A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf ε and internal resistance r. A cell C having emf ε /2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is – (a) 11/12L (b) 13/24L (c) 5/12L (d) 11/24L

Answer: (b). Current in wire: I = (ε)/(12r + r) = (ε)/(13r). Potential gradient k = (I · 12r)/(L) = (12ε)/(13L). At null point no current flows through C, so its internal resistance does not matter: (ε)/(2) = k · AJ. AJ = (ε/2)/(12ε/13L) = 13/24L. So the answer is (b) 13/24L.

A potentiometer wire AB having length L and resistance 12 r is joined… | JEE Main 2019 (10 Jan, Shift 1)

Q73JEE Main 2019 · 10 JAN · SHIFT 1Difficult

Potentiometer

A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf $\varepsilon$ and internal resistance r. A cell C having emf $\varepsilon$ /2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is –

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Answer: (b)

Current in wire: $I = \frac{\varepsilon}{12r + r} = \frac{\varepsilon}{13r}$ .
Potential gradient $k = \frac{I \cdot 12r}{L} = \frac{12\varepsilon}{13L}$ .
At null point no current flows through C, so its internal resistance does not matter: $\frac{\varepsilon}{2} = k \cdot AJ$ .
$AJ = \frac{\varepsilon/2}{12\varepsilon/13L} = \frac{13}{24}L$ .
So the answer is (b) $\frac{13}{24}L$ .

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Potentiometer question from JEE Main 2019 (10 Jan, Shift 1)

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