Maharashtra State Board Class 12th Physics Sample Paper Set 2 with Solutions Answers Pdf Download.

## Maharashtra Board Class 12 Physics Model Paper Set 2 with Solutions

**Section A**

Question 1.

Select and write the correct answers to the following questions:

(i) A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio “Rotational K.E : Translation K.E : Total K.E.” is:

(a) 1 : 1 : 2

(b) 1 : 2 : 3

(c) 1 : 1 : 1

(d) 2 : 1 : 3

Answer:

(d) 2 : 1 : 3

(ii) The ratio of emissive power of perfectly blackbody at 1327°C and 527°C is:

(a) 4 : 1

(b) 16 : 1

(c) 2 : 1

(d) 8 : 1

Answer:

(b) 16 : 1

(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 art even integral multiple of λ

Answer:

(a) Must be an odd integral multiple of λ/4

(iv) 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

(v) During refrigeration cycle, heat is rejected by the refrigerant in the:

(a) Condenser

(b) Cold chamber

(c) Evaporator

(d) Hot chamber

Answer:

(a) Condenser

(vi) In a certain unit the radius of gyration of a uniform disc about its central and transverse axis is \(\sqrt{2.5}\). Its radius of gyration about a tangent in its plane (in the same unit) must be:

(a) \(\sqrt{5}\)

(b) 2.5

(c) 2\(\sqrt{2.5}\)

(d) \(\sqrt{12.5}\)

Answer:

(b) 2.5

(vii) A charged particle is in moving having initial velocity v when it enter into a region of uniform magnetic field perpendicular to v. 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

(viii) What is the energy required to build up a current of 1A in an inductor of 20 mH?

(a) 10 mJ

(b) 20 mJ

(c) 20 J

(d) 10 J

Answer:

(a) 10 mJ

(ix) In an AC circuit, e and i given by e = 150 sin (150t) V and i = 150 sin (150t + π/3)A. The power dissipated in the circuit is:

(a) 106 W

(b) 150 W

(c) 5625 W

(d) Zero

Answer:

(c) 5625 W

(x) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for:

(a) Electron

(b) Proton

(c) α-particle

(d) Hydrogen atom

Answer:

(a) Electron

Question 2.

Answer the following questions:

(i) What is Brewster’s law? Describe the formula for Brewster angle.

Answer:

The tangent of the polarizing angle is equal to the refractive index of the reflecting medium with respect to the surrounding (_{1}n_{2}). If θ_{B} = _{1}n_{2} = \(\frac{n_1}{n_2}\)

Here n_{1} is the absolute refractive index of the surrounding and n_{2} is that of reflecting medium. The angle of θ_{B} is called the Brewster angle.

(ii) The emissive power of a sphere of area 0.02 m^{2} is 0.5 kcals^{-1} m^{-2}. What is the amount of heat radiated by the spherical surface is 20 second?

Answer:

Given: R = 0.5 kcal s^{-1} m^{-2}, A = 0.02 m^{2}, t = 20 s

Q = RAt = (0.5) (0.02) (20) = 0.2 kcal

This is the required quantity.

(iii) Give an example of some familiar process in which no heat is added to or removed from a system but the temperature of the system changes.

Answer:

Adiabatic compression is the process in which no heat is transferred to or from the system but the temperature of the system changes. When we compress gas in an adiabatic process the volume of the gas will decrease and the temperature of the gas rises as it is compressed which we have seen the warming of/a bicycle pump. Conversely, the temperature falls when the gas expands but the heat remains constant throughout the process.

(iv) A star is emitting light at the wavelength of 5000 A. Determine the limit of resolution of a telescope having an objective of diameter of 200 inch.

Answer:

Given: λ = 5000 Å = 5 × 10^{-7} m

D = 200 × 2.54 cm

= 5.08 m

θ = \(\frac{1.22 \lambda}{D}\)

= \(\frac{1.22 \times 5 \times 10^{-7}}{5.08}\)

= 1.2 × 10^{-7} rad

(v) What do you mean by electromagnetic induction?

Answer:

The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.

(vi) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.

Answer:

Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by

Hence, the equivalent inductance is less than the inductance of either coil.

(vii) What do you understand by the term wave-particle duality? Where does it apply?

Answer:

Depending upon experimental conditions of the structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = \(\frac{h}{p}\), where λ is the wavelength of matter waves, also called de Broglie waves I Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck’s constant.

(viii) State the importance of Davisson and Germer experiment.

Answer:

The Davisson 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.

Attempt any Eight of the following questions:

Question 3.

On what factors does the frequency of a conical pendulum depends? Is it independent of some factors?

Answer:

The frequency of a conical pendulum of string length L and semi-vertical angle θ is

n = \(\frac{1}{2 \pi} \sqrt{\frac{g}{L \cos \theta}}\)

Where g is the acceleration due to gravity as the place. From the above expression, we can see that

- n ∝ \(\sqrt{g}\)
- n ∝ \(\frac{1}{\sqrt{\mathrm{~L}}}\)
- n ∝ \(\frac{1}{\sqrt{\cos \theta}}\)
- The frequency is independent of the mass of the bob.

Question 4.

The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path.

Answer:

Given: Mass = m = 2 kg, Energy = E = 40 J

The speed of the body while crossing the centre of the path (mean position) is v_{max} and the total energy is kinetic energy.

∴ \(\frac{1}{2} m v_{\max }^2\) = E

∴ V_{max} = \(\sqrt{\frac{2 E}{m}}\) = \(\sqrt{\frac{2 \times 40}{2}}\) = 6.324 m/s.

Question 5.

What is a wavefront? How is it related to rays of light?

Answer:

Wavefront or wave surface: The locus of all points where waves starting simultaneously from a source reach at the same instant of time and hence the particles at the points oscillate with the same phase is called a wavefront or wave surface.

Consider a point source of light 0 in a homogeneous isotropic medium in which the speed of light is v. The source emits light in all directions. In time t, the disturbance (light energy) from the source, covers a distance Vt in all directions, i.e. it reaches out to all points which are at a distance vt from the point source.

The Locus of these points which are in the same phase is the surface of a sphere with centre O and radius vt. It is a spherical wavefront. In a given medium, a set of straight lines can be drawn which are perpendicular to the wavefront According to Huygens, these straight lines are the rays of light. Thus, rays are always normal to the wavefront.

In the case of a spherical wavefront, the rays are radial. If a wavefront has travelled a large distance away from the source, a small portion of this wavefront appears to be plane. This part is a plane wavefront.

Question 6.

A 6 µF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer:

Given: C = 6 μF = 6 × 10^{-6}F = C_{1}, V = 300, C_{2} = 3 μF

The electrostatic energy in the capacitor

= \(\frac{1}{2}\)CV^{2} = \(\frac{1}{2}\)(6 × 10^{-6}) (300)^{2}

= 3 × 10^{-6} × 9 × 10^{4} = 0.27 J

The charge on this capacitor,

Q = CV = (6 × 10^{-6}) (300) = 1.8 mC

When two capacitors of capacitances C_{1} and C_{2} are connected in parallel, the equivalent capacitance C

= C_{1} + C_{2} = 6 + 3 = 9 μF

= 9 × 10^{-6} F

By conservation of charge,

Q = 1.8 C

∴ The energy of the system

= \(\frac{Q^2}{2 C}\)

= \(\frac{\left(1.8 \times 10^{-3}\right)^2}{2\left(9 \times 10^{-6}\right)}\) = \(\frac{18 \times 10^{-8}}{10^{-6}}\) = 0.18J

The energy lost = 0.27 – 0.18 = 0.09 J.

Question 7.

State the uses of a potentiometer.

Answer:

The applications (uses) of the potentiometer:

(i) Voltage divider: The potentiometer can be used as a voltage divider to change the output voltage of a voltage supply.

(ii) Audio control- Sliding potentiometers are commonly used in modern low-power audio systems as audio control devices.

Both sliding (faders) and rotary potentiometers (knobs) are regularly used for frequency attenuation, loudness control and for controlling different characteristics of audio signals.

(iii) Potentiometer as a sensor: If the slider of the potentiometer is connected to the moving part of a machine, it can work as a motion sensor. A small displacement of the moving part causes a change in potential which is farther amplified using an amplifier circuit. The potential difference is calibrated in terms of displacement of the moving part.

(iv) To measure the emf(for this, the emf of the standard cell and potential gradient must be known).

(v) To compare the emfs to two cells.

(vi) To determine the internal resistance of a cell.

Question 8.

Calculate the value of magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5 A. (Given: μ_{0} = 4π × 10^{-7} Wb/Am).

Answer:

Given: I = 5A, a = 0.02 m, \(\frac{\mu_0}{4 \pi}\) = 10^{-7} T m/A

The magnetic induction,

B = \(\frac{\mu_0{ }^{\prime}}{2 \pi a}\) = \(\frac{\mu_0}{4 \pi} \frac{2}{a}\) = 10^{-7} × \(\frac{2(5)}{2 \times 10^{-2}}\) = 5 × 10^{-5}T

Question 9.

What should be retentivity and coercivity of permanent magnet?

Answer:

A permanent magnet should have a large zero-field magnetization and should need a vary large reverse field to demagnetize. In other words, it should have a very broad hysteresis loop with high retentivity and very high coercivity.

Question 10.

An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10^{-5} T 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^{-5} T, v = 400 m/s

The magnitude of the induced emf,

| e | = B/v = (6 × 10^{-5}) (400) (50) = 1.2 V.

Question 11.

The safest way to protect yourself from lightening is to be inside a car. Justify.

Answer:

A car is an almost deal faraday cage, when a car is truck by lightning, the charge flows on the outside surface of the car to the ground but electric field inside remains zero. This leaves the passenger inside unharmed.

Question 12.

What do you understand by the term wave-particle duality? Where does it apply?

Answer:

Depending upon experimental conditions or the structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = h/p, where λ is the wavelength of matter waves, also called de Broglie waves I Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck’s constant.

Question 13.

State the difficulties faced by Rutherford’s atomic model.

Answer:

(i) According to Rutherford, the electrons revolve in circular orbits around the atomic nucleus. The circular motion is an accelerated motion. According to the classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion should go on radiating energy.

Due to the loss of energy, the radius of its orbit should go on decreasing. Therefore, the electron should move along a spiral path and finally fall into the nucleus in a very short time, of the order of 10^{-16} s in the case of a hydrogen atom. Thus, the atom should be unstable. We exist because atoms are stable.

(ii) If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution of the electron would go on increasing. The electron, therefore, would emit radiation of continuously changing frequency and hence give rise to a continuous spectrum. However, the atomic spectrum is a line spectrum.

Question 14.

The common-base DC current gain of a transistor is 0.967. If the emitter current is 10 mA. What is the value of base current?

Answer:

Given: Current gain (α) = 0.967

Emitter current = 10 mA

To find: The value of base current of the transistor.

→ The common gain DC current us is given by

α = 0.967 = \(\frac{I_C}{I_E}\) = \(\frac{I_C}{10}\)

I_{C} = 9.67 mA

→ The base current of the transistor is given by the formula

I_{E} = I_{B} + I_{C}

10 = I_{B} + 9.67

I_{B} = 0.33 mA

The value of base current of the transistor is 0.33 mA.

**Section C**

Attempt any Eight of the following questions:

Question 15.

Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.

Answer:

Consider a symmetric rigid body, like a sphere or a wheel or a disc, rolling a plane surface with friction along a straight path. Its centre of mass (CM) moves in a straight line and if the frictional force on the body is large enough, the body rolls without slipping.

Thus, the rolling motion of the body can be treated as a translation of the CM and rotation about an axis through the CM. Hence, the kinetic energy of a rolling body is

E = E_{tran} + E_{rot} ……… (1)

Where E_{tran} and E_{rot} are the kinetic energies associated with the translation of the CM and rotation about on axis through the CM, respectively.

Let M and R be the mass and radius of the body. let ω, k and I be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its centre and v be the translation speed of the centre of

mass.

Question 16.

