A metre bridge with resistors X and Y is represented in the given figure.

(a) Balance point from end A, l₁ = 39.5 cm

Resistance of the resistor, Y = 12.5 $\mathrm{\Omega }$

Condition for the balance is given as,

$\frac{\mathrm{X}}{\mathrm{Y}}=\frac{100-{\mathrm{l}}_1}{{\mathrm{l}}_1}$
$\mathrm{X}=\frac{100-39.5}{39.5}\times 12.5=8.2 \mathrm{\Omega }$

Therefore, the resistance of resistor X is 8.2 $\mathrm{\Omega }$.

The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

(b) If X and Y are interchanged, then l₁ and 100-l₁ get interchanged.

The balance point of the bridge will be 100-l₁ from A.

$100-{\mathrm{l}}_1=100-39.5=60.5 \mathrm{cm}$

Therefore, the balance point is 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.