Question

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 k12.
How much was the resistance on the left slot
before interchanging the resistances?​

Answer 1

Case I : \frac{R}{l}=\frac{1000-R}{100-l} eq 1

Case II: \frac{1000-R}{l-10}= \frac{R}{110-l} eq 2

Multiply both equation

\frac{R(1000-R)}{l(l-10)}= \frac{R(1000-R)}{(100-l)(110-l)} \Rightarrow l^{2}-10l=11000+l^{2}-210l

\Rightarrow 200l=11000

or l=55cm

\Rightarrow \frac{R}{55}=\frac{1000-R}{45}

or 45 R=55000-55R

or R=550\Omega