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## Karnataka 2nd PUC Physics Model Question Paper 3 with Answers

Time: 3.15 Hours

Max Marks: 70

General Instructions:

- All parts are compulsory
- Answers without relevant diagram/figure/circuit wherever necessary will not carry any marks.
- Direct answers to Numerical problems without detailed solutions will not carry any marks.

Part -A

I. Answer the following questions. ( 10 × 1 = 10 )

Question 1.

What is the electric field strength inside a charged spherical conductor?

Answer:

Zero

Question 2.

How does the resistivity of a conductor vary with temperature?

Answer:

Resistivity is directly propotional to temperature

Question 3.

State Gauss’s law in magnetism.

Answer:

The net magnetic flux through any closed surface is always zero.

Question 4.

Name one application of eddy current.

Answer:

Speedometer / Induction furnace any one relevant answer.

Question 5.

What type of lens is used to correct the myopic eye?

Answer:

Concave lens of suitable focal length.

Question 6.

What is the rest mass of photon?

Answer:

Zero

Question 7.

Write one limitation of Bohr’s atom model.

Answer:

It is applicable only for hydrogen and hydrogen like atoms or any other relevant.

Question 8.

Define mean life of a radioactive element.

Answer:

The ratio oftotal life time of all the atoms of radioactive element and the total number of atoms present intially.

Question 9.

Write the logic symbol of NAND gate. A

Answer:

Question 10.

What is attenuation in communication system?

Answer:

The loss of strength of a singnal while propagating through a channel.

Part – B

II. Answer any FIVE of the following questions. ( 5 x 2 = 10 )

Question 11.

Write Coulomb’s law in vector form and explain the terms.

Answer:

\(\overrightarrow{\mathrm{f}_{12}}=\mathrm{k} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{21}\) where q_{1} and q_{2} are two points charges, r is separation between the charges and \(\begin{aligned}

&\wedge\\

&\mathbf{r}_{21}

\end{aligned}\) is the unit vector directed from q_{2} to q_{1}

Question 12.

Mention two limitations of ohm’s law.

Answer:

It is applicable only for the metallic conductor when other physical conditions

are constants..

Not applicable for semiconductor / super conductor/electrolytes.

Question 13.

Write two differences between dia and paramagnetic substances.

Diamagnetic substances

- These substances are feedly repelled by a powerful magnetic.
- Relative permeability of these substances is slightly less than one.

Paramagnetic substances.

- These substances are feedly attracted by a powerful magnetic.
- Relative permeability of these substances is slightly more than one.

Question 14.

Current in a coil falls from 5A to OA in 0.1 s, calculate the induced emf in

a coil if its self inductance is 4H.

Answer:

E = L\(\frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\)

E = 4 x \(\frac{5}{0.1}\) = 200v

Question 15.

Give two uses of UV rays.

Answer:

- In the analysis of the structure of organic compounds.
- In high resolving power microscopes.
- In the study of bacteria.

Any two uses.

Question 16.

Draw the ray diagram for the formation of image in case of a concave mirror when the object is placed at the centre of curvature of a mirror.

Answer:

Question 17.

Distinguish between intrinsic and extrinsic semiconductors.

Answer:

Intrinsic semiconductors

- These are the pure semiconductors.
- Electron density is equal to the hole density.
- Conductivity is low

Extrinsic semiconductors:

- These are the impure semiconductors.
- Electron density is not equal to the hole demsity.
- Conductivity is high.

Any two differences

Question 18.

Draw the block diagram of AM transmitter.

Answer:

Part – C

III. Answer any FIVE of the following questions. ( 5 x 3 = 15 )

Question 19.

Derive the relation between electric field and electric potential.

Answer:

Let A and B be the two points seperated by a distance dx in an electric field

+ IC in the unit positive charge palaced at A. dv in the potential difference between A and B, dx in the distance between A and B

we have, dv=work done in bringing q_{0} charge from A to B.

negative sign indicate that the field and displacement are in opposite direction.

Question 20.

Arrive at the expression for velocity selector using Lorentz force.

Answer:

we have F = qE – Bq v sinθ …..(1)

In the case, Electric and magnetic force are in opposite direction (Flemings Left handrule)

where q in the magnitude of charge

E in Electric field intensity

B in magnetic field intensity

V in the velocity of charged particle

Q in the angle between direction of B & V

In this case, θ90°sin 90° = 1

∴ equation (l)become F = qE -BqV …….(2)

whenF = 0

equation (2) become qE- BqV = 0

qE = BqV

∴ V = \(\frac{E}{B}\)

Question 21.

Mention any three salient features of Hysteresis loop.

Answer;

Hysteresis : It is the phenomenon of lagging of magnetic induction behind the magnetic in subjected to cycle of magnetization.

Retentivity: If in the amount of magnetic induction left in the specimen of ferromagnetic material when magnetic Intensity in reduced to zero.

Coercivity: It is the amount of reverse magnetic intensity required to remove the residual magnetism.

Question 22.

Derive an expression for motional.

Answer:

Consider PQRS in rectangular conducting loop placed in uniform magnetic field B, in the normal to the plane of paper and directed in ward as shown in fig.

‘1’ in the length of the conductor PQ, V in velocity of conductor PQ, ‘X’ in distance moved by the conductor PQ, G in the galvonometer.

When conductor PQ ‘x’ is distance moved by the conductor PQ, G is the Galvanometer.

When conductor PQ moves towards the left, all the free charges of the conductor PQ experience force. This producer the carrent through the loop PQRS an shown in the of sure.

Magnetic flux enclosed by the loop in given by Φ =B x Area of the loop PQRS

Φ = B x1 x x – (1)

From faradays Law e = \(\frac{d \phi}{d t}\) …..(1)

from equation (1) (2) e = \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{Blx})\)

e = \(\mathrm{Bl} \frac{\mathrm{dx}}{\mathrm{dt}}\)

e = BIV [∵ \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{V}\) ]

Question 23.

Mention three power losses in a transformer.

Answer:

power losses in a transformer is

- Loss due in a heating
- Loss due to flux leakage
- Loss due to eddy currents and
- Loss due to hysteresis

Question 24.

Using Huygen’s wave theory of light, show that angle of incidence is equal to angle of reflection in case of reflection of a plane wavefront by a plane surface.

Answer:

Law of reflection of light: Angle of incidence = angle of reflection ⇒ i = r

In above figure, MN in reflecting surface AB in the incident wave front EC in the reflected wave front, i in the angle of incidence, r in the angle of reflection.

Let V be the speed of the light wave in the medium and ‘t’ be the time taken by the secondary waveletes to move from the point B to C Then the distance, BC = vt

In order to contruct the reflected wave front, Let us draw a sphere of radium AE = vt, CE repersents the tangent drawn from the point e to this sphere, This tangent represents the position of the new reflected wave front.

From the figure it is observed that, AE = BC = vt, the triangles EAC and BAC are congruent

∴ i = r. This is the Law of reflection.

Question 25.

Explain three facts of photoelectric effect using Einstein’s photoelectric equation

Answer:

1) The photoelectirc emission is an instantaneous process without any appearent time lag (n 10^{-9} 5 or less ] even when the incident radiation in made exceedingly dim.

2) K-E= h(r-r_{0})

It r = r_{0} Then K.E = 0

ie. the electron just release but K.E = 0

when frequency of incident photon in equal to the Threshold frequency. The maximum K.E. increase as frequency of incident radiation increases.

3) r > r_{0}, KE = positive

ie. photo emission in possible when frequency of incident radiation in greater than threshold frequency.

Question 26.

Explain the working of a zener diode as a voltage regulator.

Answer:

Answer:

The circuit connection are made as shown in the figure. The circuit contain a zenerdiode an unregulated voltage source, a series resistance R_{s} and a load resistance R_{L}

If the input DC voltage increases, the current through R_{s}, voltage drop across R_{s} and the current through zener diode in unaltered.

