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Q. No.The steady-state current in the circuit shown below is:
| 1. | \(0.67~\text A\) | 2. | \(1.5~\text A\) |
| 3. | \(2~\text A\) | 4. | \(1~\text A\) |
Subtopic: Derivation of Ohm's Law |
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A uniform wire of diameter \(d\) carries a current of \(100~\text{mA}\) when the mean drift velocity of electrons in the wire is \(v.\) For a wire of diameter \({\dfrac{d}{2}}\) of the same material to carry a current of \(200~\text{mA},\) the mean drift velocity of electrons in the wire is:
| 1. | \(4v\) | 2. | \(8v\) |
| 3. | \(v\) | 4. | \(2v\) |
Subtopic: Current & Current Density |
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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 |
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The amplitude of the charge oscillating in a circuit decreases exponentially as \(\mathrm{Q=Q_{0}e}^{-Rt/2L}\) where \({Q}_0\) is the charge at \({t}=0~\text{s}.\)The time at which charge amplitude decreases to \(0.50~{Q}_0\) is nearly:
\((\text{Given that}~ {R=1.5~\Omega,~L=12~\text{mH},~\text{ln}(2)=0.693})\)
1. \(19.01~\text{ms}\)
2. \(11.09~\text{ms}\)
3. \(19.01~\text s\)
4. \(11.09~\text s\)
\((\text{Given that}~ {R=1.5~\Omega,~L=12~\text{mH},~\text{ln}(2)=0.693})\)
1. \(19.01~\text{ms}\)
2. \(11.09~\text{ms}\)
3. \(19.01~\text s\)
4. \(11.09~\text s\)
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A wire of resistance \(R\) is cut into \(8\) equal pieces. From these pieces two equivalent resistance are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
| 1. | \(\dfrac{R}{16}\) | 2. | \(\dfrac{R}{8}\) |
| 3. | \(\dfrac{R}{64}\) | 4. | \(\dfrac{R}{32}\) |
Subtopic: Combination of Resistors |
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NEET - 2025
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Arrange the following in the order of their resistance.
Choose the correct answer from the options given below:
| A. | (\(0\) to \(1~\text A\)) ranged ammeter. |
| B. | (\(0\) to \(100~\text {mA}\)) ranged milli-ammeter. |
| C. | (\(0\) to \(500~\mu\text A\)) ranged micro-ammeter. |
| D. | (\(0\) to \(100~\text V\)) ranged voltmeter. |
Choose the correct answer from the options given below:
| 1. | \(\mathrm{A > B > C > D}\) | 2. | \(\mathrm{D > C > B > A}\) |
| 3. | \(\mathrm{D > A > B > C}\) | 4. | \(\mathrm{C > B > A > D}\) |
Subtopic: Derivation of Ohm's Law |
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In an electrical circuit, the voltage is measured as \({V}=(200\pm4)\) volts and the current is measured as \({I}=(20\pm0.2)~\text A.\) The value of the resistance is:
| 1. | \((10\pm4.2)~\Omega \) | 2. | \((10\pm0.3)~\Omega \) |
| 3. | \((10\pm0.1)~\Omega \) | 4. | \((10\pm0.8)~\Omega \) |
Subtopic: Derivation of Ohm's Law |
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A uniform metal wire of length \(l\) has \(10~\Omega \) resistance. Now this wire is stretched to a length \(2l\) and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is:
| 1. | \(10~\Omega \) | 2. | \(5~\Omega \) |
| 3. | \(40~\Omega \) | 4. | \(20~\Omega \) |
Subtopic: Current & Current Density |
Level 3: 35%-60%
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The current passing through the battery in the given circuit, is:
1. \(2.5~\text{A}\)
2. \(1.5~\text{A}\)
3. \(2.0~\text{A}\)
4. \(0.5~\text{A}\)
Subtopic: Wheatstone Bridge |
Level 3: 35%-60%
NEET - 2025
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A constant voltage of \(50~\text{V}\) is maintained between the points \(A\) and \(B\) of the circuit shown in the figure. The current through the branch \(C D\) of the circuit is:
1. \(2.5~A\)
2. \(3.0~A\)
3. \(1.5~A\)
4. \(2.0~A\)
1. \(2.5~A\)
2. \(3.0~A\)
3. \(1.5~A\)
4. \(2.0~A\)
Subtopic: Kirchoff's Voltage Law |
Level 4: Below 35%
NEET - 2025
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