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Shivendra Singh

Myself Shivendra Singh worked in Kota as Physics Educator almost 10 years now I came in Sultanpur in covid pendemic in 2020. I have felt that students of Sultanpur need real guidance for JEE and NEET so I have started specialized PHYSICS coaching in Shashtri Nagar, Sultanpur for 11th and 12th (CBSE) , IIT-JEE and NEET.
(Mentor and counsellor for IIT-JEE and NEET)
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Work, Energy and Power

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Lecture 2022-03-01

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Thermodynamics,Behaviour of Perfect Gas and Kinetic Theory

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Lecture 2022-01-24

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Rotational Dynamics

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EM waves lecture 1

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NLM And WPE

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Electrostatics Sheet.pdf

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Half Yearly Physics

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AC notes.pdf

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Electric Charge And Field

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Electric charge and field- Short Notes.pdf

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Magnetism

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Study Notes - Physical World

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Electricity DPP - 1

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AC notes

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Effects of electric current, Part-1

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AC

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13.AC Circuits

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Electromagnetic induction practice set

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Full Syllabus

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Full Syllabus Mock Test

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Current Electricty

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Electric Current Previous

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Magnetism, EMI & AC (www.jeebooks.in)

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Physics 22 Sept 2021

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Alternating Current

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Alternating Current

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Alternating Current

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Plus-two-physics-previous-question-paper chapter wise

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Nov 24, 2021

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Page 2 :

