CBSE Sample Papers for Class 12 Physics Paper 6

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 6 of Solved CBSE Sample Paper for Class 12 Physics is given below with free PDF download solutions.

Answer 4.
Electric flux through plates of capacitor,

Here q is constant, the capacitor is fully charged.

As voltage becomes constant when capacitor is fully charged.

Answer 5.
The relation between the angle of incidence i, angle of prism A, and the angle of minimum deviation A , for a triangular prism is given by i = (A + A_m)/2

SECTION : B

Answer 6.
Curves 1 and 2 correspond to similar materials while curves 3 and 4 represent different materials, since the value of stopping potential for 1,2 and 3,4 are the same. For the given frequency of the incident radiation, the stopping potential is independent of its intensity.

So, the pairs of curves ( 1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation.

Answer 7.

Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by
E = 200- 10= 190 V
Hence, the current in the circuit is given by
I = E/R = 190/38 = 5 A

Answer 8.
The emf of a cell is greater than its terminal voltage because there is some potential drop across the cell due to its small internal resistance.

Answer 9.
(a) Necessary conditions for total internal reflection to occur are :

1. The incident ray on the interface should travel in optically denser medium.
2. The angle of incidence should be greater than the critical angle for the given pair of optical media.

(b)

Where a and b are the rarer and denser media respectively. C is the critical angle for the given pair of optical media.

Answer 10.
Lenz’s law states that the polarity of induced emf is such that it produces a current which opposes the change in magnetic flux that produces it. Emf will be induced in the rod as there is change in magnetic flux.

When a metallic rod held horizontally along east-west direction, is allowed to fall freely under gravity i.e. fall from north to south, the magnetic flux changes and the emf is induced in it.

SECTION : C

Answer 11.
We have focal length of convex lens,
f₁  = +25 cm = +0.25 m
Focal length of the concave lens,
f₁ = -20 cm = -0.20 m
Equivalent focal length,

∴ The focal length of the combination = -1 m = -100 cm As the focal length is negative, the system will be diverging in nature.

Answer 12.
We have, resistance of ammeter, R_A = 0.80 ohm
Maximum current across ammeter, I_A = 1.0 A.
So, voltage across ammeter, V= IR = 1 x 0.80 = 0.8 V
Let the value of shunt be x.
(i) Resistance of ammeter with shunt,

(ii) Combined resistance of the ammer and the shunt,

Answer 13.

The given figure is Common Emitter (CE) configuration of an n-p-n transistor. The input circuit is forward biased and collector circuit is reverse biased. If resistance R decreases, forward biased in the input circuit will increase, thus the base current (I_B) will decrease and the emitter current (I_E) will increase. This will increase the collector current (I_c) as
(I_E) = (I_B) + (I_c).
When I_c increases which flows through the lamp, the voltage across the bulb will also increase making the lamp brighter and the voltmeter is connected in parallel with the lamp, the reading in the voltmeter will also increase.

Answer 14.
(a) Given,
Velocity, v = v \(\overrightarrow { i }\) and E is the electric field along Y-axis and B is the magnetic field along Z-axis.
The propagation of EM wave is shown below :

(b) Speed of EM wave can be given as the ratio of magnitude of electric field (E₀) to the magnitude of magnetic field (B₀)

Answer 15.
From the given block diagram of demodulator of a typical receiver, we can conclude the following things :
(a) X represents Intermediate Frequency (IF) stage while Y , represents an amplifier.
(b) At IF stage, the carrier frequency is transformed to a lower frequency then in this process, the modulated signal is detected. The function of amplifier is to amplify the detected signal which may not be strong enough to be made use of and hence is essential.

Answer 16.
A junction diode made from light sensitive semiconductor is called a photo diode.

An electrical device that is used to detect and convert light into an energy signal with the use of a photo detector is known as a photo diode. The light that falls on it controls the function of pn-junction. Suppose, the wavelength is such that the energy of a photon hcfk is enough to break a valence bond. There is an increase in number of charge carriers and hence the conductivity of the junction also increases. New hole-electron pairs are created when such light falls on the junction. If the junction is connected in a circuit, the intensity of the incident light controls the current in the circuit.

OR

1. The reverse breakdown voltage of LEDs are very low, which is around 5 V. So enough care is to be taken while fabricating a pn-junction diode such that the high reverse voltages do not occur across them.
2. There exist very small resistance to limit the current in LED. So, a resistor must be placed in series with the LED such that no damage is occurred to the LED.The semiconductor used for fabrication of visible LEDs must atleast have a band gap of 1.8 eV.

Answer 17.
(a) Wave nature of radiation cannot explain the photoelectric effect because of :

1. The immediate ejection of photo electrons.
2. The presence of threshold frequency for a metal surface.
3. The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave nature of light.

(b) Photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based on particle nature of light. Its basic features are :

1. In interaction with matter, radiation behaves as if it is made up of particles called photons.
2. Each photon has energy (E = hv), momentum (p =hv /c) , and speed c, the speed of light.
3. All photons of light of a particular frequency v, or wavelength X, have the same energy
   
4. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity’ of radiation.
5. Photons are electrically neutral and are not deflected by electric and magnetic fields.
6. In a photon – particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, number of photons may not be conserved.

Answer 18.
(i) Initial voltage, V₁, = V volts and charge stored, Q₁ = 360 μC
Q₁ = CV₁            ………(1)
Changed potential,
V₂ = V -120
Q₂ = 120 μC
Q₂ = CV₂       ………. (2)
By dividing (1) by (2), we get

(ii) If the voltage applied had increased by 120 V, then
V₃ =180 + 120 = 300 V
Hence, charge stored in the capacitor,
Q₃ = CV₃  = 2 x 10^-6  x300 = 600 μC

OR

(i) Given,

Answer 19.


Which shows that the density is independent of mass number A.

Answer 20.
Three important factors which justify the need of modulating a message signal :

(i) Size of antenna or aerial :
For communication within the effective length of the antennas, the transmitting frequencies should be high, so modulation is required.
(ii) Effective power which is radiated by antenna :
Since the power radiated from a linear antenna is inversely proportional to the square of the transmitting wavelength. As high powers are needed for good transmission, so higher frequency is required which can be achieved by modulation.
(iii) The interference of signals from different transmitters :
To avoid the interference of the signals there is a need of high frequency which can be achieved by the modulation.

Answer 21.
Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction. Suppose the direction of the magnetic field is along +z direction.
Then, using Lorentz law, we get the following :

Thus, the direction of force on the electrons is along x-axis. So, the electrons will move towards the centre i.e., the fixed end of the rod. This movement of electrons will result in current and thus it will generate an emf in the rod between the fixed end and the point touching the ring.
Let 0 be the angle between the rod and radius of the circle at any time t.

Answer 22.
(i) Dynamic output resistance is given as :

SECTION : D

Answer 23.

1. It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage.
2. Awareness and humanity
3. Gratitude and obliged
4. Once I came across to a situation where a puppy was shuck in the middle of a busy road during rain and was not able to cross due to heavy flow, so I quickly rushed and helped him.

SECTION : E

Answer 24.
(a) A compound microscope consists of two convex lenses parallel separated by some distance. The lens nearer to the object is called the objective. The lens through which the final image is viewed is called the eyepiece. Magnifying power, when final image is at infinity: The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.
M = M_e x M₀       ……… (i)

∴ Where, M_e and M₀ are the magnifying powers of the eyepiece and objective respectively.
If u₍₎ is the distance of the object from the objective and v₀ is the distance of the image from the objective, then the magnifying power of the objective is

Where, h, h’ are object and image heights ; respectively and f₀ is the focal length of the a” objective.
L is the tube length i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece : When the final image is at infinity,

Magnifying power of compound microscope,

OR

(a) Huygens Principle :

• Each point on the primary wavefront acts as a source of secondary wavelets, transferring out disturbance in all directions in the same way as the original source of light does.
• The new position of the wavefront at any instant is the envelope of the secondary wavelets at that instant.Refraction on the basis of wave theory
• Consider any point Q on the incident wave front.
• Suppose when disturbance from point P on incident wave front reaches point P’ on the refracted wavefront, the disturbance from point Q reaches Q’ on the refracting surface XY.

Since P’A’ represents the refracted wavefront, the time taken by light to travel from a point on incident wavefront to the corresponding point on refracted wavefront should always be the same. Now, time taken by light to go from Q to Q’ will be

This is the Snell’s law for refraction of light.

(b)

1. The frequency of reflected and refracted light remains constant as the frequency of incident light only depends on the source of light.
2. Since the frequency remains same, hence there is no reduction in energy.

Answer 25.

(a) Working principle of Potentiometer :
When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

Applications of Potentiometerfor comparing emfs of two cells :
The following figure shows an application of the potentiometer to compare the emf of two cells of emf E₁ and E,. 1, 2, 3 form a two way key.
When 1 and 3 are connected, E₁ is connected to the galvanometer (G). Jockey is moved to N₁ which is at a distance l₁ from A, to find the balancing length. Applying loop rule to AN₁G31 A,

Thus, we can compare the emfs of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from equation (3).

(b)

1. The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared.
2. The positive ends of all cells are not connected to the same end of the wire.

OR

(a) Kirchhoff’s First Law – Junction Rule :
The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
Let the currents be I₁  , I_{2 ,} I₃ and I₄

Convention :
Current towards the junction – positive Current away from the junction – negative
I₃+ (- I₁) + (- I₂) +(- I₄) = 0

Kirchhoff’s Second Law – Loop Rule :
In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistance and current flowing through them.

For closed part BACB,
E₁ + E₂  =  I₁ R₁– I₃ R₃  + I₂ R₂
For closed part CADC,
E₂ =  I₃ R₃   + I₄ R₄   + I₅ R5

(b) Considering both the situations and writing them in the form of equations. Let R’ be the resistance per unit lenght of the potentiometer wire,

Answer 26.
(a) Consider a rectangular loop-ABCD carrying current I.

Case I :
The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop. No force is exerted by the magnetic field on the arms AD and BC. Magnetic field exerts a force
∴  F₁ = IbB  = F₁
Magnetic field exerts a force F₂ on arm CD.
F₂ = IbB = F₁

Net force on the loop is zero.
The torque on the loop rotates the loop in anti-clockwise direction.

τ = BIA
If there are n such turns the torque will be nIAB
Where, b →Breadth of the rectangular coil
a → Length of the rectangular coil
A = ab → Area of the coil.

Case II :

Plane of the loop is not along the magnetic field, but makes angle with it. Angle between the field and the normal is θ. Forces on BC and DA are equal and opposite and they cancel each other as they are collinear. Force on AB is F, and force on CD is F₂.
F₁= F₂ = IbB

Where A = ab
(b) We know, Lorentz force, F = Bqv sin θ
Where θ = angle between velocity of particle and magnetic field = 90
So, Lorentz force, F = Bqv
Thus, the particles will move in circular path.

Let m_p = mass of proton, m_d = mass of deuteron, v = velocity of proton and v_d = velocity of deuteron The charge of proton and deuteron are equal.

As (1) and (2) are equal, so Thus, the trajectory of both the particles will be same.

OR

(a) The torque on the needle is τ = m x B In magnitude, τ= mB sin θ
Here τ is restoring torque and 0 is the angle between m and B.

Negative sign with mB sin θ implies that restoring torque is in opposition to deflecting torque. For small values of 0 in radians, we approximate sin θ = θ and get

We hope the CBSE Sample Papers for Class 12 Physics Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physics Paper 6, drop a comment below and we will get back to you at the earliest.