12th Standard English Medium Physics Reduced syllabus Creative Five mark Question with Answerkey - 2021(Public Exam )

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:

Generic 3

2 x 3 = 6

Derive an expression for electrostatic potential due to a point charge.

Calculate the equivalent resistance between A and B in the given circuit.

Multiple Choice Question

15 x 1 = 15

The gravitational waves were theoretically proposed by _____.

(a)

Conrad Rontgen

(b)

Marie Curie

(c)

Albert Einstein

(d)

Edward Purcell

The particle size of ZnO material is 30 nm. Based on the dimension it is classified as _____.

(a)

Bulk material

(b)

Nanomaterial

(c)

Soft material

(d)

Magnetic material

The principle based on which a solar cell operates is______.

(a)

Diffusion

(b)

Recombination

(c)

Photovoltaic action

(d)

Carrier flow

Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is _____.

(a)

(b)

(c)

(d)

If a light of wavelength 330 nm is incident on a metal with work function 3.55 eV, the electrons are emitted. Then the wavelength of the wave associated with the emitted electron is _____.(Take h = 6.6 x 10^–34 Js)

(a)

< 2.75 ×10⁻⁹m

(b)

≥ 2.75 × 10⁻⁹m

(c)

≤ 2.75 × 10⁻¹²m

(d)

< 2.5 × 10⁻¹⁰m

Light transmitted by Nicol prism is, _____.

(a)

partially polarised

(b)

unpolarised

(c)

plane polarised

(d)

elliptically polarised

A simple pendulum with charged bob is oscillating with time period T and lets θ be the angular displacement. If the uniform magnetic field switched ON in a direction perpendicular to the plane of oscillation then ___________.

(a)

time period will decrease but θ will remain constant

(b)

time period remain constant but θ will decrease

(c)

both T and θ will remain the same

(d)

both T and θ will decrease

10.

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 _____.

(a)

1.00 m A

(b)

1.25 mA

(c)

1.50 mA

(d)

1.75 mA

11.

A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field \(\vec { B } \).

(a)

\(\sqrt { \frac { 2{ q }^{ 3 }BV }{ m } } \)

(b)

\(\sqrt { \frac { { q }^{ 3 }{ B }^{ 2 }V }{ 2m } } \)

(c)

\(\sqrt { \frac { 2{ q }^{ 3 }{ B }^{ 2 }V }{ m } } \)

(d)

\(\sqrt { \frac { { 2q }^{ 3 }BV }{ { m }^{ 3 } } } \)

12.

The instantaneous values of alternating current and voltage in a circuit are \(i=\frac { 1 }{ \sqrt { 2 } } \sin\left( 100\pi t \right) \) A and v \(=\frac { 1 }{ \sqrt { 2 } } \sin\left( 100\pi t+\frac { \pi }{ 3 } \right) V.\)The average power in watts consumed in the circuit is

(a)

\(\frac{1}{4}\)

(b)

\(\frac{\sqrt3}{4}\)

(c)

\(\frac{1}{2}\)

(d)

\(\frac{1}{8}\)

13.

When the current changes from +2A to −2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is

(a)

0.2H

(b)

0.4H

(c)

0.8H

(d)

0.1H

14.

A radiation of energy E falls normally on a perfectly reflecting surface. Th e momentum transferred to the surface __________________.

(a)

\(\frac{E}{c}\)

(b)

2\(\frac{E}{c}\)

(c)

(d)

\(\frac { E }{ { c }^{ 2 } } \)

15.

An electric field \(\vec { E } =10x\hat { i } \) exists in a certain region of space. Then the potential difference V = V_o – V_A, where V_o is the potential at the origin and V_A is the potential at x = 2 m is _____.

(a)

10 V

(b)

-20 V

(c)

+20 V

(d)

-10 V

16.

A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of ______.

(a)

each of them increases

(b)

each of them decreases

(c)

copper increases and germanium decreases

(d)

copper decreases and germanium increases

17.

A wire of resistance 2 ohms per meter is bent to form a circle of radius 1m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

(a)

\(\pi \Omega\)

(b)

\(\frac{\pi}{2}\Omega\)

(c)

2\(\pi \Omega\)

(d)

\(\frac{\pi}{4}\Omega\)

Generic 2

9 x 2 = 18

18.

Light of wavelength 600 nm that falls on a pair of slits producing interference pattern on a screen in which the bright fringes are separated by 7.2 mm. What must be the wavelength of another light which produces bright fringes separated by 8.1 mm with the same apparatus?

19.

What are carrier waves?

20.

What is Breakdown voltage?

21.

What is called logical variables?

22.

Explain knee voltage

23.

Define Zener effect

24.

What is rectification?

25.

If the current i flowing in the straight conducting wire as shown in the figure decreases, find out the direction of induced current in the metallic square loop placed near it.

26.

Calculate the magnetic flux coming out from closed surface containing magnetic dipole (say, a bar magnet) as shown in figure.

Generic 5

6 x 5 = 30

27.

A telescope has an objective of diameter 60 cm. The focal lengths of the objective and eyepiece are 2.0 m and 1.0 cm respectively. The telescope is directed to view two distant almost point sources of light, (e.g., two stars of a binary). The sources are roughly at the same distance = 10⁴ light years along the line of sight but are separated transversely to the line of sight by a distance of 1010 m. Will the telescope resolve the two objects i.e. will it see two distant stars?

28.

Two polaroids are set in crossed positions. A third polaroid is placed between the two making an angle e with the pass axis of the first polaroid. Write the expression for the intensity of light transmitted from the second polaroid. In what orientations will the transmitted intensity be
(i) minimum and
(ii) maximum.

29.

If the focal lengths of the objective and eyepiece of a microscope are 2 cm and 5 cm and 5 cm respectively and the distance between them is 20 cm, what is the distance of the object from the objective when the image seen by the eye is 25 cm from eyepiece? Also find the magnifying power.

30.

For a BJT, the common - base current gain α = 0.98 and the collector base junction reverse bias saturation I_CU = 0.6μA. This BJT is connected in the common emitter mode and operated in the active region with a base drive current I_D= 20 μA. The collector current I_C for this mode of operating is

31.

Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.

32.

Obtain the expression for electric field due to an infinitely long charged wire.

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Answers Key:

Generic 3

2 x 3 = 6

Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q. This is shown in Figure.

Electrostatic potential at a point P
The electric potential at the point P is
\(V=\int _{ \infty }^{ r }{ \left( -\vec { E } \right) d\vec { r } } =-\int _{ \infty }^{ r }{ \vec { E } } .d\vec { r } \) ...(1)
Electric field due to positive point charge q is
\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } \hat { r } \)
\(V=\frac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .d\vec { r } } \)
The infinitesimal displacement vector, \(d\vec { r } =dr\hat { r } \) and using \(\hat { r } \).\(\hat { r } \) = 1, we have
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .dr\hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } dr } } \)
After the integration,
\(V=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q{ \left\{ -\frac { 1 }{ r } \right\} }_{ \infty }^{ r }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)
Hence, the electric potential due to a point charge q at a distance r is
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \) ....(2)

In all the sections, the resistors are connected in parallel.
Section I
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 2 }{ 2 } \quad { R }_{ { p }_{ 1 } }=1\Omega \)

Section II
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 2 }{ 4 } ,\quad \frac { 1 }{ { R }_{ { P }_{ 2 } } } =\frac { 1 }{ 2 } ,{ R }_{ { p }_{ 2 } }=2\Omega \)

Section III
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } \)
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 3 } ,{ R }_{ { p }_{ 3 } }=3\Omega \)
Equivalent resistance is given by
R = Rp₁+ Rp₂+ Rp₃
R = 1 Ω + 2 Ω + 3 Ω = 6 Ω
The circuit became,

Equivalent resistance between A and B is

Multiple Choice Question

15 x 1 = 15

Albert Einstein theoretically proposed the existence of gravitational waves in the year 1915.

Size of the particle is between
1-100 nm - Nano.
Size of the particle is greater
than-100 nm - Bulk.

\(V_{ionisation}=\frac{13.6}{n^2}Z^2 volt\)
\(Z=\sqrt\frac{V\times n^2}{13.6}=\sqrt\frac{122.4 \times I^2}{13.6}=\sqrt{9}=3\)

(b)
≥ 2.75 × 10⁻⁹m

10.

11.

Lorentz force F = Bqv
Energy w = qV
Energy is equal to kinetic energy,
\(qV=\frac{1}{2}mv^2\)
\(v=\sqrt { \frac {2qV }{ m } } \)
\(\therefore Lorentz \ force \ F= Bq\times \sqrt \frac{2qV}{m}=\sqrt { \frac { 2{ B }^{ 2 }{ q }^{ 3 }V }{ m } } \)

12.

P_av= \(\frac{1}{2}\)V₀I₀cosΦ
\(= \frac{1}{2}\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}cos\times\frac{\pi}{3}\times \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)

13.

\(\text {emf } e=8 \mathrm{~V} \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A} \)
\(\text {dt }=0.05 \mathrm{~s} \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05} \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4} \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

14.

(b)
2\(\frac{E}{c}\)

15.

\(\vec {E}\) = 10x\(\hat{i},\) when x = 2 m
\(\vec {E}\) = 10 x 2 x \(\hat{i}\) = 20\(\hat{i}\)
Since, \(E=\frac{-dV}{dx}\therefore V=+20 V\)

16.

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

17.

Total length = 2πr = 2π
∴ Resistance of each segment = 2π/2 = π Ω

Generic 2

9 x 2 = 18

18.

ß₁ = 7.2 mm, ß₂ = 8.1 mm, λ₁ = 600 nm, λ₂=?
Dividing, we get \(\beta = \frac{D\lambda}{d}(or)\beta∝\lambda\)
\(\frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2}\)
\(\therefore, { \lambda }_{ 2 }=\frac{\beta_2\times \lambda_1}{\beta_1}=\cfrac { 8.1\times10^{-3} \times 600 \times10^{-9}}{ 7.2 \times10^{-3}} \)
\(=\cfrac { 9 \times 600\times10^{-9} }{ 8 } =675\times 10^{-9 }\Rightarrow\lambda_2= 675nm\)

19.

Waves of high frequency are called carrier waves. If an audio signal (low frequency) is transmitted as such, it does out after traversing some distance. To reach up larger distance, it is superimposed an high frequency waves called carrier wave

20.

The voltage at which the diode breaks down is called the breakdown voltage

21.

Digital electronics deals with logical operations. The variables are called logical variables

22.

At room temperature, a potential difference equal to the barrier potential is required before a reasonable forward current starts flowing across the diode. This voltage is known as threshold voltage or cut-in voltage or knee voltage.

23.

Electric field is strong enough to break (or) rep tune the covalent bonds in the lattice and there by generating electron - hole pairs. This effect is called Zener effect.

24.

The process in which alternating voltage or alternating current is converted direct voltage or direct current is called rectification.

25.

From right hand rule, the magnetic field by the straight wire is directed into the plane of the square loop perpendicularly and its magnetic flux is decreasing. The decrease in flux is opposed by the current induced in the loop by producing a magnetic field in the same direction as the magnetic field of the wire. Again from right hand rule, for this inward magnetic field, the direction of the induced current in the loop is clockwise.

26.

The total flux emanating from the closed surface S enclosing the dipole is zero. So,
\({ \Phi }_{ B }=\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero,
\(\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
This is similar to Gauss’s law in electrostatics.

Generic 5

6 x 5 = 30

27.

