2nd PUC Physics Model Question Paper 4 with Answers

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Karnataka 2nd PUC Physics Model Question Paper 4 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.
Two point charges are separated by some distance, repel each other with a force F. What will be the force if distance between them is halved?
Answer:
New force = 4F. (Since force a 1/d² and new distance = d/2)

Question 2.
In a Wheat stone’s network four resistors with resistances P, Q, R and S are connected in a cyclic order. Write the balancing condition of the network.
Answer:
P/Q = S/R

Question 3.
A current flows in a conductor from west to east. What is the direction of the magnetic field at a point below the conductor?
Answer:
Towards North

Question 4.
State Gauss law in magnetism.
Answer:
‘The net magnetic flux through any closed surface is zero”

Question 5.
Name the phenomenon in which an emf is induced in a coil due to the change of current in the same coil.
Answer:
Self – induction

Question 6.
What do you mean by dispersion of light?
Answer:
The phenomenon of splitting of light into its component colours is known as dispersion.

Question 7.
How does the de-Broglie wavelength of a Charged particle changes when accelerating potential increases?
Answer:
The de Broglie wavelength decreases when accelerating potential increases.
(Since λ ∝ \(\frac{1}{\sqrt{\mathrm{V}}}\) )

Question 8.
What is the significance of the negative total energy of an electron orbiting round the nucleus?
Answer:
The electron is bound to the nucleus.

Question 9.
A radioactive element ₉₂X²³⁸ emits one α  particle and one β⁺ particle in succession. What is the mass number of new element formed?
Answer:
Mass number of the new element = 234. (Since the mass number decreases by 4 units in alpha decay but it does not change in beta decay)

Question 10.
What is sky wave propagation?
Answer:
The radio waves emitted from the transmitter antenna reach the receiving antenna after reflection by the ionosphere of earth atmosphere. This mode of propagation is known as sky wave propagation.

Part – B

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

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

1. Field lines are continuous curves without any breaks
2. Two field lines cannot cross each other
3. Electrostatic field lines start at positive charges and end at negative charges
4. They do not form closed loops.

Question 12.
What are the limitations of ohm’s law? Ohms law fails if
Answer:

1. V depends on I non-linearly
2. The relation between V and I depends on the sign of V for the same absolute value of V.
3. The relation between V and I is non – unique.

Question 13.
Mention the expression for time period of oscillation of small compass needle in a uniform magnetic field. Explain the terms.
Answer:
Time period: T – 2 \(\pi \sqrt{\frac{\mathrm{I}}{\mathrm{m} \mathrm{B}}}\)
Where I – moment of inertia, B – magnetic field and m – magnetic moment of compass needle.

Question 14.
Give two applications of eddy currents.
Answer:
Magnetic braking in trains, Induction furnace, electromagnetic damping, electric power meters.

Question 15.
What is displacement current? Mention its need.
Answer:
The current due to changing electric field (time – varying electric field) is’called displacement current. To remove the inconsistency of Ampere’s circuital law.

Question 16.
Write the two conditions for total internal reflection.
Answer:

1. The ray must travel from a denser medium to a rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle, for the pair of media and for the given colour of light.

Question 17.
Give the circuit symbol and truth table for OR gate.
Answer:

Question 18.
Draw the block diagram of AM receiver.
Answer:

Part – C

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

Question 19.
Obtain an expression for electric potential energy of a system of two point charges in the absence of external electric field.
Consider two charges q, and q2 with position vector \(\overrightarrow{\mathrm{r}_{1}}\) and \(\overrightarrow{\mathrm{r}_{2}}\) relative to some origin. There is no external field, so work done in bringing q₁ from infinity to \(\overrightarrow{\mathrm{r}_{1}}\) is zero. This charge produces a potential in space given by V₁ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1}}{\mathrm{r}_{1 \mathrm{P}}}\)

Work done in bringing charge q₂ from infinity to the point \(\overrightarrow{\mathrm{r}_{2}}\) is q₂ times the potential \(\overrightarrow{\mathrm{r}_{2}}\) at due to q₁ i.e., W = q₂ V₁ = \(\mathrm{q}_{2}\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1}}{\mathrm{r}_{12}}\right)\) ,

Where r₁₂ is the distance between q₁ and q₂. This work gets stored in the form of potential energy of the system Thus, the potential energy of a system of two charges q₁ and q₂ is U = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}\right)\)

