Question

On interchanging, the resistances, the balance point of a meter bridge shifts to left by $10\text{cm}$. The resistance of their series combination is $1k\Omega$. How much was the resistance on the left slot before interchanging the resistances?

• A. $910\Omega$
• B. $990\Omega$
• C. $505\Omega$
• D. $550\Omega$

Answer 1

The correct option is D $550\Omega$

Initial case-
$\frac{R_1}{(1000-R_1)}=\frac{l}{(100-l)}$
$R_1(100-l)=l(1000-R_1)...\left(1\right)$
Similarly, we can write
Final case -
$\frac{R_2}{R_1}=\frac{1000-R_1}{R_1}=\frac{l-10}{110-l}$
$(1000-R_1)(110-l)=(l-10)R_1...\left(2\right)$
By solving equation (1) and equation (2), we get $l=55\text{cm}$

Putting value of $l$ in equation (1)
$R_1(100-55)=55(1000-R_1)$
$45R_1=55000-55R_1$
$100R_1=55000$
$R_1=550\Omega$

On balancing condition
$R_1(100-l)=(1000-R_2)l$
On interchanging resistances