Maharashtra Board Class 12 Physics Sample Paper Set 6 with Solutions

Maharashtra State Board Class 12th Physics Sample Paper Set 6 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Physics Model Paper Set 6 with Solutions

Time : 3 Hrs.
Max. Marks : 70

General instructions:

The question paper is divided into four sections.

1. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each. Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
2. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. IS to Q. No., 26 are Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each.’ (Attempt any Three).
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each MCQ, correct answer must be written along with its alphabet.
   e.g., (a) ……/ (b)…./ (c)…/ (d) Only first attempt will be considered for evaluation.
8. Physical constants:
   1. Charge on a proton, e = 1.6 × 10^-19 C
   2. Acceleration due to gravity, g = 9.8 m/s²

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [10 Marks]

i. A particle of mass m performs vertical motion in a circle of radius r. Its potential energy at the highest point is _____.
(g is acceleration due to gravity)
(A) 2 mgr
(B) mgr
(C) 0
(D) 3 mgr
Answer:
(B) mgr

ii. The first law of thermodynamics is concerned with the conservation of ____.
(A) momentum
(B) energy
(C) temperature
(D) mass
Answer:
(B) energy

iii. The radioactive nucleus may emit
(A) all the three α, β and γ radiations simultaneously.
(B) all the three α, β and γ, one after the other.
(C) only α and β simultaneously.
(D) only one among α or β or γ at a time.
Answer:
(D) only one among α or β or γ at a time.

iv. Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between two successive maximum intensities will be ____.
(A) \(\frac{1}{4}\)s
(B) \(\frac{1}{2}\)s
(C) 1s
(D) 4s
Answer:
(A) \(\frac{1}{4} s\)

v. Intensity of electric field at a point close and outside a charged conducting cylinder is proportional to _____.
(r is the distance of a point from the axis of cylinder)
(A) \(\frac{1}{\mathrm{r}}\)
(B) \(\frac{1}{r^2}\)
(C) \(\frac{1}{r^3}\)
(D) r³
Answer:
(A) \(\frac{1}{\mathrm{r}}\)

vi. In Young’s double slit experiment the two coherent sources have different amplitudes. If the ratio of maximum intensity to minimum intensity is 16 : 1, then the ratio of amplitudes of the two source will be ____.
(A) 4 : 1
(B) 5 : 3
(C) 1 : 4
(D) 1 : 16
Answer:
(B) 5 : 3

vii. A small piece of metal wire is dragged across the gap between the pole pieces of magnet in 0.5 second. The magnetic flux between the pole pieces is 8 × 10^-4 weber. The emf induced in the wire is ____.
(A) 1.6 millivolt
(B) 16 millivolt
(C) 1.6 volt
(D) 16 volt
Answer:
(A) 1.6 millivolt

viii. Time required for a 50 Hz alternating current to increase from zero to 70.7% of its peak value is
(A) 2.5 × 10^-3 s
(B) 0.01 s
(C) 0.2 s
(D) 0.014 s
Answer:
(A) 2.5 × 10^-3s

ix. Incident light on a photo cathode consists of two wavelengths of 3500A and 4500 A. The threshold wavelength for photo cathode is 4800A. K.E of the photoelectrons is maximum for
(A) 3500 A
(B) 4500 A
(C) either 3500 A or 4500 A
(D) no wavelength as there is no emission
Answer:
(A) 3500 A

x. A LED emits visible light when its
(A) junction is reverse biased
(B) depletion region widens
(C) holes and electrons recombine
(D) junction becomes hot
Answer:
(C) holes and electrons recombine

Question 2.
Answer the following questions: [8 Marks]

i. State an expression for moment of inertia of a solid cylinder of uniform cross section about its own axis of symmetry.
Answer:
Expression for M.I of solid cylinder about its own axis of symmetry:
Let, M = mass of cylinder,
R = radius of cross section of cylinder
L = length of cylinder,
ZZ’ = axis passing through its centre and ⊥^ar to its plane

The M. I of solid cylinder about its own axis of symmetry is given by
I = \(\frac{1}{2}\)MR²

ii. An open water tank has a hole at a distance ‘x’ m from free water surface. If the radius of the hole is 2 mm and velocity of efflux of water is 11 m/s, find ‘x’(g = 9.8 m/s²)
Answer:
v = \(\sqrt{2 g h}\)
v² = 2gh = 2 gx ……… (∵ h = x)
∴ x = \(\frac{v^2}{2 g}\) = \(\frac{(11)^2}{2 \times 9.8}\) = 6.173 m

iii. State Brewster’s law.
Answer:
The tangent of the polarizing angle is equal to the refractive index of the refracting medium at which partial reflection takes place.