Physics by Shivendra Sir, Q.1, , An alternating current changes from a complete, cycle in 1s, then the frequency in Hz will be –, (A) 10-6, (B) 50, (C) 100, (D) 106, , Q.2, , An alternating voltage voltage source is, connected, in an A.C. circuit whose maximum, value is 170 volt. The value of potential at a, phase angle of 45º will be –, (A) 120.56 Volt, (B) 110.12 Volt, (C) 240 Volt, (D) Zero, In an ac circuit, the current is given by, i = 4 sin (100t + 30º) ampere. The current, becomes maximum first time (after t = 0) at, t equal to –, (A) (1/200) sec, (B) (1/300) sec, (C) (1/50) sec, (D) None of the above, The instantaneous value of current in an ac, circuit is = 2 sin (100t + /3) A. The current, at the beginning (t = 0) will be –, , Q.3, , Q.4, , (A) 2 3 A, , 3, A, 2, , (C), Q.5, , Q.6, , , 2 2, , 3A, , (D) Zero, , (B) 2 2 , (D), , Q.13, , (C), , 5, , (C) The emf leads the current by 60º, (D) The phase difference between the current, and the emf is zero, , Q.15, , Q.16, , A mixer of 1000 resistance is connected, , (B) 0.01, (D) 1 × 10–2, , The value of alternating e.m.f. is e = 500 sin, 100t , then the frequency of this potential in, Hz is –, (A) 25, , Q.18, , (B) 100 times, (D) 500 times, , The time period of of alternating current with, frequency of one KHz one second will be –, (A) 0.10, (C) 1 × 10–3, , Q.17, , (B) one time, (D) 60 times, , If the frequency of alternating potential is 50Hz, then the direction of potential, changes in, one second by –, (A) 50 times, (C) 200 times, , 5, , to an A.C. source of 200V and 50 cycle, see the value of average potential, difference across the mixer will be (A) 308 V, (B) 264 V, (C) 220 V, (D) 0, , The direction of alternating currenting current, get changed in one cycle –, (A) two times, (C) 50 times, , 1, , (D) 5 2, , The emf and the current in a circuit are –, E = 12 sin (100t) ; = 4 sin (100t + / 3), then, (B) The current lags the emf by 60º, , 2, , (B), , (D) 3 3, , (A) The current leads the emf by 60º, , , , (C) E0 2, (D) 0, If the value of Erms is 5 volt, then the, tolerance of the component in volt is -, , If instantaneous value of current is, = 10 sin (314 t) A,, then the average current for the half cycle, will be –, (A) 10 A, (B) 7.07 A, (C) 6.37 A, (D) 3.53 A, The r.m.s. value of alternating current is 10 Amp, having frequency of 50Hz. The time taken by, the current to increase from zero to maximum, and the maximum value of current will be –, (A) 2 × 10–2 sec. and 14.14 Amp, (B) 1 × 10–2 sec. and 7.07 amp., (C) 5 × 10–3 sec. and 7.07 amp, (D) 5 × 10–3 sec. and 14.14 Amp., n a circuit an a.c. current and a d, c. current, are supplied together. The expression of the, instantaneous current is given as, i = 3 + 6 sin t, Then the rms value of the current is –, (A) 3, (B) 6, (C) 3 2, , Q.14, , The r.m.s. value of potential due to superposition, of given two alternating potentials E1 = E0, sin t and E2 = E0 cos t will be –, (A) E0, (B) 2E0, , (A) 1, , Q.9, , Q.12, , (A) peak value / 2, (B) 0, (C) peak value, (D) None of the above, The form factor for sinusoidal potential is –, , (C), , Q.8, , Q.11, , In A.C. circuit the average value per cycle, of e.m.f. or current is –, , (A) 2, , Q.7, , (B), , Q.10, , (B) 50, , (C) 75 (D) 100, , The frequency of an alternating current is, 50Hz, then the time to complete one cycle, for current vector will be–, (A) 20 ms, (C) 100 ms, , (B) 50 ms, (D) 1 s, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, Q.1, , An