If the input D.C voltage drop decreases, the current through Rs, voltage drop across Rs and current through zener diode decreases, But the voltage drop across the zener diode is unaltered

Any change (increase / decrease) in the input voltage results in change (increase / decrease) of voltage drop across Rs without any change in voltage across the zener diode, then the zener diode cut as voltage regulator

Part – D

IV. Answer any Two of the following Questions ( 2 x 5= 10 )

Question 27.

Obtain an expression for electric field for an electric dipole along its axis.

Answer:

In the above figure, -q and +q are the magnitudes of the charges of the dipole. 2a is the distance between the charges C is the point on the axcial line, r is the distance between point O and the point C.

Where electric field and dipole moment are in same direction.

Question 28.

Derive an expression for equivalent emf and equivalent internal resistance when two cells are connected in parallel.

Answer:

In the above fig – 1 P & Q are the two cells connected in paralles,

E_{1} &E_{2} are the emfs of P & Q respectively,

r_{1}, &r_{2} are the internal resistances of P & Q respectively

I_{1} and I_{2} are the currenrts sent by the cells P & Q respectively,

I is the main current sent by the two cells

I = i_{1}+ i_{2}

Let V_{AB} be the potential difference between the points A &B.

By the definition emf of the cell, E = V+Ir

⇒ V = E -Ir …… (1)

Potential difference across the terminals of second cell is given by,

(1) ⇒ V_{AB }= E_{1} – I_{1}r_{1} = E_{1} – V_{AB}

When the combination is replaced by an equivalent cell ofemfEP and an equivalent internal resistance r_{p} (figure -2)

we get, V_{AB} = EP – Irp ……. (6)

on comparing (5) & (6) we get

Question 29.

Derive an expression for the magnetic field at a point along the axis of circular current loop.

Answer:

In the figure, R is the redius of the circular loop and I is the current in a circular loop.

P is the point on the axis, \(\overrightarrow{\mathrm{dt}}\) is the length of the current element.

\(\overrightarrow{\mathrm{r}}\) is the distance between p and the current element at Q.

Q is the angle between the axis and the direction of AB. The magnetic field at p due to the current element dl^{μ} at Q is given by

The magnetic field ‘d B’ is resolved into two components along x and y axes. y components of dB will be cancelled. But x components of dB are added up.

∴ Magnetic field at p due to the loop.

From triangle POQ, Loss = =\(\frac{\mathrm{R}}{\mathrm{r}}\) …….(4)

V. Answer any two of the following Questions: ( 2 x 5 = 10 )

Question 30.

What is interference of light? Arrive at the conditions for constructive and destructive interference by assuming the expression for intensity.

Answer:

The phenomenon of redistribution of light energy due to superposition of two (or) more light waves is called interference.

I = 4a^{2}\(\left(\frac{\phi}{2}\right)\) is the intensity

conditions for constructive interference phase difference:

In the constructive interference. I maximum , I is maximum when cos \(\left(\frac{\phi}{2}\right)\) = ±1 ie

When

\(\left(\frac{\phi}{2}\right)\) = 0,π,2π, 3π

Φ = 0, 2π, 4π, 6π ……..

Φ = 2π n = 0, 1, 2, 3 ……..

For consturctive interference the phase difference should be integral multiple of 2π

Path difference – we have δx = δΦ x \(\frac{\lambda}{2 \pi} \) ⇒ δx = 2π x \(\frac{\lambda}{2 \pi}\)

⇒ δx = nλ where n =0,1,2,3 ………..

For constructive interfereace the path difference should be integral multiple of λ

Conditions for destructive interference : In the destructive interference. I minimum.

I is minimum, when Cos \(\left(\frac{\phi}{2}\right)\) = 0 i.e. when

\(\left(\frac{\phi}{2}\right)\) = π/2, 3π/2, 5π/2

Φ = π, 3π, 5π……. ⇒ Φ = (2n +1)π

where n = 0,1,2,3…………..

