HBSE Class 12th Physics Sample Paper 2024 Answer

Haryana Board Class 12th Physics Sample Paper 2024 with Answer. Haryana Board 12th Physics Model Paper with Answer. HBSE 12th Physics Model Paper 2023 Solution. HBSE 12th Physics ka Model Paper 2024. BSEH Class 12th Physics Model Paper 2024. Physics Sample Paper 12th Class HBSE Board. HBSE Class 12 Physics Sample Paper 2024 Pdf Download. HBSE Class 12th Ka Physics Model Paper Download Kare. HBSE Class 12 Physics Model Paper 2024 Pdf Download.

HBSE Class 12th Physics Sample / Model Paper 2024 Answer

Class – 12th                       Subject – Physics
Time : 3 Hours                    Total 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 these Sections are compulsory.
3. Section A contains eighteen MCQ 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  (18 Marks)
(18 MCQ of 1 Mark Each)

1. The electric potential on the axis of an electric dipole at a distance ‘r’ from its centre is V. Then the potential at a point at the same distance on its equatorial line will be :
(i) 2V
(ii) -V
(iii) V/2
(iv) Zero
Answer – (iv) Zero

2. Resistance of conductor does not depend on :
(i) Length of conductor
(ii) Nature of material
(iii) Radius of cross section of conductor
(iv) Potential difference applied across the conductor
Answer – (iv) Potential difference applied across the conductor

3. The temperature (T) dependence of resistivity of materials A and material B is represented by fig.(i) and fig.(ii) respectively. Identify material A and material B.

(i) material A is copper and material B is germanium
(ii) material A is germanium and material B is copper
(iii) material A is nichrome and material B is germanium
(iv) material A is copper and material B is nichrome
Answer – (ii) material A is germanium and material B is copper

4. Wheatstone bridge can not be used for measuring of very ………… resistances.
(i) high
(ii) low
(iii) low or high
(iv) medium value
Answer – (ii) low

5. If the magnetizing field on a ferromagnetic material is increased, its permeability is :
(i) decreases
(ii) increases
(iii) remains unchanged
(iv) first decreases and then increases
Answer – (i) decreases

6. An iron cored coil is connected in series with an electric bulb with an AC source as shown in figure. When iron piece is taken out of the coil, the brightness of the bulb will :

(i) decrease
(ii) increase
(iii) remain unaffected
(iv) fluctuate
Answer – (ii) increase

7. A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is 3/4 of the angle of prism. Speed of light in the prism is :
(i) c
(ii) c/2
(iii) c/4
(iv) none
Answer – (iv) none
A = 60°, i = ¾A = ¾(60°) = 45°
In the position of minimum deviation,
r = A/2 = 60°/2 = 30°
μ = sini/sinr = sin45°/sin30° = (1/√2)/√2 = √2
As, μ = c/v
v = c/μ = (3×10⁸)/√2 = 2 × 10⁸ m/s

8. Which of the following statement is NOT true about the properties of electromagnetic waves ?
(i)These waves do not require any material medium for their propagation.
(ii) Both electric and magnetic field vectors attain the maxima and minima at the same time.
(iii) The energy in electromagnetic wave is divided equally between electric and magnetic fields.
(iv) Both electric and magnetic field vectors are parallel to each other.
Answer – (iv) Both electric and magnetic field vectors are parallel to each other.

9. In two positions convex lens produces : magnified image of given object. The positions are :
(i) At f at 2f
(ii) Between f and 2f, between optical center and f
(iii) Beyond 2f, between c and f
(iv) At 2f, between optical centre and f
Answer – (ii) Between f and 2f, between optical center and f

10. If Young’s double slit experiment is immersed in water, then fringe width :
(i) decreases
(ii) increases
(iii) remain same
(iv) none
Answer – (i) decreases

11. The work function for a metal surface is 4.14 eV. The threshold wavelength for this metal surface is :
(i) 4125 Å
(ii) 2062.5 Å
(iii) 3000 Å
(iv) 6000 Å
Answer – (iii) 3000 Å
Work Function, w = hc/λ
hc/λ = 4.14eV
Threshold wavelength, λ = (6.6 × 10-³⁴ × 3 × 10⁸) ÷ (4.14 × 1.6 × 10-¹⁹) = 3 × 10-⁷ metres = 3000 Å

