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CBSE

Subject

Physics

Class

CBSE Class 12
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CBSE Physics 2017 Exam Questions

Short Answer Type

1.

How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. 


During charging, electric flux between the plates of capacitor keeps on
changing; this results in the production of a displacement current
between the plates.

Expression for the displacement current
contour integral stack straight B space space with rightwards arrow on top. dl with rightwards arrow on top space equals space straight mu subscript 0 left parenthesis straight I subscript straight C space plus dφ subscript straight E over dt


2.

A long straight current carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will the be an induced emf in the loop? Justify


No, Magnetic flux does not change with the change of current.


3.

A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of Ro. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire.



When the slide is in the middle of the potentiometer, only half of its total resistance i.e. R0/2 (since resistance is directly proportional to length) will be between A and point of contact (C), say R1, will be given by the following expression.
1 over straight R subscript 1 space equals 1 over straight R space plus fraction numerator 1 over denominator straight R subscript 0 divided by 2 end fraction
straight R subscript 1 space equals space fraction numerator RR subscript 0 over denominator 2 straight R plus straight R subscript 0 end fraction

The total resistance between A and B will be sum of the resistance between A & C and C&B i.e R1+R0/2
Current flowing through the potentiometer will be
straight I space equals space fraction numerator straight V over denominator straight R subscript 1 plus begin display style straight R subscript straight o over 2 end style end fraction space equals space fraction numerator 2 straight V over denominator 2 straight R subscript 1 plus straight R subscript 0 end fraction

The voltage V1 taken from the potentiometer will be the product of current I and the resistance R1
straight V subscript 1 space equals space IR subscript 1 space equals space fraction numerator 2 straight V over denominator 2 straight R subscript 1 plus straight R subscript 0 end fraction straight x space straight R subscript 1 space equals space fraction numerator 2 straight V over denominator begin display style fraction numerator 2 RR subscript 0 over denominator 2 straight R plus straight R subscript 0 end fraction end style plus straight R subscript 0 end fraction space straight x fraction numerator RR subscript 0 over denominator 2 straight R plus straight R subscript 0 end fraction space equals space fraction numerator 2 VR over denominator straight R subscript 0 plus 4 straight R end fraction


4.

Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field.


Depicting the orientation of the dipole in (i) stable equilibrium in a uniform electric field.

Net force is zero in this case as qE-qE = 0
Net torque =pEsinθ as θ = 0


(ii) Unstable equilibrium in a uniform electric field.

Net force is negative in this case as-qE-qE = -2qE
Net torque = pEsinθ asθ = 180


5.

Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give a reason for your answer.


No, Because the charge resides only on the surface of the conductor.


6.

(i) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2μF capacitance.


(ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?




1 over straight C subscript eqv space equals space 1 half space plus 1 over 6 space plus 1 half space equals space 7 over 6
straight C subscript eqv space equals space 6 over 7 space μF
ii right parenthesis space straight Q space equals space straight C subscript eqv straight V space equals space 6 over 7 space straight x space 7 space equals space 6 space μC
straight E space equals 1 half space CV squared space equals space 1 half space straight x 6 over 7 straight x space left parenthesis 49 right parenthesis space equals space 21 space straight J

7.

(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage?

(ii) Without making any other change, find the value of the additional capacitor C, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.


Given V = Vosin (1000 +Φ)
ω = 1000s-1
L = 100 mH
C =2μF
R = 400Ω
Phase difference  straight capital phi space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight X subscript straight L minus straight X subscript straight C over denominator straight R end fraction close parentheses
XL = ωL = 1000 x100 x10-3 = 100Ω
 =straight X subscript straight C space equals space 1 over ωC space equals space fraction numerator 1 over denominator 1000 space straight x 2 space straight x 10 to the power of negative 6 end exponent end fraction space equals space 500 space straight capital omega
straight capital phi space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 100 minus 500 over denominator 400 end fraction close parentheses space equals space tan to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis
straight capital phi space equals space minus space 45 to the power of straight o space and space the space current space is space leading space the space voltage.

(ii) For power factor to be unity, R = Z
or XL = XC
straight omega squared space equals space 1 over LC space left parenthesis straight C space equals space resultant space capacitance right parenthesis
10 to the power of 6 space equals space fraction numerator 1 over denominator 100 space straight x 10 to the power of negative 3 end exponent space xC end fraction
rightwards double arrow space straight C to the power of apostrophe space equals space 10 to the power of negative 5 end exponent space straight F

For two capacitance in parallel, resultant capacitance C'=C + C1
10-5 = 0.2 x 10-5 + C1
⇒ C1 = 8μF


8.

How is the speed of em-waves in vacuum determined by the electric and magnetic field?


The speed of the em-waves in a vacuum is determined by the ratio of the peak value of electric and magnetic fields.
c =E0/B0


9.

Drive the expression for electric field at a point on the equatorial line of an electric dipole.


The magnitudes of the electric field due to the two charges +q and -q are given by,
straight E subscript plus straight q end subscript space equals space fraction numerator straight q over denominator 4 space πε subscript 0 end fraction fraction numerator 1 over denominator straight r squared space plus straight a squared end fraction space..... space left parenthesis straight i right parenthesis
straight E subscript negative straight q end subscript space equals space fraction numerator straight q over denominator 4 πε subscript 0 end fraction fraction numerator 1 over denominator straight r squared plus straight a squared end fraction space space....... space left parenthesis ii right parenthesis

The directions of E+q and E-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
Therefore, Total electric field.
E = - (E+q + E-q) cosθ p [Negative sign shows that field is opposite to p]

straight E space equals space minus fraction numerator 2 qa over denominator 4 πε subscript 0 space left parenthesis straight r squared space plus straight a squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction space space space... left parenthesis iiii right parenthesis
At large distances (r>>a),this reduces to
straight E space equals negative fraction numerator 2 qa over denominator 4 πε subscript 0 straight r cubed end fraction straight p with hat on top space space.... left parenthesis iv right parenthesis
because space straight p with rightwards arrow on top space equals space straight q space straight x space 2 straight a straight p with hat on top
therefore space straight E space equals space fraction numerator negative straight p with rightwards arrow on top over denominator 4 πε subscript 0 straight r cubed end fraction space left parenthesis straight r greater than greater than straight a right parenthesis


10.

At a place, the horizontal component of earth's magnetic field is B and angle of dip are 60o. What is the value of a horizontal component of the earth's magnetic field at the equator?


The horizontal component of the electric field is given by
BH = Becos(θ)
where θ is the angle of dip at the given place
Be is teh net magnetic field
At θ = 60o
BH = Be cos(60) = Be x (1/2)
Be = 2BH = 2B (given BH =B)
Now horizontal component at equator
BH = Becos (θ)
angle of dip at equator is 0
BH = Becos(0) = Be = 2B
BH = 2B


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