alternating current changes from a complete, cycle in 1s, then the frequency in Hz will be –, (A) 10-6, (B) 50, (C) 100, (D) 106, , Q.2, , An alternating voltage voltage source is, connected, in an A.C. circuit whose maximum, value is 170 volt. The value of potential at a, phase angle of 45º will be –, (A) 120.56 Volt, (B) 110.12 Volt, (C) 240 Volt, (D) Zero, In an ac circuit, the current is given by, i = 4 sin (100t + 30º) ampere. The current, becomes maximum first time (after t = 0) at, t equal to –, (A) (1/200) sec, (B) (1/300) sec, (C) (1/50) sec, (D) None of the above, The instantaneous value of current in an ac, circuit is = 2 sin (100t + /3) A. The current, at the beginning (t = 0) will be –, , Q.3, , Q.4, , (A) 2 3 A, , 3, A, 2, , (C), Q.5, , Q.6, , , 2 2, , 3A, , (D) Zero, , (B) 2 2 , (D), , Q.13, , (C), , 5, , (C) The emf leads the current by 60º, (D) The phase difference between the current, and the emf is zero, , Q.15, , Q.16, , A mixer of 1000 resistance is connected, , (B) 0.01, (D) 1 × 10–2, , The value of alternating e.m.f. is e = 500 sin, 100t , then the frequency of this potential in, Hz is –, (A) 25, , Q.18, , (B) 100 times, (D) 500 times, , The time period of of alternating current with, frequency of one KHz one second will be –, (A) 0.10, (C) 1 × 10–3, , Q.17, , (B) one time, (D) 60 times, , If the frequency of alternating potential is 50Hz, then the direction of potential, changes in, one second by –, (A) 50 times, (C) 200 times, , 5, , to an A.C. source of 200V and 50 cycle, see the value of average potential, difference across the mixer will be (A) 308 V, (B) 264 V, (C) 220 V, (D) 0, , The direction of alternating currenting current, get changed in one cycle –, (A) two times, (C) 50 times, , 1, , (D) 5 2, , The emf and the current in a circuit are –, E = 12 sin (100t) ; = 4 sin (100t + / 3), then, (B) The current lags the emf by 60º, , 2, , (B), , (D) 3 3, , (A) The current leads the emf by 60º, , , , (C) E0 2, (D) 0, If the value of Erms is 5 volt, then the, tolerance of the component in volt is -, , If instantaneous value of current is, = 10 sin (314 t) A,, then the average current for the half cycle, will be –, (A) 10 A, (B) 7.07 A, (C) 6.37 A, (D) 3.53 A, The r.m.s. value of alternating current is 10 Amp, having frequency of 50Hz. The time taken by, the current to increase from zero to maximum, and the maximum value of current will be –, (A) 2 × 10–2 sec. and 14.14 Amp, (B) 1 × 10–2 sec. and 7.07 amp., (C) 5 × 10–3 sec. and 7.07 amp, (D) 5 × 10–3 sec. and 14.14 Amp., n a circuit an a.c. current and a d, c. current, are supplied together. The expression of the, instantaneous current is given as, i = 3 + 6 sin t, Then the rms value of the current is –, (A) 3, (B) 6, (C) 3 2, , Q.14, , The r.m.s. value of potential due to superposition, of given two alternating potentials E1 = E0, sin t and E2 = E0 cos t will be –, (A) E0, (B) 2E0, , (A) 1, , Q.9, , Q.12, , (A) peak value / 2, (B) 0, (C) peak value, (D) None of the above, The form factor for sinusoidal potential is –, , (C), , Q.8, , Q.11, , In A.C. circuit the average value per cycle, of e.m.f. or current is –, , (A) 2, , Q.7, , (B), , Q.10, , (B) 50, , (C) 75 (D) 100, , The frequency of an alternating current is, 50Hz, then the time to complete one cycle, for current vector will be–, (A) 20 ms, (C) 100 ms, , (B) 50 ms, (D) 1 s, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Page 3 :

