CBSE Class 12 Physics Question Paper 2023 (Series: GEFH1/2) with Solutions

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CBSE Class 12 Physics Question Paper 2023 (Series: GEFH1/2) with Solutions

General Instructions:

Read the following instructions very carefully and follow them:

1. This question paper contains 35 questions. All questions are compulsory.
2. Question paper is divided into FIVE sections – Section A, B, C, D and E.
3. In Section-A: question number 1 to 18 are Multiple Choice (MCQ) type questions carrying 1 mark each.
4. In Section-B: question number 19 to 25 are Short Answer-1 (SA-1) type questions carrying 2 marks each.
5. In Section-C: question number 26 to 30 are Short Answer-2 (SA-2) type questions carrying 3 marks each.
6. In Section-D: question number 31 to 33 are Long Answer (LA) type questions carrying 5 marks each.
7. In Section-E: question number 34 and 35 are case-based questions carrying 4 marks each.
8. There is no overall choice. However, an internal choice has been provided in 2 questions in Section-B, 2 questions in Section-C, 3 questions in Section-D and 2 questions in Section-E.
9. Use of calculators is NOT allowed.
   c = 3 × 10 ⁸ m/s; h = 6.63 × 10 ^-34 Js
   e = 1.6 × 10 ^-19 ; µ ₀ = 4π × 10 ^-7 T m A ^-1
   \(\varepsilon_0\) = 8.854 × 10 ^-12 C ² N ^-1 m ^-2
   \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10 ⁹ N m ² C ^-2
   Mass of electron (m _e ) = 9.1 × 10 ^-31 kg; Mass of neutron = 1.675 × 10 ^-27 kg
   Mass of proton = 1.673 × 10 ^-27 kg; Avogadro’s number = 6.023 × 1o ²³ per gram mole
   Boltzmann constant = 1.38 × 10 ^-23 JK ^-1

SET I Code No. 55/2/1
Section-A

Question 1.
The magnitude of the electric field ^due to a point chaige object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of [1]
(a) 1 m
(b) 2 m
(c) 3 m
(d) 6 m
Answer:
(c) 3 m

Question 2.
A point P lies at a distance x from the mid point of an electric dipole on its axis. The electric potential at point P is proportional to [1]
(a) \(\frac{1}{x^2}\)
(b) \(\frac{1}{x^3}\)
(c) \(\frac{1}{x^4}\)
(d) \(\frac{1}{x^{1 / 2}}\)
Answer:
(a) \(\frac{1}{x^2}\)

P ∝ \(\frac{1}{x^2}\)

Question 3.
A current of 0.8 A flows in a conductor of 40 Ω for 1 minute. The heat produced in the conductor will be [1]
(a) 1445 J
(b) 1536 J
(c) 1569 J
(d) 1640 J
Answer:
(b) 1536 J

Given, I = 0.8 A, R = 40 Ω, t = 1 min or 60 sec
H = I ² Rt
∴ H = (0.8) ² × 40 × 60 = 0.64 × 40 × 60 = 1536 J

Question 4.
A cell of emf E is connected across an external resistance R. When current ‘I’ is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance ‘r’ of the cell is [1]
(a) \(\left(\frac{E-V}{E}\right) R\)
(b) \(\left(\frac{E-V}{E}\right) R\)
(c) \(\frac{(E-V) R}{I}\)
(d) \(\left(\frac{E-V}{V}\right) R\)
Answer:
(d) \(\left(\frac{E-V}{V}\right) R\)

r = \(\left(\frac{E}{V}-1\right) R\) or \(\left(\frac{E-V}{V}\right) R\)

Question 5.
Beams of electrons and protons move parallel to each other in the same direction. They [1]
(a) attract each other.
(b) repel each other.
(c) neither attract nor repel.
(d) force of attraction or repulsion depends upon speed of beams.
Answer:
(b) repel each other.

Question 6.
A long straight wire of radius V carries a steady current T. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic field \(\overrightarrow{\mathrm{B}}_1\) at \(\frac{a}{2}\) and \(\vec{B}_2\) at distance 2a is [1]
(a) \(\frac{1}{2}\)
(b) 1
(c) 2
(d) 4
Answer:
(b) 1

Question 7.
\(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) represent the electric and the magnetic field of an electro-magnetic wave respectively. The direction of propagation of the wave is along [1]
(a) \(\vec{B}\)
(b) \(\overrightarrow{\mathrm{E}}\)
(c) \(\vec{E} \times \vec{B}\)
(d) \(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}}\)
Answer:
(c) \(\vec{E} \times \vec{B}\)

\(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

Question 8.
A ray of monochromatic light propagating in air, is incident on the surface of water. Which of the following will be the same for the reflected and refracted rays? [1]
(a) Energy carried
(b) Speed
(c) Frequency
(d) Wavelength
Answer:
(c) Frequency

Question 9.
A beam of light travels from air into a medium. Its speed and wavelength in the medium are 1.5 × 10 ⁸ ms ^-1 and 230 nm respectively. The wavelength of light in air will be [1]
(a) 230 nm
(b) 345 nm
(c) 460 nm
(d) 690 nm
Answer:
(c) 460 nm

