Q 12-03-294JEE MainJEE Main 2018 (8 Apr)Medium
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 $1\ \text{k}\Omega$. How much was the resistance on the left slot before interchanging the resistances?
Answer: (D) $550\ \Omega$
If the balance point is at $l$ cm from the left end, interchanging the resistances moves it to $100 - l$. A shift to the left by $10$ cm means
$$100 - l = l - 10 \;\Rightarrow\; l = 55\ \text{cm}$$
With $X$ on the left and $1000 - X$ on the right:
$$\frac{X}{1000 - X} = \frac{55}{45} \;\Rightarrow\; X = 550\ \Omega$$
Solution by Sreeraj P, M.Sc Physics