# Class 12 AHSEC 2020 Physics Question Paper Solved English Medium

Class 12 AHSEC 2020 Physics Question Paper Solved English Medium, AHSEC Class 12 Physics Question Paper Solved PDF Download, to each Paper is Assam Board Exam in the list of AHSEC so that you can easily browse through different subjects and select needs one. AHSEC Class 12 Physics Previous Years Question Paper Solved in English can be of great value to excel in the examination.

## Class 12 AHSEC 2020 Physics Question Paper Solved English Medium

AHSEC Old Question Paper provided is as per the 2020 AHSEC Board Exam and covers all the questions from the AHSEC Class 12 Physics Solved Question Paper 2020 English Medium. Access the detailed Class 12 Physics 2020 Previous Years Question Paper Solved provided here and get a good grip on the subject. AHSEC 2020 Physics Question Paper Solved Access the AHSEC 2020 Physics Old Question Paper Solved, AHSEC Class 12 Physics Solved Question Paper 2020 of English in Page Format. Make use of them during your practice and score well in the exams.

### PHYSICS

2020

PHYSICS OLD QUESTION PAPER SOLVED

1. (a) Mention one difference between mass and charge.

Ans: Electric charge may be positive or negative physical quantity. Mass of a body is always a positive physical quantity.

(b) The colours on a carbon resistor are yellow, violet, brown and glode respectively from left to right. If the corresponding numbers for the colours are 47,1 and 5, what will be the resistanc of the resistor.

Ans: Resistance=(1stcolour)×(2ndcolour)×10 3rd colour + Tolerance

= [47×101+5%of 47×101]ohm

= [1+5/100]×470=[470+23.5]ohm

(c) Name the beautiful natural phenomenon that occurs in the sky of polar regions of earth due to helical motion of charged particles.

Ans: An aurora is caused by the streams of electrified particles (which are emitted by the sun) trapped in the magnetic field of the earth.

(d) What is eddy current?

Ans: When a sheet of metal is placed in a changing magnetic field induced currents are set up in it, which opposes the change. These a known as eddy currents.

(e) (i) Name the portion of the electromagnetic spectrum in between ultraviolet and infrared regions.

Ans: Visible light.

Or

(i) What is the radiation pressure on earth’s surface?

Ans: The force exerted by electromagnetic wave on unit area of the surface is called radiation pressure. According to experiment,the radiation pressure of visible light is found to be the order of 7×10-6Nm-2. It means that,the force due to radiation pressure of visivle light on the surface of area 10cm2 is

= (7×10-6×10×10-4) N=7×10-9N

Whichis very small.Therefore, we do not feel when sunlight falls on us.

(1) Name the equipment which can transmit optical signal through it and are used as ‘light pipe’.

Ans: Optical fibre.

(g) (i) Guess the shape of the curve which shows the variation of V, with vin the case of phototelectric emission shown by the relation where the symbols

have their usual meaning.V0=(h/e)v-00/e

Or

(ii) The de Broglie wavelength of a heavier particle is blank)

Ans: Smaller.

(h) The loss of strength of a signal while propagating through a medium is called____ fill in the blanks)

Ans: Atltenuation.

2. (a) (i) A closed spgerical surface a charge q at its centre. Show that electric flux through the closed surface is q/e0.

Ans: Let us consider a point p at a distance r from a hollow sphere of radius R (r<R) & charge Q. To calculate the field at p we consider a Gaussian surface having radius r. According to Gauss’s law.

0=Q/E0

Or

(b) (i) In the following network, I₁ =5/₂ A, I₂ =5/A and l3= 15/8 A. Calculate the total voltage drop over the closed loopw BADEB.

(ii) Find the equivalent resistance across the terminals A and B, shown in the figer below.

Ans:

(c) (i) How would you connect resistances 1Ω,2Ω and 3Ω so as to get an equivalent resistance of 3.66Ω. Draw the required circuit diagram.

Ans:

Or

(ii) Draw a circuit diagram required to compare the e.m.fs. of two cells using potentiometer. Write also the mathematical formula required for it.

Ans:

Or

(ii) A beam of ions with velocity 2×105ms-1 enters normally into a uniform magnetic field of 0.04T. If the specific charge (i.e.q/m) of ion is 5×107 Ckg-1, find the radius of the circular path described.2

Ans: r = u/ (q/m)B = 2×10-5/5×107×0.04 = 10-1=0.1m

(e) (i) Of two metals A and B, it is found that X >>1 and -1≤ XB <0. Name the types of materials to which the metals A and B do belong. Give one example of each.

Ans:

Or

(ii) Define magnetic declination. How does it depend on latitudes?

Ans: Magnetic declination at a place is defined as the angle between geographic and magentic meridian at the place.

(f) (i) In a circuit, current decreases from 5A to 0A is 0.1s. If the average induced e.m.fis 200V, calculate the self-inductance of the circuit and write its unit also.

Ans: E = -L dl/dt or 200 = -L(0-5)/0.1 or 20=5L or 2L= -20/5 = 4 Henry

Or

(ii) When a coil of area 5m² and number of turns 100 is placed perpendicular to a magnetic field of 10T, the flux passing through it is 5×103 Wb. If the coils removed from the field in 0.1s, calculate the induced e.m.f.

Ans: E = – do/dt = – (-5×103) = 5×104 volt

(g) (i) Taking the example of a charging capacitor, name the currents that are responsible for the process of charging and also state which current flows outside and which flows inside the capacitor.

Ans: Conduction current and displacement current, Conduction current flows outside and displacement current flows inside.

Or

(ii) Match the following and rewrite:

(h) (i) A concave mirror of focal length 18cm produces 3 times magnified errect image of an object. Find the position of the object.

Ans: Here f = – 18cm

m= – v/u

Also,m= hi /ho ∴ hi /ho = -v/u

Or, 3ho/ ho =- v/u |∴  hi = 3ho or v= -3u

Mirror formula, 1/u+1/v = 1/f or, 1/u -1/3v=1/-18 or, 3-1/3u= 1/-18

Or, 2/3u=1/-18 or, u=2/3x (6/18) = – 12

∴ u= – 12cm

Or

(ii) In a combination of lenses in contact, the power of the first lens is +2.5D and the focal length of the second lens is-25cm. Calculate the power or focal length of the lens combination.

Ans: Given P1= + 2.5D

P₂ = 1/ f2=100/-25 D=- 4D

∴ P=P₁+P₂=2.5+(-4)=-1.5D

(I) (i) In 1885, a Swedish school teacher observed a series of spectral lines in the visible region of the hydrogen spectrum. Name the spectral series. If R

(Rydberg constant) =1.097×10-7 m-1 and (1/22-1/32) = 0.138.

Ans:

Or

(ii) A difference of 2.3e V separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Ans: ΔE = hv

or, 2.3×1.6×10-19 = 6.63×10-34 xv

V= 2.3×1.6×10-19/ 6.63×10-34 = 0.55505×10-19+34

= 0.55505 x 1015 = 5.55×1014 Hz

(ii) Explain in brief, how do the positively charged protons reside unitedly withen the nucleus.

Ans: Inside nucleus, nuclera force reside which is independent of charge. There is an attractive force between protons which let them to reside untiedly within the nucleus.

(k) (i) When Si is doped with B, what will be the type of resulting semiconductor? Will it possess overall charge neutrality? Under what condition ne nh = ni2?

Ans: P-type

In p-type semiconductor, no. of holes are greater than no of electrons. When nu=nn then ne nn= n21.

Or

(ii) You want to run an electric motor using a self-made full wave rectifier. Draw a neatly labelled diagram to serve your purpose. (Use the symbol Ⓞ for the electric motor).

Ans:

(I) (i) Video signals for transmission of pictures require about 4.2 MHz of bandwidth. A TV signal requires 6 MHz of bandwidth for transmission. Explain in brief. Mention the bandwidth for speech signal.

Ans: The approximate bandwidth of a video stream is a matter of multiplying the resolution of the capture image with the number of frames for the moving image. The required bandwidth for video signals is 4.2MHz.

TV Signal consists of two components voice/speech/ audio and picture.The entire TV signal will have band width ofabout 6MHz. The speech signals frequency range from 300Hz to 3100Hz. Hence bandwidth= 3100-300-2800Hz.

Or

(ii) From the diagram of a pulse shaped signal shown below indicate the following parts of it.

(a) Pulse rise

(b) Pulse duration

(C) Pulse fall

(d) Pulse amplitude

Ans:

3. (a) (i) Inthe following diagram the potential difference between the points A and B is V. Find an expression fortotal currentI. Show that.

Ans:

(ii) R1 and R2 are two resistors.Req(s) and Req9p) are their equivalent resistances when they are connected in (1 series and 2) in parallel. Draw two circuit diagrams for (1) and (2) and show that Req(s)x Req9p) =R1xR2.

(b) (i) Write Biot-Savart law in vector form and mention the direction of magnetic field. Which term in the law works as a vector source and produces the magnetic field? Mention one similarity and one dissimilarity between Biof- Savart law for magnetic field and Coulomb’s law for electrostatic field.

Ans:

Or

Ans: If T be the time period then current I = l/T

Again if v be the orbital velocity

Than V = 2πr/T

∴ I = eu/2πr

Area of the orbit A = πr2

∴ Megnetic moment

M = IA = Iu/2πr x πr2 = evr/2 = emvr/2m = eL/2m

(c) (i) Draw a ray diagram of a compund microscope forming and inverted and magnified image of an object. Which kens in the compound microscope acts as a simple microscope?

