The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 \mathrm{~cm}$…

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
  1. $9 \Omega$
  2. $7 \Omega$
  3. $3 \Omega$
  4. $5 \Omega$

Solution

$\frac{20}{80}=\frac{r_{1}}{r_{2}}=\frac{1}{4} \Rightarrow 4 r_{1}=r_{2}$ $\frac{40}{60}=\frac{r_{1}+15}{r_{2}}=\frac{2}{3} \Rightarrow\left(r_{1}+15\right) 3=2 r_{2}=2 \times 4 r_{1}=8 r_{1}$ $3 \mathrm{r}_{1}+45=8 \mathrm{r}_{1}$ $5 \mathrm{r}_{1}=45 \quad \therefore \mathrm{r}_{1}=9 \Omega$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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