Electromagnetic Induction
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A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is
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1.0 Ω
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0.5 Ω
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2.0 Ω
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zero
A.
1.0 Ω
This is a problem is a base on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential differences, then
E/V = l1/l2
where l1 and l2 are the length of potentiometer wire with and without short-circuited through a resistance
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A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction in
such a way that its axis makes an angle of 60o with.
The magnetic flux linked with the disc is
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0.02 Wb
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0.06 Wb
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0.05 Wb
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0.01 Wb
A.
0.02 Wb
The magnetic flux Φ passing through a plane surface of area A placed in a uniform magnetic field B is given by
Φ = BA Cos θ
where θ is the angle between the direction of B and the normal to the plane.
Here,
A coil of resistance 400 is placed in magnetic field. If the magnetic flux Φ (Wb) linked with the coil varies with time t (sec) as Φ = 50t2 +4.
The current in the coil at t= 2 s is
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0.5 A
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0.1 A
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2 A
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1 A
A.
0.5 A
A coil of self -inductance L is connected in series with a bulb b and an AC source. Brightness of the bulb decreases when
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frequency of the AC source is decreases
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number of turns in the coil is reduced
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a capacitance of reactance Xc=XL is included in the same circuit
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an iron is inserted in the coil
D.
an iron is inserted in the coil
A condenser of capacity C is charged to a potential difference of V1. The plates to the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is
D.
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Mock Test Series
Mock Test Series