(i) When a conductor is stretched, its resistance changes because resistance is proportional to the length of the conductor. The resistance is given by: \[ R = \rho \frac{l}{A} \] where \( l \) is the length and \( A \) is the cross-sectional area. When the length is doubled, the resistance also doubles: \[ R' = 2R \] (ii) The drift velocity is inversely proportional to the length of the conductor (as the potential difference and electric field remain the same): \[ v_d' = \frac{v_d}{2} \] Thus, the relations are: \[ R' = 2R \quad \text{and} \quad v_d' = \frac{v_d}{2} \] Thus, the final resistance is twice the initial resistance, and the final drift velocity is half the initial drift velocity.

The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. 
