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: Â Meter Bridge & Potentiometer |
Â 89%
From NCERT
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The figure given below shows a circuit when resistances in the two arms of the meter bridge are $$5~\Omega$$ and $$R$$, respectively. When the resistance $$R$$ is shunted with equal resistance, the new balance point is at $$1.6l_1$$. The resistance $$R$$ is:

1. $$10~\Omega$$
2. $$15~\Omega$$
3. $$20~\Omega$$
4. $$25~\Omega$$

Subtopic: Â Meter Bridge & Potentiometer |
Â 73%
From NCERT
AIPMT - 2014
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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: Â Meter Bridge & Potentiometer |
Â 62%
From NCERT
NEET - 2019
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A meter bridge is set up to determine unknown resistance $$x$$ using a standard $$10~\Omega$$ resistor. The galvanometer shows the null point when the tapping key is at a $$52$$ cm mark. End corrections are $$1$$ cm and $$2$$ cm respectively for end $$A$$ and $$B$$. Then the value of $$x$$ is:

1. $$10.2~\Omega$$
2. $$10.6~\Omega$$
3. $$10.8~\Omega$$
4. $$11.1~\Omega$$

Subtopic: Â Meter Bridge & Potentiometer |
Â 52%
From NCERT
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When two resistances $$X$$ and $$Y$$ are put in the left hand and right hand gaps in a Wheatstone meter bridge, the null point is at $$60$$ cm. If $$X$$ is shunted by a resistance equal to half of itself, then the shift in the null point will be:
 1 $$26.7$$ cm 2 $$33.4$$ cm 3 $$46.7$$ cm 4 $$96.7$$ cm