Question

On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10\text{cm}$. The resistance of their series combination is $1\text{k}\Omega$. How much was the resistance on the left slot before interchanging the resistance ?

• A. $910\Omega$
• B. $990\Omega$
• C. $505\Omega$
• D. $550\Omega$

Answer 1

The correct option is D $550\Omega$

Let the resistances be $R_1$ and $R_2$ in the left and right slots respectively.


Let the null point be at a distance $x$ from end $A$ of meter bridge wire $AB$.
$\frac{R_1}{R_2}=\frac{x}{100-x}..........\left(1\right)$

After interchanging the slot resistances,


After interchanging the null point becomes $(x-10)\text{cm}$ from end $A$ & other length will be,
$100-(x-10)=(110-x)\text{cm}$ from $B$.

$\therefore \frac{R_2}{R_1}=\frac{x-10}{110-x}.....\left(2\right)$

From eqs. $\left(1\right)\&\left(2\right)$

$\frac{100-x}{x}=\frac{x-10}{110-x}$

$x^2-100x-110x+11000=x^2-10x$

$200x=11000$

$\therefore x=\frac{11000}{200}=55\text{cm}$.

Given that $R_{net}$ for series combination of $R_1$ & $R_2$ is $1\text{k}\Omega$.

$\therefore R_1+R_2=1000\Omega .......\left(3\right)$

Also, from eq. $\left(1\right)$

$\frac{R_1}{R_2}=\frac{55}{45}=\frac{11}{9}$

$R_2=\frac{9R_1}{11}$

Substituting in eq. $\left(3\right)$ we get,

$R_1+\frac{9R_1}{11}=1000$

$\frac{20R_1}{11}=1000$

$\therefore R_1=\frac{11000}{20}=550\Omega$

Thus, resistance in left slot is $550\Omega$

Ans : $\left(d\right)$

Why this question ? Tip: The key concept involved here is the application of balanced bridge condition by assuming resistance $R_1$ & $R_2$ in left & right slots respectively. Also, their series equivalent is given for easy simplification of solution.