Physics
SET-I
Section-A
Q.1. Draw equipotential surfaces for an electric dipole.
Q.2. A proton is accelerated through a potential difference V, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change ?
Answer: Given, proton accelerated through a potential difference V, the direction of magnetic field is normal to velocity of protons.
As we know
Q.3. The magnetic susceptibility X of magnesium at 300 K is 1-2 × 105 . At what temperature will its magnetic susceptibility become 1-44 × 105 ?
OR
Q.3. The magnetic susceptibility X of a given material is – 0.5. Identify the magnetic material.
Answer:
Q.4. Identify the semiconductor diode whose V-I characteristics are as shown.
Answer: Photo diode.
Q.5. Which part of the electromagnetic spectrum is used in RADAR? Give its frequency range.
OR
Q.5. How are electromagnetic waves produced by accelerating charges ?
Answer : Microwaves [1GHz to 100 GHz].
Answer: An oscillating electric field in space, produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other.
Section – B
Q.6. A capacitor made of two parallel plates, each of area ‘A’ and separation W is charged by an external d.c.-source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor.
Answer:
From Ampere’s law,
Let the case-1, where a point P is considered outside the capacitor charging.
From Ampere’s law magnetic field at point P will be :
Now, take case-2 where shape of surface under consideration covers capacitor’s plate as we consider there is no current through capacitor then this value of B will be zero.
Case-2
Hence, there is a contradiction.
Therefore, this Ampere’s law was modified with addition of displacement current inside capacitor.
Where, id is displacement current.
During charging of capacitor, outside the capacitor, ic (conduction current) flows and inside ic (displacement current) flows.
Q.7. A photon and a proton have the same deBroglie wavelength X. Prove that the energy of the photon is (2mλc/h) times the kinetic energy of the proton.
Answer:
Q.8. A photon emitted during the de-excitation of electrons from a state in the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photocell, with a stopping potential of 0.55V. Obtain the value of the quantum number of the state n.
OR
Q.8. A hydrogen atom in the ground state is excited by an electron beam 12-5 eV energy. Find out the maximum number of lines emitted by atom from its excited state.
Answer:
Q.9. Draw the ray diagram of an astronomical telescope showing image formation in the normal adjustment position. Write the expression for its magnifying power.
OR
Q.9. Draw a labelled ray diagram to show the image formation by a compound microscope and write the expression for its resolving power.
Answer:
The magnifying power m is the ratio of the angle P subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye. Hence,
Q.10. Write the relation between the height of a TV antenna and the maximum range up to which signals transmitted by the antenna can be received. How is this expression modified in the case of line of sight communication by space waves? In which range of frequencies, is this mode of communication used?
Answer : The curvature of the earth limits the distance up to which a signal can be transmitted by a tower.
If the height of the transmitting antenna is ‘h’ and the radius of the earth is ‘R’, then the optimum distance ‘d’between the receiving and the transmitting antenna is given by:
d= √2Rh
Q.11. Under which conditions can a rainbow be observed? Distinguish between a primary and a secondary rainbow.
Answer : The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of Sunlight by spherical water droplets of rain. The conditions for observing a rainbow are that the Sun should be shining in one part of the sky (say near western horizontal while it is raining in the opposite part of the sky (say eastern horizon).
Difference between Primary and Secondary Rainbow:
Q.12. Explain the following :
(a) Sky appears blue.
(b) The Sun appears reddish at (i) sunset, (ii) sunrise
Answer: (a) Light from the sun reaches the atmosphere that is comprised of the tiny particles of the atmosphere. These act as a prism and cause the different components to scatter. As blue light travels in shorter and smaller waves in comparison to the other colours of the spectrum. It is scattered the most , causing the sky to appear bluish.
(b) The molecules of the atmosphere and other particles that are smaller than the longest wavelength of visible light are more effective in scattering light of shorter wavelengths than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. (Rayleigh Effect) Light from the Sun near the horizon passes through a greater distance in the Earth’s atmosphere than does the light received when the Sun is overhead. The correspondingly greater scattering of short wavelengths accounts for the reddish appearance of the Sun at rising and setting.
Section – C
Q.13. A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50 Hz. The potential difference across C and R are respectively 120 V, 90 V, and the current in the circuit is 3 A. Calculate
(i) the impedance of the circuit
(ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.