12. The radius of the inner most electron orbit of a hydrogen atom is 5.3 × 10-¹¹ m. The radius of the n = 3 orbits :
(i) 1.01 × 10-1¹⁰ m
(ii) 1.59 × 10-¹ m
(iii) 2.12 × 10-¹⁰ m
(iv) 4.77 × 10-¹⁰ m
Answer – (iv) 4.77 × 10-¹⁰ m
r3 = (n)²r1 = (3)² × 5.3 × 10-¹¹ = 4.77 × 10-¹⁰ m

13. Which of the following statements about nuclear forces is not true ?
(i) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femto meters (fm).
(ii) The nuclear force is much weaker than the Coulomb force.
(iii) The force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distance less than 0.8 fm.
(iv) Then nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same.
Answer – (ii) The nuclear force is much weaker than the Coulomb force.

14. Power of lens is 10 diopters, which of following is correct :
(i) Convex lens of focal length 10 metre
(ii) Convex lens of focal length 10 cm
(iii) Concave lens of focal length 10 metre
(iv) Concave lens of focal length 10 cm
Answer – (ii) Convex lens of focal length 10 cm
Power = 1/F (in metres)
F = 1/P = 1/10 = 0.1 m = 10 cm

Q15–Q18 Two statements are given- one labeled Assertion(A) and the other labeled 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

15. Assertion (A) : p-type semiconductors is a positive type crystal.
Reason (A) : p-type semiconductor is an negative type crystal.
Answer – (d) A is false and R is also false

16. Assertion (A) : The electrical conductivity of a semiconductor increases on doping.
Reason (A) : Doping always increases the number of electrons in the semiconductor.
Answer – (c) A is true but R is false

17. Assertion (A) : In an interference pattern observed in Young’s double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
Reason (R) : Fringe width is proportional to (d/D).
Answer – (a) Both A and R are true and R is the correct explanation of A

18. Assertion (A) : The photo electrons produced by a mono chromatic light beam incident on a metal surface have a spread in their kinetic energies.
Reason (R) : The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal.
Answer – (c) A is true but R is false

SECTION – B  (14 Marks)
(7 Questions of 2 Marks Each)

19. Electromagnetic waves with wavelength :
(i) λ1 is suitable for radar systems used in air craft navigation.
Answer – Microwave

(ii) λ2 is used to kill germs in water purifiers.
Identify and name the part of the electromagnetic spectrum to which these radiations belong.
Answer – Ultraviolet

20. A uniform magnetic field gets modified as shown in figure when two specimens A and B are placed in it.

Identify the specimen A and B.
Answer – A is diamagnetic and B is paramagnetic

21. State biot savarts law ?
Answer – The magnetic field at any point due to an element of a conductor carrying current is directly proportional to the strength of the current i, length of the element dL, sine of the angle θ between the element in the direction of current and inversely proportional to the square of the distance r of the point.
dB = μo/4π × idLsinθ/r²

OR

State ampere’s circuital law.
Answer – Ampere’s circuital law states that, the line integral of the magnetic field surrounding closed loop equals to the number of times the algebraic sum of currents passing through the loop.

22. State working principle of moving coil galvanometer ?
Answer – Moving coil galvanometers work on the principle that a current carrying coil experiences torque when placed in a magnetic field. As the electric current is passed through the coil, a torque acts on it, which deflects the coil.

23. A proton, deutron and alpha particle enter with same momentum perpendicular to same magnetic field. What is the ratio of radii of proton, deutron and alpha particle.
Answer – The masses are in the ratio mp : md : mα = 1 : 2 : 4
As the momentum is same we get the velocity in the ratio vp : vd : vα = 4 : 2 : 1
For a charged particle in uniform magnetic field, we can write,
mv²/r = Bqv
If +e is the charge on proton, then charge on deutron is also +e and charge on alpha particle is +2e.
Thus charges are in the ratio qp : qd : qα = 1 : 1 : 2
For a proton, a deutron and an alpha-particle are moving with same momentum in a uniform magnetic field
fp : fd : fα = eBv : eBv : 2eBv
As B is same we get
fp : fd : fα = 1 : 1 : 2

