2nd PUC Physics Model Question Paper 1 with Answers

Students can Download 2nd PUC Physics Model Question Paper 1 with Answers, Karnataka 2nd PUC Physics Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka 2nd PUC Physics Model Question Paper 1 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.
Draw the electric lines of force in the case of two positive point charges separated by a small distance.
Answer:

Question 2.
Name the charge carriers in metallic conductors.
Answer:
Free electrons or electrons

Question 3.
A coil perpendicular to a uniform magnetic field is rotated by 1800. What is the change in the flux through it?
Answer:
The flux becomes double.

Question 4.
Write the expression for displacement current.
Answer:
i = \(\varepsilon_{0}\left(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\right)\)

Question 5.
Mention one common method of generating X – rays.
Answer:
Bombarding a heavy metal target with high speed electrons.

Question 6.
Write the expression for magnifying power of a telescope in terms of focal lengths.
Answer:
The magnification in the case of a
compound microscope is m = \(\frac{\mathrm{L}}{\mathrm{f}_{0}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\) whereas in the case of a telescope m = \(\frac{f_{0}}{f_{e}}\) where f₀ is the focal length of object piece and f₀ is the focal length of the eye piece.

Question 7.
What is the outcome of Davisson Germer experiment?
Answer:
The theoritical and experimental values are compared. This confirms that electrons behave as a wave.

Question 8.
How does nuclear radius of an atom depend on its mass number?
Answer:
R – R₀A^1/3 i.e., R ∝ A^1/3 where ‘A’ is the mass number.

Question 9.
Who discovered the phenomenon of photoelectric effect?
Answer:
Photoelectric effect was discovered by Heinrich Hertz.

Question 10.
What is demodulation ?
Asnwer:
The process of separation of low (audio) frequency from the modulated carrier wave is called demodulation.

Part – B

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

Question 11.
Write two properties of an electric charge.
Answer:

1. Charge is conserved and universal
2. Charge is quantised
3. Charge is additive
4. Charge is invariant of speed.

Question 12.
What is electrostatic shielding? Mention its one application.
Answer:
Electric lines of force do not pass through a metallic conductor. This property is known as Electrostatic shielding.

Question 13.
State KirchholFs rules of an electrical network.
Answer:
Kirchhoff’s current law: The algebraic sum of currents entering and leaving a node of an electrical network is zero.
Kirchhoff’s voltage law: The algebraic sum of‘IR’ products is equal to the algebraic sum of emfs, in that mesh of an electrical network.

Question 14.
Mention two properties of a diamagnetic material.
Answer:
Diamagnetic substances

1. These substances are repelled by a strong magnetic field.
2. These substances move from stronger to weaker magnetic field.
3. 0 ≤ μ_r < 1
4.  χ ’ is low but negative and of the order -10^-9 to -10^-6
5. Susceptibility is independent of temperature.

Question 15.
What are eddy currents? Mention one application of eddy currents.
Answer:
If a solid metal piece is in a changing magnetic field then due to electromagnetic induction currents are induced in the metal. These currents are called eddy currents. Eddy currents are used in dead beat galvanometer, inductive furnace, odometer and electromagnetic brakes. Eddy currents produce heat and brake the electrical insulation.

Question 16.
Draw the ray diagram to construct an image when an object is placed between the principal focus and pole of a concave mirror.
Answer:
A simple microscope is a magnifier of a small object placed or held in front of it and viewed into it from the other side.
An erect and enlarged image of an object may be obtained by placing the object in between the optic centre and the focus of tile convex lens.

Question 17.
Mention two necessary conditions for doping.
Answer:

1. The size of the dopant atom should be almost the same as that of the semiconductor atom.
2. The dopant atoms should not distort the crystal lattice.

