Subject

Physics

Class

CBSE Class 12

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.

Book Store

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

Test Series

Pre-Board Question Papers are now available online for students to practice and excel the exams based on the latest syllabus and marking scheme given by respective boards.

CBSE Physics 2015 Exam Questions

Short Answer Type

21.

(a) Define the term 'intensity of radiation' in terms of photon picture of light.

(b) Two monochromatic beams, one red and the other blue, have the same intensity.

In which case:

(i) the number of photons per unit area per second is larger,

(ii) the maximum kinetic energy of the photoelectrons is more? Justify your answer.


a) The number of photons falling per unit area in unit time is defined as the intensity of radiation.

b)

i) Since, the two beams have the same intensity therefore, the number of photons emitted per unit are per unit time is the same.

ii) Maximum Kinetic energy of photoelectrons is given by, straight E straight space equals straight space hν straight space – straight space straight ϕ
We know that the frequency of blue beam is more than that of the red beam. Therefore, maximum Kinetic energy of the blue beam will be more.

2093 Views

Long Answer Type

22.

During a thunderstorm the 'live' wire of the transmission line fell down on the ground from the poles in the street. A group of boys, who passed through, noticed it and some of them wanted to place the wire by the side. As they were approaching the wire and trying to lift the cable, Anuj noticed it and immediately pushed them away, thus preventing them from touching the live wire. During pushing some of them got hurt. Anuj took them to a doctor to get them medical aid.

Based on the above paragraph, answer the following questions:

(a) Write the two values which Anuj displayed during the incident.

(b) Why is it that a bird can sit on a suspended 'live' wire without any harm whereas touching it on the ground can give a fatal shock?

(c) The electric power from a power plant is set up to a very high voltage before transmitting it to distant consumers. Explain, why. 

a) Anuj displayed concern for others lives, presence of mind and a selfless attitude.

b) When the bird perches on a live wire, body becomes charged for a moment and has a same voltage as the live wire. However, no current flows in its body. Body is a poor conductor of electricity as compared to the copper wire. So the electrons do not travel through the bird’s body.

On the other hand, if the bird touches the ground while being in contact with the high voltage live wire, then the electric circuit gets complete and high current flows through the body of the bird to the ground, giving it a fatal shock.

c) Inorder to reduce the loss of power transmission, the power plant is set up at high voltages. Power loss during the transmission is I2R. Therefore, by reducing the value of current, power loss can be minimized. So, the value of the voltage should be kept high.

1581 Views

23.

Draw a necessary arrangement for winding of primary and secondary coils in a step-up transformer. State its underlying principle and derive the relation between the primary and secondary voltages in terms of number of primary and secondary turns. Mention the two basic assumptions used in obtaining the above relation.

State any two causes of energy loss in actual transformers.


Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.

Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.

i.e.,                                                    NS > NP

                                  


Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.

Let, EP be the alternating emf applied to primary coil and np be the number of turns in it.

Consider straight ϕ as the electric flux associated with it.

Then, 

straight E subscript straight P space equals space minus straight n subscript straight p space dϕ over dt space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

where comma space fraction numerator d ϕ over denominator d t end fraction space is space the space rate space of space change space of space flux space across space
the space primary space coil. space

Now comma space Emf space across space the space secondary space coil space
be space straight E subscript straight s space and space straight n subscript straight s space be space the space number space of space turns space in space it.

straight E subscript straight S space equals space minus straight n subscript straight s space dϕ over dt space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis

Dividing space equation space left parenthesis 2 right parenthesis space by space left parenthesis 1 right parenthesis comma space

E subscript s over E subscript p equals n subscript s over n subscript p equals straight k italic colon

Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.

Two sources of energy loss in the transformer:

Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.

H = I2Rt

Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.

iii)

Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy. 

 

2204 Views

24.

(a) Use Huygens' principle to show the propagation of a plane wavefront from a denser medium to a rarer medium. Hence find the ratio of the speeds of wavefront in the two media.