Question 20.
Derive the expression for the magnetic force experienced by a current canylng conductor.
Answer:
Consider a rod of uniform cross-sectional area A and length 1.
Let the number density of mobile charge carriers (of each charge q) in it be n. Then the total number of mobile charge carriers in it is nAl. For a steady current I in this conducting rod, each mobile carrier has an average drift velocity \(\overrightarrow{\mathrm{V}_{\mathrm{d}}}\) In the presence of an external magnetic field \(\overrightarrow{\mathrm{B}}\) , the force on these carriers

\(\overrightarrow{\mathrm{F}}=(\mathrm{n} \mathrm{Al}) \mathrm{q}\left[\overrightarrow{\mathrm{v}}_{\mathrm{d}} \mathrm{x} \overrightarrow{\mathrm{B}}\right]\) But nq \(\overrightarrow{\mathrm{v}_{\mathrm{d}}}=\overrightarrow{\mathrm{j}}\) , is the current density and \(\left|\left(n q \bar{v}_{d}\right)\right|\) A = I is the current. Thus, force on the conductor

\(\left.\overrightarrow{\mathrm{F}}=\left[\left(\mathrm{n} \mathrm{q}\overrightarrow{\mathrm{v}}_{\mathrm{d}}\right) \mathrm{A}\right]\right] \times \overrightarrow{\mathrm{B}}[\overrightarrow{\mathrm{j}} \mathrm{Al}] \mathrm{x} \overrightarrow{\mathrm{B}}=\mathrm{I}(\overrightarrow{\mathrm{I}} \times \overrightarrow{\mathrm{B}})\)

Where \(\overrightarrow{\mathrm{i}}\) is a vector of magnitude 1, the length ofthe rod, and with a direction identical to the current I.

Question 21.
Write three properties of paramagnetic substances.
Answer:

1. When placed in a non-uniform magnetic field, apiece of paramagnetic substance willtend to move from weak field to strong field. i.e., they get weakly attracted to a magnet.
2. They get weakly magnetised when placed in an external magnetic field,
3. The individual atoms (or ions or molecules) of a paramagnetic material possess a permanent magnetic dipole moment oftheir own.
4. Paramagnetic substances do not posses net magnetisation.
5. The magnetisation of a paramagnetic material is inversely proportional to the absolute temperature T.
6. X is small and positive for paramagnetic materials.

Question 22.
Obtain the expression for the magnetic energy stored in a coil (solenoid).
Answer:
When a current is established in a solenoid (coil), work has to be done against the back emf. This work done is stored in the form of magnetic energy in the coil.
For a current I in the coil, the rate of work done (power) is : P = \(\frac{\mathrm{d} \mathrm{w}}{\mathrm{dt}}=|\varepsilon| \mathrm{I}\) (since power P=VI) But we know that, \(\varepsilon=\mathbf{L} \frac{\mathrm{d} \mathrm{I}}{\mathrm{d} \mathrm{t}}\)
Therefore, \(\frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}=\mathrm{L} \mathrm{I} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}} \Rightarrow \mathrm{d} \mathrm{W}=\mathrm{L} \mathrm{Id} \mathrm{I}\)
Therefore, the total work done in establishing a current I is given by

This work is stored in the coil in the form of magnetic potential energy, U =\( \frac{1}{2}\) Li²

Question 23.
What is resonance in series LCR circuit? Derive the expression for resonant angular frequency.
Answer:
In a series LCR circuit for a particular frequency of the applied ac the current in the circuit becomes maximum. This is known as electrical resonance. At resonance, inductive reactance (X_L) = capacitive reactance (X_C) ⇒ \(\omega_{0} L=\frac{1}{\omega_{0} C}\) Thus, resonant angular frequency \(\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}\)

Question 24.
Derive the expression for resultant displacement and amplitude when two waves having same amplitude and a phase difference Φ superpose.
Answer:
Let the displacement produced by source S₁ is given by y₁ = a cos (ωt) that by S₂ be y₂ = acos(ωt + Φ) where Φ is the phase difference between the waves.
The resultant displacement:

Question 25.
Give the de-Broglie’s explanation of Bohr’s second postulate.
Answer:
For an electron moving in n^th circular orbit of radius r_n, the total distance is = 2 πr_nCircumference of a stationary Bohr orbit of radius r_n is equal to integral multiple of wavelength of matter waves 2πr_n = n λ ………(1)
The de Broglie wavelength of the electron moving in the if1 orbit: \(\lambda=\frac{h}{m v}\) ….. (2)
From equation (1) & (2) , 2πr_n = \(\frac{\mathrm{nh}}{\mathrm{mv}}\) ie., \(\mathrm{mvr}_{\mathrm{n}}=\frac{\mathrm{n} \mathrm{h}}{2 \pi}\)
But, angular momentum of the electron is L = mvr_n. Hence L = \(\frac{\mathrm{n} \mathrm{h}}{2 \pi}\)

Question 26.
Distinguish between extrinsic and intrinsic semiconductors.
Intrinsic Semiconductor

1. Semiconductor without doping (pure form)
2. Number of free electrons (ne) is equal to the number of holes (nh)
3. Conductivity is low
4. Conductivity is due to both electrons and holes

Extrinsic Semiconductor

1. Semiconductor doped with impurity atoms (impure form)
2. Number of free electrons (ne) and number of holes (nh) are unequal.
3. Conductivity is high
4. Conductivity is mainly due to majority charge carriers

Part – D

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

Question 27.
Obtain the expression for the electric field at any point on the equatorial plane of an electric dipole.
Answer:

Let a point P be at a distance r from the centre of the dipole (O) Dipole length AB = 2a,
Using Pythagoras theorem, the distance of the point P from the charges is, AP = BP = d = \(\sqrt{r^{2}+a^{2}}\)
Using, the expression for electric field due to a charge q at a distance d from it, E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{d^{2}}\) , The magnitude of the electric field at P due to the charge +q ie E_+q = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}+a^{2}}\) The magnitude of the electric field at P due to the charge -q is E -q = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}+\mathrm{a}^{2}}\) The total field P, \(\overrightarrow{\mathrm{E}}_{+\mathrm{q}}\) and \(\overrightarrow{\mathrm{E}}_{-\mathrm{q}}\) are resolved in to components parallel to \(\overrightarrow{\mathrm{E}}_{+\mathrm{q}}\) and perpendicular to \(\overrightarrow{\mathrm{p}}\) (or p ) is from -q to +q.
The components normal to the dipole axis (E_+qsinθ) and (E_+qsinθ) cancel out as they are equal in magnitude and opposite in direction.
The total electric field is

Question 28.
Assuming the expression for drift velocity, derive the expression for ‘ conductivity of a material σ = ne² τ/m
Answer:

Question 29.
Using Biot Savart’s law, derive the expression for magnetic field at a point on the axis of a circular current loop.
Answer:

Let a circular loop with its centre at the origin O, radius R carrying a steady current I and having axis as X-axis. Letx be the distance of P from the centre O of the loop.
Consider a conducting element \(\overrightarrow{\mathrm{dl}}\) of the loop. The magnitude of the magnetic field dB due to \(\overrightarrow{\mathrm{dl}}\) is given by Biot-Savart’s law,
\(|\overrightarrow{\mathrm{dB}}|=\left(\frac{\mu_{0}}{4 \pi}\right)\left(\frac{1|\overrightarrow{\mathrm{d} | \mathrm{i} x}|}{r^{3}}\right)\) and wer have r² = X² + R²
Any element of the loop will be perpendicular to the displacement vector from the element to the axial point.

The direction of magnetic field d \(\vec{B})\) is perpendicular to the plane containing \((d \vec{l})\) and \(\overrightarrow{\mathbf{r}} \cdot \mathrm{d\vec{B }}\) has a X – component dB_x and a component perpendicular to x-axis dB ⊥ (in XZ plane). When the components perpendicular to the X – axis are summed over a complete loop, they cancel out. Thus, only the X-component dB cosθ remains.
The net magnetic field along X-direction can be obtained by integrating (summing)
dB_x = dB cosθ ……………..(2)
From equations (1) and (2),

The summation of elements dl over the loop gives 2πR, the circumference of the loop. i.e., Σdl =2πR
Thus, the magnetic field at P due to the entire circular loop is

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

Question 30.
Using Huygen’s wave theory of light, derive Snell’s law of refraction.
Answer:

Let pp¹ represent the surface separating medium -1 and medium – 2. Let V₁ and v₂ be the speed of light in medium -1 and medium – 2 respectively. Consider a plane wave front AB incident in medium -1 at angle T on the surface pp¹
According to Huygens principle, every point on the wave front AB is a source of secondary wavelets.

Let the secondary wavelet fromB strike the surface pp¹ at C in a time t. Then BC = v₁τ. The secondary wavelet from A will travel a distance v₂τ as radius; draw an arc in medium 2. The tangent from C touches the arc at E. Then AE=v₂τ and CE is the tangential surface touching all the spheres of refracted secondary wavelets. Hence, CE is the refracted wave front. Let r be the angle of refraction. In the above figure, ∠BAC =i = angle of incidence and ∠ECA = r = angle of refraction
BC = V₁τ and AE = v₂τ
From triangle B AC, sin i = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) and from triangle sin r = \(\frac{\mathrm{AE}}{\mathrm{AC}}\)

Since v₁ is a constant in medium -1 and v₂ is a constant in medium – 2 \(\frac{\sin i}{\sin r}=\frac{V_{1}}{v_{2}}\) = constant …….(*)
Now, refractive index (n) of a medium: n = \(\frac{c}{v}\) or v = \(\frac{c}{n}\) , where c – speed of light in vacum For the first medium: \(\mathrm{v}_{1}=\frac{\mathrm{c}}{\mathrm{n}_{1}}\) and for the second medium: \(\mathrm{v}_{2}=\frac{\mathrm{c}}{\mathrm{n}_{2}}\) ⇒ \(\frac{V_{1}}{V_{2}}=\frac{n_{2}}{n_{1}}\)
(*) becomes \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\) or n₁ sin i = n₂sin r. This is the Snell’s law of refraction.

Question 31.
Write the five experimental observations of photoelectric effect.
Answer:

1. For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light.
2. For a given photosensitive material and frequency of incident radiation, saturation current is proportional to the intensity of incident radiation but stopping potential is independent of its intensity
3. For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation called the threshold frequency, below which no photoelectrons emission.
4. Above the threshold frequency, the stopping potential or the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation, but is independent ofits intensity.
5. The photoelectric emission is an instantaneous process.

Question 32.
Explain the working of semiconductor diode when it is forward biased. Draw the I-V characteristics for both forward bias and reverse bias of semiconductor diode.
Answer:

I-V Characteristic Curve:
When an external voltage V is applied across a semiconductor diode such that p – side is connected to the positive terminal of the battery and n-side to the negative terminal it is said to be forward biased.

The direction of the applied voltage (v) is opposite to the built – in potential V₀ in the semiconductor diode. As a result, the width of depletion layer decreases and the barrier height is reduced.

If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, the current will be sma ll. If the applied voltage increase to large value, the barrier height will be reduced, the current increases.

VI. Answer Any THREE of the following questions: ( 3 x 5 = 15 )

Question 33.
A 600 pF capacitor is charged by a 200 V supply. Calculate the electrostatic energy stored in it. It is then disconnected from the supply and is connected in parallel to another uncharged 600 pF capacitor. What is the energy stored in the combination?
Answer:
Energy stored initially in the capacitor:
U = \(\frac{1}{2}\) CV² = \(\frac{1}{2}\) (600 -10^-12) (200 )² = 12 x 10^-6 J
hitial charge on the single capacitor is Q = CV = 12 x 10^-8 coulomb.
When other (second) uncharged capacitor connected to the first, charge is distributed among them Capacitors are having equal capacitances, hence charges on each capacitor are equal = Q₁= Q/2 = 6×10^-8 Now, energy stored in each capacitor :

Therefore, the total energy of the combination – 6 x 10^-6 J

Question 34.
Two cells of emf 3 V and 2V and internal resistances 1.5 Ω and 1Ωrespectively are connected in parallel across 3Ω resistor such that they tend to send current through resistor in the same direction. Calculate potential difference across 3Ω resistor.
Answer:

Potential difference across 3Ω resistor, V = IR = (2/3) (3) = 2 V
Alternate Method – Applying KirchofTs rules also this problem can be solved.

Question 35.
A 60V, 10W lamp is to be run on 100V, 60 Hz ac mains. Calculate the inductance of a choke coil required to be connected in series with it to work the bulb.
Answer:

Question 36.
A convex lens of focai length 0.24m and of refractive index 1.5 is completely immersed in water of refractive index 1.33. Find the change in the focal length of the lens.
Answer:

Dividing equation (1) by (2), f_gw = 0.939 ∴ Change in focal length = 0.699 m

Question 37.
A given coin has a mass 3.0 gram, Calculate the nuclear energy that would be required to separate all the neutrons and proteins from each other. Assume that the coil is entirely made of ₂₉Cu⁶³ atoms of mass = 62.92960u. Given Avogadro number = 6.023 x 10²³ mass of proton m_p = 1.00727 u and mass of neutron mn = 1.00866 u.
Answer:
Mass defect in each copper nucleus = 29 m_p + 34 m_n -M_cu
(29 x 1.00727) + (34 x 1.00866) – 62.92960
∆m = 0.57567u
Energy required to separate all neutrons and protons in one nucleus (is binding energy)
= ∆m x 931,5MeV
E_b =0.57567 x 931.5 = 536.2 MeV
Number of atoms in the copper coin: N = \(\frac{6.023 \times 10^{23} x^{3}}{63}\) = 2.868 x 10²²
∴ Total energy required to separate all neutrons and protons in the coin
= E_b x N
= 536.2 x 2.868 x 10²² = 1.538 x 10²⁵ MeV

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