In potentiometer experiment when terminals of the cell are at a distance of \[52cm\], then no current flows through it. When \[5\Omega \] shunt resistance is connected in it then balance length is at \[40cm\]. The internal resistance of the cell is:
A) \[5\Omega \]
B) \[1.5\Omega \]
C) \[\dfrac{200}{52}\Omega \]
D) \[\dfrac{52}{8}\Omega \]

Hint: Before proceeding with the solution, let us discuss a bit about a potentiometer. A potentiometer is an instrument for measuring voltage or 'potential difference' by comparison of an unknown voltage with a known reference voltage. The principle of a potentiometer is that the potential dropped across a segment of a wire of uniform cross-section carrying a constant current is directly proportional to its length.

Formula Used:
\[E=k.l\], \[V=E\dfrac{r}{R+r}\]

Complete step by step solution:
We know that no current flows through the circuit when the emf of the cell is balanced by the potential drop caused by the given length of the resistance.

In the given question, when no current flows through the circuit, we can say that \[E=k.52\] where \[E\] is the emf of the cell, \[52\] is the length which balances the emf, that is, \[52cm\] and \[k\] is a constant

Now, when the shunt is connected in the potentiometer, the balance length decreases. This means that now the potential drop by a lesser length can balance the emf, hence the resistance has increased. The new resistance is the sum of the shunt resistance and the internal resistance of the cell.

Let the internal resistance of the cell be \[r\text{ }\Omega \]
We have been given that, the resistance of the shunt connected \[(R)=5\Omega \]
The balance condition after the connection of the shunt can be represented as follows
\[E-E\dfrac{r}{R+r}=k.40\] where the balancing length is \[40cm\], \[E\] is the emf of the cell and \[E\dfrac{r}{R+r}\] is the potential drop caused by the length of the wire of the potentiometer
Simplifying the above equation, we get
\[\begin{align}
  & E\left( 1-\dfrac{r}{R+r} \right)=k.40 \\
 & \Rightarrow E\left( \dfrac{R+r-r}{R+r} \right)=k.40 \\
 & \Rightarrow E\left( \dfrac{R}{R+r} \right)=k.40 \\
\end{align}\]
In the beginning, when no shunt was connected, we wrote our emf value as \[E=k.52\]
Substituting the emf value in the shunt balancing equation, we will have
\[\begin{align}
  & k.52\left( \dfrac{R}{R+r} \right)=k.40 \\
 & \Rightarrow 52R=40(R+r) \\
 & \Rightarrow 52R-40R=40r \\
 & \Rightarrow 12R=40r \\
 & \Rightarrow r=\dfrac{12}{40}R \\
 & \Rightarrow r=0.3R \\
\end{align}\]
Now that we have successfully expressed internal resistance of the cell in terms of the shunt resistance, we can substitute the value of the shunt resistance
Therefore, the internal resistance of the cell \[(r)=0.3\times 5\Omega =1.5\Omega \]

Hence option (B) is the correct answer.

Note: Students often make the mistake of misinterpreting the question to mean that the shunt has been connected in parallel because the balance length decreases after the shunt connection. You must always keep in mind that a lesser balancing length implies a higher resistance and hence a series connection and not the other way around.