10Ω
15Ω
25Ω
20Ω
Step 1: Initial Balance condition for meter bridge
[Refer Fig. in Question]
As shown in the figure in Question, initial balancing length is l1
So Applying balancing condition of meter bridge:
PS = l1/(100 - l1) => 5/R = l1/(100 - l1) (1)
Step 2: New balance condition after shunting R with equal resistance R
[Ref. Fig.]
S' = R × R/(R + R) = R/2
Let the new balancing length be L = 1.6l1
So we have:
5/S' = L/(100 - L) => 5/(R/2) = 1.6l1/(100 - 1.6l1) (2)
Step 3: Solving equations
Divide equation(1) by (2)
1/2 = l1(100 - 1.6l1)/(100 - l1) × 1.6l1 => 0.8 = (100 - 1.6l1)/(100 - l1) => 80 - 0.8l1 = 100 - 1.6l1 => 0.8l1 = 20 => l1 = 25 cm
Put the value of l1 in eq(1)
5/R = 25/75 => R = 15 Ω
Hence Option B is correct.