# CBSE Sample Papers for Class 12 Physics Set 3 with Solutions

Practicing the CBSE Sample Papers for Class 12 Physics with Solutions Set 3 allows you to get rid of exam fear and be confident to appear for the exam.

## CBSE Sample Papers for Class 12 Physics Set 3 with Solutions

Time : 3 Hours
Maximum Marks: 70

General Instructions :

1. There are 35 questions in all. All questions are compulsory.
2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.
3. Section A contains eighteen MCQs of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, Section D contains three long questions of five marks each and Section E contains two case study based questions of 4 marks each.
4. There is no overall choice. However, an internal choice has been provided in Section B, C, D and E. You have to attempt only one of the choices in such questions,
5. Use of calculators is not allowed.

Section – A

The following questions are multiple-choice questions with one correct answer. Each question carries 1 mark. There is no internal choice in this section.

Question 1.
Figure shows the field lines of a positive point charge. The work done by the field in moving a small positive charge from Q to P is: [1]

(a) zero
(b) positive
(c) negative
(d) data insufficient.
(c) negative
Explanation: In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.

Question 2.
Ohm’s law deals with the relation between: [1]
(a) current and potential difference
(b) capacity and charge
(c) capacity and potential
(d) charge and potential difference
(a) current and potential difference
Explanation: Ohm’s law deals with the relation between current and potential difference.
i.e., V = IR
or V ∝ I

Question 3.
The electric potential at a point in free space due to a charge Q coulomb is Q x 1011 V. The electric field at that point is: [1]
(a) 12πε0 Q x 1022 Vm-1
(b) 4πε0 Q x 1020 Vm-1
(c) 12πε0 Q x 1020 Vm-1
(d) 4πε0 Q x 1022 Vm-1
(d) 4πε0 Q x 1022 Vm-1
Explanation: Given that,

Question 4.
The capacitor, whose capacitance is 6 μF, 6 μF and 3 μF respectively are connected in series with 20 V line. Find the charge on 3 μF. [1]

(a) 30 μC
(b) 48 μC
(c) 60 μC
(d) 120 μC
(a) 30 μC
Explanation: In series $$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$ and charge on each capacitor is same.
∴ $$\frac{1}{C}=\frac{1}{6}+\frac{1}{6}+\frac{1}{3}$$
C = $$\frac { 3 }{ 2 }$$ … (i)
∵ q = CV
∵ q = 20 x $$\frac { 3 }{ 2 }$$ [By using (i)]

Question 5.
A current of 10 A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due north on the table is : [1]
(a) 2 x 10-5 T, acting downwards
(b) 2 x 10-5 T, acting upwards
(c) 4 x 10-5 T, acting upwards
(d) 4 x 10-5 T, acting upwards
(a) 2 x 10-5 T, acting downwards
Explanation:
$$\vec{B}=\frac{\mu_0 I}{2 \pi r}$$
= $$\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}$$
= $$10^{-7} \cdot \frac{2 \times 10}{0.10}$$
$$\vec{B}$$ = 2 x 10-5 T
By using, right hand rule it should be acting downwards.

Question 6.
In the case of bar magnet, lines of magnetic induction: [1]
(a) run continuously through the bar and outside
(b) emerge in circular paths from the middle of the bar
(c) are produced only at the north pole like rays of light from a bulb
(d) start from the north pole and end at the south pole
(a) run continuously through the bar and outside
Explanation: In the bar magnet, lines of magnetic induction run continuously through the bar and outside.

Question 7.
Faraday’s laws are consequences of conservation of: [1]
(a) energy and magnetic field
(b) energy
(c) magnetic field
(d) charge
(b) energy
Explanation: Faraday’s laws involve conversion of mechanical energy into electric energy. This is in accordance with the law of conservation of energy.

Question 8.
Waves in decreasing order of their wavelength are: [1]
(a) X-rays, infrared rays, visible rays, radio waves
(b) radio waves, visible rays, infrared rays, X-rays
(c) radio waves, infrared rays, visible rays, X-rays
(d) radio waves, ultraviolet rays, visible rays, X-rays.
(c) radio waves, infrared rays, visible rays, X-rays
Explanation: The wavelength of radio waves > infrared rays > visible rays > X-rays.

