2nd PUC Physics Model Question Paper 2 with Answers

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Karnataka 2nd PUC Physics Model Question Paper 2 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. All parts are compulsory
2. Answers without relevant diagram/figure/circuit wherever necessary will not carry any marks.
3. 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 electirc dipole.
Answer:
Two equal and opposite charges separated by small distance constitutes an electric dipole.

Question 2.
State Ampers circuital law.
Answer:
The line integral of magnetic field around any closed path in free space is equal to times the net current enclosed by that path.

Question 3.
Mention any one use of electromagnet.
Answer:
Used in elecrtric cells / loud speakers / telephone.

Question 4.
Name the law which gives polarity of induced emf.
Answer:
Lenz’s law

Question 5.
Give the wavelength range of x- rays
Answer:
0.1 A⁰ to 100 A⁰

Question 6.
Mention the expression for displacement current
Answer:
Displacement current: id = \(\varepsilon_{0} \frac{\mathrm{d} \varphi_{\mathrm{E}}}{\mathrm{dt}}\)

Question 7.
How does the power of a lens vary with its focal length?
Answer:
Inversely

Question 8.
Who proposed wave theory of light ?
Answer:
Huygen

Question 9.
Write the SI unit of activity.
Answer:
Bequerel or dis / s.

Question 10.
Draw the circuit symbol of npn transistor
Answer:

Part – B

II. Answer any Five of the following questions: ( 5 × 2 = 10 )

Question 11.
Write any two properties of electric field lines.
Answer:

1. Electic field lines start from +ve charge and terminate on -ve charges
2. No two field lines intersect each other

Question 12.
Mention an expression for force on . charge moving in uniform magnetic field and explain the terms.
Answer:
F = qVB sinθ
where q’ in charge of particle
v in velocity of particle
B in magnetic field
θ is angle between velocity and direction of Magnetic field.

Question 13.
Define the terms retentivity and coercivity
Answer:
Retentivity : The value of intensity of magnetisation of a material, when the magnetising field Is redued to zero, in called retentivity of material.
Coercivity : The value of reverse magnetising field required so as to reduce residual magnetism to zero is called coersivity of the material.

Question 14.
What are eddy currents? Give one use of eddy currents.
Answer:
The eddy currents are induced currents in metallic conductor ,when the magnetic flux linked with the conductor changes use; It in used in dead beat Galvonometer

Question 15.
What is a thin prism? Write its deviation expression.
Answer:
Thin prism is prism whose refraction angle in small
It is deviation of expression in d = A(n-l)

Question 16.
Represent the plane polarised light and unpolarised light diagrammatically
Answer:

Question 17.
Define half – life of radiocative element and mention expression for it.
Answer:
It is time taken by radiactive element to disintegrate the half of its initial number of atoms It half life is T_1/2 = \(\frac{0.693}{\lambda}\)

Question 18.
Write two differences between intrinsic semiconductors.
Answer:
Intrinsic semiconductor :

1. It has equal no of holes and electron
2. Its conductivity in Low

Extrinsic semiconductor :

1. It has unequal no of holes and electron
2. Its conductivity is high

Part – C

III. Answer any five of the following questions: ( 5 × 3 = 15 )

Question 19.
Derive an expression for an electric potential energy .of a system of two point charges in the absence of electric field.

Answer:
Consider two point charges q₁ and q₂ separated by a distance ‘r’
The electric potential due to the charge q₁ at a distance ‘r’ is v = \(\frac{1}{4 \pi \mathrm{E}_{0}} \frac{\mathrm{q}_{1}}{\mathrm{r}}\)

The work done in moving charge q₂ from infinity to point B in the electric field of charge q₂ in given by
w = v₁ x q₂ = \(\frac{1}{4 \pi \mathrm{E}_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}}\)
This work done in bringing the two charges to their respective position in stored on the potential energy u = \(\frac{1}{4 \pi \mathrm{E}_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}}\)

Question 20.
State and explain ohm’s law and hence define ohm.
Answer:
ohm’s Law statement : At contant temperature, the flow of current through conductor is directly proportional to the potential difference between ends of conductor, provided other physical condition are remain constant.
Explanation :