(b) (i) Why does an unpolarized light incident on a polaroid get linearly polarized ?

(ii) Derive the expression of Brewster's law when unpolarized light passing from a rarer to a denser medium gets polarized on reflection at the interface.


a) 




As seen in the fig. above let XY be a surface separating the two media ‘1’ and ‘2’. Let v1 and v2 be the speeds of waves in these media.

A plane wavefront AB in the first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B after time-interval t.

Here, BB apostrophe space equals space straight v subscript 1 straight t.

As the wavefront AB advances, it strikes the points between A and B’ of boundary surface.

According to Huygens principle, secondary spherical wavelets emanate from these points, which travel with speed v1 in the first medium and speed v2 in the second medium.

Secondary wavelet starting from A, traverses a distance AA’ = v2t in second medium in time t. In the same time, point of wavefront traverses a distance  in first medium and reaches B’, from where the secondary wavelet starts.

So, BB' = v1 t and AA’ = v2t.

Assuming A as centre, we draw a spherical arc of radius AA’ (= v2t) and draw tangent B’A’ on this arc from B’. As the incident wavefront AB advances, the secondary wavelets start from points between A and B’, one after the other and will touch A’B’ simultaneously. 

According to Huygens principle A’B’ is the new position of wavefront AB in the second medium. Hence A’B’ will be the refracted wavefront.

Let the angle made by incident wavefront be i and angle made by the refracted wavefront A’B’ be r.

In space increment space AB apostrophe straight B comma space

space space space space space space space space space space space space space space space space space space space straight angle ABB '  equals straight space 90 to the power of 0

Hence comma space space space space space space space space space space space space space space sin straight space straight i straight space equals straight space sin  ∠ BAB ' 

space space space space space space space space space space space space space space space space space space space space space space space fraction numerator BB straight apostrophe over denominator AB straight apostrophe end fraction equals straight space fraction numerator straight v subscript 1 straight t over denominator AB straight apostrophe end fraction space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

Similarly comma space in space right minus angled space triangle comma space angle AA apostrophe straight B comma space
straight angle AA straight apostrophe straight B '  equals straight space 90 to the power of straight o

therefore space space space space space space space space space space space space space space space space space space sin space straight i straight space equals straight space sin    ∠  AB straight apostrophe straight A straight apostrophe

space space space space space space space space space space space space space space space space space space space space space fraction numerator AA straight apostrophe over denominator AB straight apostrophe end fraction equals fraction numerator straight v subscript 2 straight t over denominator AB straight apostrophe end fraction space 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

Now comma space dividing space equation space left parenthesis 1 right parenthesis space by space left parenthesis 2 right parenthesis comma space we space have space

space space space space fraction numerator sin space straight i over denominator sin space straight r end fraction straight space equals straight space straight v subscript 1 over straight v subscript 2 straight space equals straight space constant

This space is space the space Snell ’ straight s space law space of space refraction.

straight lambda subscript 1 space and space straight lambda subscript 2 space represents space the space wavelength space of space light
space in space the space rarer space medium space and space denser space medium space respectively.

Then comma space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight lambda subscript 1 over straight lambda subscript 2 equals straight space straight v subscript 1 over straight v subscript 2 

b) 

 

Polarization of light is referred to as restricting the vibration of light in a perpendicular direction perpendicular to the direction of propagation of wave.

The vibration of particles of light which is parallel to the axis of crystal passes through the Polaroid on passing an unpolarized light. All other vibrations are absorbed and that is why intensity of emerging light is reduced.

The plane of vibration here is ABCD, in which the vibrations of the polarized light is confined and the plane KLMN is called plane of polarization. KLMN is perpendicular to the plane of vibration.

Reflected light is totally polarized, when unpolarized light is incident on the glass-air interface at the Brewster angle iB. This is known as Brewster’s law.

The reflected component OB and refracted component OC are mutually perpendicular to each other, when light is incident at Brewster’s angle.

italic space italic space italic space italic space italic space italic space angle space B O y space plus space angle Y O C space equals space 90 to the power of o

space space space space space space left parenthesis 90 to the power of 0 space minus space i subscript B right parenthesis space plus space left parenthesis 90 to the power of 0 space plus space r right parenthesis space equals space 90 to the power of o
Now comma space using space Snell ’ straight s space law comma space we space have

space space space space space mu space equals space fraction numerator sin space i over denominator sin space r end fraction 

i = iB and r = (900 – iB)

Therefore, 

mu space equals space fraction numerator sin i subscript B over denominator sin space left parenthesis 90 to the power of o space minus space i subscript B right parenthesis end fraction space equals space fraction numerator sin i subscript B over denominator cos i subscript B end fraction space equals space tan i subscript B , is the expression for Brewster’s law.

3573 Views

Short Answer Type

25.

Obtain the relation between the decay constant and half-life of a radioactive sample.

The half-life of a certain radioactive material against straight alpha decay is 100 days. After how much time, will the undecayed fraction of the material be 6.25%?


In a radioactive sample, number of atoms at any instant is given by, 

straight N space equals space straight N subscript straight o space straight e to the power of negative λt end exponent

Now, when t = T, where T is the half-life of the sample.

Error converting from MathML to accessible text. 

This implies, half-life of a radioactive substance is inversely proportional to decay constant.

Numerical: 

We have,

Number of undecayed nuclei left = 6.25 % = 6.25/ 100 = 1/16

Let, t be the required time after which the undecayed fraction of the material will be 6.25%.

therefore space N space equals space N subscript o over 16

Error converting from MathML to accessible text.

3783 Views

Long Answer Type

26.

a) State Kirchhoff's rules and explain on what basis they are justified.

(b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel.

Derive the expression for the

(i) Emf and

(ii) internal resistance of a single equivalent cell which can replace this combination.


OR

 

(a) "The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement.

(b) Two identical circular loops '1' and '2' of radius R each have linear charge densities -straight lambdaand +straight lambda C/m respectively. The loops are placed coaxially with their centre distance apart. Find the:

magnitude and direction of the net electric field at the centre of loop '1'.

a)

i) Junction Rule: The algebraic sum of currents meeting at a point in an electrical circuit is always zero.

                                      sum I space equals space 0

This law is in accordance with law of conservation of charge.

ii) Loop Rule: In a closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of the resistances and the current flowing through them.

                                Error converting from MathML to accessible text.

This law is based on the conservation of energy.

b) Consider the circuit, 

 

Here, E1 and E2 are the emf of two cells,

r1 and r2 are the internal resistance of cell,

I1 and I2 current due to two cells.

Terminal potential difference across the first cell is given by, 

space space space straight V space equals space straight E subscript 1 space minus space straight I subscript 1 straight r subscript 1 space

rightwards double arrow space space straight I subscript 1 space equals space fraction numerator straight E subscript 1 space minus space straight V over denominator straight r subscript 1 end fraction

For the second cell, terminal potential difference will be equal to that across the forst cell.

So, 

V space equals space E subscript 2 space minus space I subscript 2 r subscript 2
rightwards double arrow I subscript italic 2 space equals space fraction numerator E subscript italic 2 space italic minus space V over denominator r subscript italic 2 end fraction 

Let E be the effective emf and r the resultant internal resistance.

Consider, I as the current flowing through the cell.

Therefore, 

I thin space equals space I subscript 1 space plus space I subscript 2 italic space

rightwards double arrow space I thin space equals space fraction numerator E subscript italic 1 italic minus V over denominator r subscript italic 1 end fraction plus space fraction numerator E subscript italic 2 italic minus V over denominator r subscript italic 2 end fraction italic space

rightwards double arrow space I thin space equals space fraction numerator r subscript italic 2 space italic left parenthesis E subscript italic 1 italic minus V italic right parenthesis italic plus r subscript italic 1 italic left parenthesis E subscript italic 2 italic minus V italic right parenthesis over denominator r subscript italic 1 r subscript italic 2 end fraction italic space

rightwards double arrow space I r subscript italic 1 r subscript italic 2 space equals space E subscript italic 1 r subscript italic 2 space plus space E subscript italic 2 r subscript italic 1 minus space left parenthesis r subscript italic 1 plus r subscript italic 2 right parenthesis V

rightwards double arrow space V space equals space fraction numerator E subscript italic 1 r subscript italic 2 italic plus space E subscript italic 2 r subscript italic 1 over denominator r subscript italic 1 space italic plus space r subscript italic 2 end fraction minus space fraction numerator I r subscript italic 1 r subscript italic 2 over denominator r subscript italic 1 space italic plus space r subscript italic 2 end fraction

Now, comparing the equation with V = E –Ir, we have

straight E space equals space fraction numerator E subscript 1 r subscript 2 plus space E subscript 2 r subscript 1 over denominator r subscript 1 space plus space r subscript 2 end fraction space and
Internal space resistance comma space straight r space equals space fraction numerator r subscript 1 r subscript 2 over denominator r subscript 1 space plus space r subscript 2 end fraction
 

                                               OR


a) The outward electric flux due to the charge enclosed inside a surface is the number of electric field lines coming out of the surface. Outward flux is independent of the shape and size of the surface because:

i) Number of electric field lines coming out from a closed surface is dependent on charge which does not change with the shape and size of the conductor.

ii) Number of electric lines is independent of the position of the charge inside the closed surface.

b) Magnitude of electric field at any point on the axis of a uniformly charged loop is given by, 

E space equals space fraction numerator lambda over denominator 2 epsilon subscript o end fraction fraction numerator r R over denominator left parenthesis r squared plus R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction 

 

Electric field at the centre of loop 1 due to charge present on loop 1 = 0

Electric field at the centre of loop1 due to charge present on loop 2 is given by, 

straight E '  equals straight space fraction numerator plus straight lambda over denominator 2 straight epsilon subscript straight o end fraction fraction numerator straight R square root of 3 straight R over denominator left parenthesis left parenthesis straight R square root of 3 right parenthesis squared plus straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction 

rightwards double arrow space E space apostrophe space equals space fraction numerator 3 over denominator 16 epsilon subscript o end fraction lambda over R, is the required electric field.

2585 Views

27.

Two infinitely long straight parallel wires, '1' and '2', carrying steady currents I1 and I2 in the same direction are separated by a distance d. Obtain the expression for the magnetic field  due to the wire '1' acting on wire '2'. Hence find out, with the help of a suitable diagram, the magnitude and direction of this force per unit length on wire '2' due to wire '1'. How does the nature of this force changes if the currents are in opposite direction? Use this expression to define the S.I. unit of current.


Consider a straight conductor XY lying in the plane of paper. Consider a point P at a perpendicular distance from the straight conductor.

 

Magnetic field induction (B) at a point P due to current I passing through conductor XY is given by,

straight B space equals fraction numerator straight mu subscript straight o space straight I over denominator 4 πa end fraction open square brackets sin space straight ϕ subscript 1 space plus space sin space straight ϕ subscript 2 close square brackets 

where, straight ϕ subscript 1 space and space straight ϕ subscript 2 are the angles made by point X and Y respectively.

At the centre of the infinite long wire, straight ϕ1 = straight ϕ2 = 90o

So, magnetic field is given by, 

straight B space equals space fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator 2 straight I over denominator straight a end fraction
Magnetic field produced by current I1 at any point on conductor Rs is given by, 

straight B subscript 1 space equals space fraction numerator μo space straight I space over denominator 2 πd end fraction

 


Force acting on length l of the conductor RS will be,

Error converting from MathML to accessible text.

An equal force is exerted into the wire PQ by the field of conductor RS which is given by, 

Error converting from MathML to accessible text.

Thus, the force is attractive when the current is acting along the same direction. When, current flows in opposite direction, the forces between the two conductors are repulsive.

One Ampere is that value of constant current which when flowing through each of the two parallel uniform long conductors placed in free space at a distance of 1m from each other will attract or repel with a force of 2 10-7 Newton per metre of their length.

2873 Views

28.

A biconvex lens with its two faces of equal radius of curvature R is made of a transparent medium of refractive index straight mu1. It is kept in contact with a medium of refractive index as shown in the figure.

                                                  

a) Find the equivalent focal length of the combination.

b) Obtain the condition when this combination acts as a diverging lens.

c) Draw the ray diagram for the case mu subscript 1 space greater than thin space left parenthesis mu subscript 2 space plus space 1 right parenthesis divided by 2 space comma spacewhen the object is kept far away from the lens. Point out the nature of the image formed by the system.


a) Using the Len’s maker’s formula, we have

 1 over straight f space equals space open parentheses mu space minus space 1 close parentheses space open square brackets fraction numerator 1 over denominator space R subscript 1 end fraction space minus space 1 over R subscript 2 close square brackets

L e t space f subscript 1 space a n d space f subscript 2 space b e space t h e space f o c a l space l e n g t h s space o f space t h e space
t w o space m e d i u m. space

1 over straight f subscript 1 space equals space open parentheses mu space minus space 1 close parentheses space open square brackets fraction numerator 1 over denominator space R end fraction space minus space open parentheses negative 1 over R subscript 2 close parentheses close square brackets

rightwards double arrow space 1 over straight f space equals space open parentheses mu space minus space 1 close parentheses open parentheses 2 over straight R close parentheses

space 1 over straight f subscript 2 space equals space open parentheses mu subscript italic 2 space minus space 1 close parentheses space open square brackets open parentheses negative fraction numerator 1 over denominator space R end fraction close parentheses space minus space 1 over infinity close square brackets

rightwards double arrow space 1 over straight f subscript 2 space equals space open parentheses mu subscript italic 2 space minus space 1 close parentheses space open parentheses negative 1 over straight R close parentheses

Let feq is the equivalent focal length of the combination, then

1 over f subscript e q end subscript space equals space 1 over f subscript 1 plus 1 over f subscript 2

rightwards double arrow space 1 over straight f subscript eq space equals space fraction numerator 1 space left parenthesis straight mu subscript 1 space minus space 1 right parenthesis over denominator straight R end fraction space minus space fraction numerator left parenthesis straight mu subscript 2 space minus space 1 right parenthesis over denominator straight R end fraction

rightwards double arrow space 1 over straight f subscript eq space equals space fraction numerator 2 straight mu subscript 1 space minus space straight mu subscript 2 space minus space 1 over denominator straight R end fraction

rightwards double arrow space straight f subscript eq space equals space fraction numerator straight R over denominator 2 straight mu subscript 1 space minus space straight mu subscript 2 space minus space 1 end fraction 

b) 

For, the combination of lenses to behave as diverging lens, equivalent focal length < 0.

 rightwards double arrow space fraction numerator R over denominator 2 mu subscript 1 space minus space mu subscript 2 minus 1 end fraction less than space 0

rightwards double arrow space space 2 straight mu subscript 1 space minus space straight mu subscript 2 space minus space 1 space less than 0

rightwards double arrow space straight mu subscript 1 space less than space fraction numerator left parenthesis straight mu subscript 2 space plus space 1 right parenthesis over denominator 2 end fraction 

c) 

For mu subscript 1 space greater than thin space left parenthesis mu subscript 2 space plus space 1 right parenthesis divided by 2 comma spacethe combination of lenses will behave as converging lens. An object placed far from the lens will form image at the focus of the lens.



Image formed is real and diminished in nature.

3962 Views

NCERT Solutions
Textbook Solutions | Additional Questions