A rectangular wire frame of size 2 cm × 2 cm is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm × 3 cm, calculate the work done in the process. The surface tension of soap film is 3 × 10^{-2} N/m.

Answer:

Given: A_{1} = 2 × 2 cm^{2} = 4 × 10^{-4} m^{2},

A_{2} = 3 × 3 cm^{2} = 9 × 10^{-4}m^{2}, T = 3 × 10^{-2}N/m

As the film has two surfaces, the work done is

W = 2T(A_{2} – A_{1})

= 2(3 × 10^{-2}) (9 × 10^{-4} – 4 × 10^{-4})

= 3.0 × 10^{-5}J = 30 μJ.

Question 17.

State the law of equipartition of energy and hence calculate motor specific heat of mono- and diatomic gases at constant volume and constant pressure.

Answer:

Law of equipartition of energy states that for a dynamical system in thermal equilibrium the total energy of the system is shared equally by the degrees of freedom.

The energy associated with each degree of freedom per molecule is \(\frac{1}{2}\)k_{B}T, where k_{B} is the Boltzmann’s constant.

For example, for a monatomic molecule, each molecule has 3 degrees of freedom. Accordingly to kinetic theory of gases. the mean kinetic energy of a molecule is \(\frac{3}{2}\)k_{B}T.

Specific heat capacity of monotomic gas: The molecules of a monatomic gas have 3 degrees of freedom.

The average energy of a molecule at temperature T is \(\frac{3}{2}\)k_{B}T.

The total internal energy of a mote is \(\frac{3}{2}\)N_{A}k_{B}T where N_{A} is the Avogadro number.

The molar specific heat at constant volume C_{V} is

For an idea gas,

C_{V} (monatomic gas) = \(\frac{d \mathrm{E}}{d T}\) = \(\frac{3}{2}\)RT

For an ideal gas, C_{P} – C_{V} = R

Where C_{P} is molar specific heat at constant pressure.

Thus, C_{P} = \(\frac{5}{2}\)R

Specific heat capacity of diatomic gas: The molecules of a monatomic gas have 5 degrees of freedom, 3 translation and 2 rotational.

The average energy of a molecule at temperature T is \(\frac{5}{2}\)k_{B}T.

The total internal energy of a mole is \(\frac{5}{2}\)N_{A}k_{B}T

The molar specific heat at constant volume C_{V} is For an ideal gas,

C_{V} (monatomic gas) = \(\frac{d \mathrm{E}}{d \mathrm{~T}}\) = \(\frac{5}{2}\)RT

For an ideal gas, C_{P} – Ç_{V} = R

where C_{P} is the molar specific heat at constant pressure.

Thus, C_{P} = \(\frac{7}{2}\)R

A soft or non-rigid diatomic molecule has in addition, one frequency of vibration which contributes two quadratic terms to the energy. Hence, the energy per molecule of a soft diatomic molecule is

Question 18.

A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.

Answer:

Given: P = 1 atm = 1.013 × 10^{5} Pa

V_{1} = 5 liters = 5 × 10^{-3} m^{3}

V_{2} = 10 liters = 10 × 10^{-3}m^{3}

Q = 400 J

The work done by the system (gas in this case) on its surroundings.

W = P(V_{2} – V_{1})

= (1.013 × 10^{5}) (10 × 10^{-3}m^{3} – 5 × 10^{-3}m^{3})

= 1.013 (5 × 10^{2})J

= 5.065 × 10^{2}J

The change in the internal energy of the system.

∆U = Q – W = 400J – 506.5J = -106.5 J

The minus sign shows that there is a decrease in the internal energy of the system.

Question 19.

Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.

Answer:

The expression is given as

T = 2π\(\sqrt{\frac{1}{m B}}\)

Explanation: The time period of oscillation of a magnet in a uniform magnetic field B is given by

Formula:

T = 2π\(\sqrt{\frac{1}{m B}}\)

where T = lime period

I = Moment of inertia

M = Mass of bob

B = magnetic field

Time-period of an oscillation body about a fixed point can be defined as the time taken by the body to complete one vibration around that particular point is called time period.

Question 20.

A wave of frequency 500 Hz is travelling with a speed of 350 m/s.

(i) What is the phase difference between two displacements at a certain point at times 1.0 ms apart?

(ii) What will be the smallest distance between two points which are 45° out of phase at an instant of time?

Answer:

Given: n = 500 Hz, v = 350 m/s

v = n × λ

∴ λ = \(\frac{350}{500}\) = 0.7 m

(i) t = 1.0 ms = 0.00 1 s, the path difference is the distance covered v × t= 350 × 0.001 = 0.35 m

∴ Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference

= \(\frac{2 \pi}{0.7}\) × 0.35 = π rad

(ii) Phase difference = 45° = \(\frac{\pi}{4}\) rad

∴ Path difference = \(\frac{\lambda}{2 \pi}\) × Phase difference

= \(\frac{0.7}{2 \pi} \times \frac{\pi}{4}\) = 0.0875 m.

Question 21.

Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.

Answer:

Description of Young’s doubles-slit interference experiment

(i) A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens. It is then made to pass through on opaque screen AB having two narrow and similar slits S_{1} and S_{2}. S_{1} and S_{2} are equidistant from S so that the wavefronts starting simultoneousty from S and reaching S_{1} and S_{2} at the same time are in phase.

A screen PQ is placed at some distance from AB as shown in the following figure.

(ii) S_{1} and S_{2} act as secondary sources. The crests/ troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in the above figure. The point where these lines meet the screen have high intensity and its bright

(iii) Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference. These dark and bright regions are called fringes or bands and the whole pattern is called an interference pattern.

Condition for occurrence of dark and bright fringes on the screen: Consider Young’s double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength A emerges out of two narrow and closely spaced, parallel slits S_{1} and S_{2} of equal widths. The separation S_{1}S_{2} = d is very small. The interference pattern is observed on a screen placed parallel to the plane of S_{1}S_{2} and at considerable distance. D(D >> d) from the slits. OO’ is the perpendicular bisector of a segment S_{1}S_{2}.

Geometry of the double-slit experiment: Consider, a point P on the screen at a distance y from O’ (y << 0). The two light waves from S_{1} and S_{2} reach P along paths S_{1}P and S_{2}P, respectively. If the path difference (∆I) between S_{1}P and S_{2}P is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S_{1}P and S_{2}P is a half-integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.

From figure,

The expression for the fringe width (or band width):

The distance between consecutive bright (or dark) fringes is called the fringe width (or bandwidth) W. Point P will be bright (maximum intensity), if the path difference,

∆l = y_{n}\(\frac{d}{D}\) = nλ where n = 0, 1, 2, 3, ………

Point P will be dark (minimum intensity equal to zero), if

y_{m}\(\frac{d}{D}\) = (2m – 1)\(\frac{\lambda}{2}\), where m = 1, 2, 3,………..

Thus, for bright fringes (or bands),

y_{n} = 0, λ\(\frac{\mathrm{D}}{\mathrm{~d}}\), \(\frac{2 \lambda D}{d}\) ……..

and for dark fringes (or bands),

y_{n} = \(\frac{\lambda}{2} \frac{\mathrm{D}}{d}\), 3\(\frac{\lambda}{2} \frac{D}{d}\), 5\(\frac{\lambda}{2} \frac{\mathrm{D}}{d}\) ….

These conditions show that the bright and dark fringes (or bands) occur alternately and equally spaced. For point O’, the path difference (S_{2}O’ – S_{1}O’) = 0. Hence, point O’ will be bright It corresponds to the centre of the central bright fringe (or band). On both sides of O’, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit

Let y_{n} and y_{n + 1} be the distances of the nth and (n + 1)^{th} bright fringes from the central bright fringe.

