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Published on: 28/11/2025
Download Tamil Nadu 12th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
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1.
The technology used for stopping the brain from processing pain is _____.
Precision medicine
Wireless brain sensor
Virtual reality
Radiology
2.
The blue print for making ultra durable synthetic material is mimicked from _____.
Lotus leaf
Morpho butterfly
Parrot fish
Peacock feather
3.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called ______.
Amplitude modulation
Frequency modulation
Phase modulation
Pulse width modulation
4.
The given electrical network is equivalent to ______.
AND gate
OR gate
NOR gate
NOT gate
5.
6.
A radioactive nucleus (initial mass number A and atomic number Z) emits two α-particles and 2 positons. The ratio of number of neutrons to that of proton in the final nucleus will be _____.
\(\frac{A-Z-4}{Z-2}\)
\(\frac{A-Z-2}{Z-6}\)
\(\frac{A-Z-4}{Z-6}\)
\(\frac{A-Z-12}{Z-4}\)
7.
The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as _____.
A2/3
A4/3
A1/3
A5/3
8.
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is _____.
1
2
3
4
9.
Emission of electrons by the absorption of heat energy is called ______ emission.
photoelectric
field
thermionic
secondary
10.
A light of wavelength 500 nm is incident on a sensitive metal plate of photoelectric work function 1.235 eV. The kinetic energy of the photoelectrons emitted is_____. (Take h = 6.6 x 10–34 Js)
0.58 eV
2.48 eV
1.24 eV
1.16 eV
11.
A light source of wavelength 520 nm emits 1.04 x 1015 photons per second while the second source of 460 nm produces 1.38 x 1015 photons per second. Then the ratio of power of second source to that of first source is _____.
1.00
1.02
1.5
0.98
12.
13.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is_____. [take wavelength of light, λ = 500 nm]
1 m
5 m
3 m
6 m
14.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is ______.
8 cm
10 cm
12 cm
16 cm
15.
If the velocity and wavelength of light in air is Va and λa and that in water is Vw and λw, then the refractive index of water is______.
\(\frac{V_W}{V_a}\)
\(\frac{V_a}{V_W}\)
\(\frac{\lambda_W}{\lambda_a}\)
\(\frac{{V_a}\lambda_a}{{V_W}\lambda_W}\)
16.
The speed of light in an isotropic medium depends on, ______.
its intensity
its wavelength
the nature of propagation
the motion of the source w.r.t medium
17.
A wire of length l carrying a current I along the Y direction is kept in a magnetic field is given by \(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T.\) The magnitude of Lorentz force acting on the wire is _____.
\(\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)
\(\sqrt { \frac { 1 }{ { 3 } } } \beta Il\)
\(\sqrt { 2 } \beta Il\)
\(\sqrt { \frac { 1 }{ 2 } } \beta Il\)
18.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current I = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the centre of the spiral is ______.
\(5\mu T\)
\(7\mu T\)
\(8\mu T\)
\(10\mu T\)
19.
The magnetic field at the centre O of the following current loop is
\(\frac { { \mu }_{ ° }I }{ 4r } \bigotimes \)
\(\frac { { \mu }_{ ° }I }{ 4r } \bigodot \)
\(\frac { { \mu }_{ ° }I }{ 2r } \bigotimes \)
\(\frac { { \mu }_{ ° }I }{ 2r } \bigodot \)
20.
\(\frac{20}{\pi^2}H\) inductor is connected to a capacitor of capacitance C. The value of C in order to impart maximum power at 50 Hz is
50 μF
0.5 μF
500 μF
5 μF
21.
An inductor 20 mH, a capacitor 50 μF and a resistor 40Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in AC circuit is
0.76 W
0.89 W
0.46 W
0.67 W
22.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
1.2
0.83
0.12
0.9
23.
The flux linked with a coil at any instant t is given by \(\Phi\)B = 10t2 − 50t + 250. The induced emf at t = 3s is
−190 V
−10 V
10 V
190 V
24.
In an electromagnetic wave traveling in free space the rms value of the electric field is 3 V m−1. The peak value of the magnetic field is _____.
1.414 x 10-8 T
1.0 x 10-8 T
2.828 x 10-8 T
2.0 x 10-8 T
25.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
Capacitance
Charge
Voltage
Energy density
26.
If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
Q remains the same, C is doubled
Q is doubled, C doubled
C remains same, Q doubled
Both Q and C remain same
27.
Two identical point charges of magnitude –q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?
A1 and A2
B1 and B2
both directions
No stable
28.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

1.5 Ω
2.5 Ω
3.5 Ω
4.5 Ω
29.
Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose RA = 3 RB, then what is the ratio of radius of wire A to that of B?
3
\(\sqrt3\)
\(\frac{1}{\sqrt3}\)
\(\frac{1}{3}\)
30.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

2 ohm
4 ohm
8 ohm
1 ohm
1.
Medical virtual reality is effectively used to stop the brain from processing pain and cure soreness.
2.
Parrot fish's source of bite → mimic → Ultra durable synthetic material.
Lotus leaf surface → SEM → Self Cleaning Process
The scales on the wings of a morpho butterfly → mimic → Interaction of colours.
Peacock feathers → mimic → Glowing in different colours.
3.
(b)
Frequency modulation
4.
\(Y_1=\overline{A+B}, y_2=\overline{A+B}=A+B, y=\overline{A+B}\)
5.
(b)
6.
