- Home
- Standard 12
- Physics
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 $1\ k\Omega$. How much was the resistance on the left slot before interchanging the resistances? .................. $\Omega$
$505 $
$550$
$910$
$990$
Solution
$\mathrm{R}_{1}+\mathrm{R}_{2}=1000$
$\Rightarrow \mathrm{R}_{2}=1000-\mathrm{R}_{1}$
On balancing condition
$\mathrm{R}_{1}(100-I)=\left(1000-\mathrm{R}_{1}\right) l$ …$(i)$
On Interchaanging resistance balance point shifts left by $10 \,cm$
On balancing condition $\left(1000-R_{1}\right)(110-l)=R_{1}(l-10)$
or, $\mathrm{R}_{1}(l-10)=\left(1000-\mathrm{R}_{1}\right)(110-l) \ldots(\mathrm{ii})$
Dividing eqn $(i)$ by $(ii)$
$\frac{100-l}{l-10}=\frac{l}{110-l}$
$\Rightarrow \quad(100-l)(110-l)=l(l-10)$
$ \Rightarrow 11000 – 100l – 110l + {l^2} = {l^2} – 10l$
$\Rightarrow 11000=200 l$
or, $l=55$
Putting the value of $'l'{\rm{in}}\,{\rm{eqn}}(i)$
$\mathrm{R}_{1}(100-55)=\left(1000-\mathrm{R}_{1}\right) 55$
$\Rightarrow \mathrm{R}_{1}(45)=\left(1000-\mathrm{R}_{1}\right) 55$
$\Rightarrow \mathrm{R}_{1}(9)=\left(1000-\mathrm{R}_{1}\right) 11$
$\Rightarrow 20 \mathrm{R}_{1}=11000$
$\therefore \quad \mathrm{R}_{1}=550\, \mathrm{K\Omega}$
