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TN 12th Computer Applications வலையமைப்பு வடமிடல் Sample Question Papers Study Material - QB365 Set A

Published on: 12/10/2020
12th Standard Physics English Medium Free Online Test Book Back one Mark questions
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.
Questions + Answers key
Take MCQ Physics Test1.
The blue print for making ultra durable synthetic material is mimicked from _____.
Lotus leaf
Morpho butterfly
Parrot fish
Peacock feather
2.
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
3.
The given electrical network is equivalent to ______.
AND gate
OR gate
NOR gate
NOT gate
4.
A forward biased diode is treated as ________________.
An open switch with infinite resistance
A closed switch with a voltage drop of 0V
A closed switch in series with a battery voltage of 0.7V
A closed switch in series with a small resistance and a battery
5.
The mass of a 37Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of 37Li nucleus is nearly _____.
46 MeV
5.6 MeV
3.9 MeV
23 MeV
6.
7.
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
8.
The wave associated with a moving particle of mass 3 x 10–6 g has the same wavelength as an electron moving with a velocity 6 x 106 ms-1. The velocity of the particle is _____.
1.82 x 10-18ms-1
9 x 10-2ms-1
3 x 10-31ms-1
1.82 x 10-15ms-1
9.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is _____.
red
yellow
green
violet
10.
11.
Two short bar magnets have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is (Horizontal components of Earth’s magnetic induction is 3.6 × 10–5 Wb m–2 )
3.60 × 10-5 Wb m-2
3.5 × 10-5 Wb m-2
2.56 × 10-4 Wb m-2
2.2 × 10-4 Wb m-2
12.
The BH curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is _____.
1.00 m A
1.25 mA
1.50 mA
1.75 mA
13.
A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is nearly ____.
1.0 A m2
1.2 A m2
0.5 A m2
0.8 A m2
14.
\(\frac{20}{\pi^2}H\) 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
15.
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
16.
In an electrical circuit, R, L, C, and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\). Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\). The power factor of the circuit is
1/2
1/\(\sqrt2\)
1
\(\sqrt3\)/2
17.
The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be





18.
If E = Eo sin[106 x -ωt] be the electric field of a plane electromagnetic wave, the value of ω is_____.
0.3 x 10−14 rad s−1
3 x 10−14 rad s−1
0.3 x 1014 rad s−1
3 x 1014 rad s-1
19.
During the propagation of electromagnetic waves in a medium __________________.
electric energy density is double of the magnetic energy density
electric energy density is half of the magnetic energy density
electric energy density is equal to the magnetic energy density
both electric and magnetic energy densities are zero
20.
Which of the following electromagnetic radiations is used for viewing objects through fog
microwave
gamma rays
X- rays
infrared
21.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 \(\times\) 10-2 C and 5 \(\times\) 10-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
3 \(\times\) 10-2 C
4 \(\times\) 10-2 C
1 \(\times\) 10-2 C
2 \(\times\) 10-2 C
22.
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
23.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is ______.
0.2 Ω
0.5 Ω
0.8 Ω
1.0 Ω
24.
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
\(\sqrt3\)
\(\frac{1}{\sqrt3}\)
\(\frac{1}{3}\)
25.
A carbon resistor of (47 ± 4.7 ) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be ______.
Yellow – Green – Violet – Gold
Yellow – Violet – Orange – Silver
Violet – Yellow – Orange – Silver
Green – Orange – Violet - Gold
1.
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.
2.
(b)
Frequency modulation
3.
\(Y_1=\overline{A+B}, y_2=\overline{A+B}=A+B, y=\overline{A+B}\)
4.
(d)
A closed switch in series with a small resistance and a battery
5.
\(\frac{BE}{A}=\frac{\Delta\times931MeV}{V}=\frac{0.042 \times931}{7}\)
= 5.586 = 5.6 MeV
6.
(b)
7.
\(K .E_{\max } =\mathrm{hv}-\phi \)
\(=\frac{\mathrm{hc}}{\lambda}-\phi \)
\(\mathrm{E} =\frac{6.6 \times 10^{-34} \times 3 \times 10^8-1.235}{500 \times 10^{-9} \times 1.6 \times 10^{-19}} \)
\(=2.475-1.235 \)
\(\text {K. } \mathrm{E}_{\max } =1.24 \mathrm{eV}\)
8.
\(\lambda_{\mathrm{i}} \frac{1}{\mathrm{mv}} \)
\(\frac{\lambda_p}{\lambda_e} =\frac{m_e v_e}{m_P v_P} \)
\(1 =\frac{9.1 \times 10^{-31} \times 6 \times 10^6}{3 \times 10^{-9} \times v_p} \)
\(\mathrm{v}_{\mathrm{p}} =9.1 \times 10^{-16} \times 2 \)
\(\mathrm{v}_{\mathrm{p}} =18.2 \times 10^{-16} \)
\(\mathrm{v}_{\mathrm{p}} =1.82 \times 10^{-15} \mathrm{~m} \mathrm{~s}^{-1}\)
9.
Refractive index for violet is more and wavelength for violet is very low comparing other colours. So, the letter which appears to be raised more is violet.
10.
(a)
11.
(c)
2.56 × 10-4 Wb m-2
12.
(c)
1.50 mA
13.
Dipole moment, \(\vec{p}_m=n\times I\times\vec{A}\)
\(\vec{p}_m\) = 50 x 3 x 3.14 x 25 x 10-4 ≈ 1.2 A m2
14.
\(L=\frac{20}{\pi^2} \mathrm{H}, \mathrm{f}=50 \mathrm{~Hz} \)
\(f=\frac{1}{2 \pi \sqrt{L C}} \)
\(50=\frac{1}{2 \pi \sqrt{\frac{20}{\pi^2} \times C}} \)
\(50=\frac{1}{2 \times \sqrt{20 C}} \)
\(\therefore(50)^2=\frac{1}{4 \times 20 C} \)
\(\therefore C=\frac{1}{2500 \times 4 \times 20}=5 \times 10^{-6}=5 \mu \mathrm{F}\)
15.
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
16.
\(\Phi = \frac{\pi}{3}-\frac{\pi}{3}=0 \)
Power factor = cosФ = cos 0 = 1
17.
Solution
e = -L\(\frac{dl}{dt}\)
18.
E = Eo sin(kx – ωt)
ω = ck
ω = 3 x 108 x 106
ω = 3 x 1014 rad s-1
19.
(c)
electric energy density is equal to the magnetic energy density
20.
(d)
infrared
21.
Q = q1 + q2 = 4 x 10-2C
\(q_{2f}=Q[\frac{r_2}{r_1+r_2}]\)
= 4 x 10-2 \([\frac{3}{4}]\)
q2f = 3 x 10-2 C
22.
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
23.
I = 0.2 A, R = 10 Ω, E = 2.1 V
\(I=\frac{ɛ}{R+r}\)
\(0.2=\frac{2.1}{10+r}\)
0.2 x (10 + r) = 2.1
2 + 0.2 r = 2.1
0.2 r = 2.1 - 2 = 0.1
Internal resistance, \(r=\frac{0.1}{0.2}=\frac{1}{2}\)
r = 0.5 Ω
24.
\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2} \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)
25.
Yellow - 4
Violet - 7
Orange - 103
Silver - Tolerance - 10%
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