Page 1 CBSE XII | PHYSICS Sample Paper – 3 Solution CBSE Board Class XII – Physics Sample Paper – 3 Solution A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or both are negative so the force between them will be repulsive. q 1q 2 < 0, then force will be attractive. A2. A3. The side bands are at 1000+10 = 1010kHz and 1000 -10 = 990kHz. A4. The diode is reverse biased. This is because the p side is at a lower voltage than the n side. A5. 1THz to 1000THz A6. (i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero. (ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is because the flux depends only on the charge enclosed & is independent of the shape of the Gaussian surface. A7. The neutral temperature will be 0 10 500 0.02 n a C b ? ? ? ? ? . The inversion temperature is twice the neutral temperature, so it is 1000 0 C. A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 0.048 A. Page 2 CBSE XII | PHYSICS Sample Paper – 3 Solution CBSE Board Class XII – Physics Sample Paper – 3 Solution A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or both are negative so the force between them will be repulsive. q 1q 2 < 0, then force will be attractive. A2. A3. The side bands are at 1000+10 = 1010kHz and 1000 -10 = 990kHz. A4. The diode is reverse biased. This is because the p side is at a lower voltage than the n side. A5. 1THz to 1000THz A6. (i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero. (ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is because the flux depends only on the charge enclosed & is independent of the shape of the Gaussian surface. A7. The neutral temperature will be 0 10 500 0.02 n a C b ? ? ? ? ? . The inversion temperature is twice the neutral temperature, so it is 1000 0 C. A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 0.048 A. CBSE XII | PHYSICS Sample Paper – 3 Solution A9. On introducing a dielectric in the capacitor, its capacitance will increase. Then, the total impedance of the circuit will decrease as 2 2 1 ZR C ? ?? ?? ?? ?? . Hence, the current in the circuit increases and the brightness of the lamp increases. A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if unpolarised light is incident on a polaroid the transmitted intensity is half the original intensity. The light transmitted by second Polaroid has intensity ? ? 2 0 cos 2 I ? where ? is the angle between the axes of first and second Polaroid. Thus, the light transmitted by third Polaroid has intensity ? ? ? ? 2 2 0 2 2 2 0 0 0 cos cos (90 ) cos (sin ) sin 2 2 2 8 I I I ? ? ? ? ? ? ? ? . This intensity will be maximum when 0 45 ? ? A11. NAND gate is called a universal gate because it can be used to obtain other basic gates like AND, NOT and OR gates. NAND gates can be combined as shown below to realize a basic OR gate. A12. Some characteristics of the carrier signal are varied in accordance with the modulating or message signal. This is called modulation. Amplitude modulation, frequency modulation and phase modulation of waves are the different types of modulation. A13. Nuclear density for iron will be 2.3x10 17 kg/m 3 . Nuclear density is independent of mass number A, so iron also has the same nuclear density as hydrogen. Page 3 CBSE XII | PHYSICS Sample Paper – 3 Solution CBSE Board Class XII – Physics Sample Paper – 3 Solution A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or both are negative so the force between them will be repulsive. q 1q 2 < 0, then force will be attractive. A2. A3. The side bands are at 1000+10 = 1010kHz and 1000 -10 = 990kHz. A4. The diode is reverse biased. This is because the p side is at a lower voltage than the n side. A5. 1THz to 1000THz A6. (i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero. (ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is because the flux depends only on the charge enclosed & is independent of the shape of the Gaussian surface. A7. The neutral temperature will be 0 10 500 0.02 n a C b ? ? ? ? ? . The inversion temperature is twice the neutral temperature, so it is 1000 0 C. A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 0.048 A. CBSE XII | PHYSICS Sample Paper – 3 Solution A9. On introducing a dielectric in the capacitor, its capacitance will increase. Then, the total impedance of the circuit will decrease as 2 2 1 ZR C ? ?? ?? ?? ?? . Hence, the current in the circuit increases and the brightness of the lamp increases. A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if unpolarised light is incident on a polaroid the transmitted intensity is half the original intensity. The light transmitted by second Polaroid has intensity ? ? 2 0 cos 2 I ? where ? is the angle between the axes of first and second Polaroid. Thus, the light transmitted by third Polaroid has intensity ? ? ? ? 2 2 0 2 2 2 0 0 0 cos cos (90 ) cos (sin ) sin 2 2 2 8 I I I ? ? ? ? ? ? ? ? . This intensity will be maximum when 0 45 ? ? A11. NAND gate is called a universal gate because it can be used to obtain other basic gates like AND, NOT and OR gates. NAND gates can be combined as shown below to realize a basic OR gate. A12. Some characteristics of the carrier signal are varied in accordance with the modulating or message signal. This is called modulation. Amplitude modulation, frequency modulation and phase modulation of waves are the different types of modulation. A13. Nuclear density for iron will be 2.3x10 17 kg/m 3 . Nuclear density is independent of mass number A, so iron also has the same nuclear density as hydrogen. CBSE XII | PHYSICS Sample Paper – 3 Solution A14. In amplitude modulation (AM), the amplitude of modulated (carrier) wave varies in accordance with amplitude of information (signal) wave. When amplitude of the information wave increases, the amplitude of modulated wave also increases and vice – versa. In frequency modulation (FM), the frequency of modulated wave varies in accordance with the frequency of the signal wave. In this case the amplitude of modulated wave is fixed. A15. Ampere’s circuital law states that 0 . ? ? ? Bdl i where I refers to the current passing through amperian loop S around the current element. The magnitude of the magnetic field due to a circular coil of radius R carrying a current i at its centre is 0 2 i B R ? ? Magnetic field lines due to circular current carrying coil is represented as OR The capacitive reactance is C 6 C 11 X 212 2 fC 2 (50)(15.0x10 ) The rms current is V 220 i= 1.04A (1/2) X 212 The peak current 2 i 2(1.04) 1.47A ? ? ? ? ? ?? ?? ?? If the frequency is doubled, the capacitive reactance is halved and the current gets doubled. Page 4 CBSE XII | PHYSICS Sample Paper – 3 Solution CBSE Board Class XII – Physics Sample Paper – 3 Solution A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or both are negative so the force between them will be repulsive. q 1q 2 < 0, then force will be attractive. A2. A3. The side bands are at 1000+10 = 1010kHz and 1000 -10 = 990kHz. A4. The diode is reverse biased. This is because the p side is at a lower voltage than the n side. A5. 1THz to 1000THz A6. (i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero. (ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is because the flux depends only on the charge enclosed & is independent of the shape of the Gaussian surface. A7. The neutral temperature will be 0 10 500 0.02 n a C b ? ? ? ? ? . The inversion temperature is twice the neutral temperature, so it is 1000 0 C. A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 0.048 A. CBSE XII | PHYSICS Sample Paper – 3 Solution A9. On introducing a dielectric in the capacitor, its capacitance will increase. Then, the total impedance of the circuit will decrease as 2 2 1 ZR C ? ?? ?? ?? ?? . Hence, the current in the circuit increases and the brightness of the lamp increases. A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if unpolarised light is incident on a polaroid the transmitted intensity is half the original intensity. The light transmitted by second Polaroid has intensity ? ? 2 0 cos 2 I ? where ? is the angle between the axes of first and second Polaroid. Thus, the light transmitted by third Polaroid has intensity ? ? ? ? 2 2 0 2 2 2 0 0 0 cos cos (90 ) cos (sin ) sin 2 2 2 8 I I I ? ? ? ? ? ? ? ? . This intensity will be maximum when 0 45 ? ? A11. NAND gate is called a universal gate because it can be used to obtain other basic gates like AND, NOT and OR gates. NAND gates can be combined as shown below to realize a basic OR gate. A12. Some characteristics of the carrier signal are varied in accordance with the modulating or message signal. This is called modulation. Amplitude modulation, frequency modulation and phase modulation of waves are the different types of modulation. A13. Nuclear density for iron will be 2.3x10 17 kg/m 3 . Nuclear density is independent of mass number A, so iron also has the same nuclear density as hydrogen. CBSE XII | PHYSICS Sample Paper – 3 Solution A14. In amplitude modulation (AM), the amplitude of modulated (carrier) wave varies in accordance with amplitude of information (signal) wave. When amplitude of the information wave increases, the amplitude of modulated wave also increases and vice – versa. In frequency modulation (FM), the frequency of modulated wave varies in accordance with the frequency of the signal wave. In this case the amplitude of modulated wave is fixed. A15. Ampere’s circuital law states that 0 . ? ? ? Bdl i where I refers to the current passing through amperian loop S around the current element. The magnitude of the magnetic field due to a circular coil of radius R carrying a current i at its centre is 0 2 i B R ? ? Magnetic field lines due to circular current carrying coil is represented as OR The capacitive reactance is C 6 C 11 X 212 2 fC 2 (50)(15.0x10 ) The rms current is V 220 i= 1.04A (1/2) X 212 The peak current 2 i 2(1.04) 1.47A ? ? ? ? ? ?? ?? ?? If the frequency is doubled, the capacitive reactance is halved and the current gets doubled. CBSE XII | PHYSICS Sample Paper – 3 Solution A16. Force acting on charge q moving with velocity v in magnetic field B: F q(v B) ?? (i) Kinetic energy of the particle does not change as the force acting on it is always perpendicular to the velocity. A force that is perpendicular to the velocity cannot change the magnitude of the velocity. All that the force does is to change the direction of the particle keeping the magnitude of velocity unchanged. (ii) The instantaneous power is F.v, where F and v are the force and velocity vector respectively. Since the force and velocity are perpendicular to each other, so F.v = 0. A17. Intensity is I= 4I 0 cos 2 F/2 When path difference is ?, phase difference is 2p I= 4I o cos 2 ? = 4 I o = K (given) When path difference is ? = ? /3, then the phase difference will be F'=2 ? ? / ? =2 ? × ?/3? = 2 ?/3 Hence the intensity at a point where the path difference is ?/3, is I'=4I 0 cos 2 2 ? /6 (since K = 4I 0) = K cos 2 ?/3= K x {1/2}2 = (¼) K. A18. The de Broglie wavelength is e e p p hh p 2mK For the electron, proton and -particle, is same m K m K m K cons tan t ?? ? ? ? ?? ? ? ? As mass of electron is minimum its kinetic energy will be maximum. As mass of alpha-particle is maximum its kinetic energy is minimum. A19. In any radioactive sample, the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. Radioactive decay law is dN N dt ? ? ? Integrating the above expression gives ? ? ? ? t 0 0 1/2 1/2 N t N e N Put N t and t T 2 0.693 T ?? ? ?? ? ? Page 5 CBSE XII | PHYSICS Sample Paper – 3 Solution CBSE Board Class XII – Physics Sample Paper – 3 Solution A1. q 1q 2 > 0 means that q 1 and q 2 have the same sign. Either both charges are positive or both are negative so the force between them will be repulsive. q 1q 2 < 0, then force will be attractive. A2. A3. The side bands are at 1000+10 = 1010kHz and 1000 -10 = 990kHz. A4. The diode is reverse biased. This is because the p side is at a lower voltage than the n side. A5. 1THz to 1000THz A6. (i) The flux is zero as the net charge enclosed by the Gaussian surface will be zero. (ii) If the Gaussian surface is made cubical, the flux is the same. It stays zero, this is because the flux depends only on the charge enclosed & is independent of the shape of the Gaussian surface. A7. The neutral temperature will be 0 10 500 0.02 n a C b ? ? ? ? ? . The inversion temperature is twice the neutral temperature, so it is 1000 0 C. A8. Total resistance of the circuit is 60+3 = 63 ohms. The current in the circuit is I = 3/63 = 0.048 A. CBSE XII | PHYSICS Sample Paper – 3 Solution A9. On introducing a dielectric in the capacitor, its capacitance will increase. Then, the total impedance of the circuit will decrease as 2 2 1 ZR C ? ?? ?? ?? ?? . Hence, the current in the circuit increases and the brightness of the lamp increases. A10. Two crossed polaroids are placed perpendicular to each other. Let the intensity of incident light be I 0. The light transmitted by first Polaroid has intensity I 0/2 because if unpolarised light is incident on a polaroid the transmitted intensity is half the original intensity. The light transmitted by second Polaroid has intensity ? ? 2 0 cos 2 I ? where ? is the angle between the axes of first and second Polaroid. Thus, the light transmitted by third Polaroid has intensity ? ? ? ? 2 2 0 2 2 2 0 0 0 cos cos (90 ) cos (sin ) sin 2 2 2 8 I I I ? ? ? ? ? ? ? ? . This intensity will be maximum when 0 45 ? ? A11. NAND gate is called a universal gate because it can be used to obtain other basic gates like AND, NOT and OR gates. NAND gates can be combined as shown below to realize a basic OR gate. A12. Some characteristics of the carrier signal are varied in accordance with the modulating or message signal. This is called modulation. Amplitude modulation, frequency modulation and phase modulation of waves are the different types of modulation. A13. Nuclear density for iron will be 2.3x10 17 kg/m 3 . Nuclear density is independent of mass number A, so iron also has the same nuclear density as hydrogen. CBSE XII | PHYSICS Sample Paper – 3 Solution A14. In amplitude modulation (AM), the amplitude of modulated (carrier) wave varies in accordance with amplitude of information (signal) wave. When amplitude of the information wave increases, the amplitude of modulated wave also increases and vice – versa. In frequency modulation (FM), the frequency of modulated wave varies in accordance with the frequency of the signal wave. In this case the amplitude of modulated wave is fixed. A15. Ampere’s circuital law states that 0 . ? ? ? Bdl i where I refers to the current passing through amperian loop S around the current element. The magnitude of the magnetic field due to a circular coil of radius R carrying a current i at its centre is 0 2 i B R ? ? Magnetic field lines due to circular current carrying coil is represented as OR The capacitive reactance is C 6 C 11 X 212 2 fC 2 (50)(15.0x10 ) The rms current is V 220 i= 1.04A (1/2) X 212 The peak current 2 i 2(1.04) 1.47A ? ? ? ? ? ?? ?? ?? If the frequency is doubled, the capacitive reactance is halved and the current gets doubled. CBSE XII | PHYSICS Sample Paper – 3 Solution A16. Force acting on charge q moving with velocity v in magnetic field B: F q(v B) ?? (i) Kinetic energy of the particle does not change as the force acting on it is always perpendicular to the velocity. A force that is perpendicular to the velocity cannot change the magnitude of the velocity. All that the force does is to change the direction of the particle keeping the magnitude of velocity unchanged. (ii) The instantaneous power is F.v, where F and v are the force and velocity vector respectively. Since the force and velocity are perpendicular to each other, so F.v = 0. A17. Intensity is I= 4I 0 cos 2 F/2 When path difference is ?, phase difference is 2p I= 4I o cos 2 ? = 4 I o = K (given) When path difference is ? = ? /3, then the phase difference will be F'=2 ? ? / ? =2 ? × ?/3? = 2 ?/3 Hence the intensity at a point where the path difference is ?/3, is I'=4I 0 cos 2 2 ? /6 (since K = 4I 0) = K cos 2 ?/3= K x {1/2}2 = (¼) K. A18. The de Broglie wavelength is e e p p hh p 2mK For the electron, proton and -particle, is same m K m K m K cons tan t ?? ? ? ? ?? ? ? ? As mass of electron is minimum its kinetic energy will be maximum. As mass of alpha-particle is maximum its kinetic energy is minimum. A19. In any radioactive sample, the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. Radioactive decay law is dN N dt ? ? ? Integrating the above expression gives ? ? ? ? t 0 0 1/2 1/2 N t N e N Put N t and t T 2 0.693 T ?? ? ?? ? ? CBSE XII | PHYSICS Sample Paper – 3 Solution A20. 1 amu = 931.5 MeV amu m 004 . 0 ) 009 . 1 017 . 3 ( ) 015 . 2 ( 2 ? ? ? ? ? Hence energy released per deuteron = (0.004 x 931.5)/2 = 1.863 MeV The number of deuterons in 1kg =N A/2 =6.023 × 10 26 /2 Energy released = (3.01 × 10 26 )(1.863 × 10 6 )(1.6x10 -19 ) J = 9.0×10 13 J A21. (i) If the audio signal is directly transmitted, the size of antenna will be very large which is not practically feasible. This is because, the size of the antenna required is proportional to /4 ? . (ii) Effective power radiated by antenna is proportional to square of frequency. For an audio frequency wave the radiated power will be extremely small. (iii) iIf different programmes, audio frequencies are directly transmitted then what you will hear at the receiver will be a mixture of all these signals. A22. (i) E remains same as it depends on the charge on the plates and the medium between the plates. Q remains same as charge does not change on the plates. (ii) V = Ed, so as the distance is doubled V also doubles. (iii) C = Q/V and V is doubled whereas Q remains same. So C is halved. A23. C b =0.8 = /(1- ) =0.8/(1-0.8) = 4 I = I = 6 4 = 24 mA ? ? ? ? ? ? ? ? OR Here, m = 3.2g = 3.2 10 -3 kg E = 10 10 NC -1 Let n be the number of electrons removed from the coin. Then the charge on the coin is q = +n e When the coin just floats, Upward force of electric field = weight of the coin n q E = mg n = ? ? ?? 3 7 19 10 mg 3.2x10 x9.8 1.96x10 qE 1.6x10 x10 electronsRead More

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