To find the resistance of wire Q, we begin by using the formula for resistance: \[ R = \frac{\rho L}{A} \] where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
For wire P, the resistance is given as \( R = \frac{\rho L}{A} \).
Since wire Q has twice the diameter of wire P, its cross-sectional area \( A_Q \) is four times that of P (because area \( A \propto d^2 \)). Thus, \[ A_Q = 4A \].
Wire Q also has half the length of wire P, so \( L_Q = \frac{L}{2} \).
The resistance of wire Q, \( R_Q \), is: \[ R_Q = \frac{\rho L_Q}{A_Q} = \frac{\rho \left( \frac{L}{2} \right)}{4A} = \frac{\rho L}{8A} = \frac{R}{8} \].
Therefore, the resistance of wire Q is \( \frac{R}{8} \). The correct answer is \( \frac{R}{8} \).
In the given reaction sequence, the structure of Y would be: