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NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices

NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices

Question 1.
In an n-type silicon, which of the following statement is true :
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the doplants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) ‘Holes are minority carriers and pentavalent atoms are the dopants in n type semiconductor.’

Question 2.
Which of the statements given in Exercise 1 is true for p-type semiconductors ?
Answer:
(d) Holes are majority carriers and trivalent atoms are the dopants in p-type semiconductors.

Question 3.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c, (Eg)si and (Eg)Ge-
Which of the following statements is true ?
(a) (Eg)Si < (Eg)Ge < (Eg)c
(b)(E)c<(Eg)Ge>(Eg)si
(c) (Eg)c > (Eg)si > (Eg)Ge
(d) (Eg)c = (Eg)si = (Eg)Ge
Answer:
(C) (Eg)c > (Eg)Si > (Eg)Ge. Energy band gap is maximum in carbon and least in germanium among the given elements.

Question 4.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer:
(c) hole concentration in p-region is more as compared to n-region because hole diffusion takes place from higher concentration to lower concentration.

Question 5.
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier
(b) reduces the majority carrier current to zero
(c) lowers the potential barrier
(d) none of the above.
Answer:
(c) lowers the potential barrier by cancelling the depletion layer.

Question 6.
For transistor action, which of the following statements are correct :
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Answer:
(b) and (c) : The base region must be very thin, lightly doped and the emitter junction is forward biased whereas collector junction is reverse biased to avoid unnecessary diffusion of charge carrier in the base and also for proper amplification.

Question 7.
For a transistor amplifier, the voltage gain :
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Answer:
(c) is low at high and low frequencies and constant at mid frequencies as per frequency response of a transistor.

Question 8.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Answer:
A half wave rectifier rectifies only one half cycle of input A.C.
.’. frequency of the output A.C.
= frequency of input A.C. = 50 Hz A full wave rectifier rectifies both halve cycles
of the A.C. input
.’. frequency of output A.C. = 2 x frequency of input A.C. = 2 x 50 = 100 Hz

Question 9.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kQ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Answer:


Question 10.
Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.
Answer:


Question 11.
A p-n photodiode is fabricated from a semiconductor with bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm ?
Answer:

Since the energy of the light photon is less than the bandgap energy of the p-n diode, it can not be detected.

Question 12.
The number of silicon atoms per m³ is 5 x 10²⁸. This is doped simultaneously with 5 x 10²² atoms per m³ of Arsenic and 5 x 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_i = 1.5 x 10¹⁶ m^-3. Is the material n-type or p- type ?
Answer:


Question 13.
In an intrinsic semiconductor, the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What are the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration n_i is given by

Answer:

Thus, conductivity of a semiconductor increases with rise in temperature.

Question 14.
In a P-n junction diode, the current I can be expressed as

where I₀ is reverse saturation current. V is the voltage across the diode and is positive for forwarding bias and negative for reverse bias, and I is the current through the diode, K_B is the Boltzmann constant (8.6 x 10^-5 eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 x 10^-12 A and T = 300 K, then
(a) What will be the forward current at a forwarding voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer:
The statement of the given question is incorrect. The relation should be

Question 15.
You are given the two circuits as shown in Figure. Show that circuit (a) acts as OR gate while circuit (b) acts as AND gate.

Answer:


Question 16.
Write the truth table for a NAND gate connected as given in Fig. Hence identify the exact logic operation carried out by these circuits.

Answer:
The NAND gate shown in the truth table has only one input. Therefore, the truth table is

Since Y = \( \bar { A }\) in this case, the circuit is actually a NOT gate with the truth table

Question 17.
You are given two circuits as shown in Fig., which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:

Question 18.
Write the truth table for circuit given in Fig. below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

Answer:
Let y₁ be the output which appears at the first operation of NOR gate.

Question 19.
Write the truth table for the circuits given in Fig., consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits

Answer:

We hope the NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

Question 1.
Predict the direction of induced current in the situations described by the following Fig. 2(a) to (f).


Answer:
(a) By Lenz’s law, the face of the coil towards the south pole of the magnet opposes the south pole. So this face should behave as the south pole. Hence the current flows along qrpq.
(b) Similar to the above reason the current flows along yzxy and along prqp.
(c) When the coil is energised, with a cell the increasing current produces an inverse current in the nearby coil along yzxy.
(d) Similar to the above reason the current flows along zyvz.
(e) By Lenz’s law current is along xryx.
(f) Field lines being along the plane of the loop, there is no induced current.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by Figure:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.


Answer:
When a wire of irregular shape turns into a circular (loop the magnetic flux linked with the loop increases due to an increase in area) The irregular shape. The induced e.m.f. will cause current to flow in such a direction, so that the wire forming the loop is pulled inward from all sides. It requires that the current should flow.
By applying Lenz’s Law, it follows that the current will flow in the direction a, d, c, b, a.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1s, what is the induced e.m.f. in the loop while the current is changing?
Answer:


Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut ¡s moving out of a region of the uniform magnetic field of magnitude O.3 T directed normal to the mop. What ¡s the e.m.f. developed across the cut if the velocity of the loop is 1 cm s^-2 in a direction normal to the
(a) longer side,
(b) the shorter side of the loop? For how long does the induced voltage last in each case?
Answer:
e = Blυ

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^_1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the e.m.f. developed between the center and the ring.
Answer:

Question 6.
A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 500 rad s^_1 a uniform horizontal magnetic field of magnitude 3.0 x 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:

(v) Source of power loss is the external rotor which provides the necessary torque to rotate the coil.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms^-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 x 10^-4 Wb m^-2.
(a) What is the instantaneous value of the e.m.f. induced in the wire?
(b) What is the direction of the e.m.f.?
(c) Which end of the wire is at the higher electrical potential?
Answer:

(b) Using Fleming’s Right-hand rule, the direction of induced e.m.f. is from West to East.
(c) Since the rod will act as a source, the Western end will be at a higher electrical potential.

Question 8.
Current in a circuit falls from 5.0 A to 0.0 As. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.
Answer:

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
dΦ_B = M.dl = 1.5 x (20 – 0) = 30 Wb

Question 10.
A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 x 10^-4 T and the dip angle is 30°?
Answer:

Question 11.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^-1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Answer:

The Source of this power is the external agency which brings a change in a magnetic field.

Question 12.
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^_1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^-3 T cm^-1 along the negative x-direction (that is it increases by 10^-3 T cm^-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10^-3 T s^_1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Answer:


The direction of the induced current is such that it increases the magnetic flux linking with the loop in positive 2-direction.

Question 13.
It is desired to measure the magnitude of the field between the poles of a powerful loudspeaker magnet. A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction. The total charge flew in the coil (measured by a ballistic galvanometer connected to the coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of the magnet.
Answer:

Question 14.
The figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed-loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience a magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^_1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed-circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer:
Here, B = 0.50 T; 1 = 15 cm = 15 x 10^-2 m;
R = 9.0m Q = 9.0 x 10³ fl
(a) Now, e = Bvl
Here, V = 12 cm s^{-1 =} 12 x 10^-2 ms^-1
∴ e = 0.50 x 12 x 10-²x 15 x 10 ² = 9 x 10^-3V
If q is a charge on as electron, then the electrons in the rod will experience magnetic Lorentz force \(-q[\vec { v } +\vec { B } ]\) P. Q. Hence, the end P of the rod will become positive and the end Q will become negative.

(b) When the switch K is open, the electron collects at the end Q. Therefore, excess change is built up at the end Q. However, when the switch K is closed, the accumulated charge at the end Q flows through the circuit,

(c) The magnetic Lorentz force on the electron is cancelled by the electronic force acting on it due to the electronic field set up across the two ends due to the accumulation of positive and negative charges at the ends P and Q respectively.

(d) Retarding force, F = BIl =B \(\frac { e }{ R } \)

Question 15.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^-3 s. How much is the average back emf induced across the ends of the open switch in the circuit ? Ignore the variation in magnetic field near the ends of the solenoid.
Answer:

Question 16.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Figure

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, υ = 10 m/s. Calculate the induced e.m.f. in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.
Answer:
(a) Consider a small portion of the coil of thickness dt at a distance t from the current-carrying wire. Then the magnetic field strength experienced by this portion

Question 17.
A line charge X per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Figure.) A uniform magnetic field extends over a circular region within the rim. It is given by

What is the angular velocity of the wheel after the field is suddenly switched off?

Answer:
Change in magnetic field is given by,

We hope the NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Question 1.
Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in figure). The charging current is constant and equal to 0.15 A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoffs first rule valid at each plate of the capacitor ? Explain.

Answer:

Yes. Because conduction current entering one plate is equal to the displacement current leaving that plate.

Question 2.
A parallel plate capacitor (shown in the figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c. supply with an (angular) frequency of 300 rad s^_1.
(a) What is the r .m.s. value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of \( \overrightarrow { B } \) at a point 3.0 cm from the axis between the plates.

Answer:

Question 3.
What physical quantity is the same for X-rays of g g wavelength 10^-10 m, the red light of wavelength 6800 A, and radiowaves of wavelength 500 m?
Answer:
The speed in a vacuum is the same for all are c = 3 x 10⁸ ms^-1. (Electromagnetic waves)

Question 4.
A plane electromagnetic wave travels in a vacuum along Z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
The electric field vector \( \overrightarrow { E } \) and magnetic field vector\( \overrightarrow { B } \) are in xy plane. They are normal to each other.

Question 5.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band ?
Answer:

=> The corresponding wavelength band is 40 m to 25 m.

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
The frequency of the electromagnetic waves produced is the same as that of the oscillating charged particle. Hence the frequency of the electro­magnetic waves produced is, υ= 10⁹ Hz.

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is B₀ =510 nT. What is the electromagnetic waves produced by the oscillator?
Answer:

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz.
(a) Determine, B₀, ω, k and λ.
(b) Find expressions for E and B.
Answer:

Question 9.
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Answer:
Using the relation for photon energy,


Conclusion. The above result indicates that the different wavelengths in the electromagnetic spectrum can be obtained by multiplying roughly the powers often.
The visible wavelengths are spaced by a few eV.
The nuclear energy levels (from y rays) are spaced about 1 MeV.

Question 10.
In-plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 10¹⁰ Hz and amplitude 48 Vm^_1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. [c= 3 x 10⁸ ms^-1].
Answer:

Question 11.
Suppose that the electric field part of an electromagnetic wave in a vacuum is
E = {(3.1 N/C) cos[(1.8 rad/m)y + (5.4 x 10⁶ rad/s)f]}\(\hat { i } \).
(a) What is the direction of propagation?
(b) What is the wavelength X?
(c) What is the frequency v?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Answer:

Question 12.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation?
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer:

Question 13.
Use the formula λ_mT = 029 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer:

These numbers tell us the range of temperature required to obtain the different parts of the spectrum. For example, to obtain a wavelength of 1 μm, a temperature of 2900 K is required.

Question 14.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift.)
(c) 2.7 K (temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe.)
(d) 5890 Å-5896 Å (double lines of sodium)
(e) 14.4 keV [energy of a particular transition in ⁵⁷Fe nucleus associated with a famous high-resolution spectroscopic method (Mossbauer spectroscopy).]
Answer:
(a) Radio waves (short-wavelength end)
(b) Radio waves (short-wavelength end)
(c)

(d) Given wavelength is of the order of 10^-7 m i.e visible radiations(yellow light)

Question 15.
Answer the following questions:

1. Long-distance radio broadcasts use shortwave bands. Why?  (C.B.S.E. 2005)
   
2. It is necessary to use satellites for long-distance TV transmission. Why? (C.B.S.E. 2005)
   
3. Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (C.B.S.E. 2009)
   
4. The small ozone layer on top of the stratosphere is crucial for human survival. Why ?(C.B.S.E. 2005, 2009)
   
5. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
6. Some scientists have predicted that a global nuclear war on the earth would be followed by a severe nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction? (C.B.S.E. 1995)
   

Answer:

1. The ionosphere reflects waves in these bands.
2. Television signals are of >30 MHz penetrate the ionosphere. Therefore, reflection is effected by satellites.
3. The atmosphere absorbs X-rays. while visible and radio waves can penetrate it.
4. It absorbs ultraviolet radiation from the sun and prevents it from reaching the earth’s surface and causing damage to life.
5. The temperature of the earth would be lower because the Greenhouse effect of the atmosphere would be absent.
6. The clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’. against which life on earth cannot withstand.

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NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems

NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems.

