Experimental Physics Question 2
Question 2 - 29 January - Shift 2
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 $\Omega$.
Show Answer
Answer: (5)
Solution:
Formula: Potentiometer
$ \frac{\varepsilon}{r+5} \times 5=200 x $
$\varepsilon \times 15$
$ \begin{aligned} & \frac{\varepsilon \times 1 J}{r+15}=300 x \\ & \Rightarrow r=5 \Omega \\ & \text{ Ans. } 5 \end{aligned} $
