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# CBSE Physics 2015 Exam Questions

1.

Draw a graph to show variation of capacitive-reactance with frequency in an a.c. circuit.

Here, XC is the capacitive-reactance and  is the frequency in an a.c circuit.

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2.

A series LCR circuit is connected across an a.c. source of variable angular frequency ‘’. Plot a graph showing variation of current 'i' as a function of ‘’ for two resistances R1 and R2 (R1 > R2).

Answer the following questions using this graph:

(a) In which case is the resonance sharper and why?

(b) In which case in the power dissipation more and why?

a) The variation of current with angular frequency for the two resistances R1 and R2 is shown in the graph below:

We can see that the resonance for the resistance R2 is sharper than for R1 because resistance R2 is less than resistance R1. Therefore, at resonance, the value of peak current will rise more abruptly for a lower value of resistance.

b) Power associated with resistance is given by,
From the above graph, we can say that virtual current in case of R2 is more than the virtual current in case of R1. Therefore, the power dissipation for circuit with R2 is more than that with R1.

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3.

Distinguish between emf and terminal voltage of a cell.

 Emf Terminal Voltage The maximum potential difference that can be delivered by a cell when no current flows through the circuit. Potential difference across the terminals of the load when the circuit is switched on and current flows through it. It is represented by E and remains constant for a cell. It is represented by V and depends on the internal resistance.
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4.

In a meter bridge shown in the figure, the balance point is found to be 40 cm from end A. If a resistance of 10  is connected in series with R, balance point is obtained 60 cm from A. Calculate the values of R and S.

For null point, balance length l1 = 40 cm

So,                       ... (1)

If resistance 10   is connected in series of R, balance point shifts towards AD = 60 cm.

From equations (1) and (2), we have

From equation (1), we have

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5.

State the underlying principle of a potentiometer. Write two factors by which current sensitivity of a potentiometer can be increased. Why is a potentiometer preferred over a voltmeter for measuring the emf of a cell?

Underlying principle of potentiometer: Potential difference across a uniform wire is directly proportional to the length of the part across which the potential is measured when, a steady current flow through the wire.

Sensitivity of potentiometer can be increased by:

(i) Increasing the length of the wire

(ii) Decreasing the current in the wire using a rheostat

Potentiometer is preferred over voltmeter because potentiometer used the null method. During the balanced condition of potentiometer, no current is drawn by the galvanometer. Voltmeter measures the voltage across the terminals of a cell when the cell is in closed circuit, that is, when current is flowing through the cell.

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6.

The field lines of a negative point charge are as shown in the figure. Does the kinetic energy of a small negative charge increase or decrease in going from B to A?

The electric field present due to the given point charge will be directed towards the centre. A negative charge always experiences a force in the direction opposite to that of the external electric field present; that is away from the centre. This will cause its motion to retard while moving from B to A. Hence, its kinetic energy will decrease in going from B to A.

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7.

State Lenz's law. Illustrate, by giving an example, how this law helps in predicting the direction of the current in a loop in the presence of a changing magnetic flux.

In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf induced in the coil.

OR

In what way is Gauss's law in magnetism different from that used in electrostatics? Explain briefly.

The Earth's magnetic field at the Equator is approximately 0.4 G. Estimate the Earth's magnetic dipole moment. Given: Radius of the Earth = 6400 km.

Lenz law states that,polarity of the induced emf is such that it opposes a change in magnetic flux.

The given activity demonstrates the above statement. The amount of magnetic flux linked with the coil increases, when the north pole of a bar magnet is brought near the coil. Current in the coil is induced in a so as to opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise direction, with respect to an observer. The magnetic moment  associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet.

Similarly, magnetic flux linked with the coil decreases when the north pole of the bar magnet is moved away from the coil. Inorder to oppose this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.

Given,

Self-inductance, L = 5 mH = 510-3 H

Change in current, dI = (4-1) = 3 A

Change in time, dt = 30 ms = 3010-3 s

So, emf induced in the coil is given by,

OR

i) Gauss’s law for magnetism states that magnetic flux through any closed surface is 0.

That is,

Gauss law for electrostatics states that electric flux through a closed surface is give n by,

Therefore, electric flux is zero when the surface encloses an electric dipole.

Magnetic flux is zero implies that, isolated magnetic poles do not exist.

ii) Given,

Magnetic field of Earth = 0.4 G = 0.4 10-4 G

Equatorial magnetic field of earth is given by,

where, d = 6400km = 6.4 x 106 m

Therefore, Earth’s magnetic dipole moment is given by,

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8.

Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1  F. When the ends X and Y are connected to a 6 V battery, find out

(i) the charge and

(ii) the energy stored in the network

Equivalent circuit can be drawn as,

In the above circuit, in one branch there are two capacitors in series.

Therefore, resultant capacitance is,

The circuit can be further re-arranged as,

We can see the two capacitors are in parallel.

Therefore, their resultant capacitance is,

Cres = 2 + 2 = 4

Voltage of the battery = 6V

i) Charge across the capacitor, q = CV =
ii) Energy stored in the capacitor, E is given by,

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9.

How are electromagnetic waves produced? What is the source of energy of these waves? Draw a schematic sketch of the electromagnetic waves propagating along the + x-axis.

Indicate the directions of the electric and magnetic fields. Write the relation between the velocity of propagation and the magnitudes of electric and magnetic fields.

Electromagnetic waves are produced by accelerating charged particle. When the charge moves with acceleration, both the magnetic and electric fields change continuously. This change produces electromagnetic waves.

Accelerated charge is the source of energy of these waves.

The direction of oscillating field is  .

Relation between velocity of propagation and magnitude of electric and magnetic field is given by |c| =   ; Eo is the magnitude of electric field and is the magnitude of magnetic field.

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10.

What is the function of a 'Repeater' used in communication system?

Repeater is used to increase the range of the transmission in communication systems with the help of a set of receiver and transmitter.

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