aglasem.com
Schools Admission Mock Test Playground
ClassChoose class
StateSelect state

HBSE Class 12 Sample Paper 2024 Answers Physics

Download the HBSE Class 12 Sample Paper 2024 Answers Physics PDF for free at AglaSem. Get accurate, step-by-step solutions to every question so you can check your answers, learn the correct method and see how to score full marks. More Detail
HBSE Class 12 Sample Paper 2024 Answers Physics - Page 1 of 12

Finished viewing? Save it for later —

Download HBSE Class 12 Sample Paper 2024 Answers Physics (PDF · 12 pages)
Downloaded 13 times

About HBSE Class 12 Sample Paper 2024 Answers Physics

HBSE Class 12 Sample Paper 2024 Answers Physics is available here for free download. Published by Haryana Board for Class 12, this solution can be viewed online or downloaded as a PDF (12 pages). Candidates preparing for Class 12 can use HBSE Class 12 Sample Paper 2024 Answers Physics to understand the exam pattern, the type of questions asked, and the overall difficulty level.

Frequently Asked Questions

How can I download HBSE Class 12 Sample Paper 2024 Answers Physics?

Open this page and click the Download button to save HBSE Class 12 Sample Paper 2024 Answers Physics as a PDF. It is completely free on AglaSem Docs.

Is HBSE Class 12 Sample Paper 2024 Answers Physics free to download?

Yes. HBSE Class 12 Sample Paper 2024 Answers Physics can be viewed online and downloaded as a PDF free of cost on AglaSem Docs.

How many pages does HBSE Class 12 Sample Paper 2024 Answers Physics have?

HBSE Class 12 Sample Paper 2024 Answers Physics contains 12 pages, which you can read online or download together as a single PDF.

Where can I find more Class 12 study material?

You can find more Class 12 question papers, sample papers, syllabus, and answer keys on AglaSem Docs.

HBSE Class 12 Sample Paper 2024 Answers Physics – Text

Read the full text of this solution below — useful to quickly search, copy and reference the content online without downloading the PDF.

📄 View text version (12 pages)

Page 1

HBSE
MODEL PAPER
2024
Practice Papers
MODEL PAPERS
Marking Scheme
ANSWER KEY

Page 2

Class: XII
SESSION:2023-2024
MARKING SCHEME
HBSESAMPLEQUESTIONPAPER(THEORY)
SUBJECT:PHYSICS

Q.no Marks
SECTIONA
1 (iv)ZERO 1
2 (iv)Potentialdifferenceappliedacrosstheconductor 1
3 (ii)materialAis germaniumandmaterialBiscopper 1
4 (ii)lowresistances 1
5 (i)decreases 1
6 (ii)increases 1
7 (iv)none 1
8 (iv)Both electric and magnetic field vectors are parallel to eachother. 1

9 (ii)betweenf and2f,betweenopticalcenterandf 1
10 (i)decreases 1
11 (iii)3000Å 1
12 (iv)4.77X10-10m 1
13 (ii) Thenuclearforceismuchweakerthanthe Coulombforce. 1
14 (ii)convexlensoffocallength10metre 1
15 (d)Both A and R is incorrect 1
16 c)AistruebutR isfalse 1
17 a)Both AandR aretrueandRIsthecorrectexplanationofA 1
18 c)Ais true butRisfalse 1
SECTIONB
19 λ1-Microwaveλ2 ultraviolet 1
1

20 A diamagneticB-paramagnetic 1
1

21 The magnetic field at any point due to an element of a conductor carrying current is 2
(1) directly proportional to (a) the strength of the current i
(b) length of the element dl
(c) sine of the angle θ
between the element in the direction of current
(2) inversely proportional to the square of the distance r of the point
OR
Ampere's circuital law states that ―the line integral of the magneticfield surrounding
closed-loop equals to the number of times thealgebraicsumofcurrentspassing
throughtheloop.‖

1

Page 3

22 Moving coil galvanometers work on the principle that a current-carrying 2
coil experiences torque when placed in a magnetic field. Asthe electric
current is passed through the coil, a torque acts on it,which deflectsthe coil.
23 ½

½

½

½

24 Angularwidth2φ=2λ/dGiven 1
λ=6000Å, d=2x10-2
=2x6000/2x10-2 1
=600000Å
25 1
The minimum distance between the centre of the nucleus and the alpha particle just before it gets reflected 1
back through 180° is
defined as the distance of closest approach ro (also known as contact distance).

OR
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.
SECTIONC

2

Page 4

26 2

1

27
Lawsofphotoelectricemission:(anythree)
(i) Thereisadefinitecutoffvalueoffrequencybelowwhichelectronscannotbeeje
1+1+1
ctedbyanysubstance.
(ii) Numberofemittedelectronsaredirectlyproportionalto
theintensityoflightincident.
(iii)Kineticenergyofemittedelectronsdependsonthe
frequencyofincidentlightonsubstance.
(iv)Thereisnotimeloggingbetweentheincidentoflightand
emissionofelectrons.

3

Page 5

28 1

1

1

29

1

2

OR
The instantaneous value of alternating voltage applied V=V0sinωt. ...(i)
If i is the instantaneous current in the circuit and 2
dt
dt the rate of change of current in the circuit at that instant, then instantaneous induced emf
ϵ=−Ldi/dt
According to Kirchhoff's loop rule
V+e=0⟹V−Ldi/dt=0
V=L di/dt

4

Page 6

Integrating with respect to time 't':
di=Vosinwtdt/L
=Vo/L ∫sinωtdt
=Vowcoswt/L
i=Vowsin(wt-π/2)/L
This is required expression for Current
i=iosin(wt-π/2)
io is the peak value of alternating current
Also comparing , we note that
current lags behind the applied voltage by
an angle π/2

1

5

Page 7

30
1

1

1

SECTION- D

31
p–n junction diode allows electric charges to flow in one
2
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
1.5
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 pn-junction with no external source,
electric field is from n-side to p-side and hence electrons drift
from p-side to n-side. 1.5

OR

6

Page 8

2

WorkingofFullWaveRectifier
During thepositive half cycle,diodeD1isforwardbiasedasitis connected tothe topof
thesecondarywinding whilediodeD2is reverse biased as it is connected to the bottom 3
of the secondary winding. Due tothis,diodeD1will
conductactingasashortcircuitandD2will notconduct actingasanopencircuit
During thenegative half cycle, thediodeD1isreverse biasedand the diodeD2isforward
biasedbecause the tophalf of thesecondarycircuit becomes negative and the bottom
half of the circuit becomes positive.Thus in a full wave rectifiers, DC voltage is
obtained for both positive andnegativehalf cycle.

7

Page 9

32(a) Drift velocity : It is the average velocity acquired by the free 2
electronssuperimposed over the random motion in the direction opposite
toelectric field and along the length of the metallic conductor. Let n =number of
free electrons per unit volume, vd = Drift velocity ofelectrons Total number of
free electrons passing through a crosssection in unit time N/t = Anvd So, total
charge passing through acrosssectioninunittimei.e.,current,l=Q /t=N/t=Anevd.

32(b) ½

½

1/2

1/2

1/2

1/2
OR
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 2
second rule—the loop rule: The algebraic sum of changes in potential around
any closed circuit path (loop) must be zero

1

2
Derivation of balanced equation using kirchoff’s law

2

8

Page 10

33(a)1. A=Prism angle,
2. δ=Angle of deviation,
3. i1=Angle of incidence, 1
4. i2=Angle of emergent.

1
5. In the case of minimum deviation, ∠r1 = ∠r2 = ∠r
6. A = ∠r1 + ∠r2
7. SO, A = ∠r+ ∠r = ∠2r
8. ∠r = A/2
9. Now, again
10. A + δ = i1 + i2 (∵ In the case of minimum deviation i1 = i2 = i and δ = δm)
11. So, A + δm = i + i = 2i
12. Now, i = A+δm/2
13. Now, from snell's rule, 1
14. μ= sin i/sin r
15. μ=sin((A + δm)/2)/ sin(A/2)

(b)
2

OR

3
9

Page 11

2

10

Page 12

SECTIONE
34 a) 1

1
(b)
Scalar

(c)
2

OR
Defination of Charge.
35a Self-inductance is the tendency of a coil toresist 1
changes in current in itself

b) 1
Selfinductancedependson-

1-Sizeofcoil

2-Shapeofthecoil

3-Materialofthecoil

4-Medim

c)

2
OR

Statement of Lenz’s Law.

11

Document Details

Board / OrgHaryana Board
ExamClass 12
TypeSolution
Pages12
Updated22 Jul 2026