For destructive interference the path difference should be odd multiple of π/2

We have

δx = δΦ x \(\frac{\lambda}{2 \pi} \) ⇒ δx = (2n + 1) x \(\frac{\lambda}{2 \pi}\)

⇒ δx = (2n + 1) \(\frac{\lambda}{2}\) where n = 0, 1, 2, 3………

Question 31.

Derive an expression for total energy of an electron in hydrogen like atom assuming radius of the orbit.

Answer:

In the figure +ze is the charge of the nucleus, r is the radius of the circular orbit of an electron, v is the orbital velocity of the electron.

The electron posses kinetic energy due to orbital motion and potential energy due to the presence of charge on the nucleus

Total energy , E = KE +PE → (1)

For stable orbit, centripetal force = Electrostatic force = \(\frac{m v^{2}}{r}=\frac{1}{4 \pi E_{0}} \frac{Z c \cdot c}{r^{2}}\)

Electrostatic force according to Columb’s Law:

Where m is the mass of an electron and e is the charge of an electron . On dividing the above equation by 2 on both sides we get.

Question 32.

Explain the working of npn trasitor as an amplifier in ce mode.

Answer:

The circuit connection are mode or shown in the figure – 1. The input circuit is forced biased and output circuit is recerve biased. The weak signal is superimpossed over this bias voltage V_{BB.}

In the absence of input AC signal :

The collection current flowing through the load resistance R_{c.}

Produces voltage drop V = I_{c} R_{c} in it

∵ Vout = V_{CE} = V_{CC} – I_{c} R_{c}……. (1)

In the presence of input signal :

During positive half cycle of a.c, the input circuit is more forward biased and base current I_{B} increase . As I_{B} increase. I_{c} increase B times according to the relation I_{c} = BI_{B} [where B is the amplification factor] As I_{c} increase .

I_{c} R_{L} increase. Therefore, Vout is negative from equation (1). That is the input positive half cycle is amplified in opposite direction.

During negative half cycle of a.c the base current I_{B} decrease . As I_{B} decrease, I_{c }decreases, B times. As I_{c} decreases I_{c} R_{L }decreases. Therefore, Vout is positive from equation (1). direction. The phase difference between input and output signal is 180° it is shown in the figure – 1

Voltage gain:

It is the ratio of change in output voltage to the change in input voltage

VI. Answer any three of the following. ( 3 x 5 = 15 )

Question 33.

Two point charges 5 x 10^{-8}C and -3 X 10^{-8}C are located 16cm apart. At what points on the line joining the two charges is the electric potential zero?

Answer:

Question 34.

Determine the current through the galanometer in the circuit given P = 2Ω, Q = 41Ω, R = 8Ω,G = 0Ω E = 5V& r = 0

Answer:

Applying KLR for the ABDA .We get 2I_{1} +10I_{q} – 8I_{2}

Applying KLR for the loop BCDA. We get

4I_{1} – 4OI_{q} – 4I_{2} – 4I_{q} – 10 I_{q} = 0

4I_{1 }– 18I_{q }– 4I_{2} = 0 (2)

Applying KLR for the loop

ABCA Via E, we get

6I_{1 }– 4I_{9} =5 ……..(3)

By solving the equation (1) & (2) we get

6I_{1 }– 46I_{g} …….(4)

By solve eq (3) & (4) we get

46I_{9 }– 4I_{9} =5

I_{9} =0,19A

Question 35.

Calculate the resonant frequenly in LCR circuit with inductance 2.0 H, capacitance 32MF & resistance 10Ω what is the Q value of this circuit?

Answer:

Question 36.

An object of size 3 cm is placed 14cm in front of a concave lens of focal length 21 cm. Calculate position and size of the image.

Answer:

Question 37.

Consider the fission process of 92 U by fast neutrons. In one fission event no neutrons emitted and final end products after beta decay of primary fragments are _{58}Ce^{140 }and _{44}RU^{99 }Calculate Q for this process.

Mass of _{92}U^{238} = 238.05079 U, Mass of _{58}Ce^{140} = 139.90543 u

Answer: