MHT CET · Physics · Current Electricity
Two known resistances are connected in two gaps of a meter bridge. The null point is obtained 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 resistances 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
Given \(l=20 \mathrm{~cm} \quad l^{\prime}=40 \mathrm{~cm}\)
Using \(\frac{R_1}{R_2}=\frac{l}{100-l}\)
\(\begin{aligned} & \therefore \frac{R_1}{R_2}=\frac{20}{100-20} \\ & \Rightarrow R_2=4 R_1\end{aligned}\)
Again using \(\frac{R_1}{R_2}=\frac{l^{\prime}}{100-l^{\prime}}\), where \(l^{\prime}=40 \mathrm{~cm}, R_1=R^{\prime}{ }_1+15\) and
\(\begin{aligned} & R_2^{\prime}=R_2=4 R_1 \\ & \therefore \frac{R_1+15}{4 R_1}=\frac{40}{100-40} \\ & \Rightarrow R_1=9 \Omega\end{aligned}\)
Using \(\frac{R_1}{R_2}=\frac{l}{100-l}\)
\(\begin{aligned} & \therefore \frac{R_1}{R_2}=\frac{20}{100-20} \\ & \Rightarrow R_2=4 R_1\end{aligned}\)
Again using \(\frac{R_1}{R_2}=\frac{l^{\prime}}{100-l^{\prime}}\), where \(l^{\prime}=40 \mathrm{~cm}, R_1=R^{\prime}{ }_1+15\) and
\(\begin{aligned} & R_2^{\prime}=R_2=4 R_1 \\ & \therefore \frac{R_1+15}{4 R_1}=\frac{40}{100-40} \\ & \Rightarrow R_1=9 \Omega\end{aligned}\)
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