Question

On interchanging the resistances, the balance point of a metre bridge shifts to the left by $10cm$. The resistance of their series combination is $100Î\copyright$. Initially, if $R_1$ and $R_2$ are the resistance (in $Î\copyright$ ) on the left slot and right slots respectively, then $R_1−R_2=$ ______$Î\copyright$.

Answer 1

Answer: 10

Let balancing length be $l$,
$∴\frac{R_1}{R_2}=\frac{l}{100−l}$ ... (i)
If $R_1$ and $R_2$ are interchanged,
balancing length becomes, $(l−10)$
$∴\frac{R_2}{R_1}=\frac{l−10}{[100−(l−10)]}=\frac{l−10}{110−l}$ ....(ii)
From equations (i) and (ii),
$\frac{l}{100−l}=\frac{110−l}{l−10}$
$∴l^2−10l=(110×100)+\left(l^2−210l\right)$
$∴200l=110×100$
$∴l=55cm$
Substituting in equation (i), we get,
$\frac{R_1}{R_2}=\frac{55}{45}=\frac{11}{9}$......(iii)
When $R_1$ and $R_2$ are connected in series,
$R_1+R_2=100Î\copyright$....(iv)
On solving equations (iii) and (iv), we get,
$R_1=55Î\copyright$ and $R_2=45Î\copyright$
$∴R_1−R_2=55−45=10Î\copyright$