As illustrated in the figure, the potentiometer wire AB has a total...

In a balanced meter bridge, the Wheatstone bridge principle is applicable as the galvanometer shows zero deflection. The ratio of the resistances is equal to the ratio of the corresponding balancing lengths of the wire.

Given:
Total length of the wire $AB = 50 \text{ cm}$
Balancing length $AD = 30 \text{ cm}$
Length $DB = 50 - 30 = 20 \text{ cm}$
Left resistance $= 6\ \Omega$
Right resistance $= R$

Applying the balanced Wheatstone bridge condition:
$\dfrac{6}{R} = \dfrac{AD}{DB}$

$\dfrac{6}{R} = \dfrac{30}{20}$

$\dfrac{6}{R} = \dfrac{3}{2}$

$R = \dfrac{6 \times 2}{3} = 4\ \Omega$