iv. What is the formula for electric field intensity at a point outside an infinitely long charged cylindrical conductor
Answer:
Formula for electric field intensity at a point outside an infinitely long charged cylindrical conductor is, E = \(\frac{\lambda}{2 \pi k \varepsilon_0 r}\).

v. Define Potential Gradient
Answer:
Potential gradient ¡s defined as potential difference per unit length of wire.

vi. What is the value of force on a closed circuit in a magnetic field?
Answer:
Force on a closed circuit in a magnetic field is zero.

vii. When is the wave nature of matter identified to be apparent?
Answer:
The wave nature of matter is apparent for any object only when the order of magnitude of its size and passage is same as that of its associated wavelength.

viii. In a transistor a change of 7.86 mA in emitter current produces a change of 7.2 mA in collector current, calculate the change in base current.
Answer:
∆I_E = ∆I_B + ∆I_C
∴ 7.86 = ∆I_B + 7.2
∴ ∆I_B = 0.66 mA = 660 µA

Section – B

Attempt any EIGHT of the following questions: [16 Marks]

Question 3.
A coin placed on a revolving disc, with its centre at a distance of 6 cm from the axis of rotation just slips off when the speed of the revolving disc exceeds 45 r.p.m. What should be the maximum angular speed of the disc, so that when the coin is at a distance of 12 cm from the axis of rotation, it does not slip?
Answer:
Given: r₁ = 6 cm, r₂ = 12 cm, n₁ = 45 r.p.m
To find: Maximum angular speed (n₂)
Formula: Max. C.F = mrω²

Calculation:

Ans: The maximum angular speed of the disc should be 31.8 r.p.m.

Question 4.
Draw a neat, labelled diagram for a liquid surface in contact with a solid, when the angle of contact is acute.
Answer:

Question 5.
A refrigerator with coefficient of performance releases 200 J of heat to a hot reservoir. Then find the work done on the working substance.
Answer:
Given: |Q_H| = 200 J; K = \(\frac{1}{3}\)
To find: Work done on the working substance
Formulae:
i. K = \(\frac{\left|Q_c\right|}{\left|Q_1\right|-\left|Q_c\right|}\)
ii. |W| = |Q_H| – |Q_C|
Calculation:
From formula (i),
\(\frac{1}{3}\) = \(\frac{Q_c}{200-Q_c}\)
∴ 200 – Q_C = 3Q_C
∴ 4Q_C = 200
∴ Q_C = \(\frac{200}{4}\)J = 50J
From formula (ii),
W = 200 – 50 = 150 J
Ans: Work done on the working substance is 150 J.

Question 6.
The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 7 A to 3 A in 0.2 s, calculate the value of emf induced in the other coil.
Answer:
given: M = 10 mH = 10 × 10^-3 H, I_i = 7 A, I_f = 3 A,
∆t = 0.2 s
To find: Induced emf

Formulae:
i. ∆φ = M ∆I
ii. e = \(\left|\frac{\Delta \phi}{\Delta t}\right|\)

Calculation:
From formula (i),
∆φ = 10 × 10^-3 × [3 – 7] = -4 × 10^-2 Wb
From formula (ii),
e = \(\left|\frac{-4 \times 10^{-2}}{0.2}\right|\) = 0.2 V
Ans: Induced emf is 0.2 V.

Question 7.
State any two postulates of Bohr’s theory of hydrogen atom.
Answer:
Bohr’s two postulates are:

1. In a hydrogen atom, the electron revolves round the nucleus in a fixed circular orbit with constant speed.
2. 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 is an integral multiple of \(\frac{h}{2 \pi}\), h being the Planck’s constant.

Question 8.
Obtain an expression for average power dissipated in a purely capacitive circuit.
Answer:
i. Let e = e_osinωt be applied emf across a capacitor of capacitance ‘C.

ii. In a purely capacitive circuit, the current leads the emf by a phase angle of π/2 i.e., when
e = e_o sin ωt then,
i = i_o sin (cot + π/2)

iii. Instantaneous power in the given circuit is given by,
P = ei
= (e_o sin ωt) (i_o cos ωt)
= e_o i_o sin ωt cot ωt

∴ Average power supplied to an ideal capacitor by the source over a complete cycle of AC is zero.

Question 9.
Define gyromagnetic ratio and give the necessary expression.
Answer:

1. The ratio of magnetic dipole moment with angular momentum of revolving electron is called the gyromagnetic ratio.
2. Gyromagnetic ratio is given by, \(\frac{m_{\text {orb }}}{L}\) = \(\frac{e}{2 m_e}\)

Question 10.
A mass M attached to a spring oscillates with a period of 2 second. If the mass is increased by 2 kg, the period increases by 1 second. Find the initial mass, assuming that Hooke’s law is obeyed.
Answer:
Given: T₁ = 2 s, T₂ = 3 s
To find: Mass (M)
Formula: T = 2π\(\sqrt{\frac{m}{k}}\)
Calculation:
From formula,
T₁ = 2π\(\sqrt{\frac{M}{k}}\) ….(i)
T₂ = 2π\(\sqrt{\frac{M+2}{k}}\) ….(ii)
From equations (i) and (ii),
\(\frac{2}{3}\) = \(\sqrt{\frac{M}{M+2}}\)
∴ \(\frac{4}{9}\) = \(\frac{M}{M+2}\)
∴ 9M = 4M + 8
∴ M = 1.6 kg
Ans: The initial mass attached to the spring is 1.6 kg.

Question 11.
Explain the reflection of transverse and longitudinal waves from a denser medium and rarer medium.
Answer:
i. Reflection from a denser medium:
a. In case of a longitudinal wave, a compression is reflected back as a compression and a rarefaction is reflected back as a rarefaction.
b. In case of a transverse wave, a crest is reflected back as a trough and a trough is reflected as a crest.

ii. Reflection from a rarer medium:
a. In case of a longitudinal wave, a compression is reflected as a rarefaction and a rarefaction is reflected as a compression.
b. In case of transverse wave, a crest is reflected as a crest and a trough is reflected as a trough.

Question 12.
State Kirchhoff s laws of electrical network
Answer:
i. Kirchhoff’s first law (Current law or junction law):

i. Statement: The algebraic sum of the currents at a junction is zero in an electrical network.
i.e.. \(\sum_{i=1}^n I_i\) = 0
where I_i is the current in the i^th conductor at a junction having n conductors.
Kirchhoff’s second law (Voltage law or loop theorem):

ii. Statement: The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emfs) in a closed loop is zero.
i.e., ΣIR + ΣE = 0

Question 13.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 4 A and making an angle of 30° with the direction of a uniform magnetic field of 0.30 T?
Answer:
Given: I = 4 A, θ = 30°, B = 0.30 T
To find: Force per unit length \(\left(\frac{\mathrm{F}}{l}\right)\)
Formula: \(\frac{F}{l}\) = BI sin θ
Calculation: From formula,
\(\frac{F}{l}\) = 0.30 × 4 × sin 30°
∴ \(\frac{F}{l}\) = 0.30 × 4 × \(\frac{1}{2}\) = 0.6 Nm^-1
Ans: The magnitude of magnetic force per unit length is 0.6 N/m.

Question 14.
State the conditions necessary for obtaining sharp and steady interference pattern
Answer:
Conditions for obtaining sustaining and good interference pattern are:
i. The two sources of light must be coherent:
Only two coherent sources can maintain their phase relation necessary for sustained interference pattern. For incoherent sources, emitted waves undergo rapid and random changes and steady interference cannot be obtained.

ii. The two sources of light must be monochromatic:
a. The condition for bright and dark fringes, the position of these fringes as well as the width of the fringes depend on the wavelength of light.
b. Therefore, the fringes of different colours do not coincide.
c. The resultant pattern contains coloured, overlapping bands due to their different wavelength.

iii. The two interfering waves must have the same amplitude:
Only if the amplitudes are equal, the intensity of dark fringes (destructive interference) is zero and the contrast between bright and dark fringes will be maximum.
iv. The separation between the two slits (d) must be small in comparison to the
distance between the plane containing the slits and the observing screen (D):
This is necessary as only in this case, the width of the fringes will be sufficiently large to be measurable and the fringes are well separated and can be clearly seen.

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
The difference between the two molar specific heats of a gas is 8000 J kg^-1 K^-1. If the ratio of the two specific heats is 1.65, calculate the two molar specific heats.
Answer:
\(\frac{C_p}{C_v}\) = 1.65
∴ C_P = 1.65 C_V
Given: C_P – C_V = 8000
∴ 1.65C_V – C_V = 8000
∴ 0.65C_V = 8000
∴ C_V = \(\frac{8000}{0.65}\)
∴ C_V = antilog {log (8000) – log (0.65))
= antilog {3.9031 – \(\overline{1} .8129\))
= antilog {4.0902} = 1.231 × 10⁴ J kg^-1 K^-1
Now, C_P = 8000 + C_V = 8000 + (1.231 × 10⁴) = 2.031 × 10⁴ J kg^-1 K^-1

Ans: The value C_V is 1.231 × 10⁴ Jkg^-1 K^-1 and C_P is 2.031 × 10⁴ Jkg^-1 K^-1.

Question 16.
With the help of a neat circuit diagram, explain the working of a half wave rectifier. Draw input- ” output waveforms
Answer:
i. The given figure shows the circuit of a half wave rectifier.

ii. The secondary coil AB of a transformer is connected in series with a diode D and the load resistance R_L. The AC voltage across the secondary coil AB changes its polarities after every half cycle.

iii. When the positive half cycle begins, the voltage at the point A is at higher potential with respect to that at the point B, therefore, the diode (b) is forward biased. It conducts and current flows through the circuit.

iv. When the negative half cycle begins, the potential at the point A is lower with respect to that at the point B and the diode is reverse biased, therefore, it does not conduct and no current passes through the circuit.

v. Hence, the diode conducts only in the positive half cycles of the AC input. It blocks the current during the negative half cycles. In this way, current always flows through the load R_L in the same direction for alternate positive half cycles and bC output voltage obtained across R_L in the form of alternate pulses.

Question 17.
What is de-Broglie hypothesis? Obtain the relation for de-Broglie wavelength.
Answer:
i. De Brogue proposed that a moving material particle of total energy E and momentum p has a wave associated with it (analogous to a photon).

ii. He suggested a relation between properties of wave, like frequency and wavelength,
with that of a particle, like energy and momentum.
p = \(\frac{E}{c}\) = \(\frac{h v}{c}\) = \(\frac{h}{\lambda}\)

iii. Thus frequency and wavelength of a wave associated with a material particle, of mass m moving with a velocity y, are given as
v = \(\frac{E}{h}\) and λ = \(\frac{h}{p}\) = \(\frac{h}{m v}\) ……. (1)

iv. De Broglie referred to these waves associated with material particles as matter
waves. The wavelength of the matter waves, given by equation (1), is now known as de Broglie wavelength and the equation is known as de Broglie relation.

Question 18.
Find the ratio of longest wavelength in Paschen series to shortest wavelength in Balmer series ‘
Answer:
Let, λ_S = shortest wavelength
λ_L = longest wavelength
\(\frac{1}{\lambda}\) = \(R\left(\frac{1}{n^2}-\frac{1}{m^2}\right)\)
Longest wavelength in Paschen series is obtained when n = 3, m = 4
For longest wavelength,

Shortest wavelength in Balmer series is obtained when n = 2, m = ∞
For shortest wavelength,

Ans: The ratio of longest wavelength in Paschen series to shortest wavelength in Balmer series is 5.131.

Question 19.
A 1000 mH inductor, 36 μF capacitor and 12 Ω resistor are connected in series to 120 V, 50 Hz AC source. Calculate:
i. impedance of the circuit at resonance
ii. current at resonance
iii. resonant frequency
Answer:
Given: L = 1000 mH = 1 H, R = 12 Ω, C = 36 µF = 36 × 10^-6F, e_rms = 120 V

To find:
i. Impedance of the circuit (Z)
ii. Current at resonance (i_rms)
iii. Resonant frequency (f_r)

Formulae :
i. i_rms = \(\frac{e_{\text {rms }}}{Z}\)
ii. f_r = \(\frac{1}{2 \pi \sqrt{L C}}\)

Calculation:
At resonance, Z = R
∴ Z = 12Ω
From formula (i),
i_rms = \(\frac{120}{12}\)
∴ i_rms = 10 A
From formula (ii),
f_r = \(\frac{1}{2 \times 3.142 \times \sqrt{1 \times 36 \times 10^{-6}}}\)
= \(\frac{1}{2 \times 3.142 \times 6 \times 10^{-3}}\) = \(\frac{1}{37.704 \times 10^{-3}}\)
f_r = 26.52 Hz

Ans:

i. The Impedance of the circuit is 12 Ω.
ii. The current at resonance is 10 A.
iii. The value of resonant frequency is 26.52 Hz.