Question 10.
Which one of the following metals does not exhibit emission of electrons from its surface when irradiated by visible light? [1]
(a) Rubidium
(b) Sodium
(c) Cadmium
(d) Caesium
Answer:
(c) Cadmium

Question 11.
A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is [1]
(a) \(\frac{8}{7} \lambda\)
(b) \(\frac{16}{7} \lambda\)
(c) \(\frac{24}{7} \lambda\)
(d) \(\frac{32}{7} \lambda\)
Answer:
(d) \(\frac{32}{7} \lambda\)

Question 12.
The curve of binding energy per nucleon as a function of atomic mass number has a sharp peak for helium nucleus. This implies that helium nucleus is [1]
(a) radioactive
(b) unstable
(c) easily fissionable
(d) more stable nucleus than its neighbours
Answer:
(d) more stable nucleus than its neighbours

Question 13.
In an extrinsic semiconductor, the number density of holes is 4 × 10 ²⁰ m ^-3 . If the number density of intrinsic carriers is 1.2 × 10 ¹⁵ m ^-3 . The number density of electrons in it is [1]
(a) 1.8 × 10 ⁹ m ^-3
(b) 2.4 × 10 ¹⁰ m ^-3
(c) 3.6 × 10 ⁹ m ^-3
(d) 3.2 × 10 ¹⁰ m ^-3
Answer:
(c) 3.6 × 10 ⁹ m ^-3

Given, n _i = 1.2 × 10 ¹⁵ m ^-3
n _h = 4 × 10 ²⁰ m ^-3
As we know, \(n_i^2\) = n _e . n _h
n _e = \(\frac{\left(n_i\right)^2}{n_h}\) = \(\frac{\left(1.2 \times 10^{15}\right)^2}{4 \times 10^{20}}\)
⇒ n _e = \(\frac{1.44 \times 10^{30}}{4 \times 10^{20}}\) = 3.6 × 10 ⁹ m ^-3

Question 14.
Pieces of copper and of silicon are initially at room temperature. Both are heated to temperature T. The conductivity of [1]
(a) both increases.
(b) both decreases.
(c) copper increases and silicon decreases.
(d) copper decreases and silicon increases.
Answer:
(d) copper decreases and silicon increases.

Copper decreases and silicon increases because copper is conductor and silicon is semiconductor.

Question 15.
The formation of depletion region in a p-n junction diode is due to [1]
(a) movement of dopant atoms.
(b) diffusion of both electrons and holes.
(c) drift of electrons only.
(d) drift of holes only.
Answer:
(b) diffusion of both electrons and holes.

Note: In questions number 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 Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion
(A).
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of
Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also false.

Question 16.
Assertion (A): Diamagnetic substances exhibit magnetism.
Reason (R): Diamagnetic materials do not have permanent magnetic dipole moment.
Answer:
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).

Question 17.
Assertion (A): Work done in moving a charge around a closed path, in an electric field is always zero.
Reason (R): Electrostatic force is a conservative force. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion
(A).

Question 18.
Assertion (A): In Young’s double slit experiment all fringes are of equal width.
Reason (R): The fringe width depends upon wavelength of light (λ) used, distance of screen from plane of slits (D) and slits separation (d). [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion
(A).

Section – B

Question 19.
Briefly explain why and how a galvanometer is converted into an ammeter.
Answer:
A galvanometer can detect only small currents. Thus, to measure large currents it is converted into an ammeter. It has large resistance and hence will change the value of current in circuit. It can be converted into an ammeter by connecting a low resistance in parallel to the galvanometer.

Question 20.
(a) How are infrared waves produced? Why are these waves referred to as heat waves? Give any two uses of infrared waves.
Or
(b) How are X-rays produced?’Give any two uses of these.
Answer:
(a) Infrared waves are produced by. hot bodies and molecules.
These waves are sometimes referred to as heat waves is because water molecules present in most materials readily absorb infrared wave. After absorption, their thermal motion increases, that is why they heat up and heaty their surroundings.

Uses of Infrared waves:

1. Infrared waves are used in physical therapy.
2. Electronic devices also emit infrared and are widely used in the remote switches of household electronic system.

Or

(b) X-rays are produced when high energetic electrons beam is made incident on a metallic target of high melting point and high atomic weight.

Uses:

1. Used as a diagnostic tool in the medicine.
2. Treatment of certain forms of cancer.

Question 21.
In the given figure the radius of curvature of curved face in the plano-convex and the plano-concave lens is 15 cm each. The refractive index of the material of the lenses is 1.5. Find the final position of the image formed. [2]

Answer:

Since Parallel rays are incident on the piano convex lens it will form an image at a distance of 30 cm from the lens. This will be object of the piano concave lens.
∴ Object distance, u = 30 – 20 = 10 cm
Using lens formula:

Question 22.
What happens to the interference pattern when two coherent sources are
(a) infinitely close, and
(b) far apart from each other [2]
Answer:
As we know, β = \(\frac{\mathrm{D} \lambda}{d}\)
If d is very small then fringe width (β) will be very large and a single patch will occupy the whole field of view hence pattern cannot be observed.
When sources are far apart, i.e., d is very large, then fringe width will be so small that the fringes are not resolved and cannot be seen separately.

