The Wheatstone bridge shown in the figure below is balanced when the uniform slide wire $$AB$$ is divided as shown. Value of the resistance $$X$$ is:

1. $$3~\Omega$$
2. $$4~\Omega$$
3. $$2~\Omega$$
4. $$7~\Omega$$

Subtopic:  Meter Bridge & Potentiometer |
89%
From NCERT
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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. $$\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}$$

Subtopic:  Meter Bridge & Potentiometer |
76%
From NCERT
NEET - 2015
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A resistance of 4 Ω and a wire of length 5 metres and resistance 5 Ω are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1 Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is:

1. 1.5 V

2. 3.0 V

3. 0.67 V

4. 1.33 V

Subtopic:  Meter Bridge & Potentiometer |
68%
From NCERT
PMT - 1997
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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 |
62%
From NCERT
NEET - 2019
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