Question 20.
Explain diamagnetism on the basis of paired electron orbit.
Answer:

i. Consider an electron revolving in an orbit around the nucleus. The revolving electron is
equivalent to a current loop.

ii. When the orbiting electron or current loop is brought in external magnetic field, the field induces a current according to Lenz’s law.

iii. The direction of induced current I, is such that the direction of magnetic field created by the current is opposite to the direction of applied external magnetic field.

iv. Out of pair of orbits, for one loop where the direction of induced current is the same as that of loop current, the net current ¡nc eases effectively increasing the magnetic dipole moment as shown ¡n the figure (b) above.

v. For the current loop where the direction of induced current is opposite to direction of loop current, the net current reduces effectively reducing the magnetic dipole moment as shown in the figure (a) above.

vi. This results in net magnetic dipole moment opposite to the direction of applied external magnetic field.
Examples of diamagnetic materials are copper, gold metal, bismuth and many metals, lead, silicon, glass, water, wood, plastics etc.

Question 21.
Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
i. Let ‘m’ be the mass of the bob and T’ be the tension in the string. The pendulum remains in equilibrium in the position OA, with the centre of gravity of the bob, vertically below the point of suspension O.

ii. If now the pendulum is displaced through a small angle θ, called angular amplitude, and released, it begins to oscillate on either side of the mean (equilibrium) position in a single vertical plane.

iii. In the displaced position (extreme position), two forces are acting on the bob.
a. Force T’ due to tension in the string, directed along the string, towards the support.
b. Weight mg, in the vertically downward direction.

iv. At the extreme positions, there should not be any net force along the string.

v. The component of mg can only balance the force due to tension.
Thus, weight mg is resolved into two components;
a. The component mg cose along the string, which is balanced by the tension T’.
b. The component mg sine perpendicular to the string is the restoring force acting on mass m tending to return it to the equilibrium position.
∴ Restoring force, F = -mg sinθ

vi. As θ is very small (θ < 10°).
sinθ ≈ θ^c
∴ F ≈ -mgθ
From the figure,
For small angle, θ = \(\frac{x}{L}\)
∴ F = -mg\(\frac{x}{L}\) …….. (1)
As m, g and L are constant, F ∝ -x

vii. Thus, for small displacement, the restoring force is directly proportional to the displacement and ¡s oppositely directed. Hence the bob of a simple pendulum performs linear 5.H.M. for small amplitudes.

viii. The period T of oscillation of a pendulum is given by,

This gives the expression for the time period of a simple pendulum.

Question 22.
Derive an expression for kinetic energy of a rotating body.
Answer:
i. Consider a rigid object rotating with a constant angular speed w about an axis perpendicular to the plane of paper.

ii. For theoretical simplification, let us consider the object to be consisting of N particles of masses m₁, m₂,…… m_N at respective perpendicular distances r₁, r₂,…… r_N from the axis of rotation.

iii. As the object rotates, all these particles perform UCM with the same angular speed w, but with different linear speeds,
v₁ = r₁ω = v₂ = r₂ω,……V_N = r_Nω

iv. Translational K.E. of the first particle is
(K.E.)₁ = \(\frac{1}{2} m_1 v_1^2\) = \(\frac{1}{2} m_1 r_1^2 \omega^2\)
Similar will be the case of all the other particles.

v. Rotational K.E. of the object, is the sum of individual translational kinetic energies.
Thus,

Question 23.
A sonometer wire is stretched by tension of 40 N. It vibrates in unison’ with frequency 384 Hz. How many numbers of beats get produced in two second’ wire is decreased by 1.24 N?
Answer:
Given: T = 40 N, n_fork = 384 Hz, ∆T = 1.24 N, t = 2 sec
To find: Number of beats

Formulae:
i. n_string = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
ii. n_beat = n₁ – n₂

Calculation:

From formula(ii),
n_beat = 384 – 378 = 6 Hz
Number of beats in two seconds
= 2 × 6 Hz = 12

Ans: Number of beats produced in two seconds is 12.