Question 9.
A step-down transformer is used on a 1000 V line to deliver 20 A at 120 V at the secondary coil. If the efficiency of the transformer is 80% the current drawn from the line is: [1]
(a) 0.3 A
(b) 3 A
(c) 30A
(d) 24 A
(b) 3 A
Explanation: We know that,

Question 10.
Find the value of 0 from given figure: [1]

(a) 45°
(b) 35°
(c) 30°
(d) 25°
(c) 30°
Explanation:
∠i = 90° – 60° = 30°
From figure, ∠i = ∠AOC (Angle of reflection)
∴ ∠i = ∠θ = 30°. (angle of incidence = angle of reflection)

Question 11.
The graph showing the correct variation of linear momentum (p) of a charge particle with its de-Broglie wavelength (X) is: [1]

Explanation: From de-Broglie hypothesis, the momentum of a particle, p = $$\frac { h }{ λ }$$
So the momentum versus wavelength graph is like y = $$\frac { 1 }{ x }$$ graph, which is represented in graph (c).

Question 12.
Scattering of an a-particle by an inverse square field (like produced by a charged nucleus in Rutherford’s model) is shown in figure. If charge of nucleus Z is 79 and kinetic energy of a-particle is 10 MeV, then find the impact parameter. [1]

(a) 2.1 x 10-14 m
(b) 1.1 x 10-16 m
(c) 1.1 x 10-14 m
(d) 2.2 x 10-16 m
(c) 1.1 x 10-14 m
Explanation: In Rutherford’s model, the impact parameter is given by

Question 13.
Mp and MN are masses of proton and neutron, respectively, at rest. If they combine to form deuterium nucleus. The mass of the nucleus will be : [1]
(a) less than Mp
(b) less than (Mp + MN)
(c) less than (Mp + 2MN)
(d) greater than (Mp + 2MN)
(b) less than (Mp + MN)
Explanation: We know that whenever there is fusion or fission or nucleoids and nuclei. Some mass is lost (mass defect) which converts into energy. So not mass of products is slightly less than that of substance.

Question 14.
When an electric dipole $$\vec{p}$$ is placed in a uniform electric field $$\vec{E}$$ then at what angle between $$\vec{p}$$ and $$\vec{E}$$ the value of torque will be maximum: [1]
(a) 90°
(b) 0°
(c) 180°
(d) 45°
(a) 90°
Explanation: As we know that,
$$\vec{\tau}=\vec{p} \times \overrightarrow{\mathrm{E}}$$
= pE sin θ
So, torque will be maximum at 90°.

Question 15.
What is the effective capacitance between points X and Y? [1]

(a) 24 μF
(b) 18 μF
(c) 12 μF
(d) 6 μF
(d) 6 μF
Explanation: By using

Equivalent circuit

As $$\frac{C_1}{C_3}=\frac{C_2}{C_4}$$ so it is balanced.
Hence, no charge will flow through 20 μF

C1 and C2 are in series, also C3 and C4 are in series.
Hence,
Hence, C’ = 3 μF, C” = 3 μF
C’ and C” are in parallel hence net capacitance = C’ + C” = 3 + 3 = 6 μF
(Direction: Question 16 to 18) Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, and R is also false.

Question 16.
Assertion : If the angles of the base of the prism are equal, then in the position of minimum deviation, the refracted ray will pass parallel to the base of prism.
Reason : In the case of minimum deviation, the angle of incidence is equal to the angle of emergence. [1]
(a) Both A and R are true and R is the correct explanation of A.
Explanation: In case of minimum deviation of a prism ∠i = ∠e so ∠r1 = ∠r2

Question 17.
Assertion : The kinetic energy of photoelectrons emitted from metal surface does not depend on the intensity of incident photon. [1]
Reason : The ejection of electrons from metallic surface is not possible with frequency of incident photons below the threshold frequency.
(b) Both A and R are true, but R is not the correct explanation of A.
Explanation: Kinetic energy of emitted photoelectrons does not depend on intensity of incident photon. Ejection of electrons from metallic surface is not possible with frequency of incident photons below threshold frequency.

Question 18.
Assertion : A pure semiconductor has negative temperature coefficient of resistance. [1]
Reason : In a semiconductor on raising the temperature, more charge carriers are released, conductance increases and resistance decreases.
(a) Both A and R are true and R is the correct explanation of A.
Explanation: In semiconductors, by increasing temperature, covalent bond breaks and conduction hole and electrons increase.

Section – B

Question 19.
(i) We feel the warmth of the sun light but not the pressure on our hands. Explain. [2]
(ii) Which out of wavelength, frequency and speed of an electromagnetic wave does not change on passing from one medium r .mother?
(iii) A thin ozone layer in the upper atmosphere is crucial for human survival on earth, why?
(i) Because the photons of light have only energy and do not exhibit mass.

(ii) When electromagnetic waves passes from one medium to another, frequency remains constant. Wavelength and speed both changes when medium of the wave changes.

(iii) The thin ozone layer around the earth’s atmosphere is very crucial for human survival. It protects the Earth from harmful ultraviolet (UV) rays from the sun. Without the ozone layer in the atmosphere, life on Earth would be very difficult. Plants cannot live and grow in heavy ultraviolet radiations. Ultraviolet radiations can cause skin cancer, cataracts and impaired immune systems in humans.

Question 20.
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? [2]

We know that,
V = E – Ir
Where E is the e.m.f. and r is the total internal resistance.
When I = 0,
Total emf = Terminal voltage
3E = 6 V
or E = 2 V
When V = 0
E = Ir
r = $$\frac { Total e.m.f }{ I }$$
= $$\frac { 6 }{ 1 }$$
= 6 Ω
As the cells are connected in series. So, the internal resistance of each cell = $$\frac { 6 }{ 3 }$$ = 2 Ω.

Question 21.
Write shortcomings of Rutherford atomic model. Explain how these were overcome by the postulates of Bohr’s atomic model. [2]
OR
Obtain the expression for the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of hydrogen atom.
According to Rutherford’s model, electrons revolve around the nucleus in a circular path. But particles that are in motion on a circular path would undergo acceleration and causes radiation of energy by charged particles. Eventually, electrons will loose energy and fall on to the nucleus. Thus, making nucleus unstable. Rutherford’s model cannot explain the stability of the atom.

Bohr modified this model by proposing that the electrons move in orbits of fixed size and energy. The energy of an electrons depends on the size of the orbit and is lower for smaller orbits. Radiation occur only when electron jump from one orbit to another.
OR
According to Bohr’s postulate,
mvr = $$\frac { nh }{ 2π }$$
i.e., 2πr = $$\frac { nh }{ mv }$$
or $$\frac { h }{ mv }$$ = $$\frac { 2πr }{ n }$$
Since, $$\frac { h }{ mv }$$ = $$\frac { h }{ p }$$ = λ, by de -Broglie hypothesis
Therefore, $$\frac { 2πr }{ n }$$ = λ
Now for second excited state
λ2 = $$\frac{2 \pi r_3}{3}$$
and for third excited state
λ3 = $$\frac{2 \pi r_4}{4}$$
On dividing equations (i) by (ii),
$$\frac{\lambda_2}{\lambda_3}=\frac{4 r_3}{3 r_4}=\frac{4}{3}\left(\frac{3}{4}\right)^2=\frac{3}{4}$$
This is the required expression.

Question 22.
A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? [2]
The figure shows a convex lens L placed in contact with a plane mirror M. P is the point object, kept in front of this combination at a distance of 20 cm, from it. As the image coincides with itself, the rays from the object, after refraction from lens, should fall normally on the mirror M, so that they retrace their path. For this, the rays from P, after refraction from the lens must form a parallel beam perpendicular to M. For clarity, M has been shown at a small distance from L (in diagram). As the rays from P, form a parallel beam after refraction, P must be at the focus of the lens. Hence, the focal length of the lens is 20 cm.

Question 23.
Distinguish between intrinsic and extrinsic semi-conductors. [2]
OR
Distinguish between n-type and p-type semiconductors.

 Intrinsic semiconductors Extrinsic semiconductors (i) It is a pure, natural semiconductor, such as pure Ge and pure Si. It is prepared by adding a small quantity of impurity to a pure semiconductor, such as n-type and p-type semiconductors. (ii) Intrinsic charge carriers are electrons and holes with equal concentration. In p-type semiconductor majority charge carries are holes and minority charge carriers are electrons, while vice-versa in n-type semi-conductor. (iii) Its electrical conductivity is very low. Its electrical conductivity is significantly high. (iv) Its conductivity cannot be controlled. Its conductivity can be controlled by adjusting the quantity of the impurity added. (v) Its conductivity increases exponentially with temperature. Its conductivity also increases with temperature but not exponentially.

OR

 n-type semiconductors p-type semiconductors (i) It is an extrinsic semiconductor obtained by adding a pentavalent impurity to a pure intrinsic semiconductor. It is also an extrinsic semiconductor obtained by adding a trivalent impurity to a pure intrinsic semi-conductor. (ii) The impurity atoms added provide extra free electrons to the crystal lattice and are called donor atoms. The impurity atoms added create holes in the crystal lattice and are called acceptor atoms. (iii) The electrons are majority carriers and the holes are minority carriers. The holes are majority carriers and the electrons are minority carriers. (iv) The electrons concentration is much more than the hole concentration (ne > > nh). The hole concentration is much more than the electron concentration (nh >> ne).

Question 24.
A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit. [2]
The distance of the nth minimum from the centre of the screen is,
xm = $$\frac { nDλ }{ a }$$
where, D = Distance of slit from screen,
λ = Wavelength of the light
a = Width of the slit.
For first minimum, n = 1

Question 25.
The electric field components in the given figure are Ex = αx1/2, Ey = Ez = 0, in which α = 800 N/Cm1/2. Calculate: [2]
(a) The flux through the cube
(b) The charge within the cube. (Assume that a = 0.1 m)

(a) Since the electric field x-component (E = ax1/2) only, for faces perpendicular to X-direction, the angle between E and ∆S is ± $$\frac { π }{ 2 }$$. Therefore, the flux Φ = E∆S is separately zero for each face of the cube except for two shaded faces. Magnitude of electric field at left face,
EL = αx1/2 = α a1/2
and, at right face,
ER = αx1/2 = α(2a)1/2
The flux at left face is given by,

(b) From Gauss’ law,
Φ = $$\frac{q}{\varepsilon_0}$$
where q = Charge enclosed
or q = Φεo = 1.05 x 8.854 x 10-12 C
= 9.27 x 10-12 C

Section – C

Question 26.
A short bar magnet of magnetic moment 5.25 x 10-2JT-1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on:
(i) its normal bisector, and (ii) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved. [3]
Given: Magnetic moment (M) = 5.25 x 10-2 JT-1
Earth’s magnetic field at a place (H) = 0.42 G = 0.42 x 10-4 T
(i) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B = $$\frac{\mu_0 \mathrm{M}}{4 \pi \mathrm{R}^3}$$
Where μo = Permeability of free space = 4π x 10-7 Tm A-1.
If resultant field is inclined at 45° with earth’s field, B = H

(ii) The magnetic field at a distance R’ from the centre of the magnet along its axis is given as,
B’ = $$\frac{\mu_0 2 \mathrm{M}}{4 \pi \mathrm{R}^3}$$
If resultant field is inclined at 45° with earth’s field,

Question 27.
Consider two conducting spheres of radii R1 and R2 with R1 > R2. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. [3]
Here V1 = V2
or $$\frac{q_1}{4 \pi \varepsilon_0 R_1}=\frac{q_2}{4 \pi \varepsilon_0 R_2}$$
⇒ $$\frac{q_1}{q_2}=\frac{R_1}{R_2}$$
Given R1 > R2
∴ q1 > q2
Larger sphere has more charge than the smaller sphere. Now, charge densities are,
σ1 = $$\frac{q_1}{4 \pi \mathrm{R}_1^2}$$
σ2 = $$\frac{q_2}{4 \pi \mathrm{R}_2^2}$$
$$\frac{\sigma_2}{\sigma_1}=\frac{q_2 R_1^2}{q_1 R_2^2}$$
or $$\frac{\sigma_2}{\sigma_1}=\frac{R_2 R_1^2}{R_1 R_2^2}=\frac{R_1}{R_2}$$ [Using (i)]
As R1 > R2. Therefore σ2 > σ1
Hence, charge density of smaller sphere is more than the charge density of larger sphere.

Question 28.
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also find the charge drawn from the battery in each case. [3]
OR
Use Kirchhoff’s rules to determine the potential difference between the points A and D when no current flows in the arm BE of the electric network shown in the figure.

Case I: When the capacitors are connected in series.
The equivalent capacitance is given by

As the capacitors are connected in series so, the charge remains the same.
Q = CeqV
Q = 6 x 10-12 x 50 x 50
Q = 300 pC

Case II: When the capacitors are connected is parallel.
The equivalent capacitance is given by,
Ceq = 12 + 12
= 24 pF
Electrostatic energy,
U = $$\frac { 1 }{ 2 }$$ x 24 x 10-12 x (50)²
U = 3 x 10-8 J
As the capacitors are connected in parallel so, the charge on each capacitor is different but potential difference remains the same.

Charge on C1V,
Q1 = C1V
Q1 = 12 x 10-12 x 50
= 600 x 10-12 C
Similarly, Q2 = 600 x 10-12 C
Total charge. Q = Q1 + Q2
= (600 + 600) x 10-12
= 1200 x 10-12
= 1200pC
OR
According to Kirchhoff’s junction law at B,

i3 = i1 + i2
As i2 = 0 (given)
∴ i3 = i1
Applying second law to loop AFE13,
∴ i3 x 2 + i3 x 3 + i2 R1 = 1 + 3 + 6
i3 = i1 = 2A
The potential difference between A to D along the branch AFD,
∴ VAD = 2i3 – 1 x 3 x i3
= (4 – 1 + 6) V = 9V

Question 29.
(a) Plot a graph showing variation of de-Broglie wavelength (λ) associated with a charged particle of mass m, versus $$\sqrt{V}$$, where V is the accelerating potential. [3]

(b) An electron, a proton and an alpha particle have the same kinetic energy. Which one has the shortest wavelength?
OR
A proton and an a-particle have the same de-Broglie wavelength. Determine the ratio of (z) their accelerating potentials (b) their speeds.
(a) We Know, λ = $$\frac{1.22}{\sqrt{V}}$$Å
λ$$\sqrt{V}$$ = constant
The nature of the graph between λ and $$\sqrt{V}$$ is hyperbola.

(b) According to de-Broglie wavelength
λ = $$\frac{h}{\sqrt{2 m \mathrm{E}_k}}$$
For same kinetic energy
λ ∝ $$\frac{1}{\sqrt{m}}$$
we know that
mα > mp > me
Since, alpha-particle has the maximum mass. Therefore, alpha-particle will ha ve minimum wavelength, therefore, λe > λp > λα
OR
(a) The de-Broglie wavelength of a particle is given by
λ = $$\frac{h}{\sqrt{2 m q \mathrm{~V}}}$$
where, V is the accelerating potential of the particle.
It is given that

(b) We can also write de-Broglie wavelength as λ = $$\frac { h }{ mV }$$
where, h is Planck’s constant, m is mass of the particle and v is speed of the particle.
We Know

Question 30.
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum. [3]
The Rydberg formula is
$$\frac{h c}{\lambda_{i f}}=\frac{m e^4}{8 \varepsilon_0^2 h^2}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$$
The wavelengths of the first four lines in the Lyman series correspond to transitions from ni = 2, 3, 4, 5 to nf = 1. We know that

Section – D

Question 31.
(a) Explain, using a labelled diagram, the principle and working of a moving coil galvanometer. What is the function of (i) uniform radial magnetic field, (ii) soft iron core? [5]

(b) Define the terms current sensitivity and voltage sensitivity of a galvanometer. Why does increasing the current sensitivity not necessarily increase voltage sensitivity?
OR
(a) Write two properties of a material suitable for making (i) a permanent magnet, and (ii) an electromagnet.
(b) (i) Write two characteristics of a material used for making permanent magnets.
(ii) Why is core of an electromagnet made of ferromagnetic materials ?
(a) The basic principle of a moving coil galvanometer is that when a current carrying coil is placed in a magnetic field, it experiences a torque.
When the current I is passed through the coil, the torque experienced is given by

τ = NIAB sin θ
Where N = No. of turns of the coil,
A = Area of the coil
B = Magnetic field and
θ = Angle between normal of coil and magnetic field
(i) The uniform radial magnetic field is used to make the scale linear.
(ii) The soft iron core increases the strength of the magnetic field.

(b) The current sensitivity is defined as the deflection produced in the galvanometer, while passing a current of 1 ampere.
Thus, current sensitivity
$$\left(\frac{\alpha}{\mathrm{I}}\right)=\frac{\mathrm{NBA}}{\mathrm{K}}$$
The voltage sensitivity is defined as the deflection produced in the galvanometer when a potential difference of 1 V is applied to the coil.
Thus, voltage sensitivity
$$\left(\frac{\alpha}{V}\right)=\frac{\mathrm{NBA}}{\mathrm{K}}=\frac{1}{\mathrm{R}}$$
Where R is the resistance.
Increasing the current sensitivity does not necessarily increase the voltage sensitivity as there is an increase in the resistance as well.
OR
(a) (i) Properties of a material suitable for making permanent magnet:

• High retentivity
• High coercivity

(ii) Properties of a material suitable for making electromagnet:

• High permeability
• Low retentivity

(b) (i) Two characteristics of materials used for making permanent magnets :

• High retentivity for making strong magnet.
• High coercivity for non-removal of magnetisation due to stray magnetic fields, temperature fluctuations or minor mechanical damage.

