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CBSE

Subject

Physics

Class

CBSE Class 12

CBSE Physics 2016 Exam Questions

Short Answer Type

1.

The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell?


Potential difference, E = 6/3 = 2 V

Internal resistance, r = straight r space equals space straight E over straight I equals 6 over 1 equals 6 space ohm
Given, three cells are connected in series,

r' = r/3 = 6/3 = 2 ohm

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2.

(i) Identify the part of the electromagnetic spectrum which is:

(a) suitable for radar system used in aircraft navigation,

(b) produced by bombarding a metal target by high-speed electrons.

ii) Why does a galvanometer show a momentary deflection at the time of charging or discharging a capacitor ? Write the necessary expression to explain this observation.


i) a) Microwaves are suitable for radar system used in aircraft navigation.

b) X-rays are produced by bombarding a metal target with high-speed electrons.

ii) A capacitor is connected to the battery. So, electrons start moving towards the plate connected to the negative terminal of the battery and electrons leave from the plate connected to the positive terminal of the battery.

Transfer of electrons takes place until the potential of the capacitor becomes equal to that of the battery. The whole process happens very quickly, and the charging current produces deflection.

The reverse process is repeated at the time of discharging a capacitor and again galvanometer shows a momentary deflection.

Galvanometer acts as a resistance and the circuit behaves like a R-C circuit, having time constant equal to RC.

Therefore, the required expression is given by,

straight q space equals space straight q subscript straight o space open parentheses 1 minus straight e to the power of begin inline style fraction numerator negative 1 over denominator RC end fraction end style end exponent close parentheses

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3.

Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R.

Draw the magnetic field lines due to a circular wire carrying current I.


Here,

I = current in the loop
R= radius of the loop
X= distance between O and P
dl = conducting element of the loop

According to the Biot-Savart law,

Magnetic field at point P is,

dB space equals space fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator straight I space vertical line dl space straight x space straight r vertical line over denominator straight r cubed end fraction

straight r squared space equals space space straight x squared space plus space straight R squared
Since space dl space and space straight r space is space perpendicular comma space

vertical line dl space straight x space straight r vertical line space equals space straight r space dl

Therefore comma

dB space equals space fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction. space fraction numerator straight I. space dl over denominator left parenthesis straight x squared plus straight R squared right parenthesis end fraction

dB has two components: dBx and dB subscript perpendicular.
dB subscript perpendicular is cancelled out and the x-component remains.
Therefore,

dB subscript straight x space equals space dB space cos space straight theta

cos space straight theta space equals space fraction numerator straight R over denominator left parenthesis straight x squared plus straight R squared right parenthesis to the power of bevelled 1 half end exponent end fraction
therefore space dB subscript straight x space equals space fraction numerator straight mu subscript straight o straight I space dl over denominator 4 straight pi end fraction. space fraction numerator straight R over denominator left parenthesis straight x squared space plus space straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction

Summation of dl over the loop is given by,

B = straight B subscript straight x straight i with hat on top space equals space fraction numerator straight mu subscript straight o space straight I space straight R squared over denominator 2 space left parenthesis straight x squared plus straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction space straight i with hat on top

Magnetic field lines due to a circular current carrying i is,

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4.

(i) Derive an expression for drift velocity of free electrons.

(ii) How does drift velocity of electrons in a metallic conductor vary with increase in temperature ? Explain.


i) Expression for the drift velocity of electrons:

When a potential difference is applied across a conductor, an electric field is produced and free electrons are acted upon by an electric force (=-eE).

As a result, electrons accelerate and keep colliding with each other and acquire a constant average velocity (vd).

Therefore,

Fe = -Ee

rightwards double arrow space straight F subscript straight e space equals space fraction numerator negative eV over denominator straight l end fraction
We space know comma
straight a space equals space fraction numerator negative straight F over denominator straight m end fraction equals fraction numerator negative eV over denominator straight m end fraction
Also comma
straight v space equals space straight u plus at

Here comma

straight u space equals space 0 semicolon space
Relaxation space time comma space straight t equals space straight tau

Therefore comma
straight v subscript straight d space equals space minus space aτ
straight v subscript straight d space equals space fraction numerator negative eV over denominator straight m end fraction straight tau

ii) As the temperature is increased, drift velocity of electrons in a metallic conductor increases.

