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Karnataka 2nd PUC Physics Model Question Paper 1 with Answers
Time: 3 Hrs 15 Min
Max. Marks: 70
- All parts are compulsory.
- Answers without relevant diagram/figure/circuit wherever necessary will not cany any marks
- Direct answers to the Numerical problems without detailed solutions will not carry any marks.
Part – A
I. Answer all the following questions ( 10 x 1 = 10 )
State Faraday’s law of electromagnetic induction.
I law : Whenever a magnetic flux linked with the coil, a change in emf is induced in it.
This induced emf lasts as long as the change in magnetic flux continues.
Write an expression for the displacement current.
The expression for displacement current is given by
What is an electric dipole ?
A pair of equal & opposite charges are seperated by a small distance is called electric dipole.
Draw the circuit symbol of p-n-p transistor.
How can the resolving power of a telescope be increased ?
Shorten the wavelength of light is used.
In the following nuclear reaction identify the particle X.
N → P + ē + X
In the above reaction, X – represents
Define magnetisation of a sample.
It is defined as “the net magnetic moment developed per unit volume of the material”.
How does the power of a lens vary with its focal length ?
Power of a lens is inversely proportional to its focal length.
ie. p =
What is cyclotron ?
Cyclotron is a device used to accelerate heavy charged particle like proton a – pue etc to high speed.
Give the wavelength range of X – rays.
1 x 10-10 to 3 x 10-8 m.
II. Answer any five of the following questions. ( 5 × 2 = 10 )
The current in a coil of self inductance 5 mH changes from 2.5A to 2.0A in 0.01 second. Calculate the value of self induced emf
What is a toroid ? Mention an expression for magnetic field at a point inside a toroid ?
A solenoid bent in to the form of a closed ring is called toroid.
∴ B = µrµ0nI = µnI
where n → is number of turns per unit length.
I → is current through the toroid.
What are isotopes & isobars ?
- Nuclei are having same atomic number but different mass number are called isotopes.
- Nuclei are having same mass number but different atomic number are called isobars.
Draw the variation of magnetic field (B) with magnetic intensity (H) when a ferromagnetic material is subjected to a cycle of magnetisation.
Mention two applications of polaroids.
- It is used in sunglasses, goggles etc.
- It is used in windows, head lamps of buses, cars, aeroplane etc.
Write the logic symbol & truth table of NAND gate.
Where A & B arc inputs and y is output.
Write two properties of eletric field lines.
- They never intersect each other.
- They normal to surface of the conductor.
What is myopia? how to connect it?
In this human vision defect a person can see near by object clearly but he can not see distant object clearly.
It can be corrected by using a concave less of suitable focal length.
Part – C
III. Answer any five of the following questions. (5 × 3 = 15)
What is a transformer ? Mention two sources of energy loss in a transformer.
A transformer is a device used to change AC voltage by using the principle of mutual induction.
- Loss due to flux leakage
- Loss due to eddy current
Write three characteristics of nuclear forces ?
- Nuclear forces is a strongest force in nature.
- They are short range forces
- They are charge independent
Derive the expression for energy . stored in a charged capacitor.
Charging a capacitor means transfering of eletron from one plate to other plate of capacitor.
∴ work will be done because electrons are moving against the electrostatic force of electric field. This work done is stored as a electrostatic potential energy.
Consider a capacitor of capacity is charged from DC source of ‘ V’ volts work done in adding a small charge dq to a capacitor is given by dw = v. dq
∴ Total workdone in changing a capacitor to Q is
This work done is stored and a electrostatic potential energy.
What is an amplifier? Draw the simple circuit of transistor amplifier in CE mode.
An amplifier is a device which is used for increasing the strength of a weak input ac signal.
Mention the types of transmission media.
The modes of propogation of em waves are :
- Ground wave propogation
- Sky wave propogation
- Space wave propogation
Derive an expression for drift velocity of free electrons in a conductor
Consider a conductor under the influence of electric field E. The force experienced by a free electron in a conductor is given by F = -eE, -ve sign shows that the divertion of F & E are in opposite direction.
Since the average time elapsed between successive collision is called relaxation tome (τ)
Explain briefly the coil & magnet experiment to demonstrate electromagnetic induction.
The expt. arrangement consists of coil containing ‘n’ no of turns. The free end of the coil is connected to sensitive
- When a magnet is at rest, no deflection in the galvanometer.
- When a north pole of magnet is moving towards the coil the galvanometers shown deflention in one direction. ||| ly when a north pole of the magnet is moving away from the coil the galvanometer shows deflection in opposite direction.
- The deflection of the galvanometer is large, when magnet is moving towards the coil (or) away from the coil.
From the expt it is clear that if there is a relative motion between the magnet coil, an emf is induced across the coil & induced current flows in the coil.
Write three properties of ferromagnetic materials.
- Ferromagnetic substance is strongly attracted by a magnet.
- The relative permeability of a ferromagnetic substance is very large ie. Mr >> 1.
- The magnetic susceptability of a ferromagnetic substance is high +ve value.
Part – D
IV. Answer any two of the following questions ( 2 × 5 = 10 )
Deduce the condition for balance of a wheat stone’s bridge using Kirchhoff’s rules.
Wheatstone’s bridge is a device used to determine unknown resistance. It consists of 4 arms of resistances P, Q, R & S, a galvanometer of resistance ‘G’ & a battery with plug key. Let ‘I’ be the main current & splits into branch current I1 & I2 respectively.
Applying KVL to the mesh ABDA
I1P + IgG – I2R = 0 ………..(1)
|||ly to the mesh BCDB
(I1 – Ig)Q – (I2 + Ig)S – IgG = O …… (2)
The wheatstone bridge is said to be balanced when no current flowing through the galvonometer.
i e Ig = 0
eqn (1) I1P – I2 R = 0
I1P – I2R …………(3)
eqn (2) is I1 Q – I1 S = 0
I1Q – I2S ………….. (4)
This is the balanced condition of Wheatstone bridge.
Derive an expression for the force between two parallel conductors carrying currents hence define ampere.
Consider two identical parallel conductors each of length seperated by a distance ‘r’. Let I1 is the current flowing through Ist conductor & I2 is the current flowing through IInd conductor.
The magnetic field is developed in second conductor due to flow of current in I conductor is given by B1 =
Since the second condutor carries a current, it experiences a mechanical force due to M.F _ developed in first condustor.
|||ly the first conductor experiences the same force due to presence of M.F developed by the second conductor, The divertions of there forces are obtained using Fleming left hand rule are as shown in the figure. When a current flows through a parallel conductor in same direction are attracted current flows through a opposite direction they repel each other.
Definition of ampere : Ampere is the current flowing through conductor of unit length seperated by a distance of lm, it experiences a force of 2 × 10-7N.
Derive an expression for eletric field due to an electric dipole at a point on the axial line.
Consider a dipole having dipole length ‘zl Let ‘p’ be a point at a distance ‘r’ from the centre of the dipole. Where electric field intensity is required. Imagine a unit +ve charge is placed at p. An electric field intensity is the resultant of two charges & it is given by.-
If 1 << r i.e dipole is short then l2 is neglected compared to μ2 .
V. Answer any two of the following questions ( 2 × 5 = 10 )
Write the experimental observations of photo electric effect.
The following one the other
experimental results of photoelectric effect
- Photoelectric effect is an instantaneous process.
- The minimum frequency is required to remove the electron from the metal surface is called threshold frequency. If the frequency of incident light is less than threshold frequency them their is no photoeletric effect.
- The minimum energy required to remove the electron from the metal surface is called work function.
ie. w = hυ0
- The intensity of incident light is directly proportional to photoelectric current.
- The kinetic energy of emitted electron is directly proportional to frequency of incident light.