24. A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equal to 6000 Å, separation between the slit is 2 cm. What is the angular width of the central maxima.
Answer – Given λ = 6000 Å, d = 2 cm = 2 × 10-² m
Angular width 2θ = 2λ/d = (2×6000)/(2×10-²)  = 600000 Å

25. Define distance of closest approach in Rutherford alpha scattering experiment. Write mathematical formula.
Answer – The minimum distance between the centre of the nucleus and the alpha particle just before it gets reflected back through 180° is defined as the distance of closest approach ro (also known as contact distance).
ro = 1/4πεo × 2Ze²/½mv² = 1/4πεo × 2Ze²/Ek

OR

Explain Rutherford alpha scattering experiment.
Answer – Rutherford’s alpha (α) particles scattering experiment resulted into the discovery of nucleus of an atom. That is, during his experiment, he found that, most space of an atom is empty, and he could find a small positively charged center in an atom which is called as the nucleus.

SECTION – C  (15 Marks)
(5 Questions of 3 Marks Each)

26. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.7 × 10-²² C/m². What is electric field intensity E :
(a) in the outer region of the first plate
Answer – In the outer region of first plate, electric field intensity E is Zero.

(b) between the plates ?
Answer – Electric field intensity E in between the plates is given by relation E = σ/εo
Where, εo = Permittivity of free space = 8.85 × 10-¹² N-¹ C² m-²
E = (17.7×10-²²)/(8.85×10-¹) = 2 × 10-¹⁰ N/C
Therefore, electric field between the plates is 2 × 10-¹⁰ N/C.

27. State laws of photoelectric effect ?
Answer – Laws of photoelectric emission :
(i) There is a definite cut off value of frequency below which electrons cannot be ejected by any substance.
(ii) Number of emitted electrons are directly proportional to the intensity of light incident.
(iii) Kinetic energy of emitted electrons depends on the frequency of incident light on substance.
(iv) There is no time logging between the incident of light and emission of electrons.

28. Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.
Answer –

When we move from the heavy nuclei region to the middle region of the plot, there is a gain in the overall binding energy and hence release of energy. This indicates that energy can be released when a heavy nucleus (A ≈ 240) breaks into two roughly equal fragments. This process is called nuclear fission.
Similarly, when we move from lighter nuclei to heavier nuclei, we again find that there is a gain in the overall binding energy and hence release of energy. This indicates that energy can be released when two or more lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.

29. An a.c. source generating a voltage ε = Eosinωt is connected to a capacitor of capacitance C. Find the expression for the current (I) flowing through it. Plot a graph of ε and I versus ωt to show that the current is a head of the voltage by π/2.
Answer –

For the given capacitor we can write,
v = q/C
According to Kirchhoff’s rule, we can write from the above circuit,
vm sinωt = q/C
q = vm Csinωt
The current through the circuit can be calculated using the relation,
i = dq/dt
i = d(vm Csinωt)/dt = ωCvm cosωt
i = im sin(ωt+π/2)  [Using the relation, cosωt = sin(ωt+π/2)]

OR

An a.c. voltage V = Vosinωt is applied across a pure inductor of inductance L. Find an expression for the current I, flowing in the circuit and show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of π/2. Also draw graphs of V and I versus ωt for the circuit.
Answer – The instantaneous value of alternating voltage applied V = Vosinωt.
If i is the instantaneous current in the circuit and dt the rate of change of current in the circuit at that instant, then instantaneous induced emf ε = -L di/dt
According to Kirchhoff’s loop rule
V + ε = 0
V – L di/dt = 0
V = L di/dt
Integrating with respect to time ‘t’ :
di = Vosinωtdt/L = Vo/L ∫sinωtdt = Voωcosωt/L
i = Voωsin(ωt-π/2)/L
This is required expression for Current
i = io sin(ωt-π/2)
Here, io = peak value of alternating current
Also, current lags, an angle π/2.