∴ \(\frac{y_n d}{D}\) = nλ

∴ y_{n} = \(\frac{n \lambda D}{d}\) …. (4)

and \(\frac{y_{n+1} d}{D}\) = (n + 1)λ

∴ (y_{n+1}) = \(\frac{(n+1) \lambda D}{d}\) ….. (5)

The distance between consecutive bright fringes

= y_{n+1} – y_{n} = \(\frac{\lambda D}{d}\)[(n + 1) – n]

= \(\frac{\lambda D}{d}\) …..(6)

Hence, the fringe width,

∴ W = ∆y = y_{n+1} – y_{n} = \(\frac{\lambda D}{d}\) …..(7)

(for bright fringes)

Alternatively, let y_{m} and y_{m + 1} be the distances of the m^{th} and (m + 1)^{th} dark fringes respectively from the central bright fringe.

Equations (7) and (11) show that the fringe width is the same for bright and dark fringes.

Question 22.

A potentiometer wire has a length of 1.5 m and resistance of 10Ω. It is connected in series with the cell of emf 4 volt and internal resistance 5Ω. Calculate the potential drop per centimetre of the wire.

Answer:

Given: L = 1.5 m, R = 10 Ω, E = 4 V, r = 5Ω

K = \(\frac{E R}{(R+r) L}\)

∴ K = \(\frac{4 \times 10}{(10+5) 1.5}\) = \(\frac{40}{15 \times \frac{15}{10}}\)

= \(\frac{400}{225}\) V/m = \(\frac{400}{22500}\) V/cm

= 0.0178 V/cm

The potential drop per centimetre of the wire is 0.0178 V/cm.

Question 23.

An electron is moving with a speed of 3 × 10^{-7} m/s in a magnetic field of 6 × 10^{-4} T perpendicular to its path. What will be the radius of the path? What will be frequency and the energy in keV? [Given: mass ‘ of electron = 9 × 10^{-31} kg, charge e = 1.6 × 10^{-19} C, 1 eV = 1.6 × 10^{-19}J]

Answer:

Given: v = 3 × 10^{7} m/s, B = 6 × 10^{-4} T

m_{e} = 9 × 10^{-31} kg, e = 1.6 × 10^{-19} C

1 eV= 1.6 × 10^{-19}J

The radius of the circular path,

Since the magnetic force does not change the kinetic energy of the charge,

Question 24.

Obtain and expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.

Answer:

In the Bohr model of a hydrogen atom, the electron of charge e performs a uniform circular motion around the positively charged nucleus. Let r, y and T be the orbital radius, speed and period of motion of the electron. Then,

T = \(\frac{2 \pi r}{v}\) ….. (1)

Therefore, the orbital magnetic moment associated with this orbital current loop has a magnitude,

I = \(\frac{e}{T}\) = \(\frac{e v}{2 \pi r}\) ……. (2)

Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude

M_{0} = current × area of the loop

= I(πr^{2}) = \(\frac{e v}{2 \pi r}\) × πr^{2} = \(\frac{1}{2}\)evr ……. (3)

Multiplying and dividing the righthand side of the above expression by the electron mass me.

M_{0} = \(\frac{e}{2 m_e}\)(m_{e}vr) = \(\frac{e}{2 m_e}\)L_{0} ….. (4)

where L_{0} = m_{e}vr is the magnitude of the orbital angular momentum of the electron \(\vec{M}_0\) is opposite to \(\overrightarrow{L_0}\).

∴ \(\vec{M}_0\) = –\(\frac{e}{2 m_e} \vec{L}_0\) …. (5)

which is the required expression.

According to Bohrs second postulate of stationary orbits in this theory of hydrogen atom, the angular momentum of the electron in the nth stationary orbit is equal to n\(\frac{h}{2 \pi}\), where h is the Planck constant and n is a positive integer. Thus, for an orbital electron,

L_{0} = m_{e}vr

= \(\frac{n h}{2 \pi}\) …….. (6)

Substituting for L_{0} in equation (4),

M_{0} = \(\frac{e n h}{4 \pi m_e}\)

For n = 1, M_{0} = \(\frac{e h}{4 \pi m_e}\)

The quantity \(\frac{e h}{4 \pi m_e}\) is a fundamental constant called the Bohr magneton, µ_{B} . µ_{B} = 9.2 74 × 10^{-24} J/T (or Am^{2})

= 5.788 × 10^{-5} eV/T.

Question 25.

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 an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad.

Here, for e = e_{0} sin ωt,

we have i = i_{0}sin(ωt – \(\frac{\pi}{2}\))

Instantaneous power.

P = ei

= (e_{0}sinωt) [i_{o}(sinωtcos\(\frac{\pi}{2}\) – cosωtsin\(\frac{\pi}{2}\))]

= -e_{0}i_{0}sin ωt cos ωt, as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1

Average power over one cycle,

P_{av} \(=\frac{\text { work done in one cycle }}{\text { time for one cycle }}\)

= \(\int_0^{\top} \mathrm{P} d t / \mathrm{T}\) = \(-\int_0^{\top} e_0 i_0\) sin ωt cos ωt dt/T

= -e_{0}i_{0}/T\(\int_0^{\top}\) sin ωt cos ωt dt

Now, \(\int_0^{\top}\)sin ωt cos ωt dt = 0

P_{av} = 0

Question 26.

Why is the emitter, the base and the collector of a BJT doped differently?

Answer:

A BJT being a bipolar device, both electrons and holes participate in the conduction process. Under the forward- biased condition, the majority carriers injected from the emitter into the base constitute the largest current component in a BJT.

For these carriers to diffuse across the base region with negligible recombination and reach the collector junction, these must overwhelm the majority carriers of the opposite charge in the base.

The total emitter current has two components, that due to majority carriers in the emitter and that due to minority carriers diffused from the base into the emitter. The ratio of the current component due to the injected majority carriers from the emitter to the total emitter current is a measure of the emitter efficiency.

To improve the emitter efficiency and the common-base current gain (a), it can be shown that the emitter should be much heavily doped than the base.

Also, the base width is a function of the base-collector voltage. A low doping level of the collector increases the size of the depletion region. This increases the maximum collector-base voltage and reduces the base width. Further, the large depletion region at the collector- base junction-extending mainly into the collector- corresponds to a smallest electric field and avoids avalanche breakdown of the reverse-biased collector- base junction.

**Section D**

Attempt any Three of the following questions:

Question 27.

(i) State and explain Lenz’s law in the light of principle of conservation of energy.

Answer:

Lenz’s law: The direction of the induced current is such as to oppose the change that produces it.

The change that induces a current may be:

(a) The motion of a conductor in a magnetic field or

(b) The change of the magnetic flux through a stationary circuit.

Explanation: Consider Faraday’s magnet and coil experiment. If the bar magnet is moved towards the coil with its N’-pole facing the coil, as in the shown first figure, the number of magnetic lines of induction (pointing to the left) through the coil increases.

The induced current in the coil sets up a magnetic field of its own pointing to the right (as given by Amperes right-hand rule) to oppose the growing flux due to the magnet. Hence to move the magnet towards the coil against this repulsive flux of the induced current, we must do work. The work done shows up as electric energy in the coil.

Lenz’s law and Faraday’s magnet and coil experiment. The solid curves in (a) and (b) are the lines of magnetic induction of the bar magnet and the dashed curves are those of the induced current.

When the magnet is withdrawn, with its N-pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases. The induced current reverses its direction to supplement the decreasing flux with its own, as shown in the second figure.

Facing the coil along with the magnet, the induced current is in the clockwise sense. The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force. Thus, we see that Lenz’s law is a consequence of the law of conservation of energy.

(ii) Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.

Answer:

Given: M = 4.00151 u, m_{p} = 1.00728 u, m_{n} = 1.00866 u, 1 u = 931.5 MeV/c^{2}

The binding energy of an alpha particle

= (Zm_{p} + Nm_{n} – M)c^{2}

= (2m_{p} + 2m_{n} – M)c^{2}

= [(2) (1.00728u) + 2(1.00866u) – 4.00151u]c^{2}

= (2.01456 + 2.01732 – 4.00151) (931.5) MeV

= 28.289655 MeV

= 28.289655 × 10^{6} eV × 1.602 × 10^{-10} J

= 4.532002731 × 10^{-12}J.

Question 28.

(i) State the characteristics of progressive waves.

Answer:

Progressive wave: A progressive wave is defined as the onward transmission of the vibratory motion of a body in an elastic medium from one particle to the successive particle.

Characteristics of a progressive wave:

- Energy is transmitted from particle to particle without the physical transfer of matter.
- The particles of the medium vibrate periodically about their equilibrium positions.
- In the absence of dissipative forces, every particle vibrates with the same amplitude and frequency but differs in phase from its adjacent particles. Every particle lags behind in its state of motion compared to the one before it.
- Wave motion is doubly periodic, i.e., it is periodic in time and periodic in space.
- The velocity of propagation through a medium depends upon the properties of the medium.
- A transverse wave can propagate only through solids but not through liquids and gases while a longitudinal wave can propagate through any material medium.
- Progressive waves are of two types: Transverse and longitudinal. In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. The progressively changing phase of the successive particles results in the formation of alternate crests and troughs that are periodic in space and time.

In an em wave, the electric and magnetic fields oscillate in mutually perpendicular directions, perpendicular to the direction of propagation. In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave.

The progressively changing phase of the successive particles results in the formation of typical alternate regions of compressions and rarefactions that are periodic in space and time. Periodic compression and rarefactions result in periodic pressure and density variations in the medium. There is no longitudinal em wave.

(ii) One hundred twenty-five small liquid drops, each carrying a charge of 0.5 μC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.

Answer:

n = 125, g = 0.5 × 10^{-6} C, d = 0.1 m

The radius of each small drop.

r = \(\frac{d}{2}\) = 0.05 m

The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the layer drop is

R = \(\sqrt[3]{n}\) × r = \(\sqrt[3]{125}\) × 0.05 = 5 × 0.05 = 0.25 m

The charge on the larger drop,

Q = ng = 125 × (0.5 × 10^{-6})C

∴ The electric potential of the surface of the larger drop.

V = \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\) = (9 × 10^{9}) × \(\frac{125 \times\left(0.5 \times 10^{-6}\right)}{0.25}\)

= 9 × 125 × 2 × 10^{3} = 2.25 × 10^{6}V.

Question 29.

(i) Give an example of some familiar process in which heat is added to an object, without changing its temperature.

Answer:

(a) Melting of ice

(b) Boiling of water

(ii) Find kinetic energy of 5 litre of a gas at S.T.P. given standard pressure is 1.013 × 10^{5} N/m^{2}.

Answer:

Given: P = 1.013 × 10^{5} N/m^{2}, V = 5 litres = 5 × 10^{-3} m^{3}

E = \(\frac{3}{2}\)PV

= \(\frac{3}{2}\)(1.013 × 10^{5} N/m^{2}) (5 × 10^{-3} m^{3})

= 7.5 × 1.013 × 10^{2}J

= 7.597 × 10^{2}J

This is the required energy.

Question 30.

Determine the series limit of Balmer, Paschen and Pfund series, given the limit for Lyman series is 912A.

Answer:

Given: λ_{L∞} = 912 Å

For hydrogen spectrum,

Question 31.

Derive expression for excess pressure inside a drop of liquid?

Answer:

Consider a liquid drop of radius Rand surface tension T. Due to surface tension, the molecules on the surface film experience the net force in the inward direction normal to the surface. Therefore there is more pressure inside than outside.

Let p_{1} be the pressure inside the liquid drop and p_{0} be the pressure outside the drop.

Therefore excess of pressure inside the liquid drop is,

P = p_{1} – p_{0}

Due to excess pressure inside the liquid drop the surface of the drop will experience the net force in outward direction due to which the drop will expand.

Let the free surface displace by dR under isothermal conditions.

Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.

The work done by an excess of pressure in displacing the surface is,

dW = Force × displacement

= (Excess of pressure × surface area) × displacement of the surface

= p × 4πR^{2} × dR …….. (1)

Increase in the potential energy is

dU = surface tension × increase in area of the free surface

= T[4π(R + dR)^{2} – 4πR^{2}]

= T[4π(2RdR)] ……… (2)

From (1) and (2)

p × 4πR^{2} × dR = T[4π(2RdR)]

⇒ p = \(\frac{2 T}{R}\)

The above expression gives us the pressure inside a liquid drop.