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

Question

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 kΩ . How much was the resistance on the left slot before interchanging the resistances ? (A) 990 Ω (B) 505 Ω (C) 550 Ω (D) 910 Ω

Tardigrade Admin · Accepted Answer

(R1/R2)=(l/(100-l)) (R2/R1)=((l-10)/(110-l)) (100-l)(110-l)=l(l-10) 11000+l2-2 10 l=l2-10 l ⇒ l=55 cm R1=R2((55/45)) R1+R2=1000 Ω R1=550 Ω

Q. On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10 c m$ . The resistance of their series combination is $1 k Ω$ . How much was the resistance on the left slot before interchanging the resistances ?

$990 Ω$

$505 Ω$

$550 Ω$

$910 Ω$

$\frac{R _{1}}{R _{2}} = \frac{l}{( 100 - l )}$
$\frac{R _{2}}{R _{1}} = \frac{( l - 10 )}{( 110 - l )}$
$(100 - l) (110 - l) = l (l - 10)$
$11000 + l^{2} - 210 l = l^{2} - 10 l$
$\Rightarrow l = 55 c m$
$R_{1} = R_{2} (\frac{55}{45})$
$R_{1} + R_{2} = 1000 Ω$
$R_{1} = 550 Ω$

Q. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10cm. The resistance of their series combination is 1kΩ . How much was the resistance on the left slot before interchanging the resistances ?

990Ω

505Ω

550Ω

910Ω