AZX = 242He + 201e + Ai2iY
\(\frac{N_i}{Z_i}=\frac{(A_i-Z_i)}{Z_i} =\frac{A-8-(Z-6)}{Z-6}=\frac{A-Z-2}{Z-6}\)
7.
r ∝ A1/3
Surface Area = 4πr2
Hence, Surface Area ∝ A2/3
8.
\(V_{ionisation}=\frac{13.6}{n^2}Z^2 volt\)
\(Z=\sqrt\frac{V\times n^2}{13.6}=\sqrt\frac{122.4 \times I^2}{13.6}=\sqrt{9}=3\)
9.
(c)
thermionic
10.
\(K .E_{\max } =\mathrm{hv}-\phi \)
\(=\frac{\mathrm{hc}}{\lambda}-\phi \)
\(\mathrm{E} =\frac{6.6 \times 10^{-34} \times 3 \times 10^8-1.235}{500 \times 10^{-9} \times 1.6 \times 10^{-19}} \)
\(=2.475-1.235 \)
\(\text {K. } \mathrm{E}_{\max } =1.24 \mathrm{eV}\)
11.
\(P =\frac{E}{t}=\frac{n h v}{t}=\frac{n h c}{\lambda t} \Rightarrow P \propto n / t \)
\(\frac{P_1}{P_2} =\frac{1.38 \times 10^{15}}{460} \times \frac{520}{1.04 \times 10^{15}}=1.5\)
12.
(b)
13.
λ = 500 nm = 500 x 10-9 m
x = 3 mm = 3 x 10-3 m
a = 1 mm = 1 x 10-3 m
\(d=\frac{xa}{1.22 \lambda}\)
\(d=\frac{3 \times1\times10^{-6}}{1.22 \times500\times10^{-9}}\)
\(=\frac{3 \times1\times10^{-6}}{6.10 \times 10^{-7}}\)
\(d=\frac{30}{6.1}=5 m\)
14.
Apparent depth = 3 + 5 = 8 cm
Real depth = thickness of the slab = t
n = 1.5
\(n=\frac{Real \ depth}{Apparent \ depth}\)
\(\therefore 1.5=\frac{t}{8}\)
t = 1.5 x 8
t = 12 cm
15.
Refractive index of water \(=\frac{Velocity \ of \ light \ in \ air(V_s)}{Velocity \ of \ light \ in \ water(V_w)}\)
16.
v = nג
In an isotropic medium, there is no change in the frequency of the light. So, the speed of light depends on wavelength of light.
17.
\(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T\)
Using an equation,
Lorentz force, \(\vec{F}=Il\hat{j}\times\vec B\)
We can get,
Lorentz force \(F=\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)
18.
(b)
\(7\mu T\)
19.
Magnetic filed at the centre of a circular
loop, B = \(\frac{μ_oI}{2\pi R}\)
From the figure, R =\(\frac{2r}{\pi}\)
\(\therefore B'=\frac{μ_oI}{2\pi \times\frac{2r}{\pi}}=\frac{μ_oI}{4r}\)
\(B'=\frac { { \mu }_{ ° }I }{ 4r } \bigotimes \)
20.
\(L=\frac{20}{\pi^2} \mathrm{H}, \mathrm{f}=50 \mathrm{~Hz} \)
\(f=\frac{1}{2 \pi \sqrt{L C}} \)
\(50=\frac{1}{2 \pi \sqrt{\frac{20}{\pi^2} \times C}} \)
\(50=\frac{1}{2 \times \sqrt{20 C}} \)
\(\therefore(50)^2=\frac{1}{4 \times 20 C} \)
\(\therefore C=\frac{1}{2500 \times 4 \times 20}=5 \times 10^{-6}=5 \mu \mathrm{F}\)
21.
L = 20 x 10-3H. C = 50 x 10-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340 \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340 \)
\(=6800 \times 10^{-1}=6.8 \Omega \)
\(X_C=\frac{1}{C .} \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2} \)
\(=\sqrt{(40)^2+(58.82-6.8)^2} \)
\(=\sqrt{(40)^2+(52.02)^2} \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62} \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W
22.
\(\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}, \mathrm{~V}_{\mathrm{s}}=11 \mathrm{~V} \)
\(\mathrm{I}_{\mathrm{P}}=6 \mathrm{~A}, \mathrm{I}_{\mathrm{s}}=100 \mathrm{~A} . \)
\(\text {Efficiency }=\frac{\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}} \)
\(=\frac{11 \times 100}{220 \times 6}=\frac{1100}{220 \times 6}=\frac{5}{6}=0.83\)
23.
\(\phi_B =10 t^2-50 t+250 \)
\(e =\frac{-d \phi_B}{d t} \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50)
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V
24.
\(B_o=\frac{E_o}{C}=\frac{\sqrt2E_{rms}}{C}\)
\(B_o=\frac{\sqrt2 \times 3}{3 \times10^8}=1.414 \times10^{-8}T\)
25.
Energy density uE \(=\frac{U}{volume}\)
If A' = 2A d ' = 2d
Then V ' = 2A x 2d = 4Ad = 4V
Then volume would be increased. So, energy density will change.
26.
If voltage is increased from V to 2 V
Then Q1 = CV
Q2 = C(2 V) = 2 CV
∴ Q is doubled and C remains same
27.
The charge + q will be stable between B1 and B2 with respect to the displacement.
28.
Rs = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
Rs = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω
29.
\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2} \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)
30.
Resistance, \(R=\frac{V}{I}=\frac{4}{2}=2 \ ohm\)
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