Question 23.
(a) What is meant by ionisation energy? Write its value for hydrogen atom.
Or
(b) Define the term mass defect. How is it related to stability of the nucleus?
Answer:
(a) Definition of ionization energy: “The minimum energy, required to free the electron from the ground state to outside the atom or from n = 1 to n = ∞ or from an isolated atom of an element is known as Ionization Energy.”
The ionization energy is given by: E ₀ = \(\frac{m e^4}{8 \varepsilon_0^2 h^2}\) i.e., E ₀ ∝ m
The ionisation energy for hydrogen atom is 13.6 eV.

Or

(b) The difference between the rest mass of the nucleus and the sum of the mass of the nucleons composing a nucleus is known as mass defect.
Mass defect, ∆m = [Zm _p + (A – Z) m _n ] – M _N
… where
[A is mass number, Z is atomic number, m _p is mass of proton, m _n is mass of neutron and M _N is mass of numbers.
The stability of the nucleus increases with increase in the value of mass defect.

Question 24.
Draw energy band diagram for an n-type and p-type semiconductor at T > OK. [2]
Answer:

Question 25.
Answer the following giving reasons:
(a) A p-n junction diode is damaged by a strong current.
(b) Impurities are added in intrinsic semiconductors.
Answer:
(a) If current is strong then more heat develops which results in breakdown of semiconductor. After breakdown semiconductor behaves like a conductor.
(b) Impurities are added in intrinsic semiconductor to increase the conductivity of the semiconductor.

Section – C

Question 26.
(a) Two charged conducting spheres of radii a and b are connected to each other by a wire. Find the ratio of the electric fields at their surfaces. [3]

Or

(b) A parallel plate capacitor (A) of capacitance C is charged by a battery to voltage V. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C is connected across A. Find the ratio of
(i) final charges on A and B.
(ii) total electrostatic energy stored in A and B finally and that stored in A initially. [3]
Answer:
(a) As Potential become same for both spheres after connection,

Or

(b) (i) Initially q ₁ = C ₁ V ₁ = CV
Finally q ₂ = C ₂ V ₂ = 2CV
∴ Ratio = \(\frac{q_1}{q_2}\) = \(\frac{C V}{2 C V}\) = \(\frac{1}{2}\)

(ii)

Question 27.
Define current density and relaxation time. Derive an expression for resistivity of a conductor in terms of number density of charge carries in the conductor and relaxation time. [3]
Answer:
Current density. It gives the amount of chargé flowing per second per unit area normal to the flow. The current density \(\vec{J}\) is given by
\(\vec{J}\) = nq \(\vec{V}_d\)
….. where
[n is number density of charge carriers each of charge q
V _d is the drift velocity
Relaxation time: The average time that elapses between two successive collisions of an electron is called relaxation time.
\(\vec{v}_d\) = \(\vec{a} \tau\) = \(-e \frac{\vec{E}}{m} \tau\) …where [\(\overrightarrow{\mathbf{v}}_d\) is called drift velocity of electrons.
Suppose a potential difference V is applied across a conductor of length ‘l’ and of uniform cross-section A, then electric field E set up inside the conductor is given by
E = \(\frac{\mathrm{V}}{l}\)

Under the influence of field \(\vec{E}\), the free electrons begin to drift in the opposite direction \(\vec{E}\) with an average drift velocity V _d .
Let the number of electrons per unit volume or electron density = n
Charge on an electron = e
Number of electrons in length l of the conductor = n × volume of the conductor = nAl
Total charge contained in length ¡ of the conductor, q = enAl
According to the electrons which enter the conductor at the right end will pass through the conductor at the left end in time,

Question 28.
A series CR circuit with R = 200 Ω and C = (50/π) µF is connected across an ac source of peak voltage \(\varepsilon_0\) = 100 V and frequency v = 50 Hz. Calculate—
(a) impedance of the circuit (Z), (b) phase angle (φ), and (c) voltage across the resistor. [3]
Answer:

Question 29.
Define critical angle for a given pair of media and total internal reflection. Obtain the relation between the critical angle and refractive index of the medium. [3]
Answer:
Critical angle of incidence. It is the angle of incidence for which the angle of refraction is 90° when a ray of light is incident from a denser medium to a rarer medium.
∴ n ₂₁ = sin i _c = \({ }^1 \mu_2\)
The refractive index of denser medium 2 with respect to rarer medium 1 will be
n ₁₂ = \(\frac{1}{\sin i_c}\)
Total internal reflection. When angle of incidence is greater than critical angle ‘i > i _c ’ for a ray travelling from denser medium to a rarer medium, the ray is reflected back into the same medium without any loss of intensity. This phenomenon is called total internal reflection.

