When two resistance R1 and R2 connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω is introduced into the right gap, a null point is found at 60 cm from left side. When R1 and R2 are connected in parallel and introduced into the left gap, a resistance of 3 Ω is introduced into the right-gap to get null point at 40 cm from left end. The product of R 1 R 2 is Ω2

Question

When two resistance R1 and R2 connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω is introduced into the right gap, a null point is found at 60 cm from left side. When R1 and R2 are connected in parallel and introduced into the left gap, a resistance of 3 Ω is introduced into the right-gap to get null point at 40 cm from left end. The product of R 1 R 2 is Ω2 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

( R 1+ R 2/10)=(60/40)=(3/2) ⇒ R 1+ R 2=15 Now (R1 R2/(R1+R2) × 3)=(40/60)=(2/3) ⇒ R1 R2=30

10R1​+R2​​=4060​=23​ ⇒R1​+R2​=15 Now (R1​+R2​)×3R1​R2​​=6040​=32​ ⇒R1​R2​=30

$\frac{R _{1} + R _{2}}{10} = \frac{60}{40} = \frac{3}{2}$
$\Rightarrow R_{1} + R_{2} = 15$
Now $\frac{R _{1} R _{2}}{( R _{1} + R _{2} ) \times 3} = \frac{40}{60} = \frac{2}{3}$
$\Rightarrow R_{1} R_{2} = 30$