Question

On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10cm.$ The resistance of their series combination is $1K\Omega$. 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 $550\Omega$

$R_1+R_2=1000\Rightarrow R_2=1000-R_1$


On balancing condition, $R_1(100-l)=(1000-R_1)l\dots \left(1\right)$
On lnter changing resistance


$(1000-R_1)(110-l)=R_1(l-10)$
or $R_1(l-10)=(1000-R_1)(110-l)\dots \dots \left(2\right)$

From (1) & (2)

$\frac{100-l}{l-10}=\frac{l}{110-l}$
$\Rightarrow (100-l)(110-l)=l(l-10)$
$\Rightarrow 11000-100l-110l+l=l-10l$
$\Rightarrow 11000=200l$
$l=55cm$
On putting in eq (1)
$R_1(100-55)=(1000-R_1)55$
$R_1\left(45\right)=(1000-R_1)55$
$R_1\left(9\right)=(1000-R_1)11$
$20R_1=11000$
$R_1=550\Omega$