Physics by Shivendra Sir, Q.19, , In the above question, time taken by current, to rise from zero to maximum is –, , 1, sec, (B), 200, 1, (C), sec, (D), 50, In the equation fo A.C. , current amplitude and, respectively be –, (A), , Q.20, , Q.21, , Q.22, , Q.23, , Q.24, , Q.25, , I0 , , ,, (A) 0 ,, (B), 2, 2 2, , (C) rms,, (D) 0, , 2, The sinusoidal voltage wave changes from 0, to maximum value of 100 volt. The voltage, when the phase angle is 30º will be –, , (A) 70.7 volt, (B) 50 volt, (C) 109 volt, (D) – 100 volt, If the frequency of ac is 60 Hz the time, difference corresponding to a phase difference, of 60º is –, (A) 60 s, (B) 1 s, (C) 1/60 s, (D) 1/360 s, The domestic power supply is at 220 volt., The amplitude of emf will be –, (A) 220 V, (B) 110 V, (C) 311 V, (D) None of this, The phase difference between the current, and the electromotive force in an ac circuit is, /4 radian. If the frequency is 50 Hz, then the, time difference corresponding to this phase, difference, will be –, (A) 0.25 s, (B) 0.02 s, (C) 2.5 ms, (D) 25 ms, In A.C. circuit the ratio of virtual current and, the r.m.s. current is –, (A) 0, (B) 0.5, (C) 1, , Q.26, , (D), , 2, , If the r.m.s. value of A.C. is Irms then peak, to peak value is –, (A), , Q.27, , 1, sec, 100, 1, sec, 400, = 0 sin t, the, frequency will, , 2 Irms./2, , (C), , 2 I0, , I0, 2, , (B), , Q.29, , If the instantaneous value of currents is, I = 100 sin 314t Amp. then the average of, current n Ampere for half cycle is –, (A) 100, (B) 70.7, (C) 63.7, (D) 35.3, , Q.30, , The equation of current in an ac circuit is, = 4 sin (100t + /6) ampere. The current, at the beginning (t = 0) will be –, (A) 1 A, (C) 3 A, , Q.31, , (C), Q.32, , (B) 2 A, (D) 4 A, , RMS value of ac i = i1 cos t + i2 sin t will, be –, (A), , 1, , (i + i2), 2 1, , 1, 2, , (i12 + i22)½, , 1, , (B), (D), , 2, , (i1 + i2)2, , 1, (i 2 + i22)½, 2 1, , The phase difference between the alternating, current and voltage represented by the, following equation = 0 sin t, E = E0 cos, (t + / 3), will be –, , 4, (B), 3, 3, 5, , (C), (D), 6, 2, The inductance of a resistance less coil is, 0.5 henry. In the coil the value of A.C. is 0.2, Amp whose frequency is 50Hz. The, reactance of circuit is -, , (A), , Q.33, , Q.34, , (A) 15.7 , , (B) 157, , (C) 1.57, , (D) 757, , The inductive reactance of a coil is 1000. If, its self inductance and frequency both are, increased two times then inductive reactance, will be –, (A) 1000, (C) 4000, , Q.35, , I0, , , (D) 0, , Sinusoidal peak potential is 200 volt with, frequency 50Hz. It is represented by the, equation –, (A) E = 200 sin 50t, (B) E = 200 sin 314t, (C) E = 200 2 sin 50t (D) E =200 2 sin 314t, , (B) Irms./ 2, , (C) 2 2 Irms., (D) 2 Irms., The average value or alternating current for, half cycle in terms of I0 is, (A), , Q.28, , n an L-C-R series circuit R = 10, XL = 8, and XC = 6the total impedance of the circuit, is –, (A) 10.2, (C) 10, , Q.36, , (B) 2000, (D) 16000, , (B) 17.2, (D) None of the above, , n the given figure, the potential difference is, shown on R, L and C. The e.m.f. of source, in volt is –, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, Q.19, , In the above question, time taken by current, to rise from zero to maximum is –, , 1, sec, (B), 200, 1, (C), sec, (D), 50, In the equation fo A.C. , current amplitude and, respectively be –, (A), , Q.20, , Q.21, , Q.22, , Q.23, , Q.24, , Q.25, , I0 , , ,, (A) 0 ,, (B), 2, 2 2, , (C) rms,, (D) 0, , 2, The sinusoidal voltage wave changes from 0, to maximum value of 100 volt. The voltage, when the phase angle is 30º will be –, , (A) 70.7 volt, (B) 50 volt, (C) 109 volt, (D) – 100 volt, If the frequency of ac is 60 Hz the time, difference corresponding to a phase difference, of 60º is –, (A) 60 s, (B) 1 s, (C) 1/60 s, (D) 1/360 s, The domestic power supply is at 220 volt., The amplitude of emf will be –, (A) 220 V, (B) 110 V, (C) 311 V, (D) None of this, The phase difference between the current, and the electromotive force in an ac circuit is, /4 radian. If the frequency is 50 Hz, then the, time difference corresponding to this phase, difference, will be –, (A) 0.25 s, (B) 0.02 s, (C) 2.5 ms, (D) 25 ms, In A.C. circuit the ratio of virtual current and, the r.m.s. current is –, (A) 0, (B) 0.5, (C) 1, , Q.26, , (D), , 2, , If the r.m.s. value of A.C. is Irms then peak, to peak value is –, (A), , Q.27, , 1, sec, 100, 1, sec, 400, = 0 sin t, the, frequency will, , 2 Irms./2, , (C), , 2 I0, , I0, 2, , (B), , Q.29, , If the instantaneous value of currents is, I = 100 sin 314t Amp. then the average of, current n Ampere for half cycle is –, (A) 100, (B) 70.7, (C) 63.7, (D) 35.3, , Q.30, , The equation of current in an ac circuit is, = 4 sin (100t + /6) ampere. The current, at the beginning (t = 0) will be –, (A) 1 A, (C) 3 A, , Q.31, , (C), Q.32, , (B) 2 A, (D) 4 A, , RMS value of ac i = i1 cos t + i2 sin t will, be –, (A), , 1, , (i + i2), 2 1, , 1, 2, , (i12 + i22)½, , 1, , (B), (D), , 2, , (i1 + i2)2, , 1, (i 2 + i22)½, 2 1, , The phase difference between the alternating, current and voltage represented by the, following equation = 0 sin t, E = E0 cos, (t + / 3), will be –, , 4, (B), 3, 3, 5, , (C), (D), 6, 2, The inductance of a resistance less coil is, 0.5 henry. In the coil the value of A.C. is 0.2, Amp whose frequency is 50Hz. The, reactance of circuit is -, , (A), , Q.33, , Q.34, , (A) 15.7 , , (B) 157, , (C) 1.57, , (D) 757, , The inductive reactance of a coil is 1000. If, its self inductance and frequency both are, increased two times then inductive reactance, will be –, (A) 1000, (C) 4000, , Q.35, , I0, , , (D) 0, , Sinusoidal peak potential is 200 volt with, frequency 50Hz. It is represented by the, equation –, (A) E = 200 sin 50t, (B) E = 200 sin 314t, (C) E = 200 2 sin 50t (D) E =200 2 sin 314t, , (B) Irms./ 2, , (C) 2 2 Irms., (D) 2 Irms., The average value or alternating current for, half cycle in terms of I0 is, (A), , Q.28, , n an L-C-R series circuit R = 10, XL = 8, and XC = 6the total impedance of the circuit, is –, (A) 10.2, (C) 10, , Q.36, , (B) 2000, (D) 16000, , (B) 17.2, (D) None of the above, , n the given figure, the potential difference is, shown on R, L and C. The e.m.f. of source, in volt is –, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

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Physics by Shivendra Sir, 30V, , 60V, , 100V, C, , (A), (B), (C), (D), , Curve P is for R–L and Q for R–C circuit, Curve P is for R–C and Q for R–L circuit, Both are for R–C circuit, Both are for R–L circuit, , ~ e, , Q.42, , Q.37, , Q.38, , (A) 190, (B) 70, (C) 50, (D) 40, n an L.C.R series circuit R = 1, XL =, 1000 and XC = 1000. A source of 100, m.volt is connected in the circuit the current, in the circuit is –, (A) 100 m.Amp, (B) 1 .Amp, (C) 0.1 .Amp, (D) 10 .Amp, Which of the following figure showing the, phase relationship is correct phase diagram, for an R–C circuit-, , The power factor of the following circuit, will be –, , , , , , C, , ~ e, , XL=100r = 40e-200V50Hz, R=40XC=40, (A) 0.2, (B) 0.4, (C) 0.6, (D) 0.8, Q.43, In a circuit, the reactance of a coil is 20 ., If the inductance of the coil is 50mH then, angular frequency of the current will be –, (A) 400 rad/sec., (C) 2.5 rad/sec, , I, (A) V, , (B), , Q.44, , (B) 1 rad/sec, (D) 0.2 rad/sec, , If a capacitor is connected to two different, A.C. generatiors then the value of capacitive, reactance is –, (A) directly proportion to frequency, , (C), , Q.39, , (B) inversely proportional to frequency, , (D), , A coil of inductance 0.1 H is connected to, an alternating voltage generator of voltage, E = 100 sin (100t) volt. The current flowing, through the coil will be –, , (C) independent of frequency, (D) inversely proportional to the square of, frequency, Q.45, , (A) only resistance, (B) only capacitor, , (A) = 10 2 sin (100t) A, , Q.40, , (B) = 10 2 cos (100t) A, (C) = – 10 sin (100t) A, (D) = – 10 cos (100t) A, The vector diagram of the current and voltage, in a given circuit is shown in the figure. The, components of the circuit will be –, 45º, , Q.46, , (B) L–R, (D) C–R, , Q.48, t, P, , for, , X2 for frequency n2 then, , (A) 1 : 1, (B) n1 : n2, (C) n2 : n1, (D) n12 : n22, A coil has reactance of 100 when frequency, is 50Hz. If the frequency becomes 150Hz,, then the reactance will be –, (A) 100, (C) 450, , Figure shows the variation of voltage with time, for an ac = 0 sin t flowing through a circuit Q, , The reactance of a capacitor is X1, frequency n1 and, X1 : X2 is, , Q.47, , (A) L–C–R, (C) L–C–R or L–R, Q.41, , (C) only an inductance coil, (D) capacitor and resistance both, , E=200 volts, , I=2A, , Alternating current lead the applied e.m.f. by, /2 when the circuit consists of –, , (B) 300, (D) 600, , n pure inductive circuit, the curves between, frequency f and inductive reactance 1/XL is –, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, 30V, , 60V, , 100V, C, , (A), (B), (C), (D), , Curve P is for R–L and Q for R–C circuit, Curve P is for R–C and Q for R–L circuit, Both are for R–C circuit, Both are for R–L circuit, , ~ e, , Q.42, , Q.37, , Q.38, , (A) 190, (B) 70, (C) 50, (D) 40, n an L.C.R series circuit R = 1, XL =, 1000 and XC = 1000. A source of 100, m.volt is connected in the circuit the current, in the circuit is –, (A) 100 m.Amp, (B) 1 .Amp, (C) 0.1 .Amp, (D) 10 .Amp, Which of the following figure showing the, phase relationship is correct phase diagram, for an R–C circuit-, , The power factor of the following circuit, will be –, , , , , , C, , ~ e, , XL=100r = 40e-200V50Hz, R=40XC=40, (A) 0.2, (B) 0.4, (C) 0.6, (D) 0.8, Q.43, In a circuit, the reactance of a coil is 20 ., If the inductance of the coil is 50mH then, angular frequency of the current will be –, (A) 400 rad/sec., (C) 2.5 rad/sec, , I, (A) V, , (B), , Q.44, , (B) 1 rad/sec, (D) 0.2 rad/sec, , If a capacitor is connected to two different, A.C. generatiors then the value of capacitive, reactance is –, (A) directly proportion to frequency, , (C), , Q.39, , (B) inversely proportional to frequency, , (D), , A coil of inductance 0.1 H is connected to, an alternating voltage generator of voltage, E = 100 sin (100t) volt. The current flowing, through the coil will be –, , (C) independent of frequency, (D) inversely proportional to the square of, frequency, Q.45, , (A) only resistance, (B) only capacitor, , (A) = 10 2 sin (100t) A, , Q.40, , (B) = 10 2 cos (100t) A, (C) = – 10 sin (100t) A, (D) = – 10 cos (100t) A, The vector diagram of the current and voltage, in a given circuit is shown in the figure. The, components of the circuit will be –, 45º, , Q.46, , (B) L–R, (D) C–R, , Q.48, t, P, , for, , X2 for frequency n2 then, , (A) 1 : 1, (B) n1 : n2, (C) n2 : n1, (D) n12 : n22, A coil has reactance of 100 when frequency, is 50Hz. If the frequency becomes 150Hz,, then the reactance will be –, (A) 100, (C) 450, , Figure shows the variation of voltage with time, for an ac = 0 sin t flowing through a circuit Q, , The reactance of a capacitor is X1, frequency n1 and, X1 : X2 is, , Q.47, , (A) L–C–R, (C) L–C–R or L–R, Q.41, , (C) only an inductance coil, (D) capacitor and resistance both, , E=200 volts, , I=2A, , Alternating current lead the applied e.m.f. by, /2 when the circuit consists of –, , (B) 300, (D) 600, , n pure inductive circuit, the curves between, frequency f and inductive reactance 1/XL is –, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

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Physics by Shivendra Sir, , 1, XL, , 1, XL, , (1), , Q.55, , (2), , f, , Q.50, , Q.51, , n pure capacitive circuit if the frequency of, A.C. is doubled, then the value of capacitive, reactance will become –, (A) Two times, (C) No change, , (D), , Q.56, , n an A.C. circuit, a capacitor of 1F value is, connected to a source of frequency 1000 rad/, sec. The value of capacitive reactance will be, –, (B) 100, , (C) 1000, , (D) 10,000, , 1, reactance as, . The frequency of A.C., 1000, , Q.53, , A resistance of 50, an inductance of 20/, henry and a capacitor of 5/F are connected, in series with an A.C. source of 230 volt and, 50Hz. The impedance of circuit is –, (A) 5, (B) 50, (C) 5K, (D) 500, , n an A.C. circuit the impedance is, Z = 100 30º , then the resistance of the, circuit in ohm will be –, (A) 50, (B) 100, , n an electric circuit the applied alternating, emf is given by E = 100 sin (314 t) volt, and, current flowing = sin (314t + / 3). Then the, impedance of the circuit is (in ohm) –, (A) 100 /, , Q.59, , Q.60, , n an L–C–R series circuit R =, , (B) 100, , 2, , (C) 100 2, (D) None of the above, The percentage increase in the impedance of, an ac circuit, when its power factor changes, form 0.866 to 0.5 is (Resistance constant) –, (A) 73.2%, (B) 86.6%, (C) 90.8%, (D) 66.6%, The impedance of the given circuit will be –, 150, ~, , 200, , (A) 50 ohm, (C) 200 ohm, Q.61, , (B) 150 ohm, (D) 250 ohm, , The impedance of the given circuit will be –, 150, , Q.54, , (D) 100 3, , Q.58, , (B) 100/, (D) 5000, , n an A.C. circuit XL = 300, XC = 200and, R = 100 the impedance of circuit is –, (A) 600, (B) 200, (C) 141, (D) None of the above, , (D) 5VR, , n an LCR circuit, the voltages across the, components are VL, VC and VR respectively., The voltage of source will be –, (A) [VR + VL + VC], (B) [VR2 + VL2 + VC2]1/2, (C) [VR2 + (VL + VC)2]1/2, (D) [VR2 + (VL – VC)2]1/2, , in MHz will be –, , Q.52, , (B) VR, , Q.57, , n an A.C. circuit capacitance of 5F has a, , (A) 1000/, (C) 200, , 2 VR, , (C) 50 3, , 1, times, 4, , (A) 10, , The potential difference between the ends of, , (C) VR/ 2, , f, , 1, (B), times, 2, , (D) 3 5, , (A), , f, Q.49, , (C) 2 5, , capacitor is VC = 2VR and between the ends, of inductance is VL = 3VR, then the alternating, potential of the source in terms of VR will be –, , f, , (4), , 1, XL, , (B) 3, , a resistance R is VR between the ends of, , 1, XL, , (3), , (A) 2, , 1µF, , 5 , XL = 9, + –, and XC = 7. If applied voltage in the circuit, is 50 volt then impedance of the circuit in, (A) Zero, (B) Infinite, ohm then impedance of the circuit in ohm will, (C) 55 ohm, (D) 2500 ohm, be –, Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, , 1, XL, , 1, XL, , (1), , Q.55, , (2), , f, , Q.50, , Q.51, , n pure capacitive circuit if the frequency of, A.C. is doubled, then the value of capacitive, reactance will become –, (A) Two times, (C) No change, , (D), , Q.56, , n an A.C. circuit, a capacitor of 1F value is, connected to a source of frequency 1000 rad/, sec. The value of capacitive reactance will be, –, (B) 100, , (C) 1000, , (D) 10,000, , 1, reactance as, . The frequency of A.C., 1000, , Q.53, , A resistance of 50, an inductance of 20/, henry and a capacitor of 5/F are connected, in series with an A.C. source of 230 volt and, 50Hz. The impedance of circuit is –, (A) 5, (B) 50, (C) 5K, (D) 500, , n an A.C. circuit the impedance is, Z = 100 30º , then the resistance of the, circuit in ohm will be –, (A) 50, (B) 100, , n an electric circuit the applied alternating, emf is given by E = 100 sin (314 t) volt, and, current flowing = sin (314t + / 3). Then the, impedance of the circuit is (in ohm) –, (A) 100 /, , Q.59, , Q.60, , n an L–C–R series circuit R =, , (B) 100, , 2, , (C) 100 2, (D) None of the above, The percentage increase in the impedance of, an ac circuit, when its power factor changes, form 0.866 to 0.5 is (Resistance constant) –, (A) 73.2%, (B) 86.6%, (C) 90.8%, (D) 66.6%, The impedance of the given circuit will be –, 150, ~, , 200, , (A) 50 ohm, (C) 200 ohm, Q.61, , (B) 150 ohm, (D) 250 ohm, , The impedance of the given circuit will be –, 150, , Q.54, , (D) 100 3, , Q.58, , (B) 100/, (D) 5000, , n an A.C. circuit XL = 300, XC = 200and, R = 100 the impedance of circuit is –, (A) 600, (B) 200, (C) 141, (D) None of the above, , (D) 5VR, , n an LCR circuit, the voltages across the, components are VL, VC and VR respectively., The voltage of source will be –, (A) [VR + VL + VC], (B) [VR2 + VL2 + VC2]1/2, (C) [VR2 + (VL + VC)2]1/2, (D) [VR2 + (VL – VC)2]1/2, , in MHz will be –, , Q.52, , (B) VR, , Q.57, , n an A.C. circuit capacitance of 5F has a, , (A) 1000/, (C) 200, , 2 VR, , (C) 50 3, , 1, times, 4, , (A) 10, , The potential difference between the ends of, , (C) VR/ 2, , f, , 1, (B), times, 2, , (D) 3 5, , (A), , f, Q.49, , (C) 2 5, , capacitor is VC = 2VR and between the ends, of inductance is VL = 3VR, then the alternating, potential of the source in terms of VR will be –, , f, , (4), , 1, XL, , (B) 3, , a resistance R is VR between the ends of, , 1, XL, , (3), , (A) 2, , 1µF, , 5 , XL = 9, + –, and XC = 7. If applied voltage in the circuit, is 50 volt then impedance of the circuit in, (A) Zero, (B) Infinite, ohm then impedance of the circuit in ohm will, (C) 55 ohm, (D) 2500 ohm, be –, Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

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Physics by Shivendra Sir, Q.62, , If E0 = 200 volt, R = 25 ohm. L = 0.1 H and, C = 10–5 F and the frequency is variable,, then the current at f = 0 and f = will be, respecitvely –, L, , R, , n a series resonant L–C–R circuit, if L is, increased by 25% and C is decreased by, 20%, then the resonant frequency will –, (A) Increase by 10% (B) Decrease by 10%, (C) Remain unchanged (D) Increase by 2.5%, , Q.70, , If R = 100 then the value of X and in the, given circuit will be –, , C, , ~, E, , Q.63, , Q.69, , L, , (A) 0 A, 8 A, (B) 8 A, 0 A, (C) 8 A, 8 A, (D) 0 A. 0 A, The impedance of the circuit given will be –, , I, A, , 100, , Q.71, , (A) Zero, (C) 110 ohm, , (B) Infinite, (D) 90 ohm, , A coil of resistance R and inductance L is, connected to a cell of emf E volt. The, current flowing through the coil will be –, (A) E/R, , Q.65, , E, , (D), , L2 R 2, , EL, L2 R 2, , I, , E, , (2), , , , I, 3, , (3), , Q.74, , E, , I, , E, , (4), , 3, E, , I, , n question (65) reactance X will be –, (A) 70.7 ohm, (C) 100 ohm, , Q.68, , Q.73, , n a certain circuit E = 200 cos (314t) and, = sin (314t + /4). Their vector, representation is –, , (1), , Q.67, , Q.72, , (B) E/L, , , , Q.66, , 300V, XV, 50Hz, ~, 220Hz, , (A) 800 V, 2A, (C) 220 V. 2.2A, , E, , (C), , R, , 10µF, , + –, , Q.64, , 300V, , C, , (B) 0.707 ohm, (D) 141 ohm, , n question (65) the power factor is –, (A) 0.5, (B) 0.707, (C) 0.85, (D) 1.0, The electric resonance is sharp in L-C-R, circuit if in the circuit –, (A) R is greater, (B) R is smaller, (C) R = XL or XC, (D) Does not depend on R, , n question (70) the value of inductance will, be –, (A) 0.12 H, (B) 0.24 H, (C) 0.31 H, (D) 0.43 H, n an LCR. series circuit the resonating, frequency can be decreased by –, (A) Decreasing the value of C., (B) Decreasing the value of L, (C) Decreasing both the values of L and C, (D) Increasing the value of C, Which of the following statements is correct, for L–C–R series combination in the condition, of resonance –, (A) Resistance is zero, (B) Impedance is zero, (C) Reactance is zero, (D) Resistance, impedance and reactance all, are zero, n an LCR circuit, the resonating frequency is, 500 k.Hz. If the value of L is increased two, 1, times and value of C is decreased, times,, 8, then the new resonating frequency in kHz will, be –, (A) 250, (C) 1000, , Q.75, , (B) 300 V. 2A, (D) 100 V, 2A, , (B) 500, (D) 2000, , n resonating circuit value of inductance and, capacitance is 0.1H and 200 F. For same, resonating frequency if value of inductance is, 100H then necessary value of capacitance in, F will be –, (A) 4, (C) 2, , (B) 0.2, (D) 0.3, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, Q.62, , If E0 = 200 volt, R = 25 ohm. L = 0.1 H and, C = 10–5 F and the frequency is variable,, then the current at f = 0 and f = will be, respecitvely –, L, , R, , n a series resonant L–C–R circuit, if L is, increased by 25% and C is decreased by, 20%, then the resonant frequency will –, (A) Increase by 10% (B) Decrease by 10%, (C) Remain unchanged (D) Increase by 2.5%, , Q.70, , If R = 100 then the value of X and in the, given circuit will be –, , C, , ~, E, , Q.63, , Q.69, , L, , (A) 0 A, 8 A, (B) 8 A, 0 A, (C) 8 A, 8 A, (D) 0 A. 0 A, The impedance of the circuit given will be –, , I, A, , 100, , Q.71, , (A) Zero, (C) 110 ohm, , (B) Infinite, (D) 90 ohm, , A coil of resistance R and inductance L is, connected to a cell of emf E volt. The, current flowing through the coil will be –, (A) E/R, , Q.65, , E, , (D), , L2 R 2, , EL, L2 R 2, , I, , E, , (2), , , , I, 3, , (3), , Q.74, , E, , I, , E, , (4), , 3, E, , I, , n question (65) reactance X will be –, (A) 70.7 ohm, (C) 100 ohm, , Q.68, , Q.73, , n a certain circuit E = 200 cos (314t) and, = sin (314t + /4). Their vector, representation is –, , (1), , Q.67, , Q.72, , (B) E/L, , , , Q.66, , 300V, XV, 50Hz, ~, 220Hz, , (A) 800 V, 2A, (C) 220 V. 2.2A, , E, , (C), , R, , 10µF, , + –, , Q.64, , 300V, , C, , (B) 0.707 ohm, (D) 141 ohm, , n question (65) the power factor is –, (A) 0.5, (B) 0.707, (C) 0.85, (D) 1.0, The electric resonance is sharp in L-C-R, circuit if in the circuit –, (A) R is greater, (B) R is smaller, (C) R = XL or XC, (D) Does not depend on R, , n question (70) the value of inductance will, be –, (A) 0.12 H, (B) 0.24 H, (C) 0.31 H, (D) 0.43 H, n an LCR. series circuit the resonating, frequency can be decreased by –, (A) Decreasing the value of C., (B) Decreasing the value of L, (C) Decreasing both the values of L and C, (D) Increasing the value of C, Which of the following statements is correct, for L–C–R series combination in the condition, of resonance –, (A) Resistance is zero, (B) Impedance is zero, (C) Reactance is zero, (D) Resistance, impedance and reactance all, are zero, n an LCR circuit, the resonating frequency is, 500 k.Hz. If the value of L is increased two, 1, times and value of C is decreased, times,, 8, then the new resonating frequency in kHz will, be –, (A) 250, (C) 1000, , Q.75, , (B) 300 V. 2A, (D) 100 V, 2A, , (B) 500, (D) 2000, , n resonating circuit value of inductance and, capacitance is 0.1H and 200 F. For same, resonating frequency if value of inductance is, 100H then necessary value of capacitance in, F will be –, (A) 4, (C) 2, , (B) 0.2, (D) 0.3, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

