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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2026
  • Hard
  • MCQ

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 __________ .

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Answer & Solution

Correct Answer

(C) \( R_{2}=8\Omega, R_{1}=\frac{16}{3}\Omega \)

Step-by-step Solution

Detailed explanation


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\( \frac{R_{1}}{R_{2}}=\frac{40}{60}=\frac{2}{3} \) \(\quad\)... (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)
\( \frac{2}{3}R_{2}=\frac{16R_{2}}{16+R_{2}} \)
\( \frac{32}{3}+\frac{2R_{2}}{3}=16 \)
\( \frac{2R_{2}}{3}=16-\frac{32}{3}=\frac{16}{3} \)
\( R_{2}=8\Omega \)
By equation (1)
\( R_{1}=\frac{2}{3}R_{2}=\frac{16}{3}\Omega \)
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