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Q. No.In the given circuit the potential difference across \(700~\Omega\) resistance is: (i.e., \(V_0)\)
1. \(2~\text V\)
2. \(0.5~\text V\)
3. \(1.1~\text V\)
4. zero
Subtopic: Derivation of Ohm's Law |
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A wire has a resistance of \(60~\Omega\) at temperature \(27^\circ\text{C}.\) When it is connected to a \(220 ~\text V\) DC supply, a current of \(2.75 ~\text A\) flows through it at a certain temperature. Then the value of temperature, (if the coefficient of thermal resistance \(\alpha =2 \times 10^{-4} { }^{\circ} \text{C} )\) is:
1. \(1694^\circ\text{C}\)
2. \(1500^\circ\text{C}\)
3. \(1000^\circ\text{C}\)
4. \(1200^\circ\text{C}\)
1. \(1694^\circ\text{C}\)
2. \(1500^\circ\text{C}\)
3. \(1000^\circ\text{C}\)
4. \(1200^\circ\text{C}\)
Subtopic: Derivation of Ohm's Law |
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If electric current passing through a conductor varies with time as \(I=I_0+\beta t\) where \(I_0=20~\text{A}, \beta=3~\text{A/s},\) then the charge flows through the conductor in the first \(10\) sec is:
1. \(400~\text{C}\)
2. \(500~\text{C}\)
3. \(200~\text{C}\)
4. \(350~\text{C}\)
1. \(400~\text{C}\)
2. \(500~\text{C}\)
3. \(200~\text{C}\)
4. \(350~\text{C}\)
Subtopic: Current & Current Density |
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Level 2: 60%+
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Two cells one of emf \(8~\text V,\) internal resistance \(2~\Omega\) and the other of emf \(2 ~\text V\) and internal resistance \(4~\Omega\) are connected as shown in the figure.
Then the potential difference (in \(\text V\)) across the point \(AC\) is:
1. \(5\)
2. \(2\)
3. \(0\)
4. \(4\)
Then the potential difference (in \(\text V\)) across the point \(AC\) is:
1. \(5\)
2. \(2\)
3. \(0\)
4. \(4\)
Subtopic: Grouping of Cells |
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In a given circuit, an ideal battery is connected with four resistances as shown. Find the current \(i\) as mentioned in the diagram.
1. \(2~\text A\)
2. \(1~\text A\)
3. \(4~\text A\)
4. \(0.5~\text A\)
1. \(2~\text A\)
2. \(1~\text A\)
3. \(4~\text A\)
4. \(0.5~\text A\)
Subtopic: Combination of Resistors |
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Level 1: 80%+
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Consider the circuit shown.
The ammeter reads \(0.9 ~\text A.\) Then the value of \(R\) is:
1. \(30~\Omega\)
2. \(40~\Omega\)
3. \(50~\Omega\)
4. \(60~\Omega\)
The ammeter reads \(0.9 ~\text A.\) Then the value of \(R\) is:
1. \(30~\Omega\)
2. \(40~\Omega\)
3. \(50~\Omega\)
4. \(60~\Omega\)
Subtopic: Derivation of Ohm's Law |
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Three voltmeters are connected in a circuit as shown in the diagram. Then the correct relation among their readings \(({V}_1,{V}_2~\text{and} ~{V}_3)\) is:
1. \({V}_1>{V}_2={V}_3\)
2. \({V}_1+{V}_2={V}_3\)
3. \({V}_1={V}_2={V}_3\)
4. \({V}_1+{V}_3={V}_2\)
1. \({V}_1>{V}_2={V}_3\)
2. \({V}_1+{V}_2={V}_3\)
3. \({V}_1={V}_2={V}_3\)
4. \({V}_1+{V}_3={V}_2\)
Subtopic: Kirchoff's Voltage Law |
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In a meter bridge, an unknown resistance \(X\) has a specific resistance \(S_1=\dfrac{X \pi R^2}{l},\) where \(R\) is radius and \(l\) is length. If the length and radius both are doubled, the new specific resistance is:
1. \(S_1\)
2. \(2S_1\)
3. \(4S_1\)
4. \(\frac{S_1}{4}\)
1. \(S_1\)
2. \(2S_1\)
3. \(4S_1\)
4. \(\frac{S_1}{4}\)
Subtopic: Derivation of Ohm's Law |
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In the given circuit, the reading of the voltmeter is \(1 ~\text V,\) then the resistance of the voltmeter is:
1. \(100 ~\Omega\)
2. \(200~\Omega\)
3. \(200\sqrt5~\Omega\)
4. \(50~\Omega\)
1. \(100 ~\Omega\)
2. \(200~\Omega\)
3. \(200\sqrt5~\Omega\)
4. \(50~\Omega\)
Subtopic: Combination of Resistors |
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In the given meter bridge circuit, a null point is found at \(60 ~\text{cm}\) from the end \(A.\) The unknown resistance \(S\) (in \(\Omega\)) is:
1. \(60~\Omega\)
2. \(80~\Omega\)
3. \(70~\Omega\)
4. \(90~\Omega\)
1. \(60~\Omega\)
2. \(80~\Omega\)
3. \(70~\Omega\)
4. \(90~\Omega\)
Subtopic: Meter Bridge |
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