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Physics by Shivendra Sir, Q.76, , The inductance of the motor of a fan is 1.0 H., To run the fan at 50 Hz the capacitance of, the capacitor that will cancel its inductive, reactance, will be –, (A) 10 F, (C) 0.4 F, , Q.77, , (B) 40 F, (D) 0.04 F, , (A), (C), Q.83, , n ac circuit at resonance :, (A) Impedance = R, 1 , , , (B) Impedance = L , C , , , Q.84, , (C) The voltages across L and C are in the, same phase, (D) The phase difference of current in C, relative to source voltage is , Q.78, , An ac circuit resonates at a frequency of 10, kHz. If its frequency is increased to 11 kHz,, then :, , Q.85, , (A) Impedance will increase by 1.1 times, (B) Impedance will remain remain unchanged, , Q.86, , (C) Impedance will increase and become, inductive, (D) Impedance will increase and become, capacitive, Q.79, , n an ac circuit 6 ohm resistor, an inductor of, 4 ohm and a capacitor of 12 ohm are, connected n series with an ac source of 100, volt (rms). The average power dissipated in, the circuit will be –, (A) 600 W, , Q.80, , Q.81, , (B) 500 W, , (C) 400 W, (D) 200 W, n an ac circuit emf and current are E = 5, cos t volt and = 2 sin t ampere, respectively. The average power dissipated in, this circuit will be –, (A) 10 W, (B) 2.5 W, (C) 5 W, (D) Zero, The equations of alternating e.m.f. and current, in an A.C. circuit are E = 5 cos t volt and, = 2 sin t ampere respectively. The average, power loss in this circuit will be –, (A) 1 watt, , Q.82, , Q.87, , 3, , (B), (D), , 1, 2, 1, 5, , n an L–C–R series circuit the loss of power, is in –, (A) Only R, (B) Only L, (C) Only C, (D) both L and C, n an ac circuit the readings of an ammeter, and a voltmeter are 10 A and 25 volt, respectively, the power in the circuit will be –, (A) More than 250 W, (B) Always less than 250 W, (C) 250 W, (D) Less than 250 W or 250 W, A choke coil of 100 ohm and 1 H is connected, to a generator of E = 200 sin (100t) volt. The, average power dissipated will be –, (A) Zero, (B) 200 W, (C) 141 W, (D) 100 W, A choke coil of negligible resistance carries, 5 mA current when it is operated at 220 V., The loss of power in the choke coil is –, (A) Zero, (B) 11 W, (C) 44 × 103 W, (D) 1.1 W, The ratio of apparent power and average, power in an A.C. circuit is equal to –, (A) Reciprocal of power factor, (B) Efficiency, (C) Power factor, (D) Form factor, , Q.88, , n an A.C. circuit, a resistance of 3, an, inductance coil of 4and a condenser of 8, are connected in series with an A.C. source, of 50 volt (R.M.S.). The average power loss in, the circuit will be, (A) 600 watt, (C) 400 watt, , Q.89, , (B) 2.5 watt, , (C) 3 watt, (D) Zero, The series combination of resistance R and, inductance L is connected to an alternating, source of e.m.f. e = 311 sin (100 t). If the, value of wattless current is 0.5A and the, impedance of the circuit is 311, the power, factor will be –, , 1, 2, 1, , Q.90, , (B) 500 watt, (D) 300 watt, , , ), 2, ampere an A,V, V = 200 sin (100 t) volt. The, power loss in the circuit will be –, In an A.C. circuit, i = 5 sin (100t–, , (A) 20 volt, , (B) 40 volt, , (C) 1000 watt, , (D) 0 watt, , When N identical bulbs are connected in, parallel, total power consuption is P, what, would be the power consuption when they, connected in series(A) P, (C) P/N, , (B) PN, (D) P/N2, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, Q.76, , The inductance of the motor of a fan is 1.0 H., To run the fan at 50 Hz the capacitance of, the capacitor that will cancel its inductive, reactance, will be –, (A) 10 F, (C) 0.4 F, , Q.77, , (B) 40 F, (D) 0.04 F, , (A), (C), Q.83, , n ac circuit at resonance :, (A) Impedance = R, 1 , , , (B) Impedance = L , C , , , Q.84, , (C) The voltages across L and C are in the, same phase, (D) The phase difference of current in C, relative to source voltