Maharashtra Board Class 12 Physics Sample Paper Set 1 with Solutions

Maharashtra State Board Class 12th Physics Sample Paper Set 1 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Physics Model Paper Set 1 with Solutions

Section A

Question 1.
Select and write the correct answers to the following questions:

(i) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform vertical circular motion, under gravity. Minimum speed of a particle is 5 m/s. Consider following statements.
(P) Maximum speed must be 5\(\sqrt{5}\) m/s.
(Q) Difference between maximum and minimum tensions along the string is 60 N. Select correct option.
(a) Only the statement P is correct
(b) Only the statement Q is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

(ii) If pressure of an ideal gas is decreased by 10% isothermally, then its volume will
(a) decrease by 9%
(b) increase by 9%
(c) decrease by 10%
(d) increase by 11.11%
Answer:
(d) Increase by 11.11%

(iii) A standing wave is produced on a string fixed at one end with the other end free. The length of the string:
(a) must be an odd integral multiple of \(\frac{\lambda}{4}\)
(b) must be an odd integral multiple of \(\frac{\lambda}{2}\)
(c) must be an odd integral multiple of λ
(d) must be an even integral multiple of λ
Answer:
(a) Must be an odd integral multiple of λ/4

(iv) A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential capacitance respectively are:
(a) Constant, decreases, decreases
(b) Increases, decreases, decreases
(c) Constant, decreases, increases
(d) Constant, increases, decreases
Answer:
(a) Constant, decreases, decreases

(v) Kirchhoff’s first law, i.e., ΣI = 0 at a junction, deals with the conservation of ‘ ‘
(a) charge
(b) energy
(c) momentum
(d) mass
Answer:
(a) Charge

(vi) A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
(a) BLv
(b) BLv²
(c) \(\frac{1}{2}\)BLv
(d) \(\frac{2 \mathrm{Bl}}{\mathrm{v}}\)
Answer:
(a) BLv

(vii) In a series LCR circuit, the phase difference between the voltage and the current is 450. Then the power factor will be:
(a) 0.607
(b) 0.707
(c) 0.808
(d) 1
Answer:
(b) 0.707

(viii) Which of the following properties of a nucleus does not depend on its mass number?
(a) Radius
(b) Mass
(c) Volume
(d) Density
Answer:
(d) Density

(ix) A charged particle is in motion having initial velocity V_ρ when it enters into a region of uniform magnetic field perpendicular to V_ρ. Because of the magnetic force the kinetic energy of the particle will:
(a) remain unchanged
(b) get reduced
(c) increase
(d) be reduced to zero
Answer:
(a) Remain unchanged

(x) A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth’s equator. There the magnetic flux lines of the Earth’s magnetic field are horizontal, with the field of 1.3 × 10^-4 T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east, are:
(a) 14 × 10^-4 N, downward
(b) 20 × 10^-4 N, downward
(c) 14 × 10^-4 N, upward
(d) 20 × 10^-4 N, upward
Answer:
(d) 20 × 10^-4N, upward

Question 2.
Answer the following questions:

(i) Why are curved roads banked ?
Answer:
To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is titled inward i.e., the outer side of the road is raised above its inner side. This is called banking of road.

(ii) Define athermanous substances and diathermanous substances.
Answer:
Substance that don’t allow transmission of infrared radiation through them are called athermanous substance diathermanous body in a body which transmits all the incident radiation without absorbing or reflecting.

(iii) State and explain the principle of conservation of angular momentum.
Answer:
The angular momentum of a body is conserved if the resultant external torque on body is zero. This law is used by a figure skater to increase their speed of rotation for a spin by reducing the body’s moment of inertia.

(iv) Define linear simple harmonic motion.
Answer:
The linear periodic motion of a body, in which the restoring force is always directed towards the mean position and its magnitude is directly proportional to the displacement from the mean position.