- A -ve potential is applied to anode plate just stop the photoelectric current is called stopping potential.
What is rectification ? With relevant circuit diagram & waveforms explain the working of p-n junction diode as a full wave retifier.
The process of conversion of AC in to DC is called rectification.
The p- n junction is used as a full wave rectifier are as shown in the figure.
An Ac in put is applied to primary of the transformer in first half cycle of AC, end A of secondary centre taping transformer is +ve & end B is -ve diode D1 is in forwand bias & hence current flows but diode D2 is reverse bias & hence current does not flow in next half cycle of AC end A of secondany centre taping transformer is -ve & end B is +ve diode D1 is reverse bias & hence current does not flow but diode D2 is in forward bias & hence current flows this process will continue & hence out put DC signal is obtained across the load resistance.
Derive an expression for equivalent focal length of two thin lenses kept in contact.
Consider two thin lenses of focal length f1 &f2 are in constant let ‘o’ be a point object placed on the principle axis. A ray of light from the object after passing through the combination of lenses the refronted light ray meet at I. ∴ An image is formed at I. Thus the formation of image at I can take splace in 2 stages.
Stage (i) : In the absence of lens L2 the referacted light ray is meet at I1 ∴ Thin less formula can be written as
Stage (2) : In the absence of lens L,, the refracted light ray is meet at p now it acts as a virtual object & hence read final image is formed at I ∴ Thin lens formula can be written as
VI. Answer any three of the following question ( 3 × 5 = 15 )
In young’s double slit experiment, fringes of certain.
Width are produced on the screen kept at a certain distance from the slits. When the screen is moved away from the slits by 0.1m, fringe width increases by 6 × 10-5 m. The seperation between the slits is 1mm. calculate the wavelength of the light used.
When two wapasitors are connected in series & connected across 4 kv line the enevgy stored in the system is 8j. The some capacitors, if connected in parallel across the some line, the energy stored is 36 j. Find the individual capacitances.
Given, v = 4 kv = 4 × 103 V = 4000 volts.
In series connection, energy stored
In parallel connection energy stored
C1C2 = (C1 + C2)10-6 = 4.5 × 1 μF
C1C2 = 4.5μF
We have (C1 – C2)2 = (C1 + C2)2 – 4C1C2
= (4.5)2 – 4 × 4.5 = 20.25 – 18 = 2.25 μf
C1 – C2 – 1.5µf + C2 =4.5 μf
2C1 = 6.0 μf (∴ C1 =3(μf)
also, C1 + C2 =4.5
<c2 = 4.5 – 3 = 1.5μf>
Calculate the shortest & longest wavelength of balmer series of hydrogen atom (Given r = 1.097 × 107m-1)
The longest wavelength of balmer series of H – atom is given by
The shortest wavelength of balmer series is given by
Calculate the resonant frequency & Q- factor of a series L-e-R circuit containing a pure inductor of inductance 4H, capacitor of capacitance 27 μf & resistor of resistance 8.4 Ω
Given L = 4H,C= 27μF,R = 8.4Ω, f0 = ?
Q = factor = ?
The expn for resonant frequency is
a)Three resistors of resistances 2 Ω, 3 Ω & 4 Ω are combined in series what is the total resistance of the combination.
b) If this combination is connected to a battery of emf 10V & negligible internal resistance obtain the potential drop across each resistor.
Given :R1 = 2Ω , R2 = 3Ω, R3 = 4Ω,
E = 10V,
a) In series combination effective resistance R3 is given by
Rs = R1 + R2 + R3 = 2+ 3 + 4 = 9Ω
b) Current through the circuit
1 = 1.11 A
∴Potential drop across R1 = I , R1 1.11 × 2 = 2.22 = 2.22 Volts
Potential drop across R2 = I R2
= 1.11 × 3 = 3.33 = 3.33 Volts
Potential drop across R3 = I R3 =1.11 × 4 = 4.44 = 4.44 Volts