If f = 1cm, fe=2cm and L=20cm

Respectively,

calculate the total magnification of the microscope.

Ans: Objective lens acts as a simple microscope

Given: f0=1cm

fe=2cm

L=20cm

But L=Vo+fe

Or, 20=Vo+2 or Vo=18cm

Also, 1/Vo -1/uo= 1/fo or 1/18-1/uo= – 1/1

-1/u01-1/18=17/18

uo=-18/17cm  mo=vo/ |uo|=18/18/17=17

Again,me=1+D/fe=1+25/2=1+12.5=13.5

∴ mo = me=17×13.5=229.5

Or

(ii) You know the phenomenon of scattering of light by the atmospheric particles. Write a few lines about the blue colour of sky and reddish colour of sky in the morning as well as in the evening.

Ans: The blue colour of the sky is due to the scattering of light by small particles of the atmosphere. When light is incident on particles whose side is smaller ther the wavelength of light, it is scattered. According to Rayleigh, the percentage o light scattered is inversely proportional to the fourth power of the wavelength- As the wavelength of blue colour is less than the red colour, it scattered more an the sky appears blue during morning. At evening, the sky appears red because sunlight travels a greater part of the earth’s atmospher and reaches directly the observer’s eye. This light is deprive of the blue colour due to scattering and the remaining colour is red.

(d) (i) In a prism r₁ +r ₂ = A and 6 = i+e-A. When the prism is at the position of minimum deviation Dm, show that, refractive index of the material of the prism is u = sin(A+Dm)/sinA/2 Also derive the expression for Dm for small angled prism.

Ans: When S = Sm

Then r1=r2= r

And i=e

r1=r2=A

⇒r+r =A ⇒2r= A ⇒r=A/2…..(1)

Also Sm=2i-A

or i= Sm+ A/2….(2)

From snell’s law u=sini/ sinr =sin(sm+A/2)/sin (A/2)

For Sm=Dm u=sin(A+Dm/2)

For a prism of small angel,

u= A+Dm2/A/2 = A+DmA

Or, uA=A+Dm or, A(u-1) =Dm

Or

(ii) Which optical device can produce optical dispersion and which property is responsible for dispersion of light? Draw a diagram to show that it is possible to obtain white light again after dispersion.

Ans: Spectrometer: Different components of white light have different wavelength. The wavelength of light changes with the change in refractive index of the material of the prism.

(e) (i) What did de Broglie propose regarding wavelength (A) associated with a particle of mass m? An electron of mass m and charge e is accelerated from rest through a potential V. Show that the de Broglie wavelength of the electron is 3.A=1.227 / √V (Given h √2me=1.227)

Ans:

∴ A= h/√ 2meV = 4/√2me/√v= 1.227/√v

|∴h/√ 2me=1.227

∴ A= h/√ 2meV = 4/√2me/√v= 1.227/√v

|∴h/√ 2me=1.227

Or

(ii) The threshlod frequency for a certain metal is 3.3×1014 Hz. If light of frequency 8.2×1014 Hz is incident on the metal, predict the cutoffb voltage for the photoelectric emission.

Ans: Given Vo= 3.3×1014 Hz

V = 8.2×1014 Hz

We know, V = h/e (v-vo) = 6.63×10-34/1.6×10-19 (8.2-3.3)×1014

= 4.14×10-15 ×4.9×1014 =20.304×10-1=2.03 Volt

(f) (i) Write a few lines on Bohr Model of hydrogen atom.

Ans: Bohr’s basic poztulates of Hydrogen atom:

(1) An electron cannot revolve around the nucleus in all possible orbits. The electron can revolve round the nucleus only in those allowed or permissible orbits the electron is an integral multiple of for which the angular momentum of the  electron is an integral multiple of h/2π.

(ii) An atom radiates energy only when an electron jumps from a stationary orbit of higher evergy to one of lower energy.

Or

(ii) What are the types in which radioactive decay takes place? If N is the number of nuclei in a radioactive sample and ∆N undergo decay in time ∆t, show that N(t)= Noe-At

Ans: Three types of radiactive decay takes place viz (a) a – decaym (b) B -decay (c) y-decay-

From the law of radioactivity,

• dN/dt ∝N

Or- dN/dt=AN

When is A decay constant

Ans: Let us consider a point p at a distance r from a hollow sphere of radius R (r<R) & charge Q. To calculate the field at p we consider a Gaussian surface having radius r. According to Gauss’s law.

O = Q/€0

Again O E.S

∴ E.S=Q/€o

Since charge enclosed by the Gaussian surface is zero i.e Q=10 hence E.S=O⇒E=O

Or

Or

(iii) What is a capactitor? You know that the capacitance of a parallel plate air capacitor is C = eoA/d. What will be its new capacitance C’ of the capacitor if a material d.e.c K is inserted between the plates? Calculate the capacitance of the capacitor shown below.

(b) (i) What do you mean by electromagnetic induction? Name two great experimentalists who carried a long series of experiments on electromagnetic induction. You are given two coils, one galvanometer, one battery and some connecting wires. Describe an experiment that can show the production of electromagnetic induction.

Ans: The phenomenon of producing induced e.m.f and hence induced current in a closed circuit due to the change in magentic field or change in magnetic flux

linked with the closed circuit is known as electromagnetic induction. Faraday and Henry

On closing key ‘K’, current flows in the coil and the galvanometer in coil 2 deflects whereas on opening key k, the galvanometer comes back to its initial position (i.e. pointer goes to zero). This shows that an e.m.t is produced in the coil 2 due to change in current in coil 1.

Or

(ii) What is otional e.m.f.?Deduce an expression for it from the following diagram. If R be the resistance of the loop PQRS at a given instant, what will be the induced current at that instant? A straight conductor of length 0.1m moves with a speed of 10ms perpendicular to a magnetic field of induction 1 Wbm-2. Calculate the induced c.m.f. State Lenz’s law.

Ans: O= BA cos 0=BL x,A= lx

∴ e= do/dt=BL dx/dt =B

E=Blu= 1×0.1×10=1 voly,B=1wbm-2, 1=0.1mu=10ms-1

Or

(iii) Draw a labelled diagram on an AC generator. Show that it produces sinusoidal e.m.f. pr current. What is its frequency of rotation in India?

Ans: Fig Shows the experimental arrangement when the armature rotates the magnetic flux linked with it changes & electric current is induced is it. The emf of the coil is given by

(c) (i) State Huygens’ principle in optics. Using this principle derive the law of reflection of refraction. Which quantity remains unchanged when a light wave suffers reflection of refraction?

Ans: Do your self.

Or

(ii) With the help of Young’s double-slit arrangement to produce interferenc pattern, derive an expression for fringe width B. Mention at least one differen- between the interference fringes and diffraction frings. Are two identical buk of same power and manufactured by the same company coherent sources?

Ans: At constant temperature the potential difference between two ends of a conductor is directly proportional to the current through the conductor.

Or

Write short notes on any two of the following:

(1) Diffraction

Ans: The path difference of the wavefrons reacting the point P=S2P-S1

(2) Doppler effect

Ans: Doppler’s effect: It is due to difference between true frequency and apparent frequency of light when there is relative motion between a source and an observer.

The expression frequency in case of receding is v1=v(1-v/c) with usual rotation.

(3) Polarisation of wave

Ans: Do your self.

(i) In which of the following case (s) the bulb will not glow, explain in very brief. Calculate the forward and reverse resistance of a Si diode from the following

V-I characteristics shown in figure 4.

The bulls will not glow in this case because si diodes have a forward voltage of approximately 0.7 volts.

Forward resistance,

r=ΔV/ΔΙ= 0.8-0.7/(20-10)×10-3=0.1/10×10-3=0.1/10-2=10-1+2=10Ω

Reverse resistance,

r= ΔV/ΔΙ=-10-0/(1-0)×10-6=-10/1×106=-107Ω

Or

(ii) You know that in order to get a regulated de voltage we connect a Zener diode across the output terminals of a rectifier. Draw a circuit diagram comprising a Zener diode, series resistor (Rs) and a load resistor (RL). Indicate the directions of currents Iz Is (=I, the total current) and I, passing through Zener diodem Rs and RL respectively. Choose the correct option from the following. (1) LL.=LZ (2) LL> LZ (3) LL> LZ? How can we achieve the right option?

In such a circuit if Vz= 6.0V, IL=4.0mA, Lz=20mA and unregulated input is 10V, calculate the value of series resistor to work satisfactorily.

Ans: Given Vi = 10V        IL=4.0mA

Vz = 6.0V                  Iz=20mA

VL Vz = 6.0v  and Vs=10V-+6V=4V

AIso, Rs=Vs/Is,RL=VL / IL =6/4×10-3=1.5×103Ω

Is = IL + IZ = (4.0+20mA=24mA

Rs= 4/24×10-3= 1000/6 =166.6Ω

Or

(iii) What is a logic gate? Identify the logic operation carried out by the two circuits shown below and write the truth table for each of them.

Ans: Logic gate is digital circuit consisting of diodes, transistors, ICS etc.which works according to digital relationship between input and output.

A            B          Y

1             1          1

0             1           1

0              0          0

1              0          1

A              B           Y

1               1           1

0               1           0

1                0             0

0                0             0

And gate

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