MHT CET · Physics · Current Electricity
The unknown resistances are connected in two gaps of a metre bridge. The null point is at \(20 \mathrm{~cm}\) from zero end. A resistance of \(15 \Omega\) is connected in series with the smaller of the two. The null point shifts to \(40 \mathrm{~cm}\). The smaller resistance is
- A \(9 \Omega\)
- B \(7 \Omega\)
- C \(3 \Omega\)
- D \(5 \Omega\)
Answer & Solution
Correct Answer
(A) \(9 \Omega\)
Step-by-step Solution
Detailed explanation
\(\frac{20}{80}=\frac{r_{1}}{r_{2}}=\frac{1}{4} \Rightarrow 4 r_{1}=r_{2}\)
\(\frac{40}{60}=\frac{r_{1}+15}{r_{2}}=\frac{2}{3} \Rightarrow\left(r_{1}+15\right) 3=2 r_{2}=2 \times 4 r_{1}=8 r_{1}\)
\(3 \mathrm{r}_{1}+45=8 \mathrm{r}_{1}\)
\(5 \mathrm{r}_{1}=45 \quad \therefore \mathrm{r}_{1}=9 \Omega\)
\(\frac{40}{60}=\frac{r_{1}+15}{r_{2}}=\frac{2}{3} \Rightarrow\left(r_{1}+15\right) 3=2 r_{2}=2 \times 4 r_{1}=8 r_{1}\)
\(3 \mathrm{r}_{1}+45=8 \mathrm{r}_{1}\)
\(5 \mathrm{r}_{1}=45 \quad \therefore \mathrm{r}_{1}=9 \Omega\)
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