- 1(current)
Q. No.The given circuit shows a uniform straight wire \(AB\) of \(40 ~\text{cm}\) length fixed at both ends. In order to get zero reading in the galvanometer \(G,\) the free end of \(J\) is to be placed from the end \(B\) at:
| 1. | \(32~\text{cm}\) | 2. | \(8~\text{cm}\) |
| 3. | \(16~\text{cm}\) | 4. | \(24~\text{cm}\) |
Subtopic: Wheatstone Bridge |
67%
Level 2: 60%+
NEET - 2024
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The value of \(R\) in the given circuit when there is no current in the \(5 ~\Omega\) resistor is:
| 1. | \(12~\Omega\) | 2. | \(9~ \Omega\) |
| 3. | \(3~ \Omega\) | 4. | \(2~ \Omega\) |
Subtopic: Wheatstone Bridge |
86%
Level 1: 80%+
NEET - 2024
Hints
A Wheatstone bridge is used to determine the value of unknown resistance \(X\) by adjusting the variable resistance \(Y\) as shown in the figure. For the most precise measurement of \(X\), the resistances \(P\) and \(Q\):
| 1. | do not play any significant role. |
| 2. | should be approximately equal to \(2X\). |
| 3. | should be approximately equal and are small. |
| 4. | should be very large and unequal. |
Subtopic: Wheatstone Bridge |
76%
Level 2: 60%+
NEET - 2022
Hints
The resistances of the four arms \(P,Q,R~\text{and}~S\) in a Wheatstone’s bridge are \(10~\Omega,30~\Omega,30~\Omega\) and \(90~\Omega\) respectively. The emf and internal resistance of the cell are \(7~\text{V}\) and \(5~\Omega\) respectively. If the galvanometer resistance is \(50~\Omega\) the current drawn from the cell will be:
| 1. | \(0.2~\text{A}\) | 2. | \(0.1~\text{A}\) |
| 3. | \(2.0~\text{A}\) | 4. | \(1.0~\text{A}\) |
Subtopic: Wheatstone Bridge |
78%
Level 2: 60%+
AIPMT - 2013
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- 1(current)
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