OR
Q.13. The figure shows a series LCR circuit connected to a variable frequency 230 V source.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Calculate the impedance of the circuit and amplitude of current at resonance.
(c) Show that potential drop across LC combination is zero at resonating frequency.
Answer:
Answer: (a) Source frequency will be same as resonance frequency of LC circuit,
(c) As at resonance frequency impedance of combination of L and C is 0.
Hence, the voltage drop across LC combination is zero at resonating frequency.
Q.14. Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below :
(Zener breakdown voltage is 15 V)
Answer : By heavily doping both p and n sides of the junction, depletion region formed is very thin, i.e. < 106 m. Hence, the electric field across the junction is very high (~5 × 106V/m) even for a small reverse bias voltage. This can lead to a break down during reverse biasing.
Q.15. Draw a labelled diagram of cyclotron. Explain its working principle. Show that cyclotron frequency is independent of the speed and radius of the orbit.
OR
Q.15. (a) Derive, with the help of a diagram, the expression for the magnetic field inside a very long solenoid having n turns per unit length carrying a current I.
(b) How is a toroid different from a solenoid?
Answer: Cyclotron : Cyclotron is a device by which the positively charged particles like protons, deuterons, etc. can be accelerated.
Principle : Cyclotron works on the principle that a positively charged particle can be accelerated by making it to cross the same electric field repeatedly with the help of a magnetic field.
Construction : The construction of a simple cyclotron is shown in figure above, it consists of two-semi cylindrical boxes D1 and D2, which are called Dees They are enclosed in an evacuated chamber.
Chamber is kept between the poles of a powerful magnet so that uniform magnetic field acts perpendicular to the plane of the dees. An alternating voltage is applied in the gap between the two dees by the help of a high frequency oscillator. The electric field is zero inside the dees.
Working and theory : At a certain instant, let D1 be positive and D2 be negative. A proton from an ion source will be accelerated towards D2 , it describes a semi-circular path with a constant speed and is acted upon only by the magnetic field. The radius of the circular path is given by.
From the above equation it follows that the frequency f is independent of both v and r and is called cyclotron frequency. Also if we make the frequency of applied a.c. equal to f, then every time the proton reaches the gap between the dees, the direction of electric field is reversed and proton receives a push and finally it gains very high kinetic energy. The proton follows a spiral path and finally gets directed towards the target and comes out from it.
OR
Answer: (a) Magnetic field inside the solenoid
(b) Toroid is a form in which a conductor is wound around a circular body. In this case we get magnetic field inside the core but poles are absent because circular body don’t have ends. Toroid is used in toroidal inductor, toroidal transformer.
Solenoid is a form in which conductor is wound around a cylindrical body with limbs. In this case magnetic field creates two poles N and S. Solenoids have some flux leakage. This is used in relay, motors, electro-magnetes.
Q.16. Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr’s postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state.
Answer:
Q.17. Two large charged plane sheets of charge densities and -2σ C/m2 are arranged vertically with a separation of d between them. Deduce expressions for the electric field at points
(i) to the left of the first sheet,
(ii) to the right of the second sheet, and
(iii) between the two sheets.
OR
Q.17. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Questions.
(a) A charge q is placed at the center of the shell. Find out the surface charge density on the inner and outer surfaces of the shell.
(b) Is the electric field inside a cavity (with no charge) zero; independent of the fact whether the shell is spherical or not? Explain.
Answer:
Q.18. A signal of low frequency fm is to be transmitted using a carrier wave of frequency fc . Derive the expression for the amplitude modulated wave and deduce expressions for the lower and upper side bands produced. Hence, obtain the expression for modulation index.
Q.19. Draw a plot of a-particle scattering by a thin foil of gold to show the variation of the number of the scattered particles with scattering angle. Describe briefly how the large angle scattering explains the existence of the nucleus inside the atom, Explain with the help of impact parameter picture, how Rutherford scattering serves a powerful way to determine and upper limit on the size of the nucleus.
Answer:
From the plot it is clear that Most of the a-particles passed through the foil,, only 0.14% of the incident particles scatter by more than 1% and about 1 in 8000 deflect by more than 90° a-particles deflected backward due to strong repulsive force. This force will come from positive charge concentrated at the center as most of the particles get deflected by small angles.
The α-particle’s trajectory depends on collision’s impact parameter (b) for a given beam of a-particles, distribution of impact parameters as beam gets scattered in different directions with different probabilities.
fig.2 shows a-particle close to nucleus suffers large scattering. Impact parameter is minimum for head on collision a-particles rebound by 180°. Impact parameter is high, for undeviated a-particles. With deflection angle = 0°.
As these of nucleus was 10-14 m to 10-15 m w.r.t. 10-10 m size of an atom which is 10,000 to 100,000 times larger hence most of the space is empty, only a small % of the incident particles rebound back indicates that the number of α-particle goes head on collision. Hence most of the mass of the atom is concentrated in a small volume. Thus, Rutherford scattering is a strong tool to determine upper limit to the size of the nucleus.
Q.20. A 200 µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the
(i) capacitance,
(ii) electric field between the plates,
(iii) energy density of the capacitor will change ?
As dielectric of 5 mm is inserted with spacing between the dielectric doubled then it will act as following-Fig.A and Fig-B.
Q.21. Why is it difficult to detect the presence of an anti-neutrino during β -decay ? Define the term decay constant of a radioactive nucleus and derive the expression for its mean life in terms of the decay constant.
OR
Q.21. (a) State two distinguishing features of nuclear force.
(b) Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. Mark the regions on the graph where the force is
(i) attractive, and (ii) repulsive.
Answer : The symbols v and v present antineutrino and neutrino respectively during j3 decay both are neutral particles. With very little or no mass. These particles are emitted from the nucleus along with the electron or positron during the decay process. Neutrinos interact very weakly with matter, they can even penetrate the earth without being absorbed. It is for this reason that their detection is extremely difficult and their presence went unnoticed for long.
Decay constant: Decay constant of a radioactive element is the reciprocal of time during which the number of atoms left in the sample reduces to 1/r times the number of atoms in the original sample.
Derivation of mean life : Let us consider, No be the total number of radioactive atoms present initially. After time t, total no. of atoms present (undecayed) be N. In further dt time dN be the no. of atoms disintegrated. So, the life of dN atoms ranges lies between t + dt and dt. Since, dt is very small time, the most appropriate life of aN atom is t. So the total life of N atom = f.dN
Now, substituting the value of dN and changing the limit in equation (i) from (ii) we get
This expression gives the relation between mean life and decay constant. Hence, mean life is reciprocal of decay constant.
OR
Answer: (a) Distinguish features of nuclear force are :
(i) Nuclear forces are very strong binding forces (attractive force.)
(ii) It is independent of the charges protons and neutrons (charge independent.)
(iii) It depends on the spins of the nucleons.
(b) Plot showing variation of potential energy of a pair of nucleons as a function of separation mark attractive and repulsive region.
X-axis shows separation between pair of nucleons and Y-axis shows variation of potential energy w.r.t. Separation.
(in x 10-15 m).
Q.22. A triangular prism of refracting angle 60° is made of a transparent material of refractive index 2/√3. A ray of light is incident normally on the face KL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation.
Answer: From the diagram it is clear that incidence angle at face KM is 60°.
Hence, critical angle is also 60°.
Therefore, incident light ray will not emerge from KM face due to total internal reflection at this face. Hence, it will move along face KM Angle of emergence = 90°.
Hence angle of deviation = 30° (from fig.)
Q.23. Prove that in a common-emitter amplifier, the output and input differ in phase by 180°. In a transistor, the change of base current by 30 µA produces change of 0 02 V in the base-emitter voltage and a change of 4 mA in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is 400.
Q.24. Show, on a plot, variation of resistivity of
(i) a conductor, and
(ii) a typical semiconductor as a function of temperature.
Using the expression for the resistivity in . terms of number density and relaxation time between collisions, explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of temperature.
Answer:
n → number of free electrons
t → Average time between collisions.
In metals n is not dependent on temperature to any appreciable extent and thus the decrease in the value of x with rise in temperature causes p to increases.
for semiconductors, n increases with temperature. This increases more than compensates any decrease in t, so that for such materials, p decreases with temperature.