(ii) Core of electromagnets are made of ferromagnetic materials because they have high permeability and low retentivity. This gives minimum heating losses because of narrow hysteresis curve.

Question 32.
(a) A voltage V = Vo sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit?
(b) What do you mean by the impedance of LCR circuit, derive an expression for it. [5]
OR
(a) (i) Determine the value of phase difference between the current and the voltage in the given series LCR circuit.

(ii) Calculate the value of the additional capacitor which may be joined suitably to the capacitor C that would make the power factor of the circuit unity.

(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V.
(a) Voltage V = Vo sin ωt is applied to a series LCR circuit.
Current, I = Io sin ωt(ωt + Φ)
Io = $$\frac{V_0}{Z}$$
Instantaneous power supplied by the source is
P = VI = (Vo sin ωt) x [Io sin (ωt + Φ)]
= $$\frac{V_0 I_0}{2}[(\cos \phi-\cos (2 \omega t+\phi)]$$
The average power over a cycle is average of the two terms on the R.H.S. of the above equation.
The second term is time dependent, so, its average is zero.
So, P = $$\frac{\mathrm{V}_0 \mathrm{I}_0}{2} \cos \phi$$
= $$\frac{\mathrm{V}_0 \mathrm{I}_0}{\sqrt{2} \sqrt{2}} \cos \phi=\mathrm{VI} \cos \phi$$
P = I² Z cos Φ
cos Φ is called the power factor.

Case (I) : For pure inductive circuit or pure capacitive circuit, the phase difference between current and voltage i.e., Φ is $$\frac { π }{ 2 }$$.
∴ Φ = $$\frac { π }{ 2 }$$
So cos Φ = 0
Therefore, no power is dissipated.

Case (II): For power dissipated at resonance in an LCR circuit,
Xc – XL = o, Φ = 0
∴ cos Φ = 1
So, maximum power is dissipated.

(b) Impedance : The opposition offered by the combination of a resistor and reactive components to flow of alternating current is called impedance.
It is ratio of r.m.s. voltage applied and r.m.s. current produced in the circuit.
Z = $$\frac { V }{ I }$$
Consider that an inductor of inductance L, a capacitor of capacitance C and resistor of resistance R are connected in series to an alternating source of emf (E). It is given by
E = Eoeiωt … (i)
The instantaneous complex current (I) is given by

The complex quantity R + jωL + $$\frac { 1 }{ jωC }$$ is called complex impedance of LCR circuit. It can be expressed as

is the peak value of current.
From (i) and (iv), it follows that in LCR circuit emf leads the current by phase angle Φ.
OR
(a) (i) V= Vo sin (1000 t + Φ)
Comparing this with
V = Vo sin (ωt + Φ)
we have,
ω = 1000
Given, L = 100 x 10-3 H, R = 400 Ω, C = 2µF

(ii) The power factor of the circuit is unity. It means that the given circuit is in resonance. It is possible, if another capacitor C is used in the circuit.

Since C’ > C, so an additional capacitor of 8 pF would be connected in parallel to the capacitor of capacitance, C = 2 µF.

(b) Given,
ep = 2200 V
Np = 3000
Ns = ?
es = 220 V
∴ $$\frac{e_s}{e_p}=\frac{N_s}{N_p}$$
$$\frac{220}{2200}=\frac{\mathrm{N}_s}{3000}$$
⇒ Ns = 300

Question 33.
(a) (i) Draw a ray diagram for the formation of image by a compound microscope.
(ii) You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope? [5]

 Lenses Power (D) Aperture (cm) L1 3 8 L2 6 1 L3 10 1

(b) Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Write its two important advantages over a refracting telescope.
OR
(a) An optical instrument uses a lens of power 100 D for objective lens and 50 D for its eyepiece. When the tube length is kept at 20 cm, the final image is formed at infinity.

• Identify the optical instrument.
• Calculate the magnification produced by the instrument.

(b) Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.48 x 106 m and the radius of lunar orbit is 3.8 x 108 m.
(a) (i)

(ii) Object lens → L3
Eye lens → L2

(b) Reflecting Telescope : The reflecting telescope make use of a concave mirror as objective. The rays of light coming from distant object are incident on the objective (parabolic reflector). After reflection, the rays of light meet at a point where another convex mirror is placed. This mirror focuses light inside the telescope tube. The final image is seen through the eye-piece. The images produced by the reflecting telescope is very bright and its resolving power is high.

(i) The resolving power (the ability to observe two object distinctly) is high, due to the large diameter of the objective.

(ii) There is no chromatic aberration as the objective is a mirror.
OR
(a) (i) Power of lens = $$\frac{1}{\text { Focal length }}$$
So, focal length of objective lens
= $$\frac { 1 }{ 100 }$$ m = 1 cm
Focal length of eyepiece = $$\frac { 1 }{ 50 }$$ m = 2 cm
Since, the objective has a smaller focal length than the eyepiece, the instrument is a compound microscope.

(ii) Magnification produced is formed at infinity is given by,
m = $$\left(\frac{\mathrm{L}}{f_o}\right)\left(\frac{\mathrm{D}}{f_e}\right)$$
where, tube length L = 20 cm and least distance of distinct vision, D = 25 cm
m = $$\frac { 20 }{ 1 }$$ x $$\frac { 25 }{ 2 }$$ = 250

(b) Astronomical telescope in near points adjustment:

Here, fo = 15 m, fe = 1 cm = 0.01 m
Diameter of moon, d = 3.48 x 106 m
Radius of lunar orbit uo = 3.8 x 108 m
size of the image of moon, I =?
The angle subtended by the moon at the objective lens,
tan α = $$\frac{d}{\left|u_o\right|}$$ … (i)
and the angle subtended by the image of the moon at the objective lens,
tan ß = $$\frac{\mathrm{I}}{f_o}$$ … (iii)
Both these angles in equations (i) and (ii) will be equal.

Section – E

Question 34.
Case Study: Light emitting diode
Light Emitting Diode (LED): LED is the photoelectronic device which converts electrical energy into the light energy. It has heavily doped p-n junction diode and it gives spontaneous radiation when it is connected in forward bias. In this the upper layer is of p-type semiconductor and lower layer is of n-type.

To control the brightness of light emitted by LED, a resistance is connected in the circuit with battery.
The specific materials used for making LED’s are Gallium-Arsenide- Phosphide (Ga-As-P) for yellow or red light, Gallium-phosphide (Ga-P) for red or green light etc.

LED’s are used in making calculators digital watches, burglar H alarms, computers, picture phones, remote control and traffic light etc.
(i) Which type of p-n junction diode used in LED’s
(ii) Which device is used to control brightness in LED circuit?
(iii) What kind of materials are used in making LED’s
OR
(iii) Write any two application of LED’s. [4]
(i) Heavily doped junction diode

(ii) Resistance is used to control brightness in semiconductor circuit.

(iii) The materials used for making LED’s are Gallium-Arsenide-Phosphide (Ga-As-P) for yellow or red light, Gallium-phosphide (Ga-P) for red or green light etc.
OR
(iii) LED’s are used in making calculators digital watches, burglar alarms, computers, picture phones, remote control and traffic light etc.

Question 35.
Case Study : Spherical lens
A convex or converging lens is thicker at the centre than at the edges. It converges a parallel beam of light on refraction through it. It has a real focus. Convex lens is of three types : (1) Double convex lens (2) Plano-convex lens (3) Concavo-convex lens. Concave lens is thinner at the centre than at the edges. It diverges a parallel beam of light on refraction through it. It has a virtual focus.
(i) A point object O is placed at a distance of 0.3 m from a convex lens (focal length 0.2 m) cut into two halves each of which is displaced by 0.0005 m as shown in figure.

What will be the location of the image?

(ii) Two thin lenses are in contact and the focal length of the combination is 80 cm. If the focal length of first lens is 20 cm then calculate the focal length of other.

(iii) A spherical air bubble is embedded in a piece of glass. For a ray of light passing through a bubble, how does the bubble behave, converging lens or diverging lens?
OR
(iii) At what distance an object should be placed so that the magnification of an image by a convex lens is positive? [4]
$$\frac { 1 }{ v }$$ – $$\frac { 1 }{ (-30) }$$ = $$\frac { 1 }{ 20 }$$; v = 60 cm