Book Store

Download books and chapters from book store.
Currently only available for.
CBSE

Subject

Physics

Class

CBSE Class 12

CBSE Physics 2015 Exam Questions

Short Answer Type

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.

8630 Views

2.

A series LCR circuit is connected across an a.c. source of variable angular frequency ‘straight omega’. 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, P space equals space E subscript V I subscript V
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.

2881 Views

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.

4257 Views

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, straight R over straight S equals fraction numerator 40 over denominator left parenthesis 100 minus 40 right parenthesis end fraction equals 2 over 3                      ... (1)

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

fraction numerator straight R plus 10 over denominator straight S end fraction equals fraction numerator 60 over denominator 100 minus 60 end fraction equals 3 over 2 space space space space space space space space space space... space left parenthesis 1 right parenthesis

space space space straight R over straight S plus 10 over straight S equals 3 over 2 space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis space
From equations (1) and (2), we have

space space space space space space 2 over 3 plus 10 over straight S equals 3 over 2

rightwards double arrow space 10 over straight R equals space 3 over 2 minus 2 over 3 equals space fraction numerator 9 minus 4 over denominator 6 end fraction equals 5 over 6

rightwards double arrow space italic space straight S space equals space fraction numerator 10 cross times 6 over denominator 5 end fraction equals space 12 space o h m 

From equation (1), we have

space space space space straight R over 12 equals 2 over 3

rightwards double arrow space R space equals space fraction numerator 12 space cross times 2 over denominator 3 end fraction equals space 8 space o h m

1882 Views

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. 

2318 Views

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.

3310 Views

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, 

Error converting from MathML to accessible text. 


                                                             OR


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

That is, 

                            straight ϕ subscript straight B space equals space contour integral space straight B with rightwards harpoon with barb upwards on top space. space ds with rightwards harpoon with barb upwards on top space equals space 0 space

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

                             straight ϕ subscript straight E space space end subscript equals space contour integral straight E with rightwards harpoon with barb upwards on top space. space ds with rightwards harpoon with barb upwards on top space equals space straight q over straight epsilon subscript straight o

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, 

straight B space equals space fraction numerator straight mu subscript straight o space straight M over denominator 4 πd cubed end fraction

where, d = 6400km = 6.4 x 106 m

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

Error converting from MathML to accessible text.

4187 Views

8.

Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 straight mu 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, 

1 over 1 plus 1 over 1 equals 2 space mu F

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 straight mu space straight F

Voltage of the battery = 6V

i) Charge across the capacitor, q = CV = 4 straight space cross times straight space 10 to the power of negative 6 end exponent straight space cross times straight space 6 straight space equals straight space 24 cross times straight space 10 to the power of negative 6 end exponent straight space equals straight space 24 straight space μC
ii) Energy stored in the capacitor, E is given by, 

              1 half C V squared space equals space 1 half cross times 4 space cross times space 10 to the power of negative 6 end exponent space cross times space left parenthesis 6 right parenthesis squared space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space equals space 72 space cross times space 10 to the power of negative 6 end exponent space J space equals space 72 space mu J

3372 Views

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 straight i with hat on top space equals space j with hat on top space cross times k with hat on top .

Relation between velocity of propagation and magnitude of electric and magnetic field is given by |c| = straight E subscript straight o over straight B subscript straight o  ; Eo is the magnitude of electric field and is the magnitude of magnetic field.

2454 Views

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.

2295 Views

1.png
curious learner
Do a good deed today
Refer a friend to Zigya

NCERT Solutions
Textbook Solutions | Additional Questions