Q. No.Across a metallic conductor of non-uniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is:
1.
current density
2.
current
3.
drift velocity
4.
electric field
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Two cities are \(150~\text{km}\) apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is \(8\) volts and the average resistance per km is \(0.5~\text{ohm}\). The power loss in the wire is:
1. \(19.2~\text{W}\)
2. \(19.2~\text{kW}\)
3. \(19.2~\text{J}\)
4. \(12.2~\text{kW}\)
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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. \(25~\Omega\)
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A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of \(2.0~\text{V}\) and negligible internal resistance. The potentiometer wire itself is \(4~\text{m}\) long. When the resistance, \(R\), connected across the given cell, has values of (i) infinity (ii) \(9.5\), the 'balancing lengths, on the potentiometer wire, are found to be \(3~\text{m}\) and \(2.85~\text{m}\), respectively. The value of the internal resistance of the cell is (in ohm):
1. \(0.25\)
2. \(0.95\)
3. \(0.5\)
4. \(0.75\)
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A wire of resistance \(4~\Omega\) is stretched to twice its original length. The resistance of a stretched wire would be:
1. \(4~\Omega\)
2. \(8~\Omega\)
3. \(16~\Omega\)
4. \(2~\Omega\)
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1. \(0.5~\Omega\)
2. \(0.8~\Omega\)
3. \(1.0~\Omega\)
4. \(0.2~\Omega\)
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2. \(0.1~\text{A}\)
3. \(2.0~\text{A}\)
4. \(1.0~\text{A}\)
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If the voltage across a bulb rated \((220~\text{V}\text-100~\text{W})\) drops by \(2.5\%\) of its rated value, the percentage of the rated value by which the power would decrease is:
1. \(20\%\)
2. \(2.5\%\)
3. \(5\%\)
4. \(10\%\)
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A ring is made of a wire having a resistance of \(R_0=12~\Omega.\). Find points \(\mathrm{A}\) and \(\mathrm{B}\), as shown in the figure, at which a current-carrying conductor should be connected so that the resistance \(R\) of the subcircuit between these points equals \(\frac{8}{3}~\Omega\)
1. \(\frac{l_1}{l_2} = \frac{5}{8}\)
2. \(\frac{l_1}{l_2} = \frac{1}{3}\)
3. \(\frac{l_1}{l_2} = \frac{3}{8}\)
4. \(\frac{l_1}{l_2} = \frac{1}{2}\)
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If power dissipated in the 9 resistor in the circuit shown is 36 W, the potential difference across the 2 resistor will be:
1. 8 V
2. 10 V
3. 2 V
4. 4 V
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