Explore the comprehensive NCERT solutions for Class 12 Science Physics Chapter 7 on Communication Systems. These solutions, complete with clear step-by-step explanations, are highly sought after by students. They prove invaluable for completing homework assignments efficiently and preparing for exams. All the questions and answers from the Class 12 Science Physics Chapter 7 in the NCERT Book are available here for free. These Physics Communication Systems Solutions offer a convenient resource for students to grasp concepts and enhance their understanding of the subject. Page No 530: Question 15.1: Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves? (a) 10 kHz (b) 10 MHz (c) 1 GHz (d) 1000 GHz ANSWER: (b) Answer: 10 MHz For beyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size. The high energy signal waves (1GHz − 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication. Page No 530: Question 15.2: Frequencies in the UHF range normally propagate by means of: (a) Ground waves. (b) Sky waves. (c) Surface waves. (d) Space waves. ANSWER: (d) Answer: Space waves Owing to its high frequency, an ultra high frequency (UHF) wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation. Page No 530: Question 15.3: Digital signals (i) Do not provide a continuous set of values, (ii) Represent values as discrete steps, (iii) Can utilize binary system, and (iv) Can utilize decimal as well as binary systems. Which of the above statements are true? (a) (i) and (ii) only (b) (ii) and (iii) only (c) (i), (ii) and (iii) but not (iv) (d) All of (i), (ii), (iii) and (iv). ANSWER: (c) Answer: A digital signal uses the binary (0 and 1) system for transferring message signals. Such a system cannot utilise the decimal system (which corresponds to analogue signals). Digital signals represent discontinuous values. Page No 530: Question 15.4: 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 81m tall. How much service area can it cover if the receiving antenna is at the ground level? ANSWER: Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communications it is not necessary for the transmitting and receiving antennas to be at the same height. Height of the given antenna, h = 81 m Radius of earth, R = 6.4 × 106 m For range, d = (2Rh)½, the service area of the antenna is given by the relation: A = πd2 = π (2Rh) = 3.14 × 2 × 6.4 × 106× 81 = 3255.55 × 106 m2 = 3255.55 ∼ 3256 km2 Page No 530: Question 15.5: 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: Amplitude of the carrier wave, Ac = 12 V Modulation index, m = 75% = 0.75 Amplitude of the modulating wave = Am Using the relation for modulation index: Page No 530: Question 15.6: A modulating signal is a square wave, as shown in Fig. 15.14. The carrier wave is given by  (i) Sketch the amplitude modulated waveform (ii) What is the modulation index? ANSWER: It can be observed from the given modulating signal that the amplitude of the modulating signal, Am = 1 V It is given that the carrier wave c (t) = 2 sin (8πt) ∴Amplitude of the carrier wave, Ac = 2 V Time period of the modulating signal Tm = 1 s The angular frequency of the modulating signal is calculated as: The angular frequency of the carrier signal is calculated as: From equations (i) and (ii), we get: The amplitude modulated waveform of the modulating signal is shown in the following figure. (ii)Modulation index,  Page No 531: Question 15.7: 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 volt? ANSWER: Maximum amplitude, Amax = 10 V Minimum amplitude, Amin = 2 V Modulation index μ, is given by the relation: Page No 531: Question 15.8: Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. 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 ωc and ωs be the respective frequencies of the carrier and signal waves. Signal received at the receiving station, V = V1 cos (ωc + ωs)t Instantaneous voltage of the carrier wave, Vin = Vc cos ωct At the receiving station, the low-pass filter allows only high frequency signals to pass through it. It obstructs the low frequency signal ωs. Thus, at the receiving station, one can record the modulating signal , which is the signal frequency.

NCERT Solutions for Class 12 Science Physics Chapter 6, titled “Semiconductor Electronics: Materials, Devices And Simple Circuits,” offer comprehensive explanations with step-by-step guidance. These solutions are widely favored by Class 12 Science students for Physics, as they prove invaluable for completing assignments and preparing for exams. All the questions and answers from this chapter in the NCERT Book are provided here for free, aiding students in their understanding of Semiconductor Electronics: Materials, Devices And Simple Circuits. Page No 509: Question 14.1: 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 dopants. (d) Holes are majority carriers and trivalent atoms are the dopants. ANSWER: The correct statement is (c). In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms. Page No 509: Question 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductors. ANSWER: The correct statement is (d). In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms. Page No 509: Question 14.3: 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) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge ANSWER: The correct statement is (c). Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge Page No 509: Question 14.4: 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: The correct statement is (c). The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region. Page No 510: Question 14.5: 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: The correct statement is (c). When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced. Page No 510: Question 14.6: 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: The correct statement is (b), (c). For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased. Page No 510: Question 14.7: 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: The correct statement is (c). The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies. Page No 510: Question 14.8: 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: Input frequency = 50 Hz For a half-wave rectifier, the output frequency is equal to the input frequency. ∴Output frequency = 50 Hz For a full-wave rectifier, the output frequency is twice the input frequency. ∴Output frequency = 2 × 50 = 100 Hz Page No 510: Question 14.9: For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ 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: Collector resistance, RC = 2 kΩ = 2000 Ω Audio signal voltage across the collector resistance, V = 2 V Current amplification factor of the transistor, β = 100 Base resistance, RB = 1 kΩ = 1000 Ω Input signal voltage = Vi Base current = IB We have the amplification relation as: Voltage amplification  Therefore, the input signal voltage of the amplifier is 0.01 V. Base resistance is given by the relation: Therefore, the base current of the amplifier is 10 μA. Page No 510: Question 14.13: In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration niis given by where n0 is a constant. ANSWER: Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV The temperature dependence of the intrinsic carrier-concentration is written as: Where, kB = Boltzmann constant = 8.62 × 10−5 eV/K T = Temperature n0 = Constant Initial temperature, T1 = 300 K The intrinsic carrier-concentration at this temperature can be written as:  … (1) Final temperature, T2 = 600 K The intrinsic carrier-concentration at this temperature can be written as:  … (2) The ratio between the conductivities at 600 K and at 300 …

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Explore comprehensive NCERT Solutions for Class 12 Science Physics Chapter 5 – Nuclei, featuring clear and concise step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, proving invaluable for completing homework assignments efficiently and preparing for exams. Free access to all questions and answers from the NCERT Book of Class 12 Science Physics Chapter 5 is available here, ensuring a convenient and effective study resource for students. Page No 462: Question 13.1: (a) Two stable isotopes of lithium  and have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, and. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of  and . ANSWER: (a) Mass of lithium isotope , m1 = 6.01512 u Mass of lithium isotope , m2 = 7.01600 u Abundance of , η1= 7.5% Abundance of , η2= 92.5% The atomic mass of lithium atom is given as: (b) Mass of boron isotope , m1 = 10.01294 u Mass of boron isotope , m2 = 11.00931 u Abundance of , η1 = x% Abundance of , η2= (100 − x)% Atomic mass of boron, m = 10.811 u The atomic mass of boron atom is given as: And 100 − x = 80.11% Hence, the abundance of  is 19.89% and that of is 80.11%. Page No 462: Question 13.2: The three stable isotopes of neon: and have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. ANSWER: Atomic mass of , m1= 19.99 u Abundance of , η1 = 90.51% Atomic mass of , m2 = 20.99 u Abundance of , η2 = 0.27% Atomic mass of , m3 = 21.99 u Abundance of , η3 = 9.22% The average atomic mass of neon is given as: Page No 462: Question 13.3: Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u ANSWER: Atomic mass of nitrogen, m = 14.00307 u A nucleus of nitrogen  contains 7 protons and 7 neutrons. Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn= 1.008665 u ∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307 = 7.054775 + 7.06055 − 14.00307 = 0.11236 u But 1 u = 931.5 MeV/c2 ∴Δm = 0.11236 × 931.5 MeV/c2 Hence, the binding energy of the nucleus is given as: Eb = Δmc2 Where, c = Speed of light ∴Eb = 0.11236 × 931.5  = 104.66334 MeV Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV. Page No 462: Question 13.4: Obtain the binding energy of the nuclei  and in units of MeV from the following data: = 55.934939 u = 208.980388 u ANSWER: Atomic mass of, m1 = 55.934939 u nucleus has 26 protons and (56 − 26) = 30 neutrons Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn − m1 Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939 = 26.20345 + 30.25995 − 55.934939 = 0.528461 u But 1 u = 931.5 MeV/c2 ∴Δm = 0.528461 × 931.5 MeV/c2 The binding energy of this nucleus is given as: Eb1 = Δmc2 Where, c = Speed of light ∴Eb1 = 0.528461 × 931.5  = 492.26 MeV Average binding energy per nucleon  Atomic mass of, m2 = 208.980388 u nucleus has 83 protons and (209 − 83) 126 neutrons. Hence, the mass defect of this nucleus is given as: Δm‘ = 83 × mH + 126 × mn − m2 Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm‘ = 83 × 1.007825 + 126 × 1.008665 − 208.980388 = 83.649475 + 127.091790 − 208.980388 = 1.760877 u But 1 u = 931.5 MeV/c2 ∴Δm‘ = 1.760877 × 931.5 MeV/c2 Hence, the binding energy of this nucleus is given as: Eb2 = Δm‘c2 = 1.760877 × 931.5 = 1640.26 MeV Average bindingenergy per nucleon =  Page No 462: Question 13.5: A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of atoms (of mass 62.92960 u). ANSWER: Mass of a copper coin, m’ = 3 g Atomic mass of atom, m = 62.92960 u The total number of atoms in the coin Where, NA = Avogadro’s number = 6.023 × 1023atoms /g Mass number = 63 g nucleus has 29 protons and (63 − 29) 34 neutrons ∴Mass defect of this nucleus, Δm‘ = 29 × mH + 34 × mn − m Where, Mass of a proton, mH = 1.007825 u Mass of a neutron, mn = 1.008665 u ∴Δm‘ = 29 × 1.007825 + 34 × 1.008665 − 62.9296 = 0.591935 u Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022 = 1.69766958 × 1022 u But 1 u = 931.5 MeV/c2 ∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2 Hence, the binding energy of the nuclei of the coin is given as: Eb= Δmc2 = 1.69766958 × 1022 × 931.5  = 1.581 × 1025 MeV But 1 MeV = 1.6 × 10−13 J Eb = 1.581 × 1025 × 1.6 × 10−13 = 2.5296 × 1012 J This much energy is required to separate all the neutrons and protons from the given coin. Page No 462: Question 13.6: Write nuclear reaction equations for (i) α-decay of (ii) α-decay of  (iii) β−-decay of (iv) β−-decay of  (v) β+-decay of (vi) β+-decay of  (vii) Electron capture of  ANSWER: α is a nucleus of helium and β is an electron (e− for β− and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus. For the given cases, the various nuclear reactions can be written as: Page No 462: Question 13.7: A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? ANSWER: Half-life of the radioactive isotope = T years Original amount of the radioactive isotope = N0 (a) After decay, the amount of the radioactive isotope = N It is given that only 3.125% of N0 remains after decay. Hence, we can write: Where, λ = Decay constant t = Time Hence, the isotope will take about 5T years to reduce to 3.125% of its original value. (b) After decay, the amount of the radioactive isotope = N It …

NCERT Solutions for Class 12 Science Physics Chapter 5 – Nuclei Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 4 on Atoms, featuring clear, step-by-step explanations. Widely favored by class 12 Science students, these Physics Atoms Solutions serve as a valuable resource for efficiently completing homework assignments and preparing for examinations. You can access all the questions and answers from the NCERT Book of Class 12 Science Physics Chapter 4 at no cost. These solutions are designed to assist you in grasping the concepts quickly and effectively, making your study sessions more productive. Page No 435: Question 12.1: Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a ………. but has a highly non-uniform mass distribution in ………. (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.) ANSWER: (a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude. (b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force. (c) A classical atom based on Rutherford’s model is doomed to collapse. (d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model. (e) The positively charged part of the atom possesses most of the mass in both the models. Page No 435: Question 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? ANSWER: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10−27 kg) is less than the mass of incident α−particles (6.64 × 10−27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment. Page No 436: Question 12.3: What is the shortest wavelength present in the Paschen series of spectral lines? ANSWER: Rydberg’s formula is given as: Where, h = Planck’s constant = 6.6 × 10−34 Js c = Speed of light = 3 × 108 m/s (n1 and n2 are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for values n1 = 3 and n2 = ∞. Page No 436: Question 12.4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? ANSWER: Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10−19 = 3.68 × 10−19 J Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv Where, h = Planck’s constant Hence, the frequency of the radiation is 5.6 × 1014 Hz. Page No 436: Question 12.5: The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state? ANSWER: Ground state energy of hydrogen atom, E = − 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = − E = − (− 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy = − 2 × (13.6) = − 27 .2 eV Page No 436: Question 12.6: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon. ANSWER: For ground level, n1 = 1 Let E1 be the energy of this level. It is known that E1 is related with n1 as: The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level. The amount of energy absorbed by the photon is given as: E = E2 − E1 For a photon of wavelengthλ, the expression of energy is written as: Where, h = Planck’s constant = 6.6 × 10−34 Js c = Speed of light = 3 × 108 m/s And, frequency of a photon is given by the relation, Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 1015 Hz. Page No 436: Question 12.7: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. ANSWER: (a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation, Where, e = 1.6 × 10−19 C ∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2 h = Planck’s constant = 6.62 × 10−34 Js For level n2 = 2, we can write the relation for the corresponding orbital speed as: And, for n3 = 3, we can write the relation for the corresponding orbital speed as: Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively. (b) Let T1 be the orbital period of the electron when it is in level n1 = 1. Orbital period is related to orbital speed as: Where, r1 = Radius of the orbit h = Planck’s constant = 6.62 × 10−34 Js e = Charge on an electron = 1.6 × 10−19 C ∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2 m = Mass …

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Embark on a Journey through Class 12 Science Physics Chapter 3: Exploring the Dual Nature Of Radiation And Matter with our Exclusive NCERT Solutions! Dive into straightforward explanations designed to simplify homework completion and exam preparation. Access all questions and answers from the NCERT Book of class 12 Science Physics Chapter 3 for free, and enjoy an ad-free experience, only on DD Target PMT NCERT Solutions. Page No 407: Question 11.1: Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons. ANSWER: Potential of the electrons, V = 30 kV = 3 × 104 V Hence, energy of the electrons, E = 3 × 104 eV Where, e = Charge on an electron = 1.6 × 10−19 C (a)Maximum frequency produced by the X-rays = ν The energy of the electrons is given by the relation: E = hν Where, h = Planck’s constant = 6.626 × 10−34 Js Hence, the maximum frequency of X-rays produced is (b)The minimum wavelength produced by the X-rays is given as: Hence, the minimum wavelength of X-rays produced is 0.0414 nm. Page No 407: Question 11.2: The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons? ANSWER: Work function of caesium metal,  Frequency of light,  (a)The maximum kinetic energy is given by the photoelectric effect as: Where, h = Planck’s constant = 6.626 × 10−34 Js Hence, the maximum kinetic energy of the emitted electrons is0.345 eV. (b)For stopping potential, we can write the equation for kinetic energy as: Hence, the stopping potential of the material is 0.345 V. (c)Maximum speed of the emitted photoelectrons = v Hence, the relation for kinetic energy can be written as: Where, m = Mass of an electron = 9.1 × 10−31 kg Hence, the maximum speed of the emitted photoelectrons is332.3 km/s. Page No 407: Question 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? ANSWER: Photoelectric cut-off voltage, V0 = 1.5 V The maximum kinetic energy of the emitted photoelectrons is given as: Where, e = Charge on an electron = 1.6 × 10−19 C Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10−19 J. Page No 407: Question 11.4: Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? ANSWER: Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 × 10−9 m Power emitted by the laser, P = 9.42 mW = 9.42 × 10−3 W Planck’s constant, h = 6.626 × 10−34 Js Speed of light, c = 3 × 108 m/s Mass of a hydrogen atom, m = 1.66 × 10−27 kg (a)The energy of each photon is given as: The momentum of each photon is given as: (b)Number of photons arriving per second, at a target irradiated by the beam = n Assume that the beam has a uniform cross-section that is less than the target area. Hence, the equation for power can be written as: (c)Momentum of the hydrogen atom is the same as the momentum of the photon,  Momentum is given as: Where, v = Speed of the hydrogen atom Page No 407: Question 11.5: The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm. ANSWER: Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 103 W/m2 Hence, power of sunlight per square metre, P = 1.388 × 103 W Speed of light, c = 3 × 108 m/s Planck’s constant, h = 6.626 × 10−34 Js Average wavelength of photons present in sunlight,  Number of photons per square metre incident on earth per second = n Hence, the equation for power can be written as: Therefore, every second,  photons are incident per square metre on earth. Page No 407: Question 11.6: In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10−15 V s. Calculate the value of Planck’s constant. ANSWER: The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as: Where, e = Charge on an electron = 1.6 × 10−19 C h = Planck’s constant Therefore, the value of Planck’s constant is  Page No 407: Question 11.7: A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere? ANSWER: Power of the sodium lamp, P = 100 W Wavelength of the emitted sodium light, λ = 589 nm = 589 × 10−9 m Planck’s constant, h = 6.626 × 10−34 Js Speed of light, c = 3 × 108 m/s (a)The energy per photon associated with the sodium light is given as: (b)Number of photons delivered to the sphere = n The equation for power can be written as: Therefore, every second, photons are delivered to the sphere. Page No 408: Question 11.8: The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission. ANSWER: Threshold frequency of the metal,  Frequency of light incident on the metal,  Charge on an electron, e = 1.6 × 10−19 C Planck’s constant, h = 6.626 × 10−34 Js Cut-off voltage for the photoelectric emission from the metal =  The equation for the cut-off energy is given as: Therefore, the cut-off voltage for the photoelectric emission is  Page No 408: Question 11.9: The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? ANSWER: No Work function of the metal,  Charge on an electron, e = 1.6 × 10−19 C Planck’s constant, h = 6.626 × 10−34 Js Wavelength of the incident radiation, λ = 330 nm = …

NCERT Solutions for Class 12 Science Physics Chapter 3 – Dual Nature Of Radiation And Matter Read More »

Explore comprehensive NCERT solutions for Class 12 Science Physics Chapter 2 on Wave Optics, featuring clear and concise step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students for their effectiveness in completing homework assignments and exam preparation. All the questions and answers from Chapter 2 of the NCERT Book for Class 12 Science Physics are available here at your disposal, providing valuable assistance in your studies. Access these free resources to enhance your understanding of Wave Optics and streamline your learning process. Page No 383: Question 10.1: Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33. ANSWER: Wavelength of incident monochromatic light, λ = 589 nm = 589 × 10−9 m Speed of light in air, c = 3 × 108 m/s Refractive index of water, μ = 1.33 (a) The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray. Frequency of light is given by the relation, Hence, the speed, frequency, and wavelength of the reflected light are 3 × 108 m/s, 5.09 ×1014 Hz, and 589 nm respectively. (b) Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.  Refracted frequency, ν = 5.09 ×1014 Hz Speed of light in water is related to the refractive index of water as: Wavelength of light in water is given by the relation, Hence, the speed, frequency, and wavelength of refracted light are 2.26 ×108 m/s, 444.01nm, and 5.09 × 1014 Hz respectively. Page No 383: Question 10.2: What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth. ANSWER: (a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure. (b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure. (c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane. Page No 383: Question 10.3: (a) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in vacuum is 3.0 × 108 m s−1) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism? ANSWER: (a) Refractive index of glass, μ = 1.5 Speed of light, c = 3 × 108 m/s Speed of light in glass is given by the relation, Hence, the speed of light in glass is 2 × 108 m/s. (b) The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism. Page No 383: Question 10.4: In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. ANSWER: Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m Distance between the slits and the screen, D = 1.4 m Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 × 10−2 m In case of a constructive interference, we have the relation for the distance between the two fringes as: Where, n = Order of fringes = 4 λ = Wavelength of light used ∴ Hence, the wavelength of the light is 600 nm. Page No 383: Question 10.5: In Young’s double-slit experiment using monochromatic light of wavelengthλ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ /3? ANSWER: Let I1 and I2 be the intensity of the two light waves. Their resultant intensities can be obtained as: Where, = Phase difference between the two waves For monochromatic light waves, Phase difference =  Since path difference = λ, Phase difference,  Given, I’ = K When path difference, Phase difference,  Hence, resultant intensity,  Using equation (1), we can write: Hence, the intensity of light at a point where the path difference is is units. Page No 383: Question 10.6: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? ANSWER: Wavelength of the light beam,  Wavelength of another light beam,  Distance of the slits from the screen = D Distance between the two slits = d (a) Distance of the nth bright fringe on the screen from the central maximum is given by the relation, (b) Let the nth bright fringe due to wavelength and (n − 1)th bright fringe due to wavelength  coincide on the screen. We can equate the conditions for bright fringes as: Hence, the least distance from the central maximum can be obtained by the relation: Note: The value of d and D are not given in the question. Page No 383: Question …

NCERT Solutions for Class 12 Science Physics Chapter 2 – Wave Optics Read More »

Dive into the World of Light and Lenses with our Tailored NCERT Solutions for Class 12 Science Physics Chapter 1: Ray Optics And Optical Instruments! Crafted with simplicity and precision, these solutions are the go-to resource for students seeking efficient homework completion and exam readiness. Explore the realm of optics with ease, exclusively on DD Target PMT NCERT Solutions. Page No 345: Question 9.1: A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? ANSWER: Size of the candle, h= 2.5 cm Image size = h’ Object distance, u= −27 cm Radius of curvature of the concave mirror, R= −36 cm Focal length of the concave mirror,  Image distance = v The image distance can be obtained using the mirror formula: Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image. The magnification of the image is given as: The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real. If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image. Page No 345: Question 9.2: A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. ANSWER: Height of the needle, h1 = 4.5 cm Object distance, u = −12 cm Focal length of the convex mirror, f = 15 cm Image distance = v The value of v can be obtained using the mirror formula: Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror. The image size is given by the magnification formula: Hence, magnification of the image,  The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually. Page No 345: Question 9.3: A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again? ANSWER: Actual depth of the needle in water, h1 = 12.5 cm Apparent depth of the needle in water, h2 = 9.4 cm Refractive index of water = μ The value of μcan be obtained as follows: Hence, the refractive index of water is about 1.33. Water is replaced by a liquid of refractive index,  The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the needle. Hence, we can write the relation: Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up. ∴Distance by which the microscope should be moved up = 9.4 − 7.67 Page No 345: Question 9.4: Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)]. ANSWER: As per the given figure, for the glass − air interface: Angle of incidence, i = 60° Angle of refraction, r = 35° The relative refractive index of glass with respect to air is given by Snell’s law as: As per the given figure, for the air − water interface: Angle of incidence, i = 60° Angle of refraction, r = 47° The relative refractive index of water with respect to air is given by Snell’s law as: Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as: The following figure shows the situation involving the glass − water interface. Angle of incidence, i = 45° Angle of refraction = r From Snell’s law, r can be calculated as: Hence, the angle of refraction at the water − glass interface is 38.68°. Page No 346: Question 9.5: A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) ANSWER: Actual depth of the bulb in water, d1 = 80 cm = 0.8 m Refractive index of water,  The given situation is shown in the following figure: Where, i = Angle of incidence r = Angle of refraction = 90° Since the bulb is a point source, the emergent light can be considered as a circle of radius,  Using Snell’ law, we can write the relation for the refractive index of water as: Using the given figure, we have the relation: ∴R = tan 48.75° × 0.8 = 0.91 m ∴Area of the surface of water = πR2 = π (0.91)2 = 2.61 m2 Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2. Page No 346: Question 9.6: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive …

NCERT Solutions for Class 12 Science Physics Chapter 1 – Ray Optics And Optical Instruments Read More »

Explore comprehensive solutions for Class 12 Science Physics Chapter 8 – Electromagnetic Waves in the NCERT textbook. These solutions offer step-by-step explanations, making them highly favored among students for quick completion of homework and effective exam preparation. The solutions cover all the questions from Chapter 8, providing valuable assistance to Class 12 Science Physics students. Access these free resources to enhance your understanding and excel in your studies. Page No 285: Question 8.1: Figure 8.6 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 the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. ANSWER: Radius of each circular plate, r = 12 cm = 0.12 m Distance between the plates, d = 5 cm = 0.05 m Charging current, I = 0.15 A Permittivity of free space, = 8.85 × 10−12 C2 N−1 m−2 (a) Capacitance between the two plates is given by the relation, C  Where, A = Area of each plate  Charge on each plate, q = CV Where, V = Potential difference across the plates Differentiation on both sides with respect to time (t) gives: Therefore, the change in potential difference between the plates is 1.87 ×109 V/s. (b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A. (c) Yes Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current. Page No 286: Question 8.2: A parallel plate capacitor (Fig. 8.7) 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 ac supply with a (angular) frequency of 300 rad s−1. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. ANSWER: Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F Supply voltage, V = 230 V Angular frequency, ω = 300 rad s−1 (a) Rms value of conduction current, I  Where, XC = Capacitive reactance ∴ I = V × ωC = 230 × 300 × 100 × 10−12 = 6.9 × 10−6 A = 6.9 μA Hence, the rms value of conduction current is 6.9 μA. (b) Yes, conduction current is equal to displacement current. (c) Magnetic field is given as: B  Where, μ0 = Free space permeability  I0 = Maximum value of current = r = Distance between the plates from the axis = 3.0 cm = 0.03 m ∴B  = 1.63 × 10−11 T Hence, the magnetic field at that point is 1.63 × 10−11 T. Page No 286: Question 8.3: What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m? ANSWER: The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum. Page No 286: Question 8.4: A plane electromagnetic wave travels in 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 electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x–y plane. They are mutually perpendicular. Frequency of the wave, ν = 30 MHz = 30 × 106 s−1 Speed of light in a vacuum, c = 3 × 108 m/s Wavelength of a wave is given as: Page No 286: Question 8.5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? ANSWER: A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 × 106 Hz Maximum frequency, ν2 = 12 MHz = 12 × 106 Hz Speed of light, c = 3 × 108 m/s Corresponding wavelength for ν1 can be calculated as: Corresponding wavelength for ν2 can be calculated as: Thus, the wavelength band of the radio is 40 m to 25 m. Page No 286: Question 8.6: 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 an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz. Page No 286: Question 8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? ANSWER: Amplitude of magnetic field of an electromagnetic wave in a vacuum, B0 = 510 nT = 510 × 10−9 T Speed of light in a vacuum, c = 3 × 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cB0 = 3 × 108 × 510 × 10−9 = 153 N/C Therefore, the electric field part of the wave is 153 N/C. Page No 286: Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B. ANSWER: Electric field amplitude, E0 = 120 N/C Frequency of source, ν = 50.0 MHz = 50 × 106 Hz Speed of light, c = 3 × 108 m/s (a) Magnitude of magnetic field strength is given as: Angular frequency of source is given as: ω = 2πν = 2π × 50 × 106 = 3.14 × 108 rad/s Propagation constant is given as: Wavelength of wave is given as: (b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as: And, magnetic field vector is given …

NCERT Solutions for Class 12 Science Physics Chapter 8 – Electromagnetic Waves Read More »

Explore the comprehensive NCERT solutions for Class 12 Science Physics Chapter 7 on Alternating Current, featuring easy-to-follow step-by-step explanations. Widely favored among Physics students, these solutions are a valuable resource for homework completion and exam preparation. All the questions and answers from Chapter 7 of the NCERT Book for Class 12 Science Physics are available here at no cost, providing students with a convenient tool for efficient learning. Page No 266: Question 7.1: A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? ANSWER: Resistance of the resistor, R = 100 Ω Supply voltage, V = 220 V Frequency, ν = 50 Hz (a) The rms value of current in the circuit is given as: (b) The net power consumed over a full cycle is given as: P = VI = 220 × 2.2 = 484 W Page No 266: Question 7.2: (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? ANSWER: (a) Peak voltage of the ac supply, V0 = 300 V Rms voltage is given as: (b) Therms value of current is given as: I = 10 A Now, peak current is given as: Page No 266: Question 7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. ANSWER: Inductance of inductor, L = 44 mH = 44 × 10−3 H Supply voltage, V = 220 V Frequency, ν = 50 Hz Angular frequency, ω=  Inductive reactance, XL = ω L Rms value of current is given as: Hence, the rms value of current in the circuit is 15.92 A. Page No 266: Question 7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. ANSWER: Capacitance of capacitor, C = 60 μF = 60 × 10−6 F Supply voltage, V = 110 V Frequency, ν = 60 Hz Angular frequency, ω=  Capacitive reactance  Rms value of current is given as: Hence, the rms value of current is 2.49 A. Page No 266: Question 7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. ANSWER: In the inductive circuit, Rms value of current, I = 15.92 A Rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P = VI cos Φ Where, Φ = Phase difference between V and I For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°. Hence, P = 0 i.e., the net power is zero. In the capacitive circuit, Rms value of current, I = 2.49 A Rms value of voltage, V = 110 V Hence, the net power absorbed can ve obtained as: P = VI Cos Φ For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°. Hence, P = 0 i.e., the net power is zero. Page No 266: Question 7.6: Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit? ANSWER: Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 × 10−6 F Resistance, R = 10 Ω Resonant frequency is given by the relation, Now, Q-value of the circuit is given as: Hence, the Q-Value of this circuit is 25. Page No 266: Question 7.7: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? ANSWER: Capacitance, C = 30μF = 30×10−6F Inductance, L = 27 mH = 27 × 10−3 H Angular frequency is given as: Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s. Page No 266: Question 7.8: Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time? ANSWER: Capacitance of the capacitor, C = 30 μF = 30×10−6 F Inductance of the inductor, L = 27 mH = 27 × 10−3 H Charge on the capacitor, Q = 6 mC = 6 × 10−3 C Total energy stored in the capacitor can be calculated by the relation, Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor. Page No 266: Question 7.9: A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? ANSWER: At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit. Resistance, R = 20 Ω Inductance, L = 1.5 H Capacitance, C = 35 μF = 30 × 10−6 F AC supply voltage to the LCR circuit, V = 200 V Impedance of the circuit is given by the relation, At resonance, Current in the circuit can be calculated as: Hence, the average power transferred to the circuit in one complete cycle= VI = 200 × 10 = 2000 W. Page No 266: Question 7.10: A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.] ANSWER: The range of frequency (ν) of a radio is 800 kHz to 1200 kHz. Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz Effective inductance of circuit L = 200 μH = 200 × 10−6 H Capacitance of variable capacitor for ν1 is given as: C1 Where, ω1 = Angular frequency for capacitor C1 Capacitance of variable capacitor for ν2, C2 Where, ω2 = Angular frequency for capacitor C2 Hence, the range of the variable capacitor is from 88.04 pFto 198.1pF. Page No 266: Question 7.11: Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = …

NCERT Solutions for Class 12 Science Physics Chapter 7 – Alternating Current Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Physics Chapter 6 on Electromagnetic Induction. These solutions offer clear, step-by-step explanations that are immensely popular among Class 12 Science students. Whether you’re working on your homework or gearing up for exams, the Physics Electromagnetic Induction Solutions can be a valuable resource for quick and effective preparation. All questions and answers from Chapter 6 of the NCERT Book for Class 12 Science Physics are available here for free, ensuring you have a reliable and accessible aid to support your studies. Page No 229: Question 6.1: Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ). (a) (b) (c) (d) (e) (f) ANSWER: The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively. Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows: (a) The direction of the induced current is along qrpq. (b) The direction of the induced current is along prqp. (c) The direction of the induced current is along yzxy. (d) The direction of the induced current is along zyxz. (e) The direction of the induced current is along xryx. (f) No current is induced since the field lines are lying in the plane of the closed loop. Page No 230: Question 6.2: Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire. ANSWER: According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it. (a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb. (b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along  Page No 230: Question 6.3: 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 0.1 s, what is the induced emf in the loop while the current is changing? ANSWER: Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2 Current carried by the solenoid changes from 2 A to 4 A. Change in current in the solenoid, di = 4 − 2 = 2 A Change in time, dt = 0.1 s Induced emf in the solenoid is given by Faraday’s law as: Where,  = Induced flux through the small loop = BA … (ii) B = Magnetic field =  μ0 = Permeability of free space = 4π×10−7 H/m Hence, equation (i) reduces to: Hence, the induced voltage in the loop is  Page No 230: Question 6.4: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s−1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? ANSWER: Length of the rectangular wire, l = 8 cm = 0.08 m Width of the rectangular wire, b = 2 cm = 0.02 m Hence, area of the rectangular loop, A = lb = 0.08 × 0.02 = 16 × 10−4 m2 Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s (a) Emf developed in the loop is given as: e = Blv = 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s. (b) Emf developed, e = Bbv = 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s. Page No 230: Question 6.5: 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 emf developed between the centre and the ring. ANSWER: Length of the rod, l = 1 m Angular frequency,ω = 400 rad/s Magnetic field strength, B = 0.5 T One end of the rod has zero linear velocity, while the other end has a linear velocity of lω. Average linear velocity of the rod, Emf developed between the centre and the ring, Hence, the emf developed between the centre and the ring is 100 V. Page No 230: Question 6.6: A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1 in a uniform horizontal magnetic field of magnitude 3.0×10−2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω, 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: Max induced emf = 0.603 V Average induced emf = 0 V Max current in the coil = 0.0603 A Average power loss = 0.018 W (Power comes from the external rotor) Radius of the circular coil, r = 8 cm = 0.08 m Area of the coil, A = πr2 = π × (0.08)2 m2 Number of turns on the coil, N = 20 Angular speed, ω = 50 rad/s Magnetic field strength, B = 3 × 10−2 T …

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