Final answer:
Using the principles of a potentiometer and given lengths of wire, we can conclude that the internal resistance of the cell, when shunted by a 2Ω resistor, is 2Ω. Option d is the answer.
Step-by-step explanation:
In a potentiometer experiment, a cell that initially balances at a length of 240cm, when shunted with a 2Ω resistor, balances at half this length, 120cm. This change indicates the presence of an internal resistance in the cell. By applying the potentiometer principle, which states that the potential drop along a uniform wire is directly proportional to its length, we can deduce the internal resistance of the cell using the provided lengths and the resistance of the shunt.
Let's denote the electromotive force (emf) of the cell as E and the internal resistance as r.
The original balance point gives us E = I⋅(240), where I is the current. After shunting, the cell's terminal voltage V will be equal to I' (the new current) times the new length (120) since E = I' (120 + r⋅2).
From these relations, we can solve for r, which is the internal resistance we're looking for. By setting the two equations equal due to the constant emf (E), we can isolate r and find its value.
After simplifying and solving the equations obtained from the above relations,
we find that the internal resistance r = 2Ω, which corresponds to option D.