Current Electricity Ques 36
36. 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 $\frac{8}{3} \Omega$.
(a) $\frac{l_1}{l_2}=\frac{5}{8}$
(b) $\frac{l_1}{l_2}=\frac{1}{3}$
(c) $\frac{l_1}{l_2}=\frac{3}{8}$
(d) $\frac{l_1}{l_2}=\frac{1}{2}$
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Answer:
Correct Answer: 36.(d)
Solution:
- (d) Let $x$ is the resistance per unit length then
equivalent resistance $R=\frac{R_1 R_2}{R_1+R_2} \frac{(x l_1)(x l_2)}{x l_1+x l_2}$
$\begin{aligned} & \Rightarrow \quad \frac{8}{3}=x \frac{l_1 l_2}{l_1+l_2} \\ & \frac{8}{3}=x \frac{l_1}{\frac{l_1}{l_2}+1} \quad ….(i) \\ & \text { also } R_0=x l_1+x l_2 \\ & 12=x\left(l_1+l_2\right) \\ & 12=x l_2\left(\frac{l_1}{l_2}+1\right) \quad ….(ii) \\ & \frac{(i)}{(i i)} \Rightarrow \frac{\frac{8}{3}}{\frac{12}{1}}=\frac{\frac{x l_1}{\left(\frac{l_1}{l_2}+1\right)}}{x l_2\left(\frac{l_1}{l_2}+1\right)}=\frac{l_1}{l_2\left(\frac{l_1}{l_2}+1\right)^2} \\ & \left(\frac{l_1}{l_2}+1\right)^2 \times \frac{8}{36}=\frac{l_1}{l_2} \\ & \left(y^2+1+2 y\right) \times \frac{8}{36}=y\left(\text { where } y=\frac{l_1}{l_2}\right) \\ & 8 y^2+8+16 y=36 y \\ & \Rightarrow 8 y^2-20 y+8=0 \\ & \Rightarrow 2 y^2-5 y+2=0 \\ & \Rightarrow 2 y^2-4 y-y+2=0 \\ & \Rightarrow \quad 2 y(y-2)-1(y-2)=0 \\ & \Rightarrow(2 y-1)(y-2)=0 \\ & \Rightarrow \quad y=\frac{l_1}{l_2}=\frac{1}{2} \text { or } 2 \\ & \end{aligned}$
BETA
