A meter bridge with two resistances $$R_{1}$$ and $$R_{2}$$ as shown in figure was balanced (null point) at 40 cm from the point $$P$$. The null point changed to 50 cm from the point $$P$$, when 16 $$\Omega$$ resistance is connected in parallel to $$R_{2}$$. The values of resistances $$R_{1}$$ and $$R_{2}$$ are ______

step 1: initial balance

Null point at 40 cm from P, so

R₁ / R₂ = 40 / (100 − 40) = 40 / 60 = 2 / 3

So,

R₁ = (2/3) R₂ … (1)

step 2: after adding 16 Ω in parallel with R₂

New null point = 50 cm → perfectly centered

So,

R₁ / (R₂ || 16) = 50 / 50 = 1

That means:

R₁ = R₂ || 16

Equivalent of parallel:

R₂ || 16 = (16R₂) / (R₂ + 16)

So,

R₁ = (16R₂) / (R₂ + 16) … (2)

step 3: substitute (1) into (2)

(2/3)R₂ = (16R₂) / (R₂ + 16)

Cancel R₂:

2/3 = 16 / (R₂ + 16)

Cross multiply:

2(R₂ + 16) = 48
2R₂ + 32 = 48
2R₂ = 16
R₂ = 8 Ω

step 4: find R₁

R₁ = (2/3) × 8 = 16/3 Ω