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Published on: 02/09/2022
QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Physics Subject - Retirement and Death of a Partner, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
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.
Emission of electrons by the absorption of heat energy is called ______ emission.
photoelectric
field
thermionic
secondary
2.
The work functions for metals A, B and C are 1.92 eV, 2.0 eV and 5.0 eV respectively. The metal/metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are _____.
A only
both A and B
all these metals
none
3.
Photons of wavelength λ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having magnitude B. The work function of the metal is _____.
\(\frac { hc }{ \lambda } -{ m }_{ e }+\frac { e^{ 2 }{ B }^{ 2 }{ R }^{ 2 } }{ { 2m }_{ e } } \)
\(\frac { hc }{ \lambda } +{ 2m }_{ e }{ \left[ \frac { eBR }{ { 2m }_{ e } } \right] }^{ 2 }\)
\(\\ \frac { hc }{ \lambda } -{ m }_{ e }{ c }^{ 2 }-\frac { e^{ 2 }{ B }^{ 2 }{ R }^{ 2 } }{ { 2m }_{ e } } \)
\(\frac { hc }{ \lambda } -{ 2m }_{ e }{ \left[ \frac { eBR }{ { 2m }_{ e } } \right] }^{ 2 }\)
4.
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
5.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is _____.
4125 \(\mathring { A } \)
3750\(\mathring { A } \)
6000\(\mathring { A } \)
2062.5\(\mathring { A } \)
6.
If the mean wavelength of light from sun is taken as 550 nm and its mean power as 3.8 x 1026 W, then the number of photons emitted per second from the sun is of the order of _____.
1045
1042
1054
1051
7.
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
8.
Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of emitted electrons in the two cases will be _____.
1:4
1:3
1:1
1:9
9.
10.
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ /2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the material is _____.
\(\frac{hc}{\lambda}\)
\(\frac{2hc}{\lambda}\)
\(\frac{hc}{3\lambda}\)
\(\frac{hc}{2\lambda}\)
11.
12.
When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is \(\frac{V}{4}\). The threshold wavelength for the metallic surface is _____.
4λ
5λ
\(\frac{5}{2}λ\)
3λ
13.
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
14.
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Broglie wavelength associated with the electrons would _____.
increase by 2 times
decrease by 2 times
decrease by 4 times
increase by 4 times
15.
The wavelength λe of an electron and λp of a photon of same energy E are related by _____.
λp ∝ λe
\({ \lambda }_{ p }∝ \sqrt { { \lambda }_{ e } } \)
\({ \lambda }_{ p }∝ \frac { 1 }{ \sqrt { { \lambda }_{ e } } } \)
\({ \lambda }_{ p }∝ { \lambda }_{ e }^{ 2 }\)
1.
(c)
thermionic
2.
\(E=\frac{12400 \stackrel{o}A}{4100 \stackrel{o}A}=3.02 eV\)
3.
\(\text {K.E } =\frac{B^2 q^2 r^2}{2 m} \)
\(\phi =\frac{h c}{\lambda}-K . E \)
\(=\frac{h c}{\lambda}-\frac{B^2 q^2 r^2}{2 m} \)
\(\phi =\frac{h c}{\lambda}-2 m\left(\frac{B q r}{2 m}\right)^2\)
4.
\(K .E_{\max } =\mathrm{hv}-\phi \)
\(=\frac{\mathrm{hc}}{\lambda}-\phi \)
\(\mathrm{E} =\frac{6.6 \times 10^{-34} \times 3 \times 10^8-1.235}{500 \times 10^{-9} \times 1.6 \times 10^{-19}} \)
\(=2.475-1.235 \)
\(\text {K. } \mathrm{E}_{\max } =1.24 \mathrm{eV}\)
5.
\(\lambda_0 =\frac{h c}{\phi} \)
\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.313 \times 1.6 \times 10^{-19}} \)
\( =\frac{19.8782400}{5.3} \times 10^{-7} \)
\(\lambda_0 =3.750 \times 10^{-7} \simeq 3750 \stackrel{o}A\)
6.
\(\mathrm{P} =\frac{\mathrm{n}}{\mathrm{t}} \frac{\mathrm{hc}}{\lambda} \)
\(\frac{\mathrm{n}}{\mathrm{t}} =\frac{\mathrm{P} \lambda}{\mathrm{hc}} \)
\(\frac{\mathrm{n}}{\mathrm{t}} =\frac{3.8 \times 10^{26} \times 550 \times 10^{-9}}{6.6 \times 10^{-3} \times 3 \times 10^8}=1 \times 10^{-15}\)
7.
\(P =\frac{E}{t}=\frac{n h v}{t}=\frac{n h c}{\lambda t} \Rightarrow P \propto n / t \)
\(\frac{P_1}{P_2} =\frac{1.38 \times 10^{15}}{460} \times \frac{520}{1.04 \times 10^{15}}=1.5\)
8.
K.E= hv - Φ
K.E1 = 0.9 - 0.6 = 0.3 eV
K.E2 = 3.3 - 0.6 = 2.7 ev
K.E ∝ v2
\(\frac{0.3}{2.7}=\frac{v^2_1}{v^2_2} \)
\(\frac{v^1}{v^2} =\frac{1}{3}\)
9.
(b)
10.
\(\frac{\mathrm{hc}}{\lambda} =\phi+\mathrm{K} . \mathrm{E} .....(1) \)
\(\frac{2 \mathrm{hc}}{\lambda} =\phi+3 \mathrm{~K} . \mathrm{E}......(2)\)
multiply eqn. (1) by 3, we get
\(\frac{3 \mathrm{hc}}{\lambda} =3\phi+3 \mathrm{~K} . \mathrm{E} .....(3)\)
Subtract eqn. (2) from (3), we get
\(\frac{\mathrm{hc}}{\lambda} =2\phi \)
\(\phi=\frac{ \mathrm{hc}}{2\lambda} \)
11.
(b)
12.
\(\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV} \) .....(1)
\(\frac{\mathrm{hc}}{2 \lambda}=\phi+\frac{\mathrm{eV}}{4}\) .....(2)
multiply (2) eqn by 4
\(\frac{2 h c}{\lambda}=4 \phi+\mathrm{eV}\) .....(3)
subtract eqn (1) from (3), we get
\(\frac{ h c}{\lambda}=3 \phi \Rightarrow \phi = \frac{ h c}{3\lambda}\)
\(\frac{ h c}{\lambda_o}=\frac{ h c}{3\lambda}\)
⋋o = 3⋋
13.
\(\lambda_{\mathrm{i}} \frac{1}{\mathrm{mv}} \)
\(\frac{\lambda_p}{\lambda_e} =\frac{m_e v_e}{m_P v_P} \)
\(1 =\frac{9.1 \times 10^{-31} \times 6 \times 10^6}{3 \times 10^{-9} \times v_p} \)
\(\mathrm{v}_{\mathrm{p}} =9.1 \times 10^{-16} \times 2 \)
\(\mathrm{v}_{\mathrm{p}} =18.2 \times 10^{-16} \)
\(\mathrm{v}_{\mathrm{p}} =1.82 \times 10^{-15} \mathrm{~m} \mathrm{~s}^{-1}\)
14.
\(\lambda\propto \frac{1}{\sqrt{V}}\)
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{224\times10^3}}{\sqrt{14\times 10^3}}\)
\(=\sqrt{16}=4\)
\(\lambda_{\mathrm{2}}= \frac{\lambda_1}{4}\)
15.
\(\mathrm{E}_{\mathrm{p}} =\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}} \)
\(\mathrm{E}_{\mathrm{e}} =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2} \)
\(\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}} =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2} \)
\(\lambda_{\mathrm{p}} \propto \lambda_{\mathrm{e}}^{{ }^2}\)
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