Question 24.
Two resistances X and Y in the two gaps of a metrebridge gives a null point in the ratio 2 : 3. If each resistance is increased by 30 Ω, the null point division ratio 5 : 6, calculate the value of X and Y.
Answer:
Given: \(\frac{x}{y}\) = \(\frac{2}{3}\)
\(\frac{x+30}{y+30}\) = \(\frac{5}{6}\)
To find: Resistance (X and Y)
Formula: \(\frac{X}{Y}\) = \(\frac{l_x}{l_y}\)
Calculation: From 1^st condition,
\(\frac{X}{Y}\) = \(\frac{2}{3}\)
∴ 3X = 2Y …… (i)
From 2^nd condition,
\(\frac{x+30}{y+30}\) = \(\frac{5}{6}\)
∴ 6 (X + 30) = 5(Y + 30)
∴ 6X + 180 = 5Y + 150
∴ 2(3X) + 180 = 5Y + 150
∴ 2 (2Y) + 180 = 5Y + 150 …. [From equation (1))
∴ 4Y + 180 = 5Y + 150
∴ Y = 30Ω ………. (ii)
From equations (i) and (ii), we have,
3X = 2(30)
∴ 3X = 60
∴ X = 20 Ω, Y = 30 Ω

Ans:
i. The value of resistance X is 20 Ω.
ii. The value of resistance Y is 30 Ω.

Question 25.
A wire loop of the form shown in the figure carries a current I. Obtain the magnitude and direction of the magnetic field at P. (Given: B = \(\frac{\mu_0 I}{4 \pi R} \sqrt{2}\) due to segment II)

Answer:
Magnetic induction field at P is sum of 2 sections as shown in the figure
B = B₁ + B₂ …….. (i)
Given that, B₁ = \(\frac{H_0 I}{4 \pi R} \sqrt{2}\) …… (ii)
B₂ will be magnetic field due to arc XOY subtending angle θ at P.

Question 26.
Explain Rayleigh’s criterion for the resolution of two close point objects, when their images are
i. just resolved
ii. well resolved
iii. unresolved.
Answer:
i. According to Lord Rayleigh, the ability of an optical instrument to distinguish between two closely spaced objects depends upon the diffraction patterns of the two objects (slits, point objects, stars, etc.), produced at the screen (retina, eyepiece, etc.).

ii. The two objects are said to be unresolved, if the separation between the central maximum of the two object is less than the distance between the central maximum and first minimum of any of the two objects i.e., S < ∆

iii.

Two objects are said to be just resolved when the separation between the central maxima of the two objects is just equal to the distance between the central maximum and the first minimum of any of the two objects i.e., S = ∆.

iv. The two objects are said to be well resolved, if the separation between the central maximum of the two objects is greater than the distance between the central maximum and first minimum of any of the two objects i.e., S > ∆.

v. Thus, depression in the resultant envelope is not noticeable in unresolved diffraction pattern, noticeable in just resolve diffraction pattern and clearly noticeable in well resolved pattern.

Section – D

Attempt any THREE of the following questions: [12 Marks]

Question 27.
Derive Bernoulli’s equation.
Answer:
i. Consider an ideal fluid flowing through a tube of varying cross section and height. Consider hn element of fluid lying between cross sections P and R.

ii. Let,
a. v₁, ₂ = speeds of the fluid at the lower end P and the upper end R respectively.
b. A₁, A₂ = cross section areas of the fluid at the lower end P and upper end R respectively.
c. P₁, P₂ = pressures of the fluid at the lower end P and upper end R respectively.
d. d₁, d₂ = distances travelled by the fluid at the lower end P and the upper end R during the time interval dt with velocities v₁ and v₂ respectively.
e. ρ = density of the fluid flowing through the tube
f. h₁, h₂ = mean heights of section P and R from ground or reference level.
g. P₁A₁, P₂A₂ = forces acting on the fluid at section P and R respectively.

iv. The volume dVof the fluid passing through any cross section during time interval dt is the same i.e., dV = A₁d₁ = A₂d₂ …….. (1)

v. Since, the fluid is ideal there is no internal friction in the fluid. The only non gravitational force working on the fluid element is due to the pressure of the surrounding fluid.

vi. Therefore, the net work W done on the element by the surrounding fluid during the flow from P to R is, W = P₁A₁d₁ – P₂A₂d₂ …….. (2)
The negative sign indicates that the force at R opposes the displacement of the fluid.

