ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2023
  • Medium
  • Numerical

A null point is found at \(200\,cm\) in potentiometer when cell in secondary circuit is shunted by \(5\,\Omega\). When a resistance of \(15\,\Omega\) is used for shunting null point moves to \(300\,cm\). The internal resistance of the cell is \(..............\,\Omega\).

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(B) \(5\)

Step-by-step Solution

Detailed explanation

\(\frac{\varepsilon}{ r +5} \times 5=200\,x\)

\(\frac{\varepsilon \times 15}{ r +15}=300\,x\)

\(\Rightarrow r =5\)

Same subject
Explore more questions on app