Question 18.
Draw the block diagram of a generalized communication system.
Answer:

Part – C

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

Question 19.
Arrive at an the expression for electric potential at a point due to a dipole and hence write expression for electric potential at a point on the dipole axis.
Answer:
Let \(\left|\overrightarrow{\mathrm{r}}_{1}\right|=\mathrm{r}_{1},|\overrightarrow{\mathrm{r}_{2}}|=\mathrm{r}_{2},|\overrightarrow{\mathrm{r}}|=\mathrm{r}\)
Electric potential at P is

Question 20.
Obtain the expression for effective capacitance of three capacitors connected in parallel.
Answer:
Capacitors are said to be connected in parallel when (i) the p.d. across each of these capacitors remains the same, (ii) the net charge ‘Q’ divides in the direct ratio of their capacitances.

i.e., Q = Q₁ + Q₂ + Q₃ where Q = CV
i.e., CV = C₁V + C₂V + C₃V .
or . C_p = C₁ + C₂ + C₃.
For ‘n’ capacitors, C_p = C₁ + C₂ + C₃ + ………..C_n
For ‘n’ identical capacitors, C_p = nC
Statement: ‘The equivalent capacitance of a number of capacitors connected in parallel is equal to the algebraic sum of their individual capacitances’.

Question 21.
What is a cyclotron? Draw its schematic labeled diagram.
Answer:
Cyclotron is a device to accelerate charged particles or ions to high energies. Cyclotron uses both electric field and magnetic field to increase the energy of charged particles.

Question 22.
Explain briefly the coil – magnet experiment to demonstrate the phenomenon of electromagnetic induction.
Answer:
a) Magnet and coil experiment:
If one end of a magnet is suddenly introduced into a coil connected to a sensitive galvanometer it shows a momentary deflection.

This shows that a momentary current and hence an instantaneous emf is produced in the circuit, the direction of the current depends on the pole towards the coil. The current produced is called induced current. If the magnet is withdrawn suddenly a sudden deflection is again observed but in the opposite direction. The deflection is greater if the magnet is introduced or withdrawn quickly. No induced emf isobtained when there is no relative motion between the coil and the magnet.

b) Coil and coil experiment: Consider two coils P and S placed closed to each other.
The coil P is connected to a ceE through a key K is called primary coil, the other coil S connected in series with a sensitive galvanometer G is called the secondary coil. On pressing K a momentary deflection is noticed in a galvanometer. When the key is released the galvanometer deflects in the opposite direction. However, no induced emf is obtained when steady current flows through the primary as long as secondary coil is at rest.

Question 23.
What is meant by alternating current? Define its amplitude and time period.
Answer:
The current that varies simple harmonically and reverses its direction once every half cycle is called alternating or bidirectional current.
Peak value AC is the maximum value attained by AC voltage/current in either half cycle. Period of AC is the time taken by it to complete one cycle.

Question 24.
Using Huygen’s wave theory of light, show that the angle of incidence is equal to angle of reflection in case of reflection of a plane wave by a plane surface.
Answer:

Let V be the speed of wave in the medium, τ be the time taken by the wavefront to move from the point B to C.
BC = y τ
Let AE be the radius of an are drawn at E and EC be the tangent drawn at E.
AE = BC = vτ
The triangles AEC and CBAare congruent and therefore \(\hat{i}=\hat{\mathbf{r}}\) Hence the law of reflection is proved.

Question 25.
Mention three experimental observations of photoelectric effect.
Answer:

1. Photoemission is instantaneous
2. For every photoemissive surface there is a minimum frequency of the incident radiation below which there is no photoemission. This minimum frequency is called treshold frequency.
3. For frequency greater than threshold, the strength of the photoelectric current is directly proportional to the intensity of in-cident radiation.
4. The maximum K.E of the photo electron increases linearly with the frequency of the incident radiation.
5. There exists a minimum -ve potential called stopping potential for which anode repels the electrons completely.

Question 26.
Classify metals, semiconductors and insulators based on the band theory of solids.
Answer:
In the case of conductors, CB and VB overlap with each other. Free electrons are the charge carriers. There are a number of electrons of conduct electricity at room temperature.
In the case of insulators, there is a wide gap between CB and VB. Forbidden gap energy is greater than 5 eV. There are no electrons in the CB in conduct electricity.

Part – D

IV. Answer any TWO of the following questions: ( 2 x 5 = 10 )

Question 27.
Obtain the expressions for effective emf and effective internal resistance when ‘n’ different cells are connected in parallel
Answer:

Let I₁, I₂, I₃…… In represent branch currents.
Let V be the common potential, so that

Comparing this with the terminal potential difference,
V = E_eq (2) = Ir_eq
V = E_eq – \(\frac{I}{\left(\frac{1}{r_{e q}}\right)}\) …… (2)
Comaparing (1) & (2)

Also, I = \(\frac{E_{e q}}{R+r_{e q}}\)
For two cells in parallel combination

For two cells in parallel combination

Note For ‘n’ number of identical cells
E-emf of each cell
r – internal resistance of each cell

Question 28.
State and explain Biolt-Savart’s law and write its mathematical equation in vector form.
Answer:
Biot-Savart’s law (Laplace rule) states that the magnetic flux density (dB) at a point due to a small current element (idl) of a current carrying conductor is directly proportional to (i) the current 7’, (ii) the length of the element ‘d/’, (iii) Sine ofthe angle made between the current element and the line joining the point and is inversely proportional to (iv) square of the distance between the point and the element.

I – Current
idl- Current element
θ – Current element
\(\overrightarrow{\mathrm{r}}\) Distance vector connecting dl and P

Question 29.
Assuming the expression for magnetic field at a point on the axis of a circular current loop, show that a long current carrying solenoid acts as a bar magnet.
Answer:
We know that the magnetic field at a point on the axis of a circular loop.

Where, m = IA = Magnetic moment.

Note :

1. As an analogous to the electric field intensity \(\overrightarrow{\mathrm{E}}=\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{2 \overrightarrow{\mathrm{p}_{\mathrm{e}}}}{x^{3}}\) ….. (2)

2. Comparing (1) and (2) we note that by 1 substituting μ₀ as \(\frac{1}{\varepsilon_{0}}\) and \(\overrightarrow{\mathrm{B}}\) by \(\overrightarrow{\mathrm{E}}\), \(\overrightarrow{\mathrm{m}}\) by \(\overrightarrow{\mathrm{p}_{\mathrm{e}}}\), one can relate the magnetic and electric fields due to the corresponding dipoles.

V. Answer any TWO of the following question : ( 2 x 5 = 10 )

Question 30.
Derive the expression for the refractive index of the material of a prism in terms of the angle of the prism and angle of minimum deviation.
Answer:
BC – Base of the prism
AB and AC – Refracting sides
representing the plances
ABC – Principle section of Prism.
PQ – incident ray.
RS – emergent ray
d₁ – angle of deviation d₁ = i₁ – r₁
dv – angle of deviation d₂ = i₂ – r₂
\(U \hat{T} S\) – d – total angle of deviation or net
deviation.
i.e., d = d₁ + d₂ (exterior angle = sum of two opposite interior angles)

NM is normal to AB at Q
N’M is normal to AC at R
i₁ – angle of incidence
i₂ – angle of emergence.
D = angle of minimum deviation From figure (1)
From the quadrilateral AQMR
\(A \widehat{Q} M \cong A \widehat{R} M=90^{\circ}\) (NM, N’M are normal to AB and BC)
So that A + M = 180° ……….. (1)
∴ AQMR is a cyclic quadrilateral from the triangle QMR ……… (2)
r₁ + r₂ + \(\widehat{\mathrm{M}}\) = 180° (Property of a ∆)
From (1) and (2)
A + M = r₁ + r₂ + M
i.e., A = r₁ + r₂ ……… (3)
For non-symmetric condition i₁ ≠ i₂,
d₁ ≠ d₂also net deviation d ,
d = d₁+d₂
d = i₁ – r₁+i₂-r₂
∴ d = i₁+i₂-(r₁+r₂)
using (3) we write
d = i₁ +i₂ – A ……….. (4)
from the fig. we note that for non-symmetric condition for two angles of incidence, angle of net deviation is the same. However for a symmetric condition. it = i₁ = i₂ and r₁ = r₂ = r net deviation becomes the minimum angle of deviation. In this case the refracted ray will be parallel to the base.
∴ for symmetric condition
(4) can be written as D – 2i – A
i.e., i = \(\left(\frac{A+D}{2}\right)\) …….(5)
(3) can be written as
A = 2 r
or r = \(\left(\frac{A}{2}\right)\) …….(6)
From Snell’s law of refraction at AB
n = \(\frac{\sin i_{1}}{\sin r_{1}}\)
but for symmetric refraction i₁ = i₂ = i and r₁ = r₂ = r ….. (7)
using (5), (6) and (7) we get,
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
This formula is applicable for symmetric refraction only.

