Q 12-03-280JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
In a meter bridge, resistance $X$ is in the left gap and resistance $Y$ in the right gap. It is given that resistance $Y = 12.5\ \Omega$ and that the balance is obtained at a distance $39.5\ \text{cm}$ from end $A$ (the left end of the wire) by jockey $J$. After interchanging the resistances $X$ and $Y$ a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ (in $\Omega$) and $l_2$?
Answer: (B) $8.16\ \Omega$ and $60.5\ \text{cm}$
At balance,
$$\frac{X}{Y} = \frac{l_1}{100 - l_1} = \frac{39.5}{60.5}$$
$$X = 12.5 \times \frac{39.5}{60.5} = 8.16\ \Omega$$
After interchanging, $\dfrac{Y}{X} = \dfrac{l_2}{100 - l_2}$, which is satisfied by $l_2 = 100 - 39.5 = 60.5\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics