Current Electricity Question 10
Question 10 - 29 January - Shift 2
When two resistance $R_1$ and $R_2$ connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \Omega$ is introduced into the right gap, a null point is found at $60 cm$ from left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap, a resistance of $3 \Omega$ is introduced into the right-gap to get null point at 40 $cm$ from left end. The product of $R_1 R_2$ is $\Omega^{2}$
Show Answer
Answer: (30)
Solution:
$\frac{R_1+R_2}{10}=\frac{60}{40}=\frac{3}{2} \Rightarrow R_1+R_2=15$
Now $\frac{R_1 R_2}{(R_1+R_2) \times 3}=\frac{40}{60}=\frac{2}{3} \Rightarrow R_1 R_2=30$