Section – D
Q.25.(a) Derive an expression for the induced
EMF developed when a coil of N turns, and area of cross-section A, is rotated at a constant angular speed o in a uniform magnetic field B.
(b) A wheel with 100 metallic spokes each 0-5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. If the resultant magnetic field at that place is 4 × 10-4 and the angle of dip at the place is 30°, find the emf induced between the axle and the rim of the wheel.
OR
Q.25. (a) Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain an expression for the magnetic energy density.
(b) A square loop of sides 5 cm carries a current of 0-2 A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of 1 A as shown. Calculate
(i) the resultant magnetic force, and
(ii) the torque, if any, active on the loop.
Answer: (a) As the armature coil is rotated in the magnetic field, angle 0 between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An e.m.f. is induced in the coil. According to Fleming’s right hand rule, current induced in AB is from A to B and it is from C to D in CD in the external circuit current flows from B2 to B1.
To calculate the magnitude of e.m.f. Induced:
Suppose,
A → Area of each turn of the coil
N → Number of turns in the coil
B → Strength of magnetic field
θ → Angle which normal to the coil makes
Magnetic flux linked with the coil in this position.
Answer: (a) Energy stored in an inductor : When a current flows through an inductor, a back e.m.f. is set up which opposes the growth of current. So, work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.
Let I be the current through the inductor L at any instant t.
The forces acting on all sides of the square due to current of infinite length wire are lying in the plane of coil. Thus, there is no net torque. Thus torque is zero.
Q.26. Explain, with the help of a diagram, how plane polarized light can be produced by scattering of light from the Sun.
Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity I is incident on P1. a third polaroid P3 is kept between P1 and P2 such that its pass axis makes an angle of 450 with that of P1. calculate the intensity of light transmitted P1, P2 and P3.
OR
Q.26. (a) Why cannot the phenomenon of interference be observed by illuminating two pin holes with two sodium lamps?
(b) Two monochromatic waves having displacements y1= a cos ωf and y2= a cos (ωt + Φ ) from two coherent sources interfere to produce an interference pattern. Derive the expression for the resultant in¬tensity and obtain the conditions for constructive and destructive interference.
(c) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 x 106.
Answer: Molecules behave like dipole radiators and scatter no energy along the dipole axis by this way plane polarized light can be produced during scattering of light.
Answer: (a) Phenomenon of interference can’t be observed by illuminating two pin holes with two sodium lamps because these sources are not coherent source (it means they are not in the same phase).
(b) Consider two monochromatic coherent sources A and B with waves y1 = a cos ωt and y2 = a cos (ωt + Φ ) respectively.
Q.27. (a) Describe briefly, with the help of a circuit diagram, the method of measuring the internal resistance of a cell.
(b) Give reasons why a potentiometer is preferred over a voltmeter for the measurement of emf of a cell.
(c) In the potentiometer circuit given below, calculate the balancing length l. Give reason, whether the circuit will work, if the driver cell of emf 5 V is replaced with a cell of 2 V, keeping all other factors constant.
(a) State the working principle of a meter bridge used to measure an unknown resistance.
(b)Give reasons.
(i) why the connections between the resistors in a meter bridge are made of thick copper strips.
(ii) why is it generally preferred to obtain the balance length near the midpoint of the bridge wire.
(c) Calculate the potential difference across the 4 Ω resistor in the given electrical circuit, using Kirchhoff’s rules.
Answer:
Where, r is the internal resistance of cell.
(b) Potentiometer is preferred over voltmeter for measurement of e.m.f. of cells because a voltmeter draws some current from the cell while potentiometer draws no current. Therefore, the potentiometer measures the actual e.m.f. of cell whereas voltmeter measures the terminal voltage.
Hence, balancing will not possible as it needs to cater to 300 mV.
OR
Answer: (a) Meter bridge is the practical apparatus which works on principle of Wheat-Stone bridge. It is used to measure an unknown resistance experimentally.
(b) (i) Connection between resistors are made of thick copper strips so that it will have maximum resistance and location of point of balance (D) will be more accurate which results in correct measurement of unknown resistance.
(ii) It is preferred to obtain the balance length near the midpoint of the bridge wire because it increases the sensitivity of meter bridges.
(c) From KCL (Kirchhoff’s current law) at point D.
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