CBSE Class 12 Physics 2019 Solved QP Set-I

 

Section-A

Question 1.
Draw equipotential surfaces for an electric dipole.
 

Answer:

 

Question 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 potential difference V, the direction of magnetic field is normal to velocity of proton.
As we know

Question 3.
The magnetic susceptibility X of magnesium at 300 K is \(1.2 × 10^5\) . At what temperature will its magnetic susceptibility become \(1.44 × 10^5\) ?
OR
The magnetic susceptibility X of a given material is – 0.5. Identify the magnetic material.
 

Answer:

 

Question 4.
Identify the semiconductor diode whose V-I characteristics are as shown. 
 

Answer: Photo diode.

 

Question 5.
Which part of the electromagnetic spectrum is used in RADAR? Give its frequency range.
OR
How are electromagnetic waves produced by accelerating charges ?
Answer : Microwaves [1GHz to 100 GHz].
OR
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.

 

Question 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,
\(∮Bdl=μ_0i(t)\)
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, \(i_c\) (conduction current) flows and inside \(i_d\) (displacement current) flows.

 

Question 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:

Question 8.
A photon emitted during the de-excitation of electron from a state n the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photo cell, with a stopping potential of 0.55V. Obtain the value of the quantum number of the state n.
OR
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:

 

Question 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
Draw a labelled ray diagram to show 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,

Question 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?

 

Question 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:

S.N.	Primary Rainbow (Fig-1)	Secondary Rainbow  (Fig-1)
1	Three Step process (Re­fraction Reflection and Refraction).	Four Step process (Refraction-Reflec­tion and Refraction)
2	Appearance intensity better than Second­ary.	Appearance inten­sity lesser than Pri­mary.
3.	Single reflection oc­curs.	Double reflection oc­curs.
4.	2 Degree range occurs.	3 Degree range.

Question 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 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 at setting.

 

Question 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
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:


(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, voltage drop across LC combination is zero at resonating frequency.

 

Question 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. < \(10^6\) m. Hence, electric field across the junction is very high (~\(5 × 10^6\) V/m) even for a small reverse bias voltage. This can lead to a break down during reverse biasing.

Question 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
(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, deutrons, 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 consist of two-semi cylindrical boxes \(D_1\) and \(D_2\) , 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 \(D_1\) be positive and \(D_2\) be negative. A proton from an ion source will be accelerated towards \(D_2\) , 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 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
(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 limb. In this case magnetic field creates two poles N and S. Solenoids have some flux leakage. This is used in relay, motors, electro-magnetes

 

Question 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:

 

Question 17.
Two large charged plane sheets of charge densities a and \(-2σ C/m^2\) 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
A spherical conducting shell of inner radius \(r_1\) and outer radius \(r_2\) has a charge Q.
(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:

 

Question 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 expression for the lower and upper side bands produced. Hence, obtain the expression for modulation index.

 

Question 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 a 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-particles 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 small % of the incident particles rebound back indicates that number of α-particle goes head on collision. Hence most of the mass of the atom is concentrated in small volume. Thus, Rutherford scattering is a strong tool to determine upper limit to the size of the nucleus.

 

Question 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.

 

Question 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
(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 anti neutrino 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. Neutrions 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 dN 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
(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.

 

Question 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 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.)

 

Question 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 the 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.

 

Question 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.  [5]
OR
(a) Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain the expression for the magnetic energy density.
(b) A square loop of sides 5 cm carrying a current of 0-2 A in the clockwise direction is placed at a distance of 10 cm from an indefinitely 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 \(B_2\) to \(B_1\).
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.


(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.

 

Question 26.
Explain, with the help of a diagram, how plane polarized light can be produced by scattering of light from the Sun.
Two polaroids \(P_1\) and \(P_2\) are placed with their pass axes perpendicular to each other. Unpolarised light of intensity I is incident on \(P_1\). A third polaroid \(P_3\) is kept between \(P_1\) and \(P_2\) such that its pass axis makes an angle of 45° with that of \(P_1\). Calculate the intensity of light transmitted \(P_1\), \(P_2\) and \(P_3\).
OR
(a) Why cannot the phenomenon of interference be observed by illuminating two pin holes with two sodium lamps?
(b) Two monochromatic waves having displacements \(y_1\)= a cos ωf and \(y_2\) = a cos (ωt + Φ ) from two coherent sources interfere to produce an interference pattern. Derive the expression for the resultant intensity 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 \(2×10^{-6}\) m. If the distance between the slit and the screen is 1.5 m, calculate the separation between the positions of the second maxima of diffraction pattern obtained in the two cases.
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.

(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 same phase).
(b) Consider two monochromatic coherent sources A and B with waves \(y_1\) = a cos ωt and \(y_2\) = a cos (ωt + Φ ) respectively.




 

Question 27.
(a) Describe briefly, with the help of a circuit
diagram, the method of measuring the internal resistance of a cell.
(b) Give reason 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. [5]

(a) State the working principle of a meter bridge used to measure an unknown resistance.
(b) Give reason.
(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 mid-point 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 cell because a voltmeter draws some current from the cell while potentiometer draws no current. Therefore, the potentionmeter measures the actual e.m.f. of cell whereas voltmeter measures the terminal voltage.


Hence, balancing will not possible as it needs to cater 300 mV.
OR
(a) Meter bridge is the practical apparatus which works on principle of Wheat-Stone bridge. It is used to measure 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 mid-point of the bridge wire because it increase the sensitivity of meter bridges.
(c) From KCL (Kirchhoff’s current law) at point D

 

 

CBSE Class 12 Physics 2019  Set-II

Question 1.
Write the relation for the force acting on a charged particle q moving with velocity v⃗  in the presence of a magnetic field \(\overrightarrow{B}\) 
Answer:
\(\overrightarrow{F}=q(\overrightarrow{V}×\overrightarrow{B})\)

 

Question 2.
Draw the pattern of electric field lines due to an electric dipole.
Answer:

 

Question 3.
Identify the semiconductor diode whose I-V characteristics are as shown.

Answer: Photodiode/Solar cell.

 

Question 4.
How is the equation for Ampere’s circuital law modified in the presence of displacement current? Explain.
Answer:


case-1
Let, the case-1 where a point P is considered outside capacitor charging from Ampere’s law. Magnetic field at point will be

Case-2
Now take case-2 where shape surface under consideration covers capacitor’s plate as we consider there is no current through capacitor then this value of B will be zero.
Hence, there is a contradiction. Therefore this Ampere’s law was modified with addition of displacement current inside capacitor.

Where, id displacement current.
During charging of capacitor, outside the capacitor, ic (conduction current) flows and
inside id displacement current flows.

Question 5.
Under what conditions does the phenomenon of total internal reflection take place? Draw a ray diagram showing how a ray of light deviates by 90° after passing through a right-angled isosceles prism.
Answer:
The phenomenon of total internal reflection occurs when,

• Angle of incidence is equal or greater than critical angle.
  i ≥ C
• When light travels from more denser medium to less denser medium.In case of right angle isosceles triangle if light rays fall normally on AB then light incident of face AC with angle of incidence > critical angle.

Hence, total internal reflection will occur with normal to the surface of BC.
 

Question 6.
A beam of light converges at a point P. Draw ray diagrams to show where the beam will converge if
(i) a convex lens, and
(ii) a concave lens is kept in the path of the beam.
Answer:

 

Question 7.
(a) How is the stability of hydrogen atom in Bohr model explained by de-Broglie’s hypothesis?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to n = 4 level. When it gets de-excited, find the maximum number of lines which are emitted by the atom. Identify the series to which these lines belong. Which of them has the shortest wavelength?
Answer:
(a) From Bohr’s model-An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds to a certain energy level. Electron revolves is circular orbit

The motion of an electron in circular orbits is restricted in such a manner that its angular momentum is an integral multiple h/2Π

Question 8.
What is the reason to operate photo-diodes in reverse bias?
A p-n photo diode is fabricated from a semiconductor with a band gap of range of 2.5 to 2.8 eV. Calculate the range of wavelengths of the radiation which can be detected by the photo diode.
Answer : Photo diode are reverse biased for working in photo-conductive mode. This reduces the response time because the additional reverse bias increases the width of the depletion layer, which decreases the junction capacitance. The reverse bias also increases the dark current without much change in the photo current.

The range of wavelengths of radiation which can be detected by the photo diode is 443.97 nm to 497.25 nm.

 

Question 9.
A ray of light incident on the face AB of an . isosceles triangular prism makes an angle of incidence (i) and deviates by angle P as shown in the figure. Show that in the position of minimum deviation ∠β = ∠α . Also find out the condition when the refracted ray QR suffers total internal reflection.

Answer:

 

Question 10.
A 100 μF parallel plate capacitor having plate separation of 4 mm is charged by 200 V dc. The source is now disconnected. When the distance between the plates is doubled and a dielectric slab of thickness 4 mm and dielectric constant 5 is introduced between the plates, how will
(i) its capacitance,
(ii) the electric field between the plates, and
(iii) energy density of the capacitor get affected? Justify your answer in each case.


As dielectric of 4 mm is inserted between the plates of capacitor and the spacing between the plates is doubled then it will acts as following Fig-A and Fig-B.

 

 

CBSE Class 12 Physics 2019  Set-III

Question 1.
Draw a pattern of electric field lines due to two positive charges placed a distance d apart. 
Answer : Electric field lines due to two positive charge placed to at distance d apart:

Question 2.
When a charge q is moving in the presence of electric (E) and magnetic (B) fields which are perpendicular to each other and also perpendicular to the velocity v of the particle, write the relation expressing v in terms of E and B. 
Answer : v = E/B

 

Question 3.
Draw the I-V characteristics of a Zener diode. 
Answer:

 

Question 3.
State, with the help of a ray diagram, the working principle of optical fibers. Write one important use of optical fibers. 
Answer:

Optical fiber works on principle of total internal reflection. When angle of incidence is greater than Critical angle then incident rays are totally reflected back in same media.
When, θi > θc , Total internal reflection occurs and if θi < θc , refraction occurs.
Application: Optical fiber are used for communication due to very high bandwidth of media.

 

Question 4.
How are electromagnetic waves produced by oscillating charges ? What is the source of the energy associated with the em waves. 
Answer : Oscillating charges are responsible for generation of periodically varying electric field in the space. The oscillating charges generate varying electric current which in turn is responsible for the generation of periodical varying magnetic field. This way the electromagnetic waves are generated.

 

Question 5.
The wavelength of light from the spectral emission line of sodium is 590 nm. Find the kinetic energy at which the electron would have the same deBroglie wavelength.

 

Question 6.
(a) Draw the energy level diagram for the line spectra representing Lyman series and Balmer series in the spectrum of hydrogen atom.
(b) Using the Rydberg formula for the spectrum of hydrogen atom, calculate the largest and shortest wavelengths of the emission lines of the Balmer series in the spectrum of hydrogen atom. (Use the value of Rydberg constant R = 1.1 x 107 m-1 

Answer:
(a) Energy level diagram showing lyman and balmer series :

Spectrum wavelngths of both series for hydrogen atom

Question 7.
In a network, four capacitors C1, C2, C3 and C4are connected as shown in the figure. [3]

(a) Calculate the net capacitance in the circuit.
(b) If the charge on the capacitor C1 is 6 μC,
(i) calculate the charge on the capacitors C3 and C4, and
(ii) net energy stored in the capacitors C3 and C4 connected in series.
Answer:
(a)

 

Question 8.
Draw the circuit diagram of a full wave rectifier. Explain its working principle. Show the input wave forms given to the diodes D1 and D2 and the corresponding output wave forms obtained at the load connected to the circuit.

Answer:

During first half of input sinusoidal ac signal diode D1 conducts as it is forward bias and during second half of input ac signal diode D2 conducts as it is forward bias now. D2 and D1 are inverse bias conditions during first and second half respectively and doesn’t conduct. Due to this output appears as waveform (b).

Question 9.
(a) When a convex lens of focal length 30 cm is in contact with a concave lens of focal length 20 cm, find out if the system is converging or diverging.
(b) Obtain the expression for the angle of incidence of a ray of light which is incident on the face of a prism of refracting angle A so that it suffers total internal reflection at the other face. (Given the refractive index of the glass of the prism is μ).
Answer: