In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance $$P = 4 \; \Omega$$ and the neutral point $$N$$ is at 60 cm from $$A$$. Now an unknown resistance $$R$$ is connected in series to $$P$$ and the new position of the neutral point is at 80 cm from $$A$$. The value of unknown resistance $$R$$ is -

We recall that the meter bridge works on the principle of the Wheatstone bridge. For a balanced bridge, the ratio of the resistances in the two gaps is equal to the ratio of the corresponding lengths of the wire, that is

$$\frac{\text{Resistance in left gap}}{\text{Resistance in right gap}} \;=\; \frac{\text{Length of wire from }A\text{ to }N} {\text{Length of wire from }N\text{ to the other end}}.$$

Initially only the resistance $$P$$ is placed in the left gap, while an unknown resistance, which we will call $$Q$$, is present in the right gap. We are told that $$P = 4 \;\Omega$$ and that the neutral point $$N$$ is found at 60 cm from end $$A$$. The total length of the wire is 100 cm, so the right-side length is

$$100\;\text{cm} - 60\;\text{cm} = 40\;\text{cm}.$$

Applying the Wheatstone‐bridge relation we have

$$\frac{P}{Q} = \frac{60}{40}.$$

Substituting $$P = 4\;\Omega$$ gives

$$\frac{4}{Q} = \frac{60}{40}.$$

First reduce the numerical fraction on the right:

$$\frac{60}{40} = \frac{3}{2}.$$

So

$$\frac{4}{Q} = \frac{3}{2}.$$

Cross-multiplying,

$$4 \times 2 = 3 \times Q,$$

$$8 = 3Q,$$

$$Q = \frac{8}{3}\;\Omega.$$

Now an additional resistance $$R$$ is connected in series with $$P$$, so the total resistance in the left gap becomes $$P + R$$. In the new situation the balance point shifts to 80 cm from end $$A$$. Hence the left-side length is 80 cm and the right-side length is

$$100\;\text{cm} - 80\;\text{cm} = 20\;\text{cm}.$$

Applying the Wheatstone relation again, we now write

$$\frac{P + R}{Q} = \frac{80}{20}.$$

Simplify the fraction on the right:

$$\frac{80}{20} = 4.$$

Therefore,

$$\frac{P + R}{Q} = 4.$$

Cross-multiplying gives

$$(P + R) = 4Q.$$

Substituting the known values $$P = 4\;\Omega$$ and $$Q = \dfrac{8}{3}\;\Omega$$, we get

$$4 + R = 4 \left(\frac{8}{3}\right).$$

Compute the right-hand side:

$$4 \left(\frac{8}{3}\right) = \frac{32}{3}\;\Omega.$$

Hence

$$4 + R = \frac{32}{3}.$$

To isolate $$R$$ subtract 4 Ω (that is, $$\dfrac{12}{3}\;\Omega$$) from both sides:

$$R = \frac{32}{3} - \frac{12}{3} = \frac{20}{3}\;\Omega.$$

This simplifies numerically to approximately 6.67 Ω, but the exact fractional form is already one of the options.

Hence, the correct answer is Option C.