MHT CET · Physics · Current Electricity
In meter-bridge experiment a resistance of \(18 \Omega\) is connected in left gap and an unknown resistance \(\mathrm{R}\) is connected in right gap. The null point is obtained at ' \(\ell_{1}{ }^{\circ}\) from left end. If unknown resistance is replaced by \(\left(\frac{R}{3}\right) \Omega\), the null point is obtained at \(1 \cdot 5 \ell_{1}\). The unknown resistance is
- A \(9 \Omega\)
- B \(36 \Omega\)
- C \(18 \Omega\)
- D \(27 \Omega\)
Answer & Solution
Correct Answer
(C) \(18 \Omega\)
Step-by-step Solution
Detailed explanation
(A)
\(\frac{18}{\ell_{1}}=\frac{\mathrm{R}}{100-\ell_{1}}\)...(1)
\(\frac{18}{1.5 \ell_{1}}=\frac{\mathrm{R} / 3}{100-1.5 \ell_{1}}=\frac{\mathrm{R}}{300-4.5 \ell_{1}}\)...(2)
Dividing Eq. (1) by Eq. (2) and solving, \(\ell_{1}=50 \mathrm{~cm}\)
Substituting this value of \(\ell_{1}\) in Eq. (1)
we get \(\mathrm{R}=18 \Omega\)
\(\frac{18}{\ell_{1}}=\frac{\mathrm{R}}{100-\ell_{1}}\)...(1)
\(\frac{18}{1.5 \ell_{1}}=\frac{\mathrm{R} / 3}{100-1.5 \ell_{1}}=\frac{\mathrm{R}}{300-4.5 \ell_{1}}\)...(2)
Dividing Eq. (1) by Eq. (2) and solving, \(\ell_{1}=50 \mathrm{~cm}\)
Substituting this value of \(\ell_{1}\) in Eq. (1)
we get \(\mathrm{R}=18 \Omega\)
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