Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k\(\Omega\). How much was the resistance on the left slot before interchanging the resistances? ____ \(\Omega\).

Hint:

In meter bridge problems, always use the balance condition \(\dfrac{R_1}{R_2} = \dfrac{l}{100-l}\) and apply the series/parallel constraints carefully.

Answer 1

Correct Answer: 550

Step 1: Let the resistance in the left slot be \(R\) and that in the right slot be \(S\). Given: \[ R + S = 1000 \, \Omega \quad \cdots (1) \]
Step 2: Let the initial balance point be at \(l\) cm from the left end. For a meter bridge at balance: \[ \frac{R}{S} = \frac{l}{100-l} \quad \cdots (2) \]
Step 3: After interchanging the resistances, the balance point shifts 10 cm to the left. So the new balance point is at \((l-10)\) cm. Thus, \[ \frac{S}{R} = \frac{l-10}{100-(l-10)} = \frac{l-10}{110-l} \quad \cdots (3) \]
Step 4: From equations (2) and (3), solving simultaneously: \[ l = 55 \text{ cm} \]
Step 5: Substitute \(l = 55\) in equation (2): \[ \frac{R}{S} = \frac{55}{45} = \frac{11}{9} \]
Step 6: Using equation (1): \[ R = \frac{11}{20} \times 1000 = 550 \, \Omega \] Final Answer (up to two decimal places): \[ \boxed{550.00} \]