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 $1 \mathrm{k} \Omega$. How much was the resistance on the left slot before interchanging the resistances?

JEE Main · 2018 · 08 Apr 2018 · Offline · Current Electricity

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 Ω

. How much was the resistance on the left slot before interchanging the resistances?

Answer: (D)

With the bridge balanced at

ℓ

cm,

\frac{R _{1}}{R _{2}} = \frac{ℓ}{100 - ℓ}

, and after swapping the resistances the balance moves

10

cm to the left so that

\frac{R _{2}}{R _{1}} = \frac{ℓ - 10}{110 - ℓ}

. Multiplying the two relations forces

\frac{ℓ ( ℓ - 10 )}{( 100 - ℓ ) ( 110 - ℓ )} = 1

, whose solution is

ℓ = 55

cm. Then

\frac{R _{1}}{R _{2}} = \frac{55}{45} = \frac{11}{9}

, and with

R_{1} + R_{2} = 1000 Ω

this gives

R_{1} = \frac{11}{20} \times 1000 = 550 Ω

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option