When a ray of light enters from a denser medium to rarer medium, it bends away from the normal, which means angle of refraction is larger than angle of incidence. As the angle of incidence increases, refracted ray grazes the surface at the interface between two media, as shown in the figure.
The angle of incidence corresponding to angle of refraction 90° is called critical angle.

μ _r = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin i_c}{\sin 90^{\circ}}\) = \(\frac{\sin i_c}{1}\)
⇒ \({ }^r \mu_d\) = \(\frac{\mathbf{1}}{d_{\mu_r}}\) = \(\frac{1}{\sin i_c}\)

Question 30.
(a) (i) Distinguish between nuclear fission and fusion giving an example of each.
(ii) Explain the release of energy in nuclear fission and fusion on the basis of binding energy per nucleon curve. [3]

Or

(b) (i) How is the size of a nucleus found experimentally? Write the relation between the radius and mass number of a nucleus. [3]
(ii) Prove that the density of a nucleus is independent of its mass number. [3]
Answer:
(a) (i) The breaking of heavy nucleus into smaller fragments is called nuclear fission; while the joining of lighter nuclei to form a heavy nucleus is called nuclear fusion.
Nuclear fission. It is the phenomenon of splitting of a heavy nucleus into two lighter nuclei. When \({ }_{92}^{235} \mathrm{U}\) is bombarded with thermal neutrons, it splits up into \({ }_{56}^{141} \mathrm{Ba}\) and \({ }_{36}^{92} \mathrm{Kr}\) with emission of 3 neutrons alongwith 200 MeV of energy per fission.

Nuclear fusion. It is the phenomenon of fusion of two or more lighter nuclei to form a single heavy nucleus. Mass defect in the process appears as energy.

(ii) Nuclear fission: Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e., heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.

Nuclear fusion: The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.

Or

(b)
(i) Size of nucleus: Only a small fraction of the incident α-particles rebound. This shows that the mass of the atom is concentrated in a small volume in the form of nucleus and gives an idea of the size of nucleus.
Relation between radius and mass number of nucleus = R = R ₀ A ^1/3
(ii) Radius of nucleus, R = R ₀ A ^1/3

Section – D

Question 31.
(a) (i) Use Gauss’ law to obtain an expression for the electric field due to an infinitely long thin straight wire with uniform linear charge density λ.

(ii) An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre and is perpendicular to the plane, of the circle. Find the kinetic energy of the electron in terms of magnitudes of its charge and linear charge density λ on the wire.
(iii) Draw a graph of kinetic energy as a function of linear charge density λ. [5]

Or

(b) (i) Consider two identical point charges located at points (0, 0) and (a, 0)
1. Is there a point on the line joining them at which the electric field is zero?
2. Is there a point on the line joining them at which the electric potential is zero? Justify your answers for each case.
(ii) State the significance of negative value of electrostatic potential energy of a system of charges.
Three charges are placed at the comers of an equilateral triangle ABC of side 2.0 m as shown in figure. Calculate the electric potential energy of the system of three charges.

Answer:
(i) Gauss’s law in electrostatics. It states that ” the total electric flux over the surface S in vacuum is \(\frac{1}{\varepsilon_0}\) times the total charge (q)”.
∴ φ _E = \(\oint \overrightarrow{\mathrm{E}} \cdot d \vec{S}\) = \(\frac{q}{\varepsilon_0}\)
Electric field due to an infinitely long straight wire: Consider an infinitely long straight line charge having linear charge density λ to determine its electric field at distance r. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S<sub.1 and is directed radially outward.

(ii)

(iii)

Or

(b) (i)
1. Yes, If both charges are identical then at the mid point (axial line) electric field will be zero, i.e., (\(\frac{a}{2}\), 0)

No, as here both charges are identical therefore electric potential cannot be zero at any point.

(ii) Significance of Negative value: For unlike charges (q ₁ q ₂ < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for reverse path (from infinity to the present locations). So the potential energy is negative.
The electric potential energy of the system

Question 32.
(a) (i) Define coefficient of self-induction. Obtain an expression for self-inductance of a long solenoid of length l, area of cross-section A having N turns.
(ii) Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency \(\left(\frac{200}{\pi}\right)\) Hz and a DC source is applied across the coil.

Or

(b) (i) With the help of a labelled diagram, describe the principle and working of an ac generator. Hence, obtain an expression for the instantaneous value of the emf generated.
(ii) The coil of an ac generator consists of 100 turns of wire, each of area 0.5 m ² . The resistance of the wire is 100 Ω. The coil is rotating in a magnetic field of 0.8 T perpendicular to its axis of rotation, at a constant angular speed of 60 radian per second. Calculate the maximum emf generated and power dissipated in the coil. [5]
Answer:
(a)
(i) Coefficient of self induction. Consider a coil L as shown in Figure. Suppose a current I flows through the coil at any instant then magnetic flux φ linked with the coil is directly proportional to the current passing through it at that instant,

∴ φ ∝ I
⇒ φ = LI .. .where [L is called coefficient of self induction.
If I = 1, then φ = L