Question 31.
State radioactive decay law. Show that N = N₀ ^-λt for a radioactive element
with usual notations.
Answer:
Ratio active law : The rate of disintegration of radio act ive atoms present in the sample of an element is directly proportional to the number of radioactive atoms present at that instant.
i.e., \(\frac{\mathrm{d} \mathrm{N}}{\mathrm{d}^{t}}\) ∝ N
or \(\frac{\mathrm{d} \mathrm{v}}{\mathrm{d}^{t}}\) = -λt where λ is known as the disintegration constant -ve sign indicates that with the passage of time, the no. of radio active nuclei/atoms decrease.
Hence \(\frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}\)= -λdt

Integrating both sides we get . loge = -λt + C ,
Applying the initial condition for t = 0, N = N₀
We get, C = log₀N₀
i.e., log_e \(\left(\frac{N}{N_{0}}\right)\) e^-λt
Hence N = N₀e^-λt
A graph ‘N’ v/’s time ‘t’ gives the exponential curve as shown in the figure.

Question 32.
What is a solar cell ? Explain its working with a diagram and draw the I-V characteristics.
Answer:
A solar cell is a semiconductor device which converts light energy mto electrical energy.

I_C – Short circuit current
V_oc -Open circuit Voltage

VI. Answer any “THREE of the following questions ; ( 3 x 5 = 15 )

Question 33.
The elctrostatic force on a metal sphere of charge 0,4 µ C due to another identical metal sphere of charge – 0.8 µ C in air is 0.2N. Find the distance between the two spheres and also the force between the same two spheres when they are brought into contact and then replaced in their initial positions.
Answer:

Question 34.
In the given circuit, calculate the (i) effective resistance between A and B (ii) current through the circuit and (iii) current through 3 Ω resistor.
Answer:
Equivalent circuit may be written as


Branch current in 3Ω resistor is 4.0A

Question 35.
A resistor, an inductor and a capacitor are connected in series with a 120V, 100Hz acsource. Voltage leads the current by 35° in the circuit. If the resistance of the resistor is 10 Ω and the sum of inductive and capacitive reactances is 17 Ω, calculate the self inductance of the inductor.
Answer:
Given R= 10 Ω
X_L + X_C=17Ω
given Φ = 35°
L = ?
We know that tan Φ = \(\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}\)
Hence X_L – X_C = R tan Φ
i.e.,X_L – X_C = R tan Φ
= 10 x tan 35°
= 10 x 0.7002
= 7.002
i.e., X_L – X_C 17Ω
Since X_L + X_C = 17
and X_L – X_C = 7
Solving we get 2X_L = 24
∴ X_L =12Ω
But X_L = 2 πfL = 12Ω
∴ L = \(\frac{12}{2 \times 3.142 \times 100}\)= 0.019H
Self inductance of the coil = 0.019H

Question 36.
A beam of light consisting of two wavelengths 500nm and 400nm is used to obtain interference fringes in Young’s double slit experiment. The distance between the slits is 0.3mm and the distace between the slits and the screen is 1.5m. Compute the least distance of the point from the central maximum, where the bright fringes due to both the wavelengths coincide.
Answer:
Given λ₁ = 500 x 10^-9 m
X2 = 400 x 10^-9m
d = 0.3 x 10^-3m
D= 1.5m
W.K.that x = \(\frac{n \lambda D}{d}\)
So that x₁ = x₂
i.e., n₁λ₁ = n₂λ₂

Here 5th bright fringe of λ₂, coincides with 4th bright fringe of λ₁.

= 10⁴ x 10^-9 x 10³m = 10^-2m
x = 0.01m

Question 37.
The first member of the Balmer series of hydrogen atom has wavelength of 656.3nm. Calculate the wavelength and frequency of the second member of the same series. Given, c = 3 x 10⁸m/s.
Given λ = 656.3 x 10^-9m
c = 3 x 10⁸ ms^-1
For balner series
n₁ = 2 First member n₂ = 3