vii. Substituting equation (1) in equation (2), we get,
W = P₁dV – P₂dV
∴ W = (P₁ – P₂) dV …….. (3)

viii. At the beginning of the time interval dt,
a. mass of the fluid between P and Q = ρA₁d₁,
b. Kinetic energy of the fluid between P and Q = \(\frac{1}{2} \rho\left(A_1 d_1\right) v_1^2\)
c. Potential energy of the fluid between P and Q = mgh₁ = ρdVgh₁

ix. At the nd of the time interval dt,
a. mass of the fluid between R and S = ρA₂d₂
b. Kinetic energy of the fluid between R and S = \(\frac{1}{2} \rho\left(A_2 d_2\right) v_2^2\)
c. Potential energy of the fluid between R and S = mgh₂ = ρdVgh₂

x. The net change in the kinetic energy ∆ K.E, during time interval dt is,

xi. The net change in the gravitational potential energy during time interval dt is,
∆P.E. = ρdVgh₂ – ρdVgh₁
∴ ∆P.E. = ρdVg(h₂ – h₁) …….(5)
As the work done W is due to forces other than the conservative force of gravity, ¡t equals the change in the total mechanical energy
∴ W = ∆K.E. + ∆P.E.
Substituting equation (3), (4) and (5) in the above equation,

This is known as Bernoulli’s equation.
Bernoulli’s equation can be written as

xiii. Bernoulli’s equation can be written as,

Question 28.
i. Explain, on the basis of kinetic theory, how the pressure of gas changes if its volume is reduced at constant temperature.
ii. The energy of 6000 J is radiated in 5 minutes by a body of surface area 100 cm2. Find emissive power of the body.
Answer:
i. a. Suppose, P = pressure exerted by the gas
V = volume of the gas, N = number of molecules of the gas
m = mass of each molecule of the gas
∴ Total mass of the gas, M = mN
b. From kinetic theory of gases, P = \(\frac{1}{3} \frac{\mathrm{mN}}{\mathrm{~V}} \overline{\mathrm{v}^2}\)
∴ Pressure exerted by o gas in an enclosed vessel, P = \(\frac{2}{3} \frac{N}{V}\left(\frac{1}{2} m \overline{v^2}\right)\)
c. But, \(\frac{1}{2} m \overline{v^2}\) = K.E constant for all the gases at a given temperature.
N = number of molecules which is constant for a given mass of the gas.
∴ P\(=\frac{\text { constant }}{V}\) ∴ P ∝ \(\frac{1}{V}\)
Hence, at constant temperature, pressure of the gas is increased if its volume is reduced.

ii.
Solution:
Given: Q = 6000 J, t = 5 minutes = 5 × 60 s = 300 g
A = 100 cm² = 100 × 10^-4m² = 10^-2m²
To find: Emissive power (R)
Formula: R = \(\frac{Q}{A t}\)
Calculation: From formula,
R = \(\frac{6000}{10^{-2} \times 300}\) = \(\frac{20}{10^{-2}}\) = 20 × 10²
∴ R = 2000 J/m²s
Ans: The emissive power of the body is 2000 J/m²s.

Question 29.
i. Distinguish between Isothermal process and Adiabatic process
ii. A heat engine works between two reservoirs. The working substance used in the engine absorbs 2.8 × 10⁵ J of heat. If the efficiency of the engine is 70%, find the heat rejected by the working substance.
Answer:
Given: 2.8 × 10⁵ J, η = 70% = 0.7
To find: Heat rejected (Q_C)
Formula:
η = \(\frac{Q_H-Q_c}{Q_H}\)
Calculation: From formula,
0.7 = \(\frac{\left(2.8 \times 1.0^5\right)-Q_c}{2.8 \times 10^5}\)
Q_C = 28 × 0.3 × 10⁵
= 8.4 × 10⁴J
Ans: The heat rejected by working substance will be 8.4 × 10⁴ J.

