Q128 / 200 · NEET 2012 (Screening) · Physics

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

  • A
    $\frac{l_1}{l_2}=\frac{5}{8}$
  • B
    $\frac{l_1}{l_2}=\frac{1}{3}$
  • C
    $\frac{l_1}{l_2}=\frac{3}{8}$
  • Correct answer
    D
    $\frac{l_1}{l_2}=\frac{1}{2}$

Solution

We knows $R \propto l$ Here, $\quad R_1+R_2=12 \Omega$ and $\quad \frac{R_1 \times R_2}{R_1+R_2}=\frac{8}{3} \Omega$ $\Rightarrow \quad R_1 R_2=32 \Omega$ We get, $R_1=8$ and $R_2=4$ Again, $R_1=\frac{12 l_1}{l_1+l_2}$ and $R_2=\frac{12 l_2}{l_1+l_2}$ Hence, $\quad \frac{l_1}{l_2}=\frac{1}{2}$

Hence, the correct answer is option: 'D'.

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