Question

On interchanging the resistances, the balance point of a metre bridge shifts to the left by $10 \,cm$. The resistance of their series combination is $100\, \Omega$. Initially, if $R_{1}$ and $R_{2}$ are the resistance (in $\Omega$ ) on the left slot and right slots respectively, then $R _{1}- R _{2}=$ ______$\Omega$.

Answer 1

Let balancing length be $l$,
$\therefore \frac{ R _{1}}{ R _{2}}=\frac{l}{100-l}$ ... (i)
If $R_{1}$ and $R_{2}$ are interchanged,
balancing length becomes, $(l-10)$
$\therefore \frac{ R _{2}}{ R _{1}}=\frac{l-10}{[100-(l-10)]}=\frac{l-10}{110-l}$ ....(ii)
From equations (i) and (ii),
$ \frac{l}{100-l}=\frac{110-l}{l-10} $
$\therefore l^{2}-10 l=(110 \times 100)+\left(l^{2}-210 l\right) $
$\therefore 200 l=110 \times 100$
$\therefore l=55 \,cm$
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}=100 \,\Omega $....(iv)
On solving equations (iii) and (iv), we get,
$R _{1}=55\, \Omega $ and $ R _{2}=45\, \Omega$
$\therefore R _{1}- R _{2}=55-45=10\, \Omega$