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Published on: 12/10/2020
12th Standard Physics English Medium Free Online Test Book Back 1 Mark Questions with Answer Key Part Three
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 given electrical network is equivalent to ______.
AND gate
OR gate
NOR gate
NOT gate
2.
If a positive half-wave rectified voltage is fed to a load resistor, for which part of a cycle there will be current flow through the load?
00–900
900–1800
00–1800
00–3600
3.
4.
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
5.
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
6.
A non-conducting charged ring carrying a charge of q, mass m and radius r is rotated about its axis with constant angular speed ω. Find the ratio of its magnetic moment with angular momentum is _____.
\(\\ \frac { q }{ m } \)
\(\\ \frac { 2q }{ m } \)
\(\\ \frac { q }{ 2m } \)
\(\\ \frac { q }{ 4m } \)
7.
The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be
8.
Consider an oscillator which has a charged particle oscillating about its mean position with a frequency of 300 MHz. The wavelength of electromagnetic waves produced by this oscillator is _____.
1 m
10 m
100 m
1000 m
9.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is _______.
8.80 x 10-17 J
-8.80 x 10-17 J
4.40 x 10-17 J
5.80 x 10-17 J
10.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be ______.
R
2R
\(\frac{R}{4}\)
\(\frac{R}{2}\)
1.
\(Y_1=\overline{A+B}, y_2=\overline{A+B}=A+B, y=\overline{A+B}\)
2.
(c)
00–1800
3.
(b)
4.
\(P =\frac{E}{t}=\frac{n h v}{t}=\frac{n h c}{\lambda t} \Rightarrow P \propto n / t \)
\(\frac{P_1}{P_2} =\frac{1.38 \times 10^{15}}{460} \times \frac{520}{1.04 \times 10^{15}}=1.5\)
5.
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
6.
Magnetic moment,
μ = IA
Angular momentum,
L = Iω
Ratio \(\frac{p_m}{L}=\frac{(q/T)\pi r^2}{mr^2\omega}=\frac{q}{2m}\)
7.
Solution
e = -L\(\frac{dl}{dt}\)
8.
ג = \(\frac{c}{f}=\frac{3 \times10^8}{300 \times 10^6}=1 m\)
9.
W = qV
V = 7 - (-4) = 11 V
q = ne = 50 x 1.6 x 10-19 = 8 x 10-18C
W = 8 x 10-18 x 11 = 88 x 10-18 = 8.8 x 10-17]
10.
\(\mathrm{V}_1 =220 \mathrm{~V}, \quad \mathrm{P}_1=60 \mathrm{~W} \)
\(\mathrm{~V}_{\mathrm{U}} =110 \mathrm{~V}, \mathrm{P}_{\mathrm{U}}=60 \mathrm{~W} \)
\(P =\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P} \)
\(\therefore R_l =\frac{V_I^2}{P_l} \text { Similarly, } \quad \mathrm{R}_U=\frac{V_U^2}{P_U} \)
\(R_l =\frac{220 \times 220}{60} \quad \mathrm{R}_{\mathrm{U}}=\frac{110 \times 110}{60} \)
\(R_I =\frac{48400}{60} \quad R_U=\frac{12100}{60} \)
\(\frac{R_U}{R_l} =\frac{12100}{60} \times \frac{60}{48400}=\frac{1}{4} \)
\(R_U =\frac{R_l}{4}=\frac{R}{4}\)
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