CBSE Sample Papers for Class 12 Physics Set 7 with Solutions

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Set 7 with Solutions

Time : 3 Hours
Maximum Marks: 70

General Instructions :

1. There are 33 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.
3. Air the sections are compulsory.
4. Section A contains sixteen questions, twelve MCQ and four Assertion-Reasoning based questions of 1 mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains two case study-based questions of four marks each and Section E contains three long answer questions of five marks each.
5. There is no overall choice. However, an internal choice has been provided in one question in Section B, one question in Section C, one question in each CBQ in Section D and all three questions in Section E. You have to attempt only one of the choices in such questions.
6. Use of calculators is not allowed.
7. You may use the following values of physical constants wherever necessary
   (i) c = 3 ×10⁸ m/s
   (ii) m_e = 9.1 × 10^-31 kg
   (iii) e = 1.6 × 10^-19 C 
   (iv) μ₀= 4π × 10^-17 TmA^-1
   (v) h = 6.63 × 10^-34 Js  
   (vi) ε₀ = 8.854 ×10^-12 C²N^-1m^-2
   (vii) Avogadro’s number = 6.023 × 10²³ per gram mole

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:

(a) zero
(b) positive
(c) negative
(d) data insufficient.
Answer:
(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:
(a) current and potential difference
(b) capacity and charge
(c) capacity and potential
(d) charge and potential difference
Answer:
(a) current and potential difference

Explanation: Ohm’s law deals with the relation between current and potential difference.
i.e., V = IR
or V ∝ l

Question 3.
The electric potential at a point in free space due to a charge Q coulomb is Q * 1011 V. The electric field at that
point is:
(a) 12πε_oQ × 10²² Vm^-1
(b) 4πε_oQ × 10²⁰ Vm^-1
(c) 12πε_oQ × 10²⁰ Vm^-1
(d) 4πε_oQ × 10²² Vm^-1
Answer:
(d) 4πε_oQ × 10²² Vm^-1

Explanation: Given that,

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

(a) 30 μC
(b) 48 μC
(c) 60 μC
(d) 120 μC
Answer:
(a) 30 μC

Question 5.
A current of 10A 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:
(a) 2 × 10^-5 T, acting downwards
(b) 2 × 10^-5 T, acting upwards
(c) 4 × 10^-5 T, acting downwards
(d) 4 × 10^-5 T, acting upwards
Answer:
(a) 2 × 10^-5 T, acting downwards

Explanation:

By using, right hand rule it should be acting downwards.

Question 6.
In the case of bar magnet, lines of magnetic induction:
(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
Answer:
(a) run continuously through the bar and outside

Explanation: In the bar magnet, lines of magnetic induction runcontinuously through the barand outside.

Question 7.
Faraday’s laws are consequences of conservation of:
(a) energy and magnetic field
(b) energy
(c) magnetic field
(d) charge
Answer:
(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:
(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.
Answer:
(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 at120 V at the secondary coil. If the efficiency of the transformer is 80% the current drawn from the line is:
(a) 0.3 A
(b) 3 A
(c) 30 A
(d) 24 A
Answer:
(b) 3 A

Explanation: We know that,

Question 10.
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 :
(a) less than Mp
(b) less than (Mp + MN)
(c) less than (Mp + 2Mn)
(d) greater than (Mp + 2Mn)
Answer:
(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 11.
The graph showing the correct variation of linear momentum (p) of a charge particle with its de-Broglie wavelength (λ) is :

Answer:

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

Question 12.
Scattering of an ∝-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.

(a) 2.1 × 10^-14 m
(b) 1.1 × 10^-16 m
(c) 1.1 × 10^-14 m
(d) 2.2 × 10^-16 m
Answer:
(c) 1.1 × 10^-14 m

Explanation: In Rutherford’s model, the impact parameter is given by

(Direction: Question 13 to 16) 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 13.
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.
Answer:
(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 14.
Assertion: The kinetic energy of photoelectrons emitted from metal surface does not depend on the intensity of incident photon.
Reason: The ejection of electrons from metallic surface is not possible with frequency of incident photons below the threshold frequency.
Answer:
(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 15.
Assertion: A pure semiconductor has negative temperature coefficient of resistance.
Reason: In a semiconductor on raising the temperature, more charge carriers are released, conductance increases and resistance decreases.
Answer:
(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.

Question 16.
Assertion: At a fixed temperature, silicon will have a minimum conductivity when it has a smaller acceptor doping.
Reason: The conductivity of an intrinsic semiconductor is slightly higher than that of a lightly doped p-type semiconductor.
Answer:
(c) A is true, but R is false.

Explanation: Conductivity of an intrinsic semiconductor is less than that of a lightly doped p-type semiconductor.

Section – B

Question 17.
(i) Which out of wavelength, frequency and speed of an electromagnetic wave does not change on passing from one medium to another?
(ii) A thin ozone layer in the upper atmosphere is crucial for human survival on earth, why?
Answer:
(i) When electromagnetic waves passes from one medium to another, frequency remains constant. Wavelength and speed both changes when medium of the wave changes.

(ii) 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 18.
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?

Answer:

Question 19.
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?
Answer:
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 20.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

	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.

Question 21.
Write shortcomings of Rutherford atomic model. Explain how these were overcome by the postulates of Bohr’s atomic model.
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.
Answer:
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

Section – C

Question 22.
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.
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.

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

Question 23.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (i) FI, (ii) B, (iii) M and (iv) the magnetising current I_m.
Answer:
(i) The field H is dependent of the material of the core, and is
H = nl
= 1000 × 2.0 = 2 × 10³ A/m.

(ii) The magnetic field B is given by
B = μ_rμ_oH
= 400 × 4π × 10^-7 (N/A²) × 2 × 10³ (A/m)
= 1.0T

(iii) Magnetisation is given by

\(M=\frac{\left(B-\mu_0 H\right)}{\mu_0}=\frac{\left(\mu_r \mu_0 H-\mu_0 H\right)}{\mu_0}\)

= (μ_r-1)H
= 399 × H
≅ 8 × 10⁵ A/m

(iv) The magnetising current I_m is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core.
Thus,

B = μ_rμ_o (I + I_m).
Using I = 2A, B = 1 T,
B = μ_rμ_o nl
=μ_o n (I + Im)
μ_rI = I + Im
I_m = (μ_r-l)I
= 798 A

Question 24.
An electromagnetic wave has a wavelength of 3 × 10^-3 m and electric field associated with it has an amplitude of 30 V/m. Calculate the amplitude and frequency of oscillation of the magnetic field in vacuum.
Answer:
Velocity of propagation (c) = 3 × 10⁸ m/s
The amplitudes of electric and magnetic fields are related as

Frequency of oscillations of magnetic field and frequency of the wave are equal.
Therefore λ = 3 × 10^-3 m.

Question 25.
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Answer:
The Rydberg formula is

The wavelengths of the first four lines in the Lyman series correspond to’ transitions from n_i  = 2, 3, 4, 5 to n_ƒ = 1. We know that

Question 26.
Draw V-I characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage?
Answer:
V-I characteristics of p-n junction diode:

(i) Under the reverse bias condition, the holes of p-side are attracted towards the negative terminal of the battery and the electrons of the n-side are attracted towards the positive terminal of the battery. This increases the depletion layer and the potential barrier.

However, the minority charge carriers are drifted across the junction producing a small current. At any temperature, the number of minority carriers is constant and very small so there is the small current at any applied potential. This is the reason for the current under reverse bias to be almost independent of applied potential. At the critical voltage, avalanche break down takes place which results in a sudden flow of large current.

(ii) At the critical voltage, the holes in the n-side and conduction electrons in the p-side are accelerated due to the reverse-bias voltage. These minority carriers acquire sufficient kinetic energy from the electric field and collide with valence electrons. Thus, the bond is finally broken and the valence electrons move into the conduction band resulting in enormous flow of electrons and thus result in formation of hole-electron pairs. Thus, there is a sudden increase in the current at the critical voltage.

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.

As R₁ > R₂. Therefore, σ₂ > σ₁.
Hence, charge density of smaller sphere is more than the charge density of larger sphere.

Question 28.
(a) Plot a graph showing variation of de-Broglie wavelength (λ) associated with a charged particle of mass m, versus √V, where V is the accelerating potential.
(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-partide have the same de-Broglie wavelength. Determine the ratio of (a) their accelerating potentials (b) their speeds.
Answer:

Section – D

Question 29.
Read the passage given below and answer the following questions:
Sometimes we notice that the ceiling fan does not start rotating soon as it is switched on. But when we rotate the blades, it starts to rotate as usual. Why it is so? We know that to rotate any object, there must be a torque applied on the object. For the ceiling fan, the initial torque is given by the capacitor widely known as a condenser, if the condenser is faulty, it will not give sufficient initial torque to rotate the blades when the fan is switched on.
(i) Capacitance (in F) of a spherical capacitor of radius 1 m is :
(a) 1.1 × 10^-10
(b) 10^-6
(c) 9 × 10^-9
(d) 10^-3
Answer:
(a) For a spherical conductor

(ii) A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor:
(a) decreases
(b) remains unchanged
(c) becomes infinite
(d) increases
Answer:
(b) As aluminium is a metal and its thickness is negligible, the potential difference between the plates remain unchanged. Thus, the capacitance will remain unchanged.

OR

When a capacitor is connected to a battery:
(a) a current flows in the circuit for some time, then decreases to zero.
(b) no current flows in the circuit at all
(c) an alternating current flows in the circuit
(d) none of the above
Answer:
(a) A current flows in the circuit for sometime, then decrease to zero’s capacitor gets fully changed.

(iii) If potential difference across a capacitor is changed from 15 V to 30 V, work done is W. What will be the work done when potential difference is changed from 30 V to 60 V?
(a) W
(b) 4W
(c) 3 W
(d) 2 W
Answer:

(iv) When 1.0 ×  10¹² electrons are transferred from one capacitor to another of a capacitor, a potential difference of 10 V develops between the two conductors. What is the capacitance of the capacitor?
(a) 1.6 × 10^-4 F
(b) 1.6 × 10^-8 F
(c) 1.6 × 10^-9 F
(d) 11.6 × 10^-8 F
Answer:
(b) q = ne = 1.0 × 10¹² × 1.6 × 10^-19 = 1.6 × 10^-7 C

Question 30.
Read the passage given below and answer the following questions:
The discovery of Infrared waves was done by Herschell. The frequency of these waves is in between 3 × 10¹¹ Hz to 4×10¹⁴ Hz. These are also known as heat waves because due to absorption of these waves heating effect in the bodies is observed. These waves are produced by hot bodies. The infrared waves are used in physical therapy to reduce or treat muscular strain. These waves also give electrical energy to satellites with the help of solar cells.

These waves are also helpful in taking photographs during fog, smoke etc. These also keep warm the plants in green houses. Weather forecasting is also done by infrared photography. These are also helpful in the study of molecular structure.

(i) These waves give
(a) Mechanical
(b) Magnetic
(c) Electrical
(d) None of these
Answer:
(c) Electrical

(ii) Waves in decreasing order of their wavelength are:
(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, UV rays, visible rays, X-rays
Answer:
(c) radio waves, infrared rays, visible rays, X-rays

(iii) The e.m. waves used to treat muscular pain is:
(a) VHF radio waves
(b) UV rays
(c) X-rays
(d) infrared rays
Answer:
(d) infrared rays

(iv) Find the ratio of the speed of the microwaves and infrared rays in the vaccum.
(a) 1:4
(b) 4:1
(c) 1:1
(d) 1:2
Answer:
(c) 1:1

OR
An infrared waves is incident on a material surface. The wave delivers momentum P and energy E. Then:
(a) P≠0, E≠0
(b) P≠0, E≠0
(c) P≠0, E≠0
(d) P≠0, E≠0
Answer:
(a) P≠0, E≠0

Section – E

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?
(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?
(c) Define the following terms:
(i) Magnetic domains
(ii) Coercive force
Answer:
(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
0 = 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

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

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 :
(1) High retentivity
(2) High coercivity

(ii) Properties of a material suitable for making electromagnet :
(1) High permeability
(2) Low retentivity

(b) (i) Two characteristics of materials used for making permanent magnets :
(1) High retentivity for making strong magnet.
(2) 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.

(c) (i) A magnetic domain is a region in which the magnetic fields of atoms are grouped together and aligned. In unmagnetised objects, all of the magnetic domains point in different directions. When it becomes magnetised all magnetic domains lined up and align in same direction.
(ii) Coercive force: It is the intensity of the applied magnetic field required to reduce the magnetisation of a given material to zero.

Question 32.
(a) A voltage V = V_o 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, and (ii) maximum power dissipated in the circuit?
(b) What do you mean by the impedance of LCR circuit, derive an expression for it.
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.
Answer:
(a) Voltage V_o sin ωt is applied to a series LCR circuit.
Current, I = Io sin (of + <(>)

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.

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 \(\frac{\pi}{2}\)

Therefore, no power is dissipated.
Case (II): For power dissipated at resonance in an LCR circuit,
X_C-X_L = 0, 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

(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.

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?

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.
(i) Identify the optical instrument.
(ii) 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 * 106 m and the radius of lunar orbit is 3.8 × 10⁸ m.
Answer:
(a) (i)

(ii) Objective 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.

Advantages:
(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

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,

(b) Astronomical telescope in near points adjustment:

Here, ƒ₀ = 15 m, ƒ_e =1 cm = 0.01 m
Diameter of moon, d = 3.48 × 10⁶ m
Radius of lunar orbit μ₀ = 3.8 × 10⁸ m
size of the image of moon, I =?
The angle subtended by the moon at the objective lens,