From the above relation,

straight v subscript straight d space proportional to space straight tau
Therefore, as the temperature of the metallic conductor increases, the collision between the electrons and ions increases, resulting in decrease in the relaxation time.

Thus, the drift velocity decreases.

4127 Views

5.

A battery of emf 12V and internal resistance 2 ohm is connected to a 4 ohm resistor as shown in the figure.

a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.

b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit?


a) Emf, E = 12 V

Internal resistance, r = 2V

Now, using the formula,

E = V + Ir

space space space space space space straight E space equals space straight V space plus space Ir
rightwards double arrow space straight V space equals space straight E space minus space Ir

straight E space equals space 12 space straight V space semicolon space straight r space equals space 2 space straight capital omega

straight V space equals space 12 space minus space 2 straight I

When the voltmeter is connected across the cell,

I = fraction numerator 12 over denominator 4 plus 2 end fraction space equals space 2 space A
V1 = 12 - 2(2) = 8 V

When the voltmeter is connected across the resistor,

V2 = IR

    = 2 x 4 = 8 V

That is, V1 = V2

Hence proved.

b) Voltmeter has very high resistance to ensure that it is connection does not alter the flow of current in the circuit. Current chooses the low resistance path. Therefore, voltmeter is connected in parallel to the load across which potential difference is to be measured.

Ammeter measures the value of current flowing through the circuit. Ammeter has a very low value of resistance to ensure that all the current flows through it. Hence, it should be connected in series.

8986 Views

6.

Define an equipotential surface. Draw equipotential surfaces:

(i) in the case of a single point charge and in a constant electric field in Z-direction.

(ii) Why the equipotential surfaces about a single charge are not equidistant?

(iii) Can electric field exist tangential to an equipotential surface? Give reason.


i) Equipotential surface for a single point charge is:       

Equipotential surface in a constant field in Z- direction.


ii) The equipotential surface about a single charge is not equidistant because V is inversely proportional to r.
Also, the equipotential surfaces about a single charge are not equidistant because electric field due to a  single charge is not constant.

iii) Electric field cannot exist tangential to an equipotential surface because if the field lines are tangential, work will be done in moving a charge on the surface, which is against the theory of equipotential surface.
7668 Views

7.

In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field?


Difference in the behaviour of a diamagnetic material and paramagnetic material:

Diamagnetic Paramagnetic
1. Diamagnetic substance would move towards the weaker region of the magnetic field. 1. Paramagnetic substance move towards the stronger region of the magnetic field.
2. Diamagnetic substance is repelled by a magnet. 2. Paramagnetic substance moves towards the magnet.
3. Diamagnetic substance get aligned perpendicular to the field. 3. Paramagnetic substance get aligned along the magnetic field.
3455 Views

8.

Ram is a student of class X in a village school. His uncle gifted him a bicycle with a dynamo fitted in it. He was very excited to get it. While cycling during night, he could light the bulb and see the objects on the road. He, however, did not know how this device works. He asked this question to his teacher. The teacher considered it an opportunity to explain the working to the whole class.

Answer the following questions:

(a) State the principle and working of a dynamo.
(b) Write two values each displayed by Ram and his school teacher.


Dynamo:

a) Principle: Whenever a coil is rotated in a magnetic field, an emf is induce in it due to change in magnetic flux linked with the coil.

Working:

As the coil in the dynamo rotates, its inclination (straight theta) with respect to the field changes. Therefore, a varying emf is obtained which is given by,

straight e space equals space straight e subscript straight o space sin space straight omega space straight t
b) Values displayed by:

Ram: curiosity, scientific aptitude, keenness to learn.

Teacher: Depth of knowledge, motivational approach, generous, dedicated.

2120 Views

9.

A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.


Work done, W = q x VAB = q X 0 = 0

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10.

(i) When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.

(ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor ? Explain.


Average power supplied by the source over a complete cycle is,

P = VI cos straight phi;

cos straight phi is called as the power factor.

For a pure inductive circuit,

Phase difference between current and voltage = straight pi over 2
And, cos straight pi over 2 equals 0
Therefore,

Average power dissipated = 0

ii) The brightness of the lamp decreases because when an iron rod is inserted in the inductor, the value of inductance L increases. Therefore, the current flowing across the bulb will decrease, thus decreasing the brightness of the bulb.

3265 Views

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