30. Write Bohr’s postulates for the hydrogen atom model.
Answer : First Postulate – Electron revolves around the nucleus in discrete circular orbits called stationary orbits without emission of radiant energy. These orbits are called stable orbits or non-radiating orbits.
• Second Postulate – Electrons revolve around the nucleus only in orbits in which their angular momentum is an integral multiple of h/2π.
• Third Postulate – When an electron makes a transition from one of its non-radiating orbits to another of lower energy, a photon is emitted having energy equal to the energy difference between the two states. The frequency of the emitted photon is then given by, ν = (Ei-Ef)/h

SECTION – D  (15 Marks)
(3 Long Questions of 5 Marks Each)

31. What is p-n junction diode. Explain the process involved in formation of p-n junction diode with the help of suitable diagram.
Answer – A p-n junction diode is a basic semiconductor device that controls the flow of electric current in a circuit. It has a positive (p) side and a negative (n) side created by adding impurities to each side of a silicon semiconductor. The symbol for a p-n junction diode is a triangle pointing to a line. A p-n junction diode allows electric charges to flow in one direction, but not in the opposite direction; negative charges (electrons) can easily flow through the junction from n to p but not from p to n, and the reverse is true for holes.
The processes that follow after forming a p-n junction are of two types– diffusion and drift.
There is a difference in the concentration of holes and electrons at the two sides of a junction. The holes from the p-side diffuse to the n-side, and the electrons from the n-side diffuse to the p-side.
Drift is the process of movement of charge carriers due to the net electric field. In a p-n junction with no external source, electric field is from n-side to p-side and hence electrons drift from p-side to n-side.

OR

Explain principle and working of p-n junction diode in fullwave rectifier ?
Answer –

Full-wave rectifier circuit

Working of Full Wave Rectifier :
During the positive half cycle, diode D1, is forward biased as it is connected to the top of the secondary winding while diode D2 is reverse biased as it is connected to the bottom of the secondary winding. Due to this, diode D1 will conductacting as a short circuit and D2, will not conductacting as an open circuit.
During the negative half cycle, the diode D1 is reverse biased and the diode D2 is forward biased because the top half of the secondary circuit becomes negative and the bottom half of the circuit becomes positive. Thus in a full wave rectifiers, DC voltage is obtained for both positive and negative half cycle.

32. (a) Explain the term drift velocity of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of drift velocity.
Answer : Drift velocity – It is the average velocity acquired by the free electrons super imposed over the random motion in the direction opposite to electric field and along the length of the metallic conductor. Let n = number of free electrons per unit volume, vd = Drift velocity of electrons, Total number of free electrons passing through a cross section in unit time N/t = Anvd
So, total charge passing through across section in unit time. i.e., current (l) = Q/t = N/t = Anevd

(b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 respectively are connected in parallel as shown in the figure.

Deduce the expression for the :
(i) Equivalent emf of the combination
Answer : Eeq = (E1r2+E2r1)/(r1+r2)

(ii) Equivalent internal resistance of the combination
Answer : req = r1r2/(r1+r2)

(iii) Potential difference between the points A and B.
Answer : V = Eeq – Ireq

OR

(a) State the two Kirchhoff’s rules used in then analysis of electric circuits and explain them.
Answer : Kirchhoff’s first rule (the junction rule) – The sum of all currents entering a junction must equal the sum of all currents leaving the junction.
Kirchhoff’s second rule (the loop rule) – The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

(b) Derive the equation of the balanced state in a Wheatstone bridge using Kirchhoff’s laws.
Answer – In balanced condition, Ig = 0
So, VB = VD or P/Q = R/S. This is called condition of balance.

33. (a) Prove prism formula.
Answer –

A = Prism angle
δ = Angle of deviation
i1 = Angle of incidence
i2 = Angle of emergent
In the case of minimum deviation, ∠r1 = ∠r2 = ∠r
A = ∠r1 + ∠r2
So, A = ∠r + ∠r = 2∠r
∠r = A/2
Now, again
A + δ = i1 + i2  (:. In the case of minimum deviation i1 = i2 = i and δ = δm)
So, A + δm = i + i = 2i
Now, i = A + δm/2
Now, from snell’s rule,
μ = sini/sinr
μ = sin[(A+δm)/2] ÷ sin(A/2)

(b) A ray PQ is incident normally on the face AB of a triangular prism of refracting angle 60° as shown in figure.

The prism is made of a transparent material of refractive index 2/√3. Trace the path of the ray as it passes through the prism. Calculate the angle of emergence and the angle of deviation.
Answer –

The incident ray is normal to the face AB.
Thus we get is i1 = 0° and r1 = 0°
Now this ray would refract at the face AC with the incident angle here as r2 and the emergence angle as e.
We know that Angle of prism A is given as r1 + r2. Thus we get r2 = 60°
Now Refractive index of the prism is given as
μ = sine/sinr2
2/√3 = sine/(√3/2)
sine = 1 = sin90°
Thus we get e = 90°
And angle of deviation is given as
δ = i1 + e – A = 30°

OR


(i) A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.
Answer – Given, angle of minimum deviation, δm = 30°
Angle of prism, A = 60°
By prism formula, reflected index
μ = [sin(δm+A)/2] ÷ [sin(A/2)]
= [sin(30°+60°)/2] ÷ [sin(60°/2)]
= sin45°/sin30° = (1/√2) ÷ (1/2) = 2/√2 = √2
Also, μ = speed of light in vaccum(c) ÷ speed of light in prism(v)
v = c/μ = (3×10⁸)/√2 m/s
Hence, speed of light through prism is (3×10⁸)/√2 m/s

(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.
Answer – Critical angle ic, is given as,
sinic = 1/μ = 1/√2
ic = 45°
A = r + ic = 60°
r = 60° – 45° = 15°
sini/sinr = √2  (using Snell’s law)
sini = √2sinr = √2sin15°
i = sin-¹(√2sin15°)

SECTION – E  (8 Marks)
(2 Case Study Questions of 4 Marks Each)

34. Case Based Study :
Read the following paragraph and answer the questions.
Smallest charge that can exist in nature is the charge of an electron. During friction it is only the transfer of electron which makes the body charged. Hence net charge on any body is an integral multiple of charge of an electron (1.6 × 10-¹⁹C) i.e., q = ±ne where r = 1, 2, 3, 4, ……
Hence no body can have a charge represented as 1.8e, 2.7e, 2e/5, etc. Recently, it has been discovered that elementary particles such as protons or neutrons are elemental units called quarks
(I) If a charge on a body is 1nC, then how many electrons are present on the body ?
Answer – Charge (Q) = Ne, where, N is number of electrons present on the body, e is the charge on an electron.
-1×10-⁹C = N × (-1.6×10-¹⁹C)
N = (10-⁹)/(1.6×10-¹⁹) = 6.25 × 10⁹ electrons

(II) Charge is scalar or vector ?
Answer – Scalar

(III) A polythene piece rubbed with wool is found to have a negative charge of 3.2 × 10-⁷ C. Calculate the number of electrons transferred.
Answer – Given, Charge (Q) = 3.2 × 10-⁷ C
Charge on the electron, e = 1.6 × 10-¹⁹ C
Therefore,
Number of electron transferred is given by,
N = Q/e = (3.2×10-⁷)/(1.6×10-¹⁹) = 2 × 10¹² electrons

OR

What is charge ?
Answer – Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charges; positive and negative.

35. Case Based Study : Electromagnetic Induction.
Read the following paragraph and answer the questions.
When a current I flows through a coil, flux linked with it is Φ = LI, where L is a constant known as self inductance of the coil.

Any charge in current sets up an induced emf in the coil. Thus, self inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth or the decay of current flowing through the coil. Also value of self inductance depends on the number of turns in the solenoid, its area of cross-section and the permeability of its corematerial.
(I) What is self inductance ?
Answer – Self inductance is the tendency of a coil to resist changes in current in itself.

(II) State the factors on which self inductance of a long solenoid depends ?
Answer – Self inductance depends on: Size of coil, Shape of the coil, Material of the coil and Medium.

(III) What is the induced emf in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2 second.
Answer – Induced emf, e = -di/dt
Given, L = 10 H, ∆i = 9A – 4A = 5A, dt = 0.2s
emf, e = -L(di/dt) = 10 × (5/0.2) = 250 V

OR

What is Lenz’s law ?
Answer – Lenz’s law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux and to exert a mechanical force opposing the motion.

Leave a Comment

error: