\( A \propto d^2 \)). Therefore, \( A_Q = 4A \).

Furthermore, wire Q's length, \( L_Q \), is half the length of wire P, so \( L_Q = \frac{L}{2} \).

The resistance of wire Q, denoted as \( R_Q \), is calculated as follows: \[ R_Q = \frac{\rho L_Q}{A_Q} = \frac{\rho \left( \frac{L}{2} \right)}{4A} = \frac{\rho L}{8A} = \frac{R}{8} \].

Consequently, the resistance of wire Q is \( \frac{R}{8} \). The final answer is \( \frac{R}{8} \).