The emf of a cell having internal resistance \(1~ \Omega\) is balanced against a length of \(330~\text{cm}\) on a potentiometer wire. When an external resistance of \(2 ~\Omega\) is connected across the cell, the balancing length will be: 
1. \(220~\text{cm}\) 2. \(330~\text{cm}\)
3. \(115~\text{cm}\) 4. \(332~\text{cm}\)
Subtopic:  Meter Bridge & Potentiometer |
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In a meter bridge experiment, the null point is at a distance of \(30~\text{cm}\) from \(\mathrm{A}\). If a resistance of \(16~\Omega\) is connected in parallel with resistance \(Y\), the null point occurs at \(50~\text{cm}\) from \(\mathrm{A}\). The value of the resistance \(Y\) is:
   

1. \(\dfrac{112}{3}~\Omega\) 2. \(\dfrac{40}{3}~\Omega\)
3. \(\dfrac{64}{3}~\Omega\) 4. \(\dfrac{48}{3}~\Omega\)
Subtopic:  Meter Bridge & Potentiometer |
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The sliding contact \(C\) is at one fourth of the length of the potentiometer wire (\(AB\)) from \(A\) as shown in the circuit diagram. If the resistance of the wire \(AB\) is \(R_0\), then the potential drop (\(V\)) across the resistor \(R\) is:
1. \(\dfrac{4V_0R}{3R_0+16R}\) 2. \(\dfrac{4V_0R}{3R_0+R}\)
3. \(\dfrac{2V_0R}{4R_0+R}\) 4. \(\dfrac{2V_0R}{2R_0+3R}\)
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In a potentiometer circuit, a cell of emf \(1.5~\text{V}\) gives a balance point at 36 cm length of wire. If another cell of emf 2.5 V replaces the first cell, then at what length of the wire, the balance point occur? 
1. 64 cm 
2. 62 cm 
3. 60 cm 
4. 21.6 cm

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A resistance wire connected in the left gap of a meter bridge balances a \(10~\Omega\) resistance in the right gap at a point which divides the bridge wire in the ratio \(3:2\). lf the length of the resistance wire is \(1.5~\text{m}\), then the length of \(1~\Omega\) of the resistance wire will be:
1. \(1.0\times 10^{-1}~\text{m}\) 
2. \(1.5\times 10^{-1}~\text{m}\)
3. \(1.5\times 10^{-2}~\text{m}\)
4. \(1.0\times 10^{-2}~\text{m}\)

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The metre bridge shown is in a balanced position with \(\frac{P}{Q} = \frac{l_1}{l_2}\). If we now interchange the position of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?

    
1. Yes, \(\frac{P}{Q}=\frac{l_1-l_2}{l_1+l_2}\)
2. No, no null point
3. Yes, \(\frac{P}{Q}= \frac{l_2}{l_1}\)
4. Yes, \(\frac{P}{Q}= \frac{l_1}{l_2}\)

Subtopic:  Meter Bridge & Potentiometer |
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A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:

1. the potential gradients.
2. a condition of no current flow through the galvanometer.
3. a condition of cells, galvanometer, and resistances.
4. the cells.

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A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite directions. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of the emf of the two cells is:
1. 5 : 4
2. 3 : 4
3. 3 : 2
4. 5 : 1
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A potentiometer wire of length \(L\) and a resistance \(r\) are connected in series with a battery of EMF \(E_{0 }\) and resistance \(r_{1}\). An unknown EMF is balanced at a length l of the potentiometer wire. The EMF \(E\) will be given by:
1. \(\frac{L E_{0} r}{l r_{1}}\)
2. \(\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}\)
3. \(\frac{E_{0} l}{L}\)
4. \(\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}\)

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A potentiometer wire has a length of \(4​​~\text{m}\) and resistance  \(8~\Omega.\)  The resistance that must be connected in series with the wire and an energy source of emf \(2~\text{V}\), so as to get a potential gradient of \(1~\text{mV}\) per cm on the wire is:
1. \(32~\Omega\)
2. \(40~\Omega\)
3. \(44~\Omega\)
4. \(48~\Omega\)

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