(v) If the density of oxygen is 1.44 kg/m³ at a pressure of 10⁵ N/m², find the root mean square velocity of oxygen molecules.
Answer:
p = 1.44 kg/m³, P = 10⁵ N/m²
∴ The root mean square velocity of oxygen molecules,
V_rms = \(\sqrt{\frac{3 P}{\rho}}\) = \(\sqrt{\frac{3 \times 10^5}{1.44}}\) m/s
= \(\sqrt{2.083 \times 10^5}\) = \(\sqrt{20.83 \times 10^4}\)
= 4.564 × 10² m/s

(vi) Which property of soft iron makes it useful for preparing electromagnet?
Answer:
An electromagnet should become magnetic when a current is passed through its coil but electromagnet lose its magnetism once the current is switched off Hence ferromagnetic core used for an electromagnet should have high permeability and low retentivity i.e., it should be magnetically soft.

(vii) A ceiling fan having moment of inertia 2 kg-m² attains its maximum frequency of 60 rpm in 2π seconds. Calculate its power rating.
Answer:
Given: ω₀ = 0, ω = 2πn = 2π × 2 = 4π rad/s
α = \(\frac{\left(\omega-\omega_0\right)}{t}\) = \(\frac{(4 \pi-0)}{2 \pi}\) = 2 rad/s²
P = r.ω = lα.ω = 2 × 2 × 4π = 16π watt ≈ 50 watt.

(viii) Write ideal gas equation for a mass of 7g of nitrogen gas.
Answer:
Ideal gas equation, PV = nRT
Here, n \(=\frac{\text { mass of the gas }}{\text { molar mass }}\) = \(\frac{7}{28}\) = \(\frac{1}{4}\)
Therefore, the corresponding ideal gas equation is
PV = \(\frac{1}{4}\)RT

Section B

Attempt any Eight of the following questions:

Question 3.
Why is the surface tension of paints and lubricating oils kept low?
Answer:
The surface tension of paints and lubricating oils kept low for better surface coverage.

Question 4.
Why should a Carnot cycle have two isothermal two adiabatic processes?
Answer:
With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.

Question 5.
Mention the conditions under which a real gas obeys ideal gas equation.
Answer:
A real gas obeys the ideal gas equation when:

1. Temperature is very high.
2. Pressure is very low.

Question 6.
What are primary and secondary sources of light?
Answer:
The source that emit tight on their own are called primary source. Some sources are not self-luminous, i.e., they do not emit tight on their own but reflect or scatter the light incident on them. Such sources of light are called secondary sources.

Question 7.
Why two or more mercury drops from a single drop when brought in contact with each other?
Answer:
A spherical shape has the minimum surface area to volume ratio of all geometric forms. When two drops of a liquid are brought ¡n contact, the cohesive forces between their molecules coalesce the drops into a single
larger drop.

Question 8.
What do you mean by electromagnetic induction? State Faraday’s Law of induction.
Answer:
The phenomenon of production of emf in circuit by a changing magnetic flux through the circuit is called electromagnetic induction.
Faraday’s first law: Whenever there is a change in magnetic flux associated with a circuit, an emf is induced in the circuit.
Faroes’s second law: The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.

Question 9.
State the importance of Davisson and Germer experiment.
Answer:
The Dbvisson and Germer experiment are probably one of the most important experiments ever since it verified that de Broglie’s “matter wave” hypothesis applied to matter (electrons) as well as light. From this emerged modern quantum theory, the most stupendous revolution in physics of all time.

Question 10.
A system releases 125 kJ of heat while 104 kJ of work is done on the system. Calculate the change in internal energy.
Answer:
Given: Q = -125 kJ, W = – 104 Id
∆U = Q – W = – 125 kJ – (-104 kJ)
= (-125 + 104) kJ = – 21 kJ

Question 11.
A violin string vibrates with fundamental frequency of 440 Hz. What are the frequencies of first and second overtones?
Answer:
Given, n = 440 Hz
The first overtone,
n₁ = 2n = 2 × 440
= 880 Hz
The second overtone,
n₂ = 3n = 3 × 440 = 1320 Hz.

Question 12.
White light consists of wavelengths from 440 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55?
Answer:
Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let λ_a be the wavelength of light in vacuum.
λ = \(\frac{\lambda_a}{n}\) = \(\frac{400 \times 10^{-9} \mathrm{~m}}{1.55}\) = 258.06 × 10^-9 m
λ = \(\frac{\lambda_a}{n}\) = \(\frac{700 \times 10^{-9} \mathrm{~m}}{1.55}\) = 451.61 × 10^-9 m
The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Question 13.
A galvanometer has a resistance of 25 Ω and its full scale deflection current is 25 μA. What resistance should be added to it to have a range of 0-10 V?
Answer:
Given: G = 25 mA
Maximum voltage to be measured is V = 10 V.
The galvanometer resistance G = 25 Ω
The resistance to be added in series,
X = \(\left(\frac{V}{I_G}\right)\) – G = (\(\frac{10}{25}\) × 10^-6) – 25 = 399.975 × 10³ Ω

Question 14.
Calculate the value of magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5A. (Given: µ₀ = 4π × 10^-7 Wb/Am)
Answer:
Given: I = 5A, a = 0.02 m, \(\frac{\mu_0}{4 \pi}\) = 10^-7T\(\frac{\mathrm{m}}{\mathrm{~A}}\)
The magnetic induction,
B = \(\frac{\mu_0{ }^{\prime}}{2 \pi a}\) = \(\frac{\mu_0 21}{4 \pi a}\) = \(\frac{10^{-7} \times 2(5)}{2 \times 10^{-2}}\) = 5 × 10^-5T

Section C

Attempt any Eight of the following questions:

Question 15.
Obtain an expression for conservation of mass starting from the equation of continuity.
Answer:

Consider a fluid in steady or streamline flow, that is its density is constant. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude. Suppose the velocity is v₁, at the point P and v₂ at point Q. If A₁ and A₂ are the cross-sectional areas of the tube at these two points.
The volume flux across A₁, \(\frac{d}{d t}\)(V₁) = A₁v₁ and that across A₂, \(\frac{d}{d t}\)(V₂) = A₂V₂
By the equation of continuity of the flow for a fluid,
A₁v₂ = A₂v₂
i.e., \(\frac{d}{d t}\)(V₁) = \(\frac{d}{d t}\)(V₂)
If ρ₁ and ρ₂ are the densities of the fluid at P and Q, respectively, the mass flux across A₁.
\(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(ρ₁V₁) = A₁ρ₁V₁ and that across A₂ = \(\frac{d}{d t}\)(m₂) = \(\frac{d}{d t}\)(ρ₂V₂) = A₂ρ₂V₂
Since no fluid can either can enter or leave through the boundary of the tube, the conservation of mass requires the mass fluxes to be equal, i.e..
\(\frac{d}{d t}\)(m₁) = \(\frac{d}{d t}\)(m₂)
i.e., A₁ρ₁V₁ = A₂ρ₂V₂
i.e., Aρv = constant which is the required expression.

Question 16.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Linear SHM is defined as the linear periodic motion of a body, in which the restoring force (or acceleration) is always directed towards the mean position and its magnitude is directly proportional to the displacement from the mean position.
cos (ωt + α) = \(\frac{x}{a}\)
x = a cos (ωt + α) ….(1)
This is the expression for displacement of particle M at time t.
As velocity of the particle is the time rate of change of displacement then we have
v = \(\frac{d x}{d t}\) = \(\frac{d}{d t}\) [a cos (ωt + α)]
v = -aωsin (ωt + α) ….(2)
As acceleration of particle is the time rate of change of velocity, we have
a = \(\frac{d v}{d t}\) = \(\frac{d}{d t}\) [- aω sin (ωt + α)]
a = -aω² cos (ωt + α)
a = ω²x
It shows that acceleration of particle M is directly proportional to its displacement and its direction is opposite to that of displacement. Thus, particle M performs simple harmonic motion but M is projection of particle performing UCM hence SHM is projection of UCM along a diameter of circle.

Question 17.
State the characteristics of stationary waves.
Answer:
Characteristics of stationary waves:

1. Stationary waves are produced by the interference of two identical. progressive waves travelling in opposite directions, under certain condition.
2. The overall appearance of standing wave is of alternate intensity maximum and minimum.
3. The distance between adjacent node \(\frac{\lambda}{2}\)
4. The distance between successive node \(\frac{\lambda}{4}\).
5. The stationary wave does not propagate in on direction.
6. The stationary wove does not transport energy through the medium.
7. There is no progressive change of phase from particle to particle.

Question 18.
A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.
Answer:
Assume parollet-plate capacitor of plate area A and plate separation d is filled with a dielectric of relative permittivity (dielectric constant) k. Its capacitance is
C = \(\frac{k \varepsilon_0 A}{d}\) …… (1)
If it is charged to a voltage (potential) V, the charge on its plates is
Q = CV
Since the battery is disconnected after it is charged, the charge Q on its plates and consequently the product CV remain unchanged.
On removing the dielectric completely, its capacitance becomes from equation (1).
C = \(\frac{\varepsilon_0 A}{d}\) = \(\left(\frac{1}{k}\right) c\) ……… (2)
That is, its capacitance decreases by the factor k. Since C’V’ = CV, its new voltage is
V’ = \(\left(\frac{C}{C^{\prime}}\right) V\) = kV ….. (3)
So that its voltage increases by the factor k.
The stored potential energy. u = \(\frac{1}{2}\)QV, so that Q remaining constant, u increases by the factor k. The electric field, E = \(\frac{V}{d}\), so that E also increases by a factor k.

Question 19.
When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer:
In on AC circuit containing only an ideal inductor, the current i logs behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad.
Here, for e = e₀ sin ωt,
we have, i = i₀ sin (ωt – \(\frac{\pi}{2}\))
Instantaneous power,
P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) – cos ωt sin \(\frac{\pi}{2}\))]
= -e₀i₀sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1
Average power over one cycle,
P_av = Work done in one cycle/time for one cycle

Question 20.
Is it always possible to see photoelectric effect with red light?
Answer:
No, it is not possible but due to present technology it may be possible.

Explanation: The photons in red light to not have the necessary energy required to rip an electron out of its orbital (this needed energy is equal to the electron’s work function. Because light behaves like particles rather than a continuous stream, even very high-intensity red light will never be able to overcome an electron’s work function (in this situation), as every individual photon fails to do so. This shows the particle behaviour of light, because of light behaved like a wave, the red light would be able to overcome the electron’s work function with high intensity or a long time.

Question 21.
Explain the construction and working of solar cell.
Answer:
Construction: A simple pn-junction solar cell consist of a p-type semiconductor substrate backed with a metal electrode. A thin layer of silicon is grown over the p-type of substrate by doping with suitable donor impurity. Metal finger electrode are prepared on top of n-layer so that there is enough space between the fingers for sunlight to reach the n-layer and underlying pn-junction:

Working: When exposed to sunlight, the absorption of incident radiation creates electron hole pairs in and near the depletion layer. The photo-generated electrons and holes moves towards the n-side and p-side, respectively. If no external load is connected, these photo-generated charges get collected at the two sides of the junction and give rise to forward photo-voltage. If a closed circuit a current I passes through external load as long as the solar cell is exposed to sunlight.

A solar cell consist several solar cells which are connected in series for higher output.

Question 22.
Disintegration rate of a sample of 1010 per hour at 20 hrs from the start. It reduces to 6.3 × 10⁹ per hour after 30 hours. Calculate its half-life and the initial number of radioactive atoms in the sample.
Answer:
A(t₁) = 10¹⁰ per hour, where t₁ = 20 h
A(t₁) = 6.3 × 10⁹ per hour, where t₂ = 30 h
A(t) = A₀e^-λt
A(t₁) = A₀\(\mathrm{A}_0 e^{-\lambda t_1}\) and A(t₂) = A₀\(e^{-\lambda t_2}\)
\(\frac{\mathrm{A}\left(t_1\right)}{\mathrm{A}\left(t_2\right)}\) = \(\left(\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}\right)\) = \(e^{\lambda\left(t_2-t_1\right)}\)
\(\frac{10^{10}}{6.3 \times 10^9}\) = \(e^{\lambda(30-20)}\) = e^10λ
1.587 = e^10λ
log_e 1.587 = 10λ
10λ = 2.303 log₁₀ (1.587)
λ = (0.2303) (0.2007) = 0.04622 per hour
The half life of the material

Question 23.
Calculate the wavelength associated with an electron, its momentum and speed. When it is accelerated through a potential of 54 V?
Answer:
Given: V = 54 V, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, h = 6.63 × 10^-13 J.s, K.E. = 150 eV.
We assume that the electron is initially at rest
V_e = \(\frac{1}{2}\)mv²
V = \(\sqrt{2 V_e / m}\)
= \(\sqrt{2(54)\left(1.6 \times 10^{-19}\right) / 9.1 \times 10^{-31}}\)
= \(\sqrt{19 \times 10^{12}}\)
= 4.359 × 10⁶ m/s
This is the speed of the electron.
Now, p = mv= (9.1 × 10^-31) (4.359 × 10⁶)
= 3.967 × 10^-24 kg m/s
This the momentum of the electron.
The wavelength associated with the electron.
λ = \(\frac{h}{p}\) = \(\frac{\left(6.63 \times 10^{-34}\right)}{\left(3.367 \times 10^{-24}\right)}\)
= 1.671 × 10^-10m
= 1.671 Å
= 0.1671 nm.

Question 24.
The distance between two consecutive bright fringes in a biprism experiment using Light of wavelength 6000 A is 0.32 mm by how much will the distance change if light of wavelength 4800 A is used?
Answer:
Given: λ₁ = 6000 Å = 6 × 10^-7 m, λ₂ = 4800 Å = 4.8 × 10^-7 m, W₁ = 0.32 mm = 3.2 × 10^-4 m.
Distance between consecutive bright fringes,

Question 25.
One mole of an ideal gas is initially kept in a cylinder with a movable frictionless and massless piston at pressure of 1.0 mPa and temperature 27°C. It is then expanded till its volume is doubled. How much work is done if the expansion is isobaric?
Answer:
Work done in isobaric process given by
W = p∆V = (V_f – V_i)
V_f = 2V_i
∴ W = 2pV_i
V_i can be found by using the ideal gas equation for initial state.
= 24.9 × 10^-4 m³
W = 2 × 10⁶ × 24.9 × 10^-4
W = 4.9 kJ.

Question 26.
The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1 µV/mm?
Answer:
Given: R = 8 Ω, L = 8 m, E = 2V
K = 1 V/mm
= \(=\frac{1 \times 10^{-6} \mathrm{~V}}{10^{-3} \mathrm{~m}}\) = 10^-3V/m
K = \(\frac{V}{L}\) = ER/(R + R_B)L
where R_B is the resistance in the box.
10^-3 = 2 × \(\frac{8}{\left(8+R_B\right) 8}\)
8 + R_B = \(\frac{2}{10^{-3}}\) = 2 × 10³
R_B = 200 – 8 = 1992 Ω

Section D

Attempt any Three of the following questions:

Question 27.
(i) What are eddy currents? State applications of eddy currents.
Answer:
Answer:
Whenever a conductor or part of it is moved in magnetic field cutting magnetic field lines, the free electrons in the bulk of metal starts circulating in closed path equivalent to current carrying loops. These current resemble to eddies in fluid stream and hence called as eddy current.

Application:

1. Electric brakes,
2. Galvanometer.

(ii) Magnetic field at a distance 2.4 cm from a long straight wire is 16 µT. What must be current through the wire?
Answer:

Question 28.
(i) State the postulates of Bohr’s atomic model.
Answer:
(a) The electron revolves with a constant speed in circular orbit around the nucleus.
(b) The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits in
an integral multiple of \(\frac{h}{2 \pi}\), h being the Planck’s constant
(c) An electron can make a transition from one of its orbit to another orbit having lower energy. In doing so, it emits a photon of energy equal to the difference in its energies in the two orbits.

(ii) A short bar magnet is placed in an external magnetic field of 700 gauss. When its axis makes an angle of 30° with the external magnetic field, it experiences a torque of 0.014 Nm. Find the magnetic moment of the magnet and the work done in moving it from its most stable to most unstable position.
Answer:
Given: B = 700 gauss = 0.07 tesla, θ = 30°, \(\tau\) = 0.014 Nm, \(\tau\) = MB sin θ
The magnetic moment of the magnet is
M = \(\frac{\tau}{\mathrm{B} \sin \theta}\) = \(\frac{0.014}{(0.07)\left(\sin 30^{\circ}\right)}\) = 0.4 A.m²
The most stable state of the bar magnet is for θ = 0°. It is in the most unstable state when θ = 180° is
W = MB (cos θ₀ – cos θ)
= MB (cos 0° – cos 180°)
= MB[1 – (- 1)] = 2 MB = (2) (0.4) (0.07)
= 0.056 J.

Question 29.
(i) On what factors do the degrees of freedom depend?
Answer:
The degree of freedom depend upon
(a) the number of atoms forming a molecule,
(b) the structure of the molecule,
(c) the temperature of the gas.

(ii) An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10^-5T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.
Answer:
Given: I = 50 m, B = 6 × 10^-5T, v = 400 m/s
The magnitude of the induced emf
| e | = Blv = (6 × 10^-5) (50) (400) = 1.2 V.

Question 30.
(i) A dipole with its charges, -q and +q located at the points (0, -b, 0) and (0, b, 0) is present in a uniform electric field E. The equipotential surfaces of this field are planes parallel to the YZ planes
(a) What is the direction of the electric field E?
(b) How much torque would the dipole experience in this field?
Answer:
(a) Given, the equipotentials of the external uniform electric field are planes parallel to the yz-plane, the electric field \(\vec{E}\) = ± E \(\hat{i}\) that is. \(\vec{E}\) is parallel to the x.

(b) The above diagram, the dipole moment

So that the magnitude of the torque is \(\tau\) = 2qbE.
If \(\overrightarrow{\mathrm{E}}\) is in the direction of the + x-axis, the torque \(\vec{\tau}\) is in the direction of – z-axis, while if \(\vec{E}\) is in the direction of the x-axis, the torque \(\vec{t}\) is in the direction of + z-axis.

(ii) A 25 µF capacitor, a 0.10 H inductor and a 25 Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Given: C = 25 µF = 25 × 10^-6 F, L = 0.10 H, R = 25 Ω, e = 310 sin (314t) [volt]
Comparing e = 310 sin (314t) with e = e₀ sin (2πft), we get
The frequency of the alternating emf as

Question 31.
(i) What happens to ferromagnetic material when its temperature increases above Curie temperature?
Answer:
A ferromagnetic material is composed of small regions called domains. Within each domain, the atomic magnetic moments of nearest-neighbour atoms interact strongly through exchange interaction, a quantum mechanical phenomenon and align themselves parallel to each other even in the absence of an external magnetic field. A domain is, therefore, spontaneously magnetized to saturation.

The material retains its domain structure only up to a certain temperature. On heating, the increased thermal agitation works against the spontaneous domain magnetization. Finally, at a certain critical temperature, called the Curie point or Curie temperature, thermal agitation overcomes the exchange forces and a certain temperature.

On heating, the increased thermal agitation works against the spontaneous domain magnetization. Finally, at a certain critical temperature, called the Curie point or Curie temperature, thermal agitation overcomes the exchange forces and exchange forces and keeps the atomic magnetic moments randomly oriented. Thus, above the Curie point, the material becomes paromagnetic. The ferromagnetic to paramognetic transition is an order to disorder transition. When cooled below the Curie point the material becomes ferromagnetic again.

(ii) In a common-base connection, the emitter current is 6.28 mA and collector current is 6.20 mA. Determine the common base DC current gain.
Answer:

Maharashtra Board Class 12 Physics Previous Year Question Papers