Question 30.
i. Obtain an expression for potential energy of an electric dipole in an external field.
ii. Two charged particles having equal charge of 3 × 10^-5 C each are brought from infinity to a separation of 30 cm. Find the increase in electrostatic potential energy during the process.
Answer:
i. Consider a dipole with charges -q and +q separated by a finite distance 2l, placed in a uniform electric field \(\vec{E}\)

ii. It experiences a torque \(\vec{\tau}\) which tends to rotate it as shown in figure below, \(\vec{\tau}\) = \(\vec{p} \times \vec{E}\) = pE sinθ

iii. To neutralize this torque, let us assume an external torque \(\vec{\tau}_{\text {ext }}\) be applied, which rotates it in the plane of the paper from angle θ₀ to angle θ, without angular acceleration and at an infinitesimal angular speed.

iv. Work done by the external torque

This work done is stored as the potential energy of the system in the position when the dipole makes an angle θ with the electric field.
Thus, potential energy of electric dipole in external electric field is,
U(θ) – U(θ₀) = pE(cosθ_o – cosθ)

Special cases:
Choosing U(θ₀) = 0, we get,

a. If initially the dipole is perpendicular to the field \(\overrightarrow{\mathrm{E}}\) i.e., θ₀ = \(\frac{\pi}{2}\) then
U(θ) = pE(cos\(\frac{\pi}{2}\) – cosθ) = -pE cosθ
U(θ) = \(-\vec{p} \cdot \vec{E}\)

b. If initially the dipole is parallel to the field \(\overrightarrow{\mathrm{E}}\) then θ₀ = 0
U(θ) = pE(cos0 – cosθ)
U(θ) = pE(1 – cosθ)

ii. Solution:
Given: q₁ = q₂ = q = 3 × 10^-5 C,
r = 30 cm = 0.3 m
To find: Increase in Potential Energy
Formula: U_A – U_B = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\)
Calculation: Assuming the potential energy (PE) at ∞ to be zero,
Increase in energy = Energy of system at present
From formula,
U_A – U_B = \(\frac{9 \times 10^9 \times\left(3 \times 10^{-5}\right)^2}{0.3}\) = 9 × 3 × 10^9-10+1 = 27J
Ans Increase in electrostatic potential energy of the system is 27 J

Question 31.
State the principle of working of transformer. Explain the construction and working of a transformer. Derive an expression for e.m.f. and current in terms of turns ratio.
Answer:
Transformer:
Transformer is an electrical device which converts low alternating voltage at high current to high alternating! voltage at low current and vice-versa.

Principle:
It is based on the principle of mutual induction i.e., whenever the magnetic flux linked with a coil changes, an e.m.f is induced in the neighbouring coil.

Construction:

1. A transformer consists of two sets of coils primary P and secondary S insulated from each other. The coil P is called the input coil and coil S is called the output coil.
2. The two coils are wound separately on a laminated soft iron core.

Working:

i. When an alternating voltage is applied to the primary coil the current through the coil goes on changing. Hence, the magnetic flux through the core also changes.

ii. As this changing magnetic flux is linked with both the coils, an e.m.f is induced in each coil.

iii. The amount of the magnetic flux linked with the coil depends upon the number of turns of the coil.

iv. Let, ‘φ’ be the magnetic flux linked per turn with both the coils at certain instant Y. Let

v. ‘N_p‘ and ‘N_s‘ be the number of turns of primary and secondary coil,
N_pφ = magnetic flux linked with the primary coil at certain instant ‘t’
N_sφ = magnetic flux linked with the secondary coil at certain instant ‘t’

vi. Induced e.m.f produced in the primary and secondary coil is given by,
e_p = \(-\frac{\mathrm{d} \phi_p}{\mathrm{~d} t}\) = \(-N_p \frac{d \phi}{d t}\) ……. (1)
e_s = \(-\frac{d \phi_s}{d t}\) = \(-N_s \frac{d \phi}{d t}\) …….. (2)

vii. Dividing equation (2) by (1),
∴ \(\frac{e_s}{e_p}\) = \(\frac{N_s}{N_p}\) …… (3)
Equation (3) represents equation of transformer.
The ratio \(\frac{N_s}{N_p}\) is called turns ratio (transformer ratio) of the transformer.

viii. For an ideal transformer,
Input power = Output power
∴ e_pI_p = e_sI_s
∴ \(\frac{e_s}{e_p}\) = \(\frac{I_p}{I_s}\) …. (4)

ix. From equation (3) and (4), \(\frac{e_s}{e_p}\) = \(\frac{N_s}{N_p}\) = \(\frac{I_p}{I_s}\)

Case I: When N_s > N_p
Then e_s > e_p (step up transformer) and i_p> i_s. Current in the primary coil is more than that in the secondary coil.

Case II: When N_s < N_p
Then e_s < e_p (step down transformer) and i_p < i_s is. Current in primary coil is less than that in secondary coil

Maharashtra Board Class 12 Physics Previous Year Question Papers