Thus, self inductance of a coil is numerically equal to the magnetic flux linked with the coil, when a unit current flows through it.
The SI unit of self inductance is henry (H).
Expression for self-inductance: Consider a long solenoid of length l and cross-sectional area A having n turns per unit length.
The magnetic field due to a current flowing in the solenoid is B = µ ₀ nl
Total flux linked with the solenoid is,
φ = (nl) (µ ₀ nI) A = µ ₀ n ² AlI . …(i) …where [nl is the total number of turns.
Thus, the self inductance is, L = \(\frac{\phi}{\mathrm{I}}\)
= \(\frac{\mu_0 n^2 \mathrm{~A} l \mathrm{I}}{\mathrm{I}}\) …. [From (i)
= µ ₀ n ² Al
If we fill the inside of the solenoid with a material of relative permeability pr then, L = µ ₀ µ ₀ n ² Al.

(ii)

Or

(b) (i) (a) Principle of A.C. generator: The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.
(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field \(\vec{B}\) at any instant t
ω = the angular velocity with which coil rotates

The magnetic flux linked with the coil at any instant t will be,
φ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
ε = –\(\frac{d \phi}{d t}\) = \(\frac{-d}{d t}\) NAB (cos ωt)
ε = NAB (sin ωt). ω
⇒ ε = ε ₀ sin ωt

(ii) Maximum emf generated, ε ₀ = NBAω
= 100 × 0.8 × 0.5 × 60 = 2400 V
Power dissipated, P = \(\frac{\varepsilon_{\text {rms }}^2}{\mathrm{R}}\)
= \(\frac{\left(\frac{2400}{\sqrt{2}}\right)^2}{100}\) = \(\frac{2400 \times 2400}{2} \times \frac{1}{100}\) = 28.8 kW

Question 33.
(a) (i) State Huygen’s principle. With the help of a diagram, show how a plane wave is reflected from a surface. Hence verify the law of reflection.
(ii) A concave mirror of focal length 12 cm forms a three times magnified virtual image of an object. Find the distance of the object from the mirror. [5]
Or
(b) (i) Draw a labelled ray diagram showing the image formation by a refracting telescope. Define its magnifying power. Write two limitations of a refracting telescope over a reflecting telescope.
(ii) The focal lengths of tho-objective and the eye-piece of a compound microscope are 1.0 cm and 2.5 cm respectively. Find the tube length of the microscope for obtaining a magnification of 300. [5]
Answer:
(a) (i) Huygen’s principle: Huygen’s principle is based on two assumptions:
(ii) Each point on the primary wavefront is a source of a new disturbance called secondary wavelets which travel in all directions with same velocity as that of original waves.
(ii) A surface tangential to the secondary wavelets gives the position and shape of new wavefront at any instant. This is called secondary wavefront.
The principle leads to the well known laws of reflection and refraction.

Since time taken by waves from point B to C and from A to E is same
∴ BC = AE = vt
In ∆ABC and ∆AEC,
AC AC …[Common
∠ABC = ∠AEC …[90° each
AE = BC
∴ ∆ABC \(\cong\) ∆AEC ‘
Hence ∠BAC = ∠ECA
∠i = ∠r
i.e., Angle of incidence = Angle of reflection

(ii)

Or

(b)
(i) Refracting telescope:

Magnifying power. It is defined as the ratio of angle (β) subtended by the final image on the eye to the angle (α) subtended by object on eye.
M = \(\frac{\tan \beta}{\tan \alpha}\) = \(\left(\frac{\beta}{\alpha}\right)\)
Magnifying power, M = \(\frac{-f_0}{f_e}\) (for comfortable view)
= \(\frac{-f_0}{f_e}\left(1+\frac{f_e}{\mathrm{D}}\right)\) (for strained view)

Limitations of refracting telescope over a reflecting type telescope:

1. It suffers from chromatic aberration due to refraction and hence the image obtained is multicoloured and blurred.
2. As a lens of large apparatus can’t be manufactured easily, its light gathering power is low and hence can’t be used to see faint stars.

(ii) Given. f ₀ = 1 cm, f _e = 2.5 cm, m = 300, D = 25 cm, L = ?
|m| = \(\frac{\mathrm{L}}{f_0} \cdot \frac{\mathrm{D}}{f_e}\)
⇒ 300 = \(\frac{\mathrm{L}}{1} \cdot \frac{25}{2.5}\) ⇒ 30 = 10 L ∴ L = 3 cm

Section – E

Note: Questions number 34 and 35 are Case Study based questions.

Read the following paragraph and answer the questions.

Question 34.
(a) Consider the experimental set up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting nonmagnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off.

Answer the following questions:
(i) Explain the reason of jumping of the ring when the switch is closed in the circuit.
(ii) What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
(iii) Explain the two laws that help us understand this phenomenon. [4]

Or

(b) Briefly explain various ways to increase the strength of magnetic field produced by a given solenoid.
Answer:
(a)
(i) The direction of induced current in the ring is such that the polarity developed in the ring is same as that of the polarity on the face of the coil, hence it will jump up due to repulsive force.
(ii) The polarity of the induced current in the ring will get reversed on changing the – terminals of the battery, so the ring will jump again.
(iii) Two laws Lenz’s law. It states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it.
Faradav’s law of EMI. Whenever there is change in magnetic flux through a coil an emf is induced.

Or

(b) Various ways to increase the magnetic field produced by a solenoid:

1. By increasing the number of turns per unit length.
2. By increasing the strength of current.

Question 35.
(a) Figure shows the variation of photoelectric current measured in a photo cell circuit as a function of the potential difference between the plates of the photo cell when light beams A, B, C and D of different wavelengths are incident on the photo cell. Examine the given figure and answer the following questions:

(i) Which light beam has the highest frequency and why? [1]
(ii) Which light beam has the longest wavelength and why?
(iii) Which light beam ejects photoelectrons with maximum momentum and why? [4]

Or

(b) What is the effect on threshold frequency and stopping potential on increasing the frequency of incident beam of light? Justify your answer.
Answer:
(a)
(i) As we know the frequency increases, work function remaining same, the stopping potential increases.
In the given case, for light beam B stopping potential is maximum. Hence, it has the highest frequency.
(ii) Light beam C has the lowest frequency. Hence, this beam has the longest wavelength.
(iii) Light beam B has the highest frequency. So, electrons ejected due to this radiation will have maximum kinetic energy and hence maximum momentum.

Or

• Stopping potential increases with increase of frequency of incident beam of light. From Einsten’s equation,
  hv = hv ₀ + eVs
  v ₀ is constant for a particular substance. So, as v increases, Vs increases.
• There is no effect on threshold frequency. Threshold frequency (v ₀ ) is a constant for a substance. If frequency of incident light is lower than this, no photoemission will take place. If frequency of incident light is higher than this, photoemission will take place.

SET II Code No. 55/2/2
Section – A

Except for the following questions, all the remaining questions have been asked in Set I.

Question 1.
The ratio of the magnitudes of the electric field and magnetic field of a plane electromagnetic wave is [1]
(a) 1
(b) \(\frac{1}{c}\)
(c) c
(d) \(\frac{1}{c^2}\)
Answer:
(c) c

\(\frac{E}{B}\) = c = velocity

Question 2.
Specify the transition of electron in the wavelength of the line in the Bohr model of hydrogen atom which gives rise to the spectral line of highest wavelength. [1]
(a) n = 3 to n = 1
(b) n = 3 to n = 2
(c) n = 4 to n = 1
(d) n = 4 to n = 2
Answer:
(b) n = 3 to n = 2

Question 5.
An isolated point charge particle produces an electric field \(\overrightarrow{\mathbf{E}}\) at a point 3 m away from it. The distance of the point at which the field is \(\frac{\vec{E}}{4}\) will be [1]
(a) 2 m
(b) 3 m
(c) 4 m
(d) 6 m
Answer:
(d) 6 m

Question 7.
A steady current of 8 mA flows through a wire. The number of electrons passing through a cross-section of the wire in 10 s is [1]
(a) 4.0 × 1o ¹⁶
(b) 5.0 × 10 ¹⁷
(c) 1.6 × 10 ¹⁶
(d) 1.0 × 10 ¹⁷
Answer:
(b) 5.0 × 10 ¹⁷

Question 8.
Which one of the following elements will require the highest energy to take out an electron from them? [1]
(a) Ge
(b) C
(c) Si
(d) Pb
Answer:
(b) C

Carbon

Question 9.
A conductor of 10 Ω is connected across a 6 V ideal source. The power supplied by the source to the conductor is [1]
(a) 1.8 W
(b) 2.4 W
(c) 3.6 W
(d) 7.2 W
Answer:
(c) 3.6 W

P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) = \(\frac{6 \times 6}{10}\) = 3.6 W

Question 12.
A photon of wavelength 663 nm is incident on a metal surface. The work function of the metal is 1.50 eV. The maximum kinetic energy of the emitted photo electrons is [1]
(a) 3.0 × 10 ^-20 J
(b) 6 0 × 10 ^-20 J
(c) 4 5 × 10 ^-20 J
(d) 9 0 × 10 ^-20 J
Answer:
(b) 6 0 × 10 ^-20 J

Question 14.
A ray of light of wavelength 600 nm propagates from air into a medium. If its wavelength in the medium becomes 400 nm, the refractive index of the medium is [1]
(a) 1.4
(b) 1.5
(c) 1.6
(d) 1.8
Answer:
(b) 1.5

Given. λ _m = 400 and λ _a = 600 mn μ = \(\frac{\lambda_a}{\lambda_m}\) = \(\frac{600 \mathrm{~nm}}{400 \mathrm{~nm}}\) = 1.5

Section – B

Question 19.
In a Young’s double slit experiment, the separation between the two slits is d and distance of the screen from the slits is 1000 d. If the first minima falls at a distance d from the central maximum, obtain the relation between d and λ. [2]
Answer:
Given. D = 1000d; n = 1; x ₁ = d
for n ^th dark fringe (minima) from the central maxima, x _n = \(\frac{(2 n-1) \lambda D}{2 d}\)
x ₁ = (2 × 1 – 1) \(\frac{\lambda 1000 d}{2 d}\)
∴ x = 500λ

Question 25.
A point object in air is placed symmetrically at a distance of 60 cm in front of a concave spherical surface of refractive index 1.5. If the radius of curvature of the surface is 20 cm, find the position of the image formed. [2]
Answer:
For refraction from concave spherical mirror, \(\frac{\mu_1}{-u}+\frac{\mu_2}{v}\) = \(\frac{\mu_2-\mu_1}{R}\)

Section – C

Question 27.
A ray of light is incident on a glass prism of refractive index µ and refracting angle A. If it just suffers total internal reflection at the other face, obtain a relation between the angle of incidence, angle of prism and critical angle. [3]
Answer:

Question 29.
Two cells of emf E ₁ and E ₂ and internal resistance r ₁ and r ₂ are connected in parallel, with their terminals of the same polarity connected together. Obtain an expression for the equivalent emf of the combination. [3]
Answer:
Let I ₁ and I ₂ be the current due to the two cells

Since the voltage across each resistor connected in parallel is the same,
Voltage across the first cell,

SET I Code No. 55/2/3
Section – A

Except for the following questions, all the remaining questions have been asked in Set I & Set II.

Question 2.
The energy of a photon of wavelength 663 nm is [1]
(a) 6.64 × 10 ^-20 J
(b) 5.18 × 10 ^-19 J
(c) 3.0 × 10 ^-19 J
(d) 2.0 × 10 ^-20 J
Answer:
(c) 3.0 × 10 ^-19 J

Question 3.
An electromagnetic wave is produced by a charge [1]
(a) moving with a constant velocity
(b) moving with a constant speed parallel to a magnetic field
(c) moving with an acceleration
(d) at rest
Answer:
(c) moving with an acceleration

Accelerated charge produces EM waves.

Question 4.
A semiconductor device is connected in series with a battery, an ammeter and a resistor. A current flows in the circuit. If the polarity of the battery is reversed, the current in the circuit almost becomes zero. The device is a/an [1]
(a) intrinsic semiconductor
(b) p-type semiconductor
(c) n-type semiconductor
(d) p-n junction diode
Answer:
(d) p-n junction diode

p-n diode does not conduct in reverse bias situation.

Question 6.
The radius of \({ }_{13}^{27} X\) nucleus is R. The radius of \({ }_{53}^{125} Y\) nucleus will be [1]
(a) \(\frac{5}{3} R\)
(b) \(\left(\frac{13}{53}\right)^{1 / 3} \mathrm{R}\)
(c) \(\left(\frac{5}{3} R\right)^{1 / 3}\)
(d) \(\left(\frac{13}{53} R\right)^{1 / 3}\)
Answer:
(a) \(\frac{5}{3} R\)

By using, R = R ₀ A ^1/3
⇒ \(\frac{\mathrm{R}_y}{\mathrm{R}_x}\) = \(\frac{(125)^{1 / 3}}{(27)^{1 / 3}}\)
⇒ \(\frac{R_y}{R}\) = \(\frac{5}{3}\)

Question 7.
An electric dipole of dipole moment 2 × 10 ^-8 C-m in a uniform electric field experiences a maximum torque of 6 × 10 ^-1 N-m. The magnitude of electric field is [1]
(a) 2.2 × 10 ³ Vm ^-1
(b) 1.2 × 10 ⁴ Vm ^-1
(c) 3.0 × 10 ⁴ Vm ^-1
(d) 4.2 × 10 ³ Vm ^-1
Answer:
(c) 3.0 × 10 ⁴ Vm ^-1

By using: \(\vec{\tau}\) = \(\vec{p} \times \vec{E}\)
For max. torque θ = 90°
\(\tau\) = p E sin 90°
⇒ E = \(\frac{\tau}{p}\) = \(\frac{6 \times 10^{-4}}{2 \times 10^{-8}}\) = 3 × 10 ⁴
∴ E = 3 × 10 ⁴
∴ E = 3 × 10 ⁴ Vm ^-1

Question 9.
A point charge q ₀ is moving along a circular path of radius a, with a point charge -Q at the centre of the circle. The kinetic energy of q ₀ is [1]
(a) \(\frac{q_0 \mathrm{Q}}{4 \pi \epsilon_0 a}\)
(b) \(\frac{q_0 \mathrm{Q}}{8 \pi \epsilon_0 a}\)
(c) \(\frac{q_0 \mathrm{Q}}{4 \pi \epsilon_0 a^2}\)
(d) \(\frac{q_0 \mathrm{Q}}{8 \pi \epsilon_0 a^2}\)
Answer:
(b) \(\frac{q_0 \mathrm{Q}}{8 \pi \epsilon_0 a}\)

If a particle is revolving in circular orbit.

Question 11.
The current in a device varies with time t as I = 6 t, where I is in mA and t is in s. The amount of charge that passes through the device during t Os to t = 3s is [1]
(a) 10 mC
(b) 18 mC
(c) 27 mC
(d) 54 mC
Answer:
(c) 27 mC

By using, I = \(\frac{d \mathrm{Q}}{d t}\)
dQ = Idt
Q = \(\int d \mathrm{Q}\) = \(\int_0^3 I d t\)
= \(\int_0^3 6 t . d t\) = \(\int_0^3 6 t . d t\)
= \(\left[\frac{6 t^2}{2}\right]_0^3\) = \(\left[3 t^2\right]_0^3\) ∴ Q = 3 × 9 = 27 mC

Question 12.
A ray of light travels a distance of 12.0 m in a transport sheet in 60 ns. The refractive index of the sheet is [1]
(a) 1.33
(b) 1.50
(c) 1.65
(d) 1.75
Answer:
(b) 1.50

Section – B

Question 21.
With the help of a circuit diagram, explain how a full wave rectifier gives output rectified voltage corresponding to both halves of the input ac voltage. [2]
Answer:
p—n junction diode as full wave rectifier. A full wave rectifier consists of two diodes and special type of transformer known as centre tap transformer as shown in the circuit. The secondary of transformer gives the desired a,c. voltage across A and B.
During the positive half cycle of a.c. input, the diode D ₁ is in forward bias and conducts cur
rent while D ₂ is in reverse biased and does not conduct current. So we get an output voltage across the load resistor R _L .

During the negative half cycle of a.c. input, the diode D ₁ is in reverse biased and does not conduct current while diode D ₂ in forward biased and conducts current. So we get an
output voltage across the load resistor R _L .
Note. This is a more efficient circuit for getting rectified voltage or current.

Let I = I ₀ sin ωt be the input curre1t to be rectified
∴ The average current = \(\frac{2 \mathrm{I}_0}{\pi}\)
Hence output voltage = \(\frac{2 \mathrm{I}_0}{\pi}\) RL

Question 23.
The power of a thin lens is +5D. When it is immersed in a liquid, it behaves like a concave lens of focal length 100 cm. Calculate the refractive index of the liquid. Given refractive index of glass = 1.5. [2]
Answer:

Section – C

Question 26.
A ray of light is refracted by a glass prism. Obtain an expression for the refractive index of the glass in terms of the angle of prism A and the angle of minimum deviation 8m. [3]
Answer:
A ray PQ is incident on the face AB of prism at ∠i and effected along QR at ∠r. The angle of incidence (from glass to air) to the second face is ‘r’ and the angle of refraction or emergence is i’.
The angle between the emergent ray RS and incident ray in the direction PQ is called the angle of deviation δ.

In the quadrilateral AQNR,
∠Q and ∠R are right angles
∴ ∠A + ∠QNR = 180° …… (i)
and in ∆QNR
r + r’ + ∠QNR = 180° ……(ii)
Comparing these equations, we get
r + r’ = ∠A
Also δ = ∠MQR + ∠MRQ = i – r + i’ – r’ = i + i – (r + t’)
∴ 6 = i + i ’ – ∠A
In the minimum deviation position,
i = i’, r = r’ and δ = δ _m
∴ r + r = A ∴ r = \(\frac{\mathrm{A}}{2}\)
and δ _m = i + i – A ∴ i = \(\frac{A \delta_m}{2}\)
Put the value of (iii) in Snell’s law
µ = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin \left(\mathrm{A}+\frac{\delta m}{2}\right)}{\sin \frac{\mathrm{A}}{2}}\)

Question 29.
A potential difference V is applied across a conductor of length l and cross-sectional area A. Briefly explain how the current density j in the conductor will be affected if
(a) the potential difference V is doubled,
(b) the conductor were gradually stretched to reduce its cross-sectional area to \(\frac{\mathrm{A}}{2}\) and then the same potential difference V is applied across it. [3]
Answer:
(a) As we know, current density, j = \(\frac{n e^2 E \tau}{m}\)
j = \(\frac{n e^2 v \tau}{m l}\)
So, j ∝ v
If v is doubled, j is also doubled.

(b) From the above formula, j × \(\frac{1}{l}\)
As, Area is reduced to A/2, it’s length will become double (2l) as volume of conduc-tor remains constant.
So, j will become half.

Question 30.
A resistor of 50 Ω, a capacitor of \(\left(\frac{25}{\pi}\right)\)µF and an inductor of \(\left(\frac{4}{\pi}\right)\)H are connected in series across an ac source whose voltage (in volt) is given by V = 70 sin (100 π t). Calculate:
(a) the net reactance of the circuit,
(b) the inpedence of the circuit
(c) the effective value of current in the circuit. [3]
Answer:
Given. R = 50Ω; L = \(\frac{4}{\pi}\)H, C = \(\frac{25}{\pi}\) × 10 ^-6 F; V ₀ = 70 V; ω = 100 π rad/sec