Data on focal lengths is not required, The limit of resolution of the telescope is given by
\(\theta =\cfrac { 1.22\lambda }{ D } =\cfrac { 1.22\times 6\times { 10 }^{ -7 } }{ 0.60 } rad=1.22\times { 10 }^{ -6 }\)

Here, the value chosen corresponds roughly to the wavelength of yellow light. The transverse separation between the two sources subtends an angle equal to
\(\cfrac { { 10 }^{ 10 }m }{ 9.46\times { 10 }^{ 19 } } ={ 10 }^{ -10 }red\)
Here, we have used 1 light-year = 9.46 x 10¹⁵ m This angle is much too small compared to e above. The two stars of the binary cannot be resolved by the given telescope.

28.

Let polaroids P₁ and P₃ be in a crossed positions.
Let the polaroid P₂ make an angle e with the pass axis of polaroid Pr
Let I₁ be the intensity of polarised light emerging our of P₁ Then intensity of light after passing through P₂ will I₂= I₁cos²θ
Since P₃ and P, are in crossed position, therefore, the angle made by P₂ with P₃ is \(\left( \cfrac { \pi }{ 2 } -\theta \right) \)
∴ The intensity of light coming out of P₃ is
\({ I }_{ 3 }={ I }_{ 2 }{ cos }^{ 2 }\left( \cfrac { \pi }{ 2 } -\theta \right) \)
or \({ I }_{ 3 }={ I }_{ 2 }{ cos }^{ 2 }{ sin }^{ 2 }\theta =\theta ={ I }_{ 1 }\left( \cfrac { 1 }{ 2 } sin20 \right) ^{ 2 }\)
If I₀ is the intensity of the unpolarised light falling on P_I' then \({ I }_{ 1 }=\cfrac { { I }_{ o } }{ 2 } \)
\(\therefore { I }_{ 3 }=\cfrac { { I }_{ o } }{ 2 } \left( \cfrac { 1 }{ 2 } sin20 \right) ^{ 2 }\)
(i) Minimum outcoming intensity is zero.
(ii) Maximum outcoming intensity is received
when \(\theta =\cfrac { \pi }{ 4 } \)
\(\therefore \left( { I }_{ 3 } \right) _{ max }=\cfrac { { I }_{ o } }{ 2 } \left( \cfrac { 1 }{ 2 } \right) ^{ 2 }=\cfrac { { I }_{ o } }{ 8 } \)

29.

\(\cfrac { 1 }{ -25 } -\cfrac { 1 }{ { u }_{ e } } =\cfrac { 1 }{ 5 } \)
or \({ u }_{ e }=-\cfrac { 25 }{ 6 } cm\)
\({ v }_{ o }=20-\cfrac { 25 }{ 6 } =\cfrac { 95 }{ 6 } cm\)
\(\cfrac { 1 }{ { v }_{ 0 } } =\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ { f }_{ 0 } } \)
\(\cfrac { 6 }{ 95 } -\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ 2 } ,{ u }_{ 0 }=2.29cm\)
\(m=-\cfrac { 95/6 }{ 190/83 } \left( 1+\cfrac { 25 }{ 5 } \right) =-41.5\)

30.

α = 0.98
I_Cu = 0.6 μA
\(\alpha =\frac { \beta }{ 1+\beta } \)
(or)
\(0.98=\frac { 1 }{ \frac { 1 }{ \beta } +1 } \)
β = 49
I_CEO = (1 + β)I_CBO
= (1 + 49) x 0.6μA
I_CEO = 30 μA
= (1 + 49) x 0.6 μA
I_CEO = 30 μA
I_C = βI_B+ I_CEO
= 49 x 20 μA + 30 μA
I_C = 1.01 mA

31.

