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NEET 2012
Exam
NEET
Year
2012
Subject
Physics
Chapter
Current Electricity

NEET 2012 · Physics · Current Electricity

A ring is made of a wire having a resistance R_0=12 . Find the...

A ring is made of a wire having a resistance R0=12ΩR_0=12 \Omega. Find the points AA and BB, as shown in the figure, at which a current carrying conductor should be connected so that the resistance RR of the sub circuit between these points is equal to 8/3Ω8 / 3 \Omega A ring is made of a wire having a resistance R_0=12 . Find the points A and B , as shown in the figure, at which a

Options

  1. Al1l2=58\frac{l_1}{l_2}=\frac{5}{8}
  2. Bl1l2=13\frac{l_1}{l_2}=\frac{1}{3}
  3. Cl1l2=38\frac{l_1}{l_2}=\frac{3}{8}
  4. Dl1l2=12\frac{l_1}{l_2}=\frac{1}{2}

Correct answer

D. l1l2=12\frac{l_1}{l_2}=\frac{1}{2}

Step-by-step solution

We knows R∝lR \propto l Here, R1+R2=12Ω\quad R_1+R_2=12 \Omega and R1×R2R1+R2=83Ω\quad \frac{R_1 \times R_2}{R_1+R_2}=\frac{8}{3} \Omega ⇒R1R2=32Ω\Rightarrow \quad R_1 R_2=32 \Omega We get, R1=8R_1=8 and R2=4R_2=4 Again, R1=12l1l1+l2R_1=\frac{12 l_1}{l_1+l_2} and R2=12l2l1+l2R_2=\frac{12 l_2}{l_1+l_2} Hence, l1l2=12\quad \frac{l_1}{l_2}=\frac{1}{2}
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