In a meter bridge, initially the balance point is found at 50 cm. When the resistance in the left gap is increased by 5 ohms, the balance point shifts to 60 cm. Calculate the original resistances in t

Step 1: Understand the meter bridge principle. In a meter bridge, the balance condition is given by the ratio of resistances being equal to the ratio of the corresponding lengths: (R_1)/(R_2) = (L_1)/(L_2) Since the total length of the wire is 100 cm, if L_1 is the length from one end, then L_2 = (100 - L_1) cm. So, the formula becomes: (R_1)/(R_2) = (L_1)/(100 - L_1) Step 2: Apply the initial condition. Initially, the balance point is at L_1 = 50 cm. Substituting this into the formula: (R_1)/(R_2) = (50)/(100 - 50) (R_1)/(R_2) = (50)/(50) (R_1)/(R_2) = 1 This implies that R_1 = R_2. Let's call this original resistance R. So, R_1 = R and R_2 = R. Step 3: Determine which resistance was increased. The balance point shifts from 50 cm to 60 cm. This means L_1 increased. For L_1 to increase, the ratio (R_1)/(R_2) must increase. If R_1 is increased, then (R_1 + 5)/(R_2) would be greater than (R_1)/(R_2), causing L_1 to increase. If R_2 is increased, then (R_1)/(R_2 + 5) would be less than (R_1)/(R_2), causing L_1 to decrease. Since the balance point shifted from 50 cm to 60 cm (an increase in L_1), it means the resistance R_1 (the one corresponding to L_1) was increased by 5. Step 4: Apply the final condition. The new resistance in the first gap is R_1' = R + 5. The resistance in the second gap remains R_2' = R. The new balance point is L_1' = 60 cm. Substituting these values into the meter bridge formula: (R_1')/(R_2') = (L_1')/(100 - L_1') (R + 5)/(R) = (60)/(100 - 60) (R + 5)/(R) = (60)/(40) Step 5: Solve for R. Simplify the ratio (60)/(40): (R + 5)/(R) = (3)/(2) Cross-multiply: 2(R + 5) = 3R 2R + 10 = 3R Subtract 2R from both sides: 10 = 3R - 2R R = 10 \, The original resistances were R_1 = R and R_2 = R. Therefore, the original resistances are 10 \, and 10 \, . The original resistances are 10 and 10 .