© All Rights reserved @ LearnWithDash
Step-by-Step Solution
1. Understanding the Problem
We have two resistances placed on the left and right gaps of a meter bridge. Their series combination is given as 1 kΩ (i.e., 1000 Ω). The balance point of the meter bridge shifts 10 cm to the left upon interchanging these two resistances. We need to find the original resistance on the left side before interchanging.
2. Setting Up the Equations
Let the resistance on the left slot be $R_1$ and the one on the right slot be $R_2$. Since their series combination is 1 kΩ, we have:
$R_1 + R_2 = 1000 \, \Omega.$
If the initial balance point is at $L$ cm (measured from the left), the meter bridge principle tells us:
$ \frac{R_1}{R_2} \;=\; \frac{L}{(100 - L)}. $
After interchanging the two resistances, the new balance point is at $(L - 10)$ cm. Then:
$ \frac{R_2}{R_1} \;=\; \frac{L - 10}{100 - (L - 10)} \;=\; \frac{L - 10}{110 - L}. $
3. Relating the Ratios
Notice that:
$ \frac{R_2}{R_1} \;=\; \frac{1}{\,R_1 / R_2\,} \;=\; \frac{110 - L}{L - 10}. $
But from the first balance condition, we also have:
$ \frac{R_1}{R_2} \;=\; \frac{L}{100 - L}. $
Thus, equating the two forms for the ratio leads to:
$ \frac{L}{100 - L} \;=\; \frac{R_1}{R_2} \quad\text{and}\quad \frac{110 - L}{L - 10} \;=\; \frac{R_2}{R_1}. $
So,
$ \frac{R_1}{R_2} \;=\; \frac{L}{100 - L}
\quad\Longrightarrow\quad
\frac{R_2}{R_1} \;=\; \frac{100 - L}{L}.
$
By comparing this to $ \frac{110 - L}{L - 10}$, we get a condition for $L$:
$ \frac{100 - L}{L} \;=\; \frac{L - 10}{110 - L}. $
4. Solving for the Balance Length $L$
Cross-multiplying yields:
\[
(100 - L)\,(110 - L) \;=\; L\,(L - 10).
\]
Expand both sides:
\[
(100 \times 110) \;-\; 100L \;-\; 110L \;+\; L^2
\;=\; L^2 - 10L.
\]
\[
11000 \;-\; 210L \;+\; L^2
\;=\; L^2 - 10L.
\]
The $L^2$ terms cancel out, leaving:
\[
11000 - 210L = -10L
\quad\Longrightarrow\quad
11000 = 200L
\quad\Longrightarrow\quad
L = \frac{11000}{200} = 55 \,\text{cm}.
\]
5. Finding the Unknown Resistance $R_1$
Using the ratio from the first balance condition:
\[
\frac{R_1}{R_2} = \frac{L}{100 - L} = \frac{55}{45} = \frac{11}{9}.
\]
Also, $R_1 + R_2 = 1000\,\Omega$. Let $R_2 = 1000 - R_1$. Substituting into the ratio:
\[
\frac{R_1}{1000 - R_1} = \frac{11}{9}.
\]
Cross-multiply and solve for $R_1$:
\[
9\,R_1 = 11\,(1000 - R_1).
\]
\[
9R_1 = 11000 - 11R_1.
\]
\[
9R_1 + 11R_1 = 11000.
\]
\[
20R_1 = 11000
\quad\Longrightarrow\quad
R_1 = \frac{11000}{20} = 550\,\Omega.
\]
6. Final Answer
The resistance on the left slot before interchanging was 550 Ω.
