Question

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

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

Answer 1

The correct option is A

$550\Omega$

Step 1. Given data:

1. The balance point is shifted by $\mathrm{x}=10\mathrm{cm}$
2. The resistance of the series combination${\mathrm{R}}_{\mathrm{s}}=1\mathrm{K\Omega }$

Step 2. Formula used:

1. For the balanced Wheatstone bridge,
   $\begin{array}{rcl}\frac{\mathrm{P}}{\mathrm{R}}& =& \frac{\mathrm{Q}}{\mathrm{S}}\left[\mathrm{P}=\mathrm{L},\mathrm{Q}=100-\mathrm{L}\right]\\ & \Rightarrow & \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}\\ & \Rightarrow & \frac{\mathrm{L}}{100-\mathrm{L}}=\frac{\mathrm{R}}{\mathrm{S}}\end{array}$(where $\mathrm{L}=$Balancing length)

Step 3: Calculation for the required resistance before interchanging.

$\mathrm{R},\mathrm{S}$are in a series combination.

According to the given data,

${\mathrm{R}}_{\mathrm{s}}=\mathrm{R}+\mathrm{S}$

Before interchanging the resistances,

$\begin{array}{rcl}\frac{\mathrm{L}}{100-\mathrm{L}}& =& \frac{\mathrm{R}}{\mathrm{S}}\end{array}$ --------------------(1)

After interchanging the resistances,

the balancing length becomes$\left(\mathrm{L}-10\right)$.

Now the equation becomes,

$\frac{L-10}{100-\left(L-10\right)}=\frac{S}{R}$

$\Rightarrow \frac{S}{R}=\frac{L-10}{110-L}$ ------------------(2)

From equations (1) and (2) we can get,

$$\begin{gathered} \frac{\mathrm{L}}{100-\mathrm{L}}=\frac{110-\mathrm{L}}{\mathrm{L}-10} \\ \Rightarrow {\mathrm{L}}^2-10\mathrm{L}=11000-100\mathrm{L}-110\mathrm{L}+{\mathrm{L}}^2 \\ \Rightarrow 200\mathrm{L}=11000 \\ \Rightarrow \mathrm{L}=55\mathrm{cm} \end{gathered}$$

Then,

$\begin{array}{rcl}\frac{\mathrm{R}}{\mathrm{S}}& =& \frac{\mathrm{L}}{100-\mathrm{L}}\\ & =& \frac{55}{45}\\ & =& \frac{11}{9}\\ \mathrm{then},\mathrm{S}& =& \left(\frac{9}{11}\right)\mathrm{R}\\ \mathrm{and}\mathrm{R}+\mathrm{S}& =& 1\mathrm{K\Omega }=1000\mathrm{\Omega }\\ & & \mathrm{By}\mathrm{simplifying},\\ & \Rightarrow & \mathrm{R}+\left(\frac{9}{11}\right)\mathrm{R}=1000\\ & \Rightarrow & \left(\frac{20}{11}\right)\mathrm{R}=1000\\ \mathrm{then},\mathrm{R}& =& \left(\frac{11000}{20}\right)=550\mathrm{\Omega }\\ & & \end{array}$

Thus, before interchanging the resistances the resistance on the left slot was$550\Omega$

Hence, option A is the correct answer.