The resistance of the meter bridge AB in given figure is \(4 \Omega\). With a cell of emf \(\varepsilon=0.5 \mathrm{~V}\) and rheostat resistance \(\mathrm{R}_{\mathrm{h}}=2 \Omega\) the null point is obtained at some point J . When the cell is replaced by another one of emf \(\varepsilon=\varepsilon_{2}\) the same null point J is found for \(\mathrm{R}_{\mathrm{h}}=6 \Omega\). The emf \(\varepsilon_{2}\) is:

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Current Electricity

The resistance of the meter bridge AB in given figure is

4Ω

. With a cell of emf

ε = 0.5 V

and rheostat resistance

R_{h} = 2Ω

the null point is obtained at some point J . When the cell is replaced by another one of emf

ε = ε_{2}

the same null point J is found for

R_{h} = 6Ω

. The emf

ε_{2}

is:

(A)0.4 V
(B)0.3 V✓ Correct
(C)0.6 V
(D)0.5 V

Answer: (B)

Solution

At the null point no current flows through the detector, so the potential drop across the section

A J

of the bridge wire is what balances the cell. Keeping the same point

J

means that same length, and hence the same section resistance, is balanced in both trials, and the loop relation for the driving cell through the rheostat and the wire makes the balanced emf proportional to

(R_{h} + R_{A B})

. Hence

\frac{ε _{2}}{ε _{1}} = \frac{R _{h}^{(1)} + R _{A B}}{R _{h}^{(2)} + R _{A B}} = \frac{2 + 4}{6 + 4} = \frac{6}{10}

, so

ε_{2} = 0.5 \times 0.6 = 0.3 V

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Current Electricity

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