Question 1.
At which of the following frequency/frequencies the communication will not be reliable for a receiver situated beyond the horizon:
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) is correct. Here (c) and (d) frequencies have high penetration power so the earth will absorb them. Radiation (a) of 10 kHz will suffer from the problem of the size of the antenna.

Question 2.
Frequencies in the UHF range normally propagate by means of
(a) ground waves
(b) sky waves
(c) surface waves
(d) space waves.
Answer:
(d) space waves.

Question 3.
Digital signals (i) do not provide a continuous set of values, (ii) represent values as discrete steps, (Hi) can utilize the only binary system, and (iv) can utilize decimal as well as a binary system. Which of the following options is true :
(a) Only (i) and (ii).
(b) Only (ii) and (iii).
(c) Only (i), (ii) and (iii), but not (iv).
(d) AH the above (i) to (iv).
Answer:
(c) is correct because the decimal system is concerned with continuous values (i) to (iii).

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer:
For line-of-sight communication, it is necessary that the transmitting antenna and receiving antenna should be eye to eye but it is not necessary that they should be at the same height.

Question 5.
A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?
Answer:

Question 6.
A modulation signal is a square wave as shown in the figure. The carrier wave is given by
C(t) = 2 sin(8πt) V

(a) Sketch the amplitude modulated waveform.
(b) What is the modulation index?
Answer:


Accordingly, the amplitude modulated waveform is shown ahead:

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index μ. What would be the value of μ if the minimum amplitude is zero V?
Answer:

Question 8.
Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let there be two signals represented by
A_c cos ω_ct and A₀ cos(ω_c + ω_m)t where A_c is the
amplitude, ω_c is the angular frequency of a carrier wave at the receiving end and A₀ is the amplitude, (ω_c+ ω_m) is the angular velocity of the modulated wave.
Multiplying these signals, we get

We hope the NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Question 1.
Two charges 5 x 10^-8 C and -3 x 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Ans. Let the potential be zero at 0, then
Answer:
Let the potential be zero at 0, then v_A + V_B= 0, where V_A is electric potential due to charge q_A and V_B is the electric potential due to charge q_B.

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center of the hexagon.
Answer:
Total potential at O is given by,

Question 3.
Two charges 2μC and -2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system. % What is the direction of the electric field at every through the mid-point. On this plane, the potential is zero everywhere.
(b) The direction of the electric field is from positive to negative charge i.e. A to B, which is in fact perpendicular to the equipotential plane.
Answer:
(a) A surface containing an equatorial line and a perpendicular line.
(b) Towards the side of – ve charge, parallel to the axis.

Question 4.
A spherical conductor of a radius of 12 cm has a charge of 1.6 x 10^-7 C distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the center of the sphere?
Answer:
(a) inside a conductor, the electric field is zero because the charge resides on the surface of a conductor.
(b) Electric field just outside the sphere is given by

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF
(1 pF = 10^-12 F.) What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer:

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Answer:

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer:
(a) Total capacitance
= c₁ + c₂ + c₃
= 2 + 3+ 4 = 9pF.
(b) Using C = \(\frac { q }{ v } \) we get q = CV
∴ q_x = C₁V = 2 x 10-¹² x 100
= 2 x 10-¹⁰ C = 200 pC
q₂ = c₂V
= 3 x 10^-12 x 100
= 3 x 10-¹⁰ C = 300 pC
q₃ = c₃v
= 4 x 10-¹² x 100
= 4 x 10-¹⁰ C = 400 pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of
6 x 10^-3 in² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?
Answer:

Question 9.
Explain what would happen if in the capacitor given in Q. 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
Answer:

Question 10.
A 12 pF capacitor is connected to a 50 V battery.How much electrostatic energy is stored in the capacitor ?
Answer:
E = \(\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 12 x 10-¹² x 50 x 50 = 1.5 2 2 x 10-⁸J.

Question 11.
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from die supply and is connected  to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?
Ans.
Here, C₁ = 600pF = 6 x 10^-10 F, C₂ =
6 x 10-¹⁰ F, V₁ = 200V, V₂ = 0

Question 12.
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10^-9 C from a point P(0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point
R(0, 6 cm, 9 cm.)
Answer:
The work done by electrostatic force on a charge is independent of the path followed by the charge. It depends only on the initial and final positions of the charge.

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the center of the cube.
Answer:
(1) Distance of the center of the cube from vertex is half of the diagonal of the cube

(2) From symmetry, it is clear that electric field at center of the cube is zero.

Question 14.
Two tiny spheres carrying charges 1.5 μc and 2.5 μc are located 30 cm apart. Find the potential and electric field :
(a) at the mid point of the line joining the two charges,
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid­point.
Answer:

Question 15.
A spherical conducting shell of inner radius r_l and outer radius r₂ has a charge Q.
(a) A charge q is placed at the center of the shell. What is 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, even if the shell is not spherical, but has any irregular shape? Explain.
Answer:
(a) Charge Q appears on the outer surface.

When charge q is placed at the center, it induces – q charge on the inner surface and +q on the outer surface.
.’. charge density of the inner surface,

and charge density of the outer surface,

Consider a cavity of an irregular shape with the net charge to be zero inside it. Let a closed-loop be partially inside and the rest outside the cavity. The field inside the conductor is zero, so some work is done by the field to carry a test charge in the closed-loop, but this is against the provisions of an electrostatic field because as per Gauss’s law, the net charge inside a Gaussian surface must be zero. Thus, there cannot be field lines inside the cavity irrespective of its shape.

Question 16.
(a) Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by

where \(\hat { n } \) is a unit vector normal to the surface at a point and a is the surface charge density at that point. (The direction of h is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ\(\hat { n } \) /ε₀
(b) Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another.[Hint. For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Answer:
Consider a sheet of charge having charge density a. E on either side of the sheet, perpendicular to the plane of sheet, has same magnitude at all points equidistant from the sheet.
Electric field intensity on the left side of the sheet,

The electric field tangential to the plate is continuous throughout.

Question 17.
a long charged cylinder of linear charged density A. is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Answer:
A cylinder P has linear charge density, λ, length l, and radius r₁
The charge on cylinder P, q = XL A hollow co-axial conducting cylinder of length / and radius r₂ surrounds the cylinder P. Charge on cylinder Q = – q.

Consider a Gaussian surface in the form of a cylinder of radius r and length l. The electric flux through the curved surface of the Gaussian surface,

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential Energy of the system in eV, taking the zero of the potential energy at an infinite separation of the electron from the proton.
(b) What is the minimum work required to free the electron, given that it’s kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?
Answer:

Question 19.
If one of the two electrons of an H₂ molecule is removed, we get a hydrogen molecular ion H⁺₂. In the ground state of an H⁺₂, the two protons are separated by roughly 1.5 Å, and the electron is roughly Å from each proton. Determine the potential energy of the system. Specify your choice of zero potential energy.
Answer:

Question 20.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of the electric field at the surfaces of the two spheres? Use the result obtained to explain why the charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer:
Two charged conducting spheres of radii a and b connected by a wire will reach to the same potential.

Clearly, electric charge density for the pointed surface will be more because a flat surface can be equated to a spherical surface of large radius and a pointed portion to a spherical surface of small radius.

Question 21.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a > > 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Answer:

The point (x, y, 0) is perpendicular to Z-axis, there­fore, the potential at (x, y, 0) is zero.
(b) Consider P to be the point of observation at a distance r from the center (O) of the electric dipole.
Let OP make an angle 0 with the dipole moment \( \overrightarrow { p } \)
and r_1, r₂ be the distances of point P from – q charge and + q charge respectively. Potential at P due to – q charge,



The answer does not change because, in electrostatics, the work done does not depend upon the actual path, it simply depends upon the initial and final positions.

Question 22.
The figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a > > l, and contrast your results with that due to an electric dipole, and an electric monopole
(i.e., a single charge.)

Answer:

(2) Due to electric dipole, the potential is of 1/r² type.
(3) Due to an electric monopole, the potential is of 1/r type.

Question 23.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Let N capacitors be used in m rows when each row has n capacitor i.e. N = mn
In series

Question 24.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm ? (You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance ((0.1 F) because of very minute separation between the conductors.)
Answer:

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor. (C.B.S.E. 2008)

Answer:
The equivalent circuit is as shown below :

Potential difference across C₄ is in the ratio 2:1
i.e., 200 V
.’. Charge on C₄ = C₄V₄
= 100 x 200 x 10^-12 = 2 x 10^-8 C
Potential difference across C₁= 100 V
Charge on C₁ = C₁x V₁
=100 x 100 x 10-¹² = 1 x 10^-8 C
Potential difference across C₂ and C₃ is 50 v each
∴ Charge on C₂ or C₃ = C₂V₂
= 200 x 50 x 10^-12 = 10-⁸ C.

Question 26.
The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:

Question 27.
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Answer:

Question 28.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor and E is the magnitude of the electric field between the plates. Explain the origin of the factor (1/2).
Answer:
Let F be the force on each plate of the capacitor. If the distance between the plates of the capacitor is increased by dx, then work done = F dx. This work done is stored as the potential energy of the capacitor. The increase in the volume of capacitor = A dx

Question 29.
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports (Figure.) Show that the capacitance of a spherical capacitor is given by

where r_x and r₂ are the radii of outer and inner spheres, respectively.
Answer:
It consists of two concentric spherical shells A and B of radii a and b with charge +q and charge -q respectively. (Outer sphere is grounded)

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 pC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer:

Question 31.
Answer carefully:
(a) Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of the electrostatic force between them exactly given by Q₁Q₂/4Πε₀r², where r is the distance between their centers?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that the electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (=6).
Answer:
(a) No. The given relation is Coulomb’s law which is true for point charges. In the present case, as the spheres are brought closer, the distribution of charge on them becomes nonuniform.

(b) No. The surface area in space varies as r₂ so that field varies as \(\frac { 1 }{ { r }^{ 2 } }\). Hence \(\frac { 1 }{ { r }^{ 2 } }\) dependence is essential.

(c) Not necessarily. The motion of charged particles need not be along the line of the field. It does so in the uniform field. The field gives the direction of acceleration and not that of velocity in general.

(d) Zero. For any complete path in the electrostatic field (the shape does not matter), it is zero.

(e) No. Potential is continuous there.

(f) The single conductor can form a condenser with the other conductor at infinity. Hence the meaning of storage of charge retains.

(g) Water molecules are polar molecules.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 pC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects, (i.e., bending of field lines at the end)
Answer:
The capacitance of a cylindrical capacitor is given by


Question 33.
A parallel plate capacitor is to be designed with a voltage rating of 1 kV, using a material of dielectric constant 3 and dielectric strength of about 10⁷ Vm^-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionization.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Answer:
10% of the given field i.e. 10⁷ V m¹ gives E = 0.1 X 10⁷ Vnr¹

Question 34.
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
(a) A plane parallel to XY plane.
(b) Plane parallel to XY plane but the planes having different fixed potential will become closer with the increase in field intensity.
(c) Concentric spheres with origin as the center.
(d) A time-dependent changing shape nearer to the grid which slowly becomes planar and parallel to the grid at far off distances from the grid.

Question 35.
In a van de Graaff type generator a spherical metal shell is to be a 15 X 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 x 10⁷ vm^-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.) (C.B.S.E. 2008)
Answer:
The minimum radius of the shell of the van de Graaff generator is given by the relation

Question 36.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂ . Show that if q₁ is positive, the charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge
q₂ on the shell is.
Answer:
Charge resides on the outer surface of a conductor. So the charge on the inner sphere will flow towards the shell through the conducting wire. Moreover, from Gauss’s law, no electric field exists inside a Gaussian surface, and also the charges enclosed by a closed surface only contribute towards the field. So q₂ does not matter in this case. It is positive, a potential difference is also positive.

Question 37.
Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decrases with altitude. Near the surface of the earth, the field is about 100 Vm^_1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a stell cage so there is no field inside !)
(b) A man fixes outside his house one evening a two meter high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning ?
[Hint. The earth has an electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9 C m^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about +1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.]
Answer:
(a) Our body and the ground form an equipotential surface. As we step out into the open, the original equipotential surfaces of open-air change, keeping our head and the ground at the same potential.

(b) Yes. The steady discharging current in the atmosphere charges up the aluminum sheet gradually and raises its voltage to an extent depending on the capacitance of the capacitor (formed by the sheet, slab, and the ground).

(c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing currents are, on average, in equilibrium.

(d) Light energy involved in lightning; heat and sound energy in the accompanying thunder.

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