JEE Mains · Physics · STD 12 - 3. current electricity
A meter bridge with two resistances \( R_{1} \) and \( R_{2} \) as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 \( \Omega \) resistance is connected in parallel to \( R_{2} \). The values of resistances \( R_{1} \) and \( R_{2} \) are __________ .

- A \( R_{2}=16\Omega, R_{1}=\frac{16}{3}\Omega \)
- B \( R_{2}=4\Omega, R_{1}=\frac{4}{3}\Omega \)
- C \( R_{2}=8\Omega, R_{1}=\frac{16}{3}\Omega \)
- D \( R_{2}=12\Omega, R_{1}=\frac{12}{3}\Omega \)
Answer & Solution
Correct Answer
(C) \( R_{2}=8\Omega, R_{1}=\frac{16}{3}\Omega \)
Step-by-step Solution
Detailed explanation
\( \frac{R_{1}}{R_{2}}=\frac{40}{60}=\frac{2}{3} \) \(\quad\)... (1) \( \frac{R_{1}}{(\frac{R_{2}\times16}{R_{2}+16})}=\frac{50}{50}\Rightarrow R_{1}=\frac{16R_{2}}{16+R_{2}} \) \(\quad\)...(2) \( \frac{2}{3}R_{2}=\frac{16R_{2}}{16+R_{2}} \)…
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