Consider a conductor AB, Let V be the potential difference applied across in ends It be the flow of current through conductor. According to ohms Law
I ∝ V or V ∝ I
V = RI
where R in proportionality constant called Resistance of conductor

Ohm: The Resistance of conductor is Said to be one ohm, if its potential difference across conductor is 1 volt when 1 ampre of current flows through it

Question 21.
Write any three properties of ferromagnetic materials.
Answer:

1. Ferromagnetic materials are strongly attracted by the magnets
2. They obey curies Law above curie temperature
3. Their magnetisation in large and positive

Question 22.
Derive an expression for motional emf, induced across the end of a conduction rod moving in a uniform perpendicular magnetic field
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 fiq.
‘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 × Area of the loop PQRS
Φ = B × 1 × x – (1)
From faradays Law e = \(\frac{d \phi}{d t}\) …..(1)
from equation (a) (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 any three sources of energy losses in transformer.
Answer:
power losses in a transformer is

1. Loss due in aheating
2. Loss due to flux leakage
3. Loss due to eddy currents and
4. Loss due to hysteresis

Question 24.
Mention three properties of nuclear forces.
Answer:

1. Nuclear forces are strongest force in nature
2. Nuclear forces are charge independent
3. Nuclear force are spin dependent.

Question 25.
Distinguish between conductors, insulators and semiconductors on the basis of band theory.
Answer:
Conductors: In conductor the conduction band and fore if a small potential difference applied across conductor, valence electron can move from valence band to conduction band.

Insulators : on insulators, there is a large energy gap between conduction band and valence band If large potential difference applied across Insulator, electron cannot Jump from valence band to conduction band.

Semiconduction : In semiconduction, there in a small energy gap between conduction band and valence band Conduction band | Conduction band Conduction band

Question 26.
Draw the block diagram of amplitude modulation transmitter.
Answer:

Part – D

IV. Answer any two of the following questions: ( 2 × 5 = 10 )

Question 27.
Obtain an expression for electric field due to an infinitely long straight uniformly charged wire using Gauss law.
Answer:
Consider an infinity long thin straight wire having a uniform change density λ. Let ‘p’ be a point at a distance v from the wire, the wire have axis of symmetry.

To find the total electric field at p imagine a cylindrical Gaussian surface with radium ‘r’ and length 1. The electric flux passes only through the curved surface of the cylinder. The electriflux through only the curved surfaced of the cylinder, At the curved surface of the cylinder the electric field \(\overrightarrow{\mathrm{E}}\) in normal at every point on the cylinder.
The surface area of the curved point in = 2πrl
∴ The flux through the Gaussian surface = flux through the curved surface of the cylinder
i .e Φ = E × 2πrl – (1)
By Gauss Law, the total charge enclosed by the Gaussian surface is Φ  = \(\frac{1}{\mathrm{E}_{0}}, \mathrm{q}\) – (2)
from eqn (1) & (2) E × 2πrl = \(\frac{1}{\mathrm{E}_{0}}, \mathrm{q}\)
but q = λl
∴ E × 2πrl = \(\frac{1}{\mathrm{E}_{0}}\) λl
E = \(\frac{\lambda}{2 \pi \epsilon_{0} r}\)
In vector form \(\overrightarrow{\mathrm{E}}=\frac{\lambda}{2 \pi \epsilon_{0} \mathrm{r}} \hat{\mathrm{n}}\)
where n is the radial unit vector in the plane normal to the wire.

Question 28.
Derive an expression of effective resistance connected in parallel.

Answer:
Consider two resistors R₁ and R₂ are connected in parallel, across A and B. Let V in the common potential across the combination. Let I be the main current and I₁ & I₂ are the current through R₁ & R₂respectively
Then by definition I = I₁ + I₂ — (1)
From ohm’s law , V = R₁ or I = \(\frac{\mathrm{v}}{\mathrm{R}}\)

It R_p in equivelent resistance of the parallel combination then
V = R_pI or I = \(\frac{\mathrm{v}}{\mathrm{R}_{\mathrm{p}}}\) … (3)
From equ(2) & (3) \(\frac{\mathcal{V}}{\mathrm{R}_{\mathrm{P}}}=\mathcal{V}\left[\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right]\)

\(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\left[\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right]\)
Then for parallel combination, the reciprocal of equivelent resistance equal to the sum of reciprocal of the individual resistance.

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 ofdB will be cancelled. But x components of dB are added up.
∴ Magnetic field at p due to the loop.

V. Answer any two of the following questions:  (2 × 5 = 10 )

Question 30.
Derive the expression for the refrective index of the material of the prism in terms of angle of prism A and angle of minimum deviation D.

ABC → principle section of the prism
A → angle of prism
n → R . I of the prism
l₁ and r₁ → are the angle of incidence and angle of refraction at Q

r₂ & I₂ → are the angle of incidence and angle of refraction at R
PQ →incident ray, QR → refracted Ray RS → emergent ray
Total deviation by the prism in
d=(i₁ -r₁ ) + (i₂ – r₂ )
d = (i₁ + i₂ ) – (r₁ + r₂) …… (1)
From the quadricateral AQNR
\(A + ∠QOR = 180……(2)
From the triangle
QOR r₁ + r₂ + ∠QOR = 180° …. (3)
From eqn (2) and (3) we get
A = r₁ + r₂…….(4)
From eqn (1) & (2) we get
d= i₁ + i₂ – A ……(5) .
At minimum deviation d = D, i₁ + i₂ = i and r₁ + r₂ = r
equation (5) becoms d= i + i – A =2i – A
∴ I = [latex]\frac{A+D}{2}\) …. (6)
equation (4) becoms A = 2r
∴r = A/2 ……(7)
From Snell’s Law n = \(\frac{\sin i}{\sin r}\)
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin r A / 2}\)

Question 31.
What in photoelectric effect? using Einstein’s photoelectric equation, Explain three experiments results

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.
A 400 pF capacitor by a 100 V d c supply is disconnected from the supply and connected to another uncharged 400pF capacitor calculate the loss of energy.
Answer:
Initial energy stored in capacitor
∪i = \(\frac{1}{2}\) CV2 = \(\frac{1}{2}\) qv = \(\frac{1}{2}\) × 400 x 10^-12 ×(100)²
∪i = 200 × 10^-12 × 10⁴
∪i = 2 × 10^-6 J
After connecting charged capacitor to uncharged capacitor of same capacity \(\frac{1}{2}\) of the charge will flows.

Question 34.
Three resistors 1Ω,2Ω and are connected in series, what is the total resistors of the combination? if the combination is connected to a battery of emf 12 v and negligible internal resistance then obtain the potential drop across each resistor.
Given:Rj = 1Ω, R2 = 2Ω, R3 =3Ω
emf = 12v, R_s = ? v₁=? v₂ = ? v₃ =?
effective resistance of 3 Resistorn are connected in serin
R_s =R₁ + R₂ + R₃ = 1+2+3 = 6Ω
I = \(\frac{E}{R+r}\) = \(\frac{12}{6}\) = 2A

Potential difference across
R₁,V₁ = IR₁ =2X1 =2V
Potential difference across
R₂,V₂ = IR₂ =2X2 =4V
Potential difference across
R₃,V₃=IR₃ =2X3 =6V

Question 35.
A serin L C R circuit with R = 20Ω, L = 1.5 HandC= =35 μF is connected to a variable frequency of 200 AC supply when the frequency of the supply is equal to the natural frequency of the circuit, what is the average power transferred to the circuit, in one complete cycle.
Answer:
Given:
R =20 μ,<=1.5H and
R =20 μ,<=1.5HVrmn =200v p =?
Impedence of the circuit

Question 36.
Monochromatic light of wavelength 5000 A0from a narrow slit is incident on the double slit. If the separation of 10 fringes on the screen 1cm away is 2 cms. Find the slit separation.
Answer:
Given: A 5000 Å ,D =2cm nB = 1cm d = ?

Question 37.
Calculate the shortest and longest wavelength of Balmer series of hydrogen atom. Given R = 1.097 x = 10⁷m
Answer:

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