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TN 12th Question Paper 2025 Physics

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Page 1

TAMIL NADU

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Q apeion
P r
DOWNLOAD

Page 2

No. of Printed Pages : 11
8317
£vÄ Gs
A Register Number

!8317Physics!
PART - III
C¯Ø¤¯À / PHYSICS

( uªÌ ©ØÖ® B[Q» ÁÈ / Tamil & English Version )

Põ» AÍÄ : 3.00 ©o ÷|µ® ] [ ö©õzu ©v¨ö£sPÒ : 70

Time Allowed : 3.00 Hours ] [ Maximum Marks : 70

AÔÄøµPÒ : (1) AøÚzx ÂÚõUPЮ \›¯õP¨ £vÁõQ EÒÍuõ GߣuøÚa
\›£õºzxU öPõÒÍÄ®. Aa_¨£vÂÀ SøÓ°¸¨¤ß, AøÓU
PsPõo¨£õÍ›h® EhÚi¯õPz öu›ÂUPÄ®.
(2) }»® AÀ»x P¸¨¦ ø©°øÚ ©mk÷© GÊxÁuØS®
AiU÷PõikÁuØS® £¯ß£kzu ÷Ásk®. £h[PÒ ÁøµÁuØS
ö£ß]À £¯ß£kzuÄ®.

Instructions : (1) Check the question paper for fairness of printing. If there is any lack of fairness,
inform the Hall Supervisor immediately.

(2) Use Blue or Black ink to write and underline and pencil to draw diagrams.

£Sv & I / PART - I
SÔ¨¦ : (i) AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 15x1=15
(ii) öPõkUP¨£mkÒÍ |õßS ©õØÖ ÂøhPÎÀ ªPÄ® Hئøh¯
Âøhø¯z ÷uº¢öukzxU SÔ±mkhß Âøh°øÚ²® ÷\ºzx
GÊuÄ®.
Note : (i) Answer all the questions.
(ii) Choose the most appropriate answer from the given four alternatives and write
the option code and the corresponding answer.

[ v¸¨¦P / Turn over

Page 3

8317 2

1. RÌUPõq® »õâU ÷PmkPÎß ÷\ºUøP°À, EÒÏkPÒ A, B, C öPõsk öÁαk
Y&UPõÚ §¼¯ß \©ß£õk GÊxP.

(A) A B C (B) AB C (C) A+B+C (D) A+B+C

In the combination of the following gates, write the Boolean equation for output Y in terms
of input A, B, C.

(a) ABC (b) AB C (c) A+B+C (d) A+B+C

2. 500 Am −1 Põ¢u©õUS ö\ÔÄ öPõsh ö£õ¸ÎøÚ 1000 Am −1 ©v¨¦øh¯
Põ¢u©õUS ¦»zvÀ øÁUS®÷£õx A¨ö£õ¸Îß Põ¢u HئzvÓß ©v¨¦ ¯õx ?
(A) 0.2 (B) 0.8 (C) 0.7 (D) 0.5
If a material having intensity of magnetisation 500 Am−1 is placed in a magnetising field of
1000 Am−1, then the susceptibility of the material is :
(a) 0.2 (b) 0.8 (c) 0.7 (d) 0.5

3. J¸ ªß©õØÔ°À •ußø© ©ØÖ® xøna _ØÖPÎÀ •øÓ÷¯ 410 ©ØÖ®
1230 _ØÖPÒ EÒÍÚ. •ußø©a_¸ÎÀ EÒÍ ªß÷Úõmh® 6 A GÛÀ, xøna
_¸Îß ªß÷Úõmh©õÚx :
(A) 12 A (B) 2 A (C) 1 A (D) 18 A
In a transformer, the number of turns in the primary and the secondary are 410 and 1230
respectively. If the current in primary is 6 A, then that in the secondary coil is :
(a) 12 A (b) 2A (c) 1A (d) 18 A

A

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3 8317

4. áü¼ß öÁ¨£ Âv°À, R ©ØÖ® t ©õÔ¼PÍõP EÒÍx. H &I y Aa]¾® I2 &I
x Aa]¾® öPõsk Áøµ¯¨£mh Áøµ£h® J¸ :
(A) Ámh® (B) ÷|ºU÷Põk
(C) }ÒÁmh® (D) £µÁøÍ¯®
In Joule’s heating law, when R and t are constant, if the H is taken along the y-axis and I2
along the x-axis, the graph is :
(a) circle (b) straight line
(c) ellipse (d) parabola

1
5. µ0 ε0 &ß £›©õn® :

(A) [L−1 T] (B) [L T−1] (C) [L−2 T2] (D) [L2 T−2]
1
The dimension of µ ε is :
0 0

(a) [L−1 T] (b) [L T−1] (c) [L−2 T2] (d) [L2 T−2]

6. "ìQ ö©ÊS' Gߣx |õ÷Úõ ö£õ¸Îß £¯ß£õk BS®. Ax £¯ß£k® xøÓ :
(A) ÂøÍ¯õmk (B) ©¸zxÁ®
(C) ÁõPÚ öuõÈØ\õø» (D) áÄÎ
“Ski wax” is an application of nano product in the field of :
(a) Sports (b) Medicine
(c) Automotive industry (d) Textile

7. 2×105 NC−1 ©v¨¦ÒÍ ªß¦»zvÀ 308 J¸[Pø©Ä ÷PõnzvÀ ªß C¸•øÚ
JßÖ øÁUP¨£mkÒÍx. Auß «x ö\¯À£k® v¸¨¦ Âø\°ß ©v¨¦ 8 Nm.
ªß C¸•øÚ°ß }Í® 1 cm GÛÀ Av¾ÒÍ J¸ ªß xPÎß ªßÞmh ©v¨¦ :
(A) 5 mC (B) 4 mC (C) 7 mC (D) 8 mC
An electric dipole is placed at an alignment angle of 30 o with an electric field of
2×105 NC−1. It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole
length is 1 cm is :
(a) 5 mC (b) 4 mC (c) 7 mC (d) 8 mC

A [ v¸¨¦P / Turn over

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8317 4

8. PõØÔ¼¸¢x JλPÀ Gs 2 öPõsh Psnõi¨ £mhPzvß «x JÎ
ÂÊQÓx GÛÀ, \õzv¯©õÚ ö£¸© »S÷Põnzvß ©v¨¦ GßÚ ?
(A) 608 (B) 308 (C) 908 (D) 458
For light incident from air on a slab of refractive index 2, the maximum possible angle of
refraction is :
(a) 608 (b) 308 (c) 908 (d) 458

9. t=0 ÷|µzvÀ Pv›¯UP uÛ©® JßÔ¾ÒÍ AqUP¸UPÎß GsoUøP N 0.
 1 
Aøµ B²mPõ»zvÀ £õv¯ÍÄ Põ»®  t = T1  BS®÷£õx EÒÍ
 2 2
AqUP¸UPÎß GsoUøP :
N0 N0 N0 N0
(A) (B) (C) (D) 2
4 2 8
A radioactive element has N0 number of nuclei at t = 0. The number of nuclei remaining after
 1 
half of a half-life  that is, at time t = T1  is :
 2 2

N0 N0 N0 N0
(a) (b) (c) (d)
4 2 8 2

10. 500 nm Aø»}Í® Eøh¯ JÎ 0.5 mm AP»•øh¯ xøÍ°ß ÁÈ÷¯ ö\À¾®
÷£õx Âή¦ ÂøÍÄ AøhQÓx. C¢{PÌÂÀ Pvº Jΰ¯ø»¨ £¯ß£kzx®
öuõø»ÂøÚU PõsP.
(A) 20 cm (B) 25 m (C) 25 cm (D) 30 cm
Calculate the distance upto which ray optics is a good approximation for light of wavelength
500 nm falls on an aperture of width 0.5 mm.
(a) 20 cm (b) 25 m (c) 25 cm (d) 30 cm

11. 1.0×10−5 cm AP»® öPõsh JØøÓ¨ ¤ÍÂÚõÀ HØ£k® Âή¦ ÂøÍÂß
•uÀ ]Ö©® 308 GÛÀ, £¯ß£kzu¨£k® J롧 Aø»}Í® GßÚ ?
(A) 600 Å (B) 400 Å (C) 700 Å (D) 500 Å
First diffraction minimum due to a single slit of width 1.0×10 −5 cm is at 308.
Then wavelength of light used is :
(a) 600 Å (b) 400 Å (c) 700 Å (d) 500 Å

A

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5 8317

12. R Bµ•øh¯ ªßPhzx¨ ö£õ¸Íõ»õÚ, ö©À¼¯ ÷PõÍPU Tmiß £µ¨¤À
Q ªßÞmh AÍÄÒÍ ªßxPÒPÒ ^µõP¨ £µÂ²ÒÍÚ GÛÀ, AuÚõÀ HØ£k®
{ø» ªßÚÊzuzvØPõÚ \›¯õÚ Áøµ£h® Gx ?

(A) (B)

(C) (D)

A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed
on its surface. The correct plot for electrostatic potential due to this spherical shell is :

(a) (b)

(c) (d)

A [ v¸¨¦P / Turn over

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8317 6

13. J뻧 EªÌÄ {PÌÂÀ, J¸ SÔ¨¤mh E÷»õPzvß £¯ßöuõhUP
AvºöÁsøn Âh 4 ©h[S AvºöÁs öPõsh PvºÃa_ A¢u
E÷»õP¨£µ¨¤À £k®÷£õx, öÁΨ£k® G»UmµõÛß ö£¸© vø\÷ÁP©õÚx,

h ν0 h ν0 h ν0 6h ν0
(A) 2 (B) (C) (D)
m m 2m m

In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a
certain metal is incident on the metal. Then, the maximum possible velocity of the emitted
electron will be :

h ν0 h ν0 h ν0 6h ν0
(a) 2 (b) (c) (d)
m m 2m m

14. Kº Aø» C¯ØÔ°À öuõhºa]¯õÚ Aø»ÄPÒ HØ£h :
(A) ¤ßÞmh ©õÔ¼ JßÓõP C¸UP ÷Ásk®.
(B) Pmh ©õØÓ® _È AÀ»x 2π &¯õP C¸UP ÷Ásk®.
(C) ÷|º ¤ßÞmh® C¸UP ÷Ásk®.
(D) ÷©ØTÔ¯ AøÚzx®.
To obtain sustained oscillation in an oscillator,

(a) Feedback factor must be unity

(b) Phase shift must be 0 or 2π

(c) Feedback should be positive

(d) All the above

15. J롧 SÖUPø»¨ £s¤øÚ öÁΨ£kzx® {PÌÄ :
(A) JÎa]uÓÀ (B) SÖURmk ÂøÍÄ
(C) uÍÂøÍÄ (D) Âή¦ ÂøÍÄ
The transverse nature of light is shown in :

(a) scattering (b) interference

(c) polarisation (d) diffraction

A

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7 8317

£Sv & II / PART - II
SÔ¨¦ : GøÁ÷¯Ý® BÖ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 24 &US
Pmhõ¯® Âøh¯ÎUPÄ®. 6x2=12
Note : Answer any six questions. Question No. 24 is Compulsory.
16. Cønzumk ªß÷uUQ JßÖ 5 cm £UP® öPõsh C¸ \xµ umkPøÍ 1 mm
CøhöÁΰÀ öPõskÒÍx. ªß÷uUQ°ß ªß÷uUSzvÓøÚU PnUQkP.
A parallel plate capacitor has two square plates of side 5 cm and separated by a distance of
1 mm. Calculate the capacitance of this capacitor.

17. ö»ßì Âvø¯U TÖP.
State Lenz’s law.

18. ø\U÷ÍõmµõÛß Áµ®¦PøÍ GÊxP.
Write the limitations of Cyclotron.

19. {øÓ SøÓ£õk GßÓõÀ GßÚ ? Auß ÷PõøÁø¯ GÊxP.
What is mass defect ? Give its expression.

20. Jΰ¯À £õøu GßÓõÀ GßÚ ?
What is optical path ?

21. ¤ßÁ¸ÁÚÁØÔØS Aø»•P¨¤ß ÁiÁ[PÒ ¯õøÁ ?
(A) DÔ»õz öuõø»ÂÀ ‰»® (B) ¦ÒÎ ‰»® (C) ÷|›¯À ‰»®
What are the shapes of wavefront for a :
(a) Source at infinite (b) Point source (c) Line source

22. p-n \¢v øh÷¯õk •ß÷ÚõUS \õº¤ØPõÚ ªß_ØÖ¨ £h® ÁøµP.
Draw the circuit diagram for a forward biased p-n junction diode.

23. {Özx ªßÚÊzu® – Áøµ¯Ö.
Define stopping potential.

24. J¸ ªßP»® 2Ω ªßuøh ÁȯõP 0.9 A ªß÷Úõmhzøu²®, 7Ω ªßuøh ÁÈ÷¯
0.3 A ªß÷Úõmhzøu²® HØ£kzxQÓx GÛÀ, ªßP»zvß AP ªß uøhø¯U
PnUQkP.
A cell supplies a current of 0.9 A through a 2Ω resistor and a current of 0.3 A through a 7Ω
resistor. Calculate the internal resistance of the cell.

A [ v¸¨¦P / Turn over

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8317 8

£Sv & III / PART - III
SÔ¨¦ : GøÁ÷¯Ý® BÖ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 33 &US
Pmhõ¯©õP Âøh¯ÎUPÄ®. 6x3=18
Note : Answer any six questions. Question No. 33 is Compulsory.

25. QºUPõL¤ß ªß÷Úõmh Âv ©ØÖ® ªßÚÊzu ÷ÁÖ£õmk ÂvPøÍU TÖP.
State Kirchhoff’s current rule and voltage rule.

26. J¸ PõÀ]¯® E÷»õP¨ £µ¨¤ß «x 300 nm Aø»}Í® öPõsh PvºÃa_ £k®
÷£õx JÎ G»Umµõß öÁΨ£k©õ ? EÚx Âøhø¯ {¯õ¯¨£kzxP.
(PõÀ]¯zvß öÁÎ÷¯ØÖ BØÓÀ 3.20eV)
A radiation of wavelength 300 nm is incident on calcium surface. Will photoelectrons be
observed ? Justify your answer.
(Work function of calcium = 3.20eV)

27. Põ¢u¯À »õµß_ Âø\ø¯¨ £ØÔ SÔ¨¦ ÁøµP.
Give an account of magnetic Lorentz force.

28. ÷|ºzvø\ ªß÷Úõmh Aø©¨ø£ Âh, ©õÖvø\ ªß÷Úõmh Aø©¨¤ß
|ßø©PÒ ©ØÖ® SøÓ£õkPÒ ¯õx ?
What are the advantages and disadvantages of Alternating Current System over Direct
Current System ?

29. ªßPõ¢u Aø»PÎß £s¦PøÍU TÖP. (H÷uÝ® BÖ)
Write down the properties of electromagnetic waves. (Any six)

30. Pmh ÷ÁÖ£õmiØS®, £õøu ÷ÁÖ£õmiØS® EÒÍ öuõhºø£ Á¸Â.
Obtain the relation between phase difference and path difference.

31. T¿® Âø\US®, ¦ÂDº¨¦ Âø\US® Cøh÷¯¯õÚ ÷ÁÖ£õkPøÍU TÖP.
What are the differences between coulomb force and gravitational force ?

A

Page 10

9 8317

32. EÒÍõº¢u ©ØÖ® ¦Ó¯»õÚ SøÓ PhzvPøÍ ÷ÁÖ£kzxP.
Distinguish between intrinsic and extrinsic semiconductors.

33. 13.25 Å Bµ® öPõsh øímµáß AqÂß 5 &Áx _ØÖ¨ £õøu°ß

(i) ÷Põn E¢u® ©ØÖ®
(ii) Av¾ÒÍ G»UmµõÛß vø\÷ÁP® BQ¯ÁØøÓU PnUQkP.
Find the (i) Angular momentum (ii) Velocity of the electron revolving in the 5th orbit of
hydrogen atom of radius 13.25 Å.

£Sv & IV / PART - IV
SÔ¨¦ : AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 5x5=25

Note : Answer all the questions.

34. (A) ªßÞmh® ö£ØÓ •i»õ }Í•ÒÍ P®¤°ÚõÀ HØ£k® ªß¦»zvØPõÚ
\©ß£õmøh¨ ö£ÖP.
AÀ»x
(B) j ©õºPÛß •uÀ ©ØÖ® CµshõÁx ÷uØÓ[PøÍ TÔ {¹¤.
(a) Obtain the expression for electric field due to an infinitely long charged wire.

OR

(b) State and prove DeMorgan’s first and second theorems.

35. (A) ªßÚÊzu©õÛø¯ £¯ß£kzv C¸ ªßP»[PÎß ªßÛ¯US Âø\PÒ
GÆÁõÖ J¨¤h¨£kQßÓÚ ?
AÀ»x
(B) J롧 ÷ÁPzøuU PshÔ²® L¤ë² (Fizeau) •øÓø¯ ÂÁ›.
(a) How the emf of two cells are compared using potentiometer ?

OR

(b) Describe the Fizeau’s method to determine the speed of light.

A [ v¸¨¦P / Turn over

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8317 10

36. (A) Põ¢u¨¦»zvÀ EÒÍ ªß÷Úõmha_ØÔß «x ö\¯À£k® v¸¨¦Âø\UPõÚ
÷PõøÁø¯ Á¸Â.
AÀ»x
(B) JÎ EªÌÄ ªßP»zvß Aø©¨¦ ©ØÖ® ÷Áø» ö\´²® Âuzøu
ÂÍUSP. ÷©¾® JÎ ªßP»zvß £¯ßPÒ (H÷uÝ® Cµsk) u¸P.
(a) Derive the expression for the torque on a current - carrying coil in a magnetic field.
OR
(b) Explain the construction and working of photo emissive cell and give any two
applications of photo cell.

37. (A) (i) J¸ _¸Ò EÒÍhUQ¯ £µ¨ø£ ©õØÖÁuß ‰»®, J¸ ªßÛ¯US
Âø\ø¯ GÆÁõÖ ysh»õ® ?
(ii) ^µõÚ Põ¢u¨¦»® 0.4 T CÀ 0.03 m2 £µ¨¦ öPõsh Ámh E÷»õP Ámk
JßÖ _ǾQÓx. _ÇØ] Aa\õÚx Ámiß ø©¯® ÁȯõPÄ®, Auß
uÍzvØS ö\[SzuõPÄ® Aø©¢xÒÍx. ÷©¾® _ÇØ] Aa\õÚx
Põ¢u¨¦»zvß vø\US Cøn¯õP EÒÍx. Ámk J¸ ÂÚõi ÷|µzvÀ
20 _ÇØ]PøÍ {øÓÄ ö\´QÓx. Ámiß ªßuøh 4 Ω GÛÀ, Auß
Aa_US® Âή¦US® Cøh÷¯ ysh¨£k® ªßÛ¯US Âø\ø¯U
PnUQkP.
AÀ»x
(B) (i) ¦¸ìhº Âvø¯U TÖP.
(ii) QøhzuÍzvØS Cøn¯õPa ö\À¾® uÍÂøÍÁØÓ JÎUPØøÓ,
JλPÀ Gs 1.65 öPõsh Psnõi¨ £µ¨¤ß «x £mk
GvöµõΨ¦ AøhQÓx. GvöµõΨ¦ Aøh¢u JÎUPØøÓ •ÊÁx®
uÍÂøÍÄ Aøh¯ ÷Áskö©ÛÀ, Psnõi£µ¨¦ QøhzuÍzxhß
G¢uU ÷PõnzvÀ \õ´zx øÁUP¨£h ÷Ásk® ?
(a) (i) How will you induce an emf by changing the area enclosed by a coil ?
(ii) A circular metallic disc of area 0.03 m2 rotates in a uniform magnetic field of
0.4 T. The axis of rotation passes through the centre and perpendicular to its
plane and is also parallel to the magnetic field. If the disc completes 20 revolutions
in one second and the resistance of the disc is 4 Ω, calculate the induced emf
between the axis and the rim.
OR

A

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11 8317

(b) (i) State Brewster’s Law.

(ii) What is the angle at which a glass plate of refractive index 1.65 is to be kept with
respect to the horizontal surface so that an unpolarised light travelling horizontal
after reflection from the glass plate is found to be plane polarised ?

38. (A) Pv›¯UP ]øuÄ Âv°øÚz u¸ÂUP.
AÀ»x
(B) EmPÁº {Ó©õø» GßÓõÀ GßÚ ? EmPÁº {Ó©õø»°ß ÁøPPøÍ
ÂÍUPÄ®.
(a) Obtain the law of radioactive decay.

OR

(b) What is absorption spectrum ? Explain the types of absorption spectrum.

-oOo-

A [ v¸¨¦P / Turn over

Page 13

Study Materials
Notes

Model Papers Class 6 Notes

Sample Papers Class 7 Notes
Half Yearly Sample Papers Class 8 Notes

Class 9 Notes
Important Resources
Class 10 Notes
Periodic Table
Class 11 Notes
Writing Skills / Formats

Maps of India / World Class 12 Notes

Books and Solutions

NCERT Books
NCERT Book Solutions
HC Verma Chapter Wise Solutions
RD Sharma Solutions
CGBSE Solutions

Document Details

Board / OrgTamil Nadu Board
ExamClass 12
TypeQuestion Paper
Pages13
Updated30 Apr 2026