CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 25
Total: 26
(a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A , rotating with a constant angular speed 'ω ' in a magnetic field B , directed perpendicular to the axis of rotation.
(b) An aeroplane is flying horizontally from west to east with a velocity of 900kmhour . Calculate the potential difference developed between the ends of its wings having a span of 20m . The horizontal component of the Earth's magneticfield is 5×104T and the angle of dip is 30 .
OR
A device X is connected across an ac source of voltage V=V0sinωt . The current through X is given as I=I0sin(ωt+
π
2
)
.
(a) Identify the device X and write the expression for its reactance.
(b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.
(c) How does the reactance of the device X vary with frequency of the ac? Show this variation graphically.
(d) Draw the phasor diagram for the device X.
Solution:  
(a) Principle of ac generator
working
Mark labelled diagram
Derivation of the expression for induced emf
(b) Calculation of potential difference
(a) The AC Generator works on the principle of electromagnetic induction.
When the magnetic flux through a coil changes, an emf is induced in it.
As the coil rotates in magnetic field the effective area of the loop, (i.e. Acosθ) exposed to the magnetic field keeps on changing, hence magnetic flux changes and an emf is induced.

When a coil is rotated with a constant angular speed ' ω ', the angle ' θ ' between the magnetic field vector B and the area vector A , of the coil at any instant ' t ' equals ωt ; (assuming θ=0 at t=0 ) As a result, theeffective area of the coil exposed to the magnetic field changes with time; The flux at any instant ' t ' is given by
ϕB=NBAcosθ=NBAcosωt
The induced emf e=N
dϕ
dt

=NBA
dϕ
dt
(cosωt)

e=NBAωsinωt
(b) Potential difference developed between the ends of the wings ' e ' =Blv
Given Velocity v=900kmhour
=250m s
Wing span (l)=20m
Vertical component of Earth's magnetic field
Bv=BHtanδ
=5×104(tan30) Tesla
Potential difference
=5×104(tan30)×20×250
=
5×20×250×104
3

=1.44 volt
OR
(a) Identification of the device X Expression for reactance
(b) Graphs of voltage and current with time
(c) Variation of reactance with frequency (Graphical variation)
(d) Phasor Diagram
(a) X : capacitor
Reactance Xc=
1
ωC
=
1
2πvC

(b)

(c) Reactance of the capacitor varies in inverse proportion to the frequency i.e., Xcα
1
v

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