Consider a solenoid of length L having N turns. The diameter of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

Consider a rectangular loop abcd. Then from Ampere's circuital law,
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } \) = μ₀ I_enclosed= μ₀ x (total current enclosed by Amperian loop)
The left hand side of the equation is
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } =\int _{ a }^{ b }{ \vec { B } .\vec { dl } } +\int _{ b }^{ c }{ \vec { B } .\vec { dl } } +\int _{ c }^{ d }{ \vec { B } .\vec { dl } } +\int _{ d }^{ a }{ \vec { B } .\vec { dl } } \)
Elemental lengths bc and da are perpendicular to magnetic field.
\(\therefore \int _{ b }^{ c }{ \vec { B } .\vec { dl } } =\int _{ b }^{ c }{ |\vec { B } ||\vec { dl } | } cos 90^o = 0\)
similarly,
\(\int _{ d }^{ a }{ \vec { B } .\vec { dl } } = 0\)
Since the magnetic field outside the solenoid is zero,
\(\int _{ c }^{ d }{ \vec { B } .\vec { dl } } =0\) and
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } } =BL \quad \quad (\because \theta =0^o)\)
Let I be the current passing through the solenoid of N turns, then
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } } = BL = μ_o NI ⇒ B = μ_o\frac {NI}{L}\)
The number of turns per unit length is given by \(\frac { N }{ L } \) = n, Then
B = \(\mu_o \frac { nLI }{ L } \) = μ₀nI
Since n is a constant for a given solenoid and μ₀ is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

32.

(i) Consider an infinitely long straight wire having uniform linear charge density λ(charge per unit length).
(ii) Let P be a point located at a perpendicular distance r from the wire.
(iii) The electric field at the point P can be found using Gauss law. we choose two small charges elements A₁ and A₂ on the wire which are at equal distance from the point P.
(iv) The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r.
This is shown in the Figure(b). From this property, the charged wire possesses a cylindrical symmetry.
(v) Let us choose a cylindrical Gaussian surface of radius r and length L .
The total electric flux through this closed surface is calculated as follows:
\(\Phi_{E}=\oint \vec{E} \cdot d \vec{A} \)
\(=\int_{\substack{\text { Curved } \text { surface }}} \vec{E} \cdot d \vec{A}+\int_{\substack{\text { top } \text { sufface }}} \vec{E} \cdot d \vec{A}+\int_{\substack{\text { bottom } \text { surface }}} \vec{E} \cdot d \vec{A} \) .....(1)
It is seen from Figure that for the curved surface, \(\vec{E}\) is parallel to \(\vec{A}\) and \(\vec{E} \cdot d \vec{A}=E d A\) . For the top and bottom surfaces, \(\vec{E}\) is perpendicular to \(\vec{A}\)and \(\vec{E} \cdot d \vec{A}=0\)
Substituting these values in the equation (1) and applying Gauss law to the cylindrical surface, we have
\(\Phi_{E}=\int_{\substack{\text { Curved } \\ \text { surface }}} E d A=\frac{Q_{\text {encl }}}{\varepsilon_{o}}\) ...........(2)

Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration and Q_encl is given by Q_encl= λL where, λ is linear charge density (charge per unit length).
\(E\int _{ Curved \ surface }^{ }{ dA } =\frac { \lambda L }{ { \varepsilon }_{ 0 } } \quad ....(3)\)
Here,= \(\int _{ Curved \ surface }^{ }{ dA } \) total area of the curved surface = 2πrL
Substituting this in equation (3), we get
E.2πrL = \(\frac { \lambda L }{ { \varepsilon }_{ 0 } } \)
\(E=\frac { 1 }{ 2\pi { \varepsilon }_{ 0 } } \frac { \lambda }{ r } \quad \quad ...(4)\)
In vector form, \(\vec { E } =\frac { 1 }{ 2\pi { \varepsilon }_{ 0 } } \frac { \lambda }{ r } \hat { r } \quad \quad \quad \quad \quad ...(5)\)
The electric field due to the infinite charged wire depends on \(\frac{1}{r}\) rather than \(\frac{1}{r^2}\) for a point charge.
Equation (5) indicates that the electric field is always along the perpendicular direction \((\hat { r } )\) to wire. In fact, if λ > 0 then \(\vec { E } \)points perpendicular outward \((\hat { r } )\) from the wire and if λ < 0, then \(\vec { E } \) points perpendicular inward \((\hat { -r } )\).