Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1758: NEET 2012 Physics - Class 12 AIPM-2012-Ph-1758 Solution: A ring is made of a wire having a resistance \(R_0 = 12 \).…

A ring is made of a wire having a resistance R0=12ΩR_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 point is equal to 83Ω\frac{8}{3}\Omega. l1l2=58\frac{l_1}{l_2} = \frac{5}{8} l1l2=13\frac{l_1}{l_2} = \frac{1}{3} l1l2=38\frac{l_1}{l_2} = \frac{3}{8} l1l2=12\frac{l_1}{l_2} = \frac{1}{2}

Multiple Choice Options

  • A f1+f2{f_1} + {f_2} f 1 _ + f 2 _
  • B −- _ f1+f2{f_1} + {f_2} f 1 _ + f 2 _
  • C 2 f1+f2{f_1} + {f_2} f 1 _ + f 2 _
  • D −- _ 2 f1+f2{f_1} + {f_2} f 1 _ + f 2 _

Step-by-Step Verified Solution

_ d = 2f 1 + f 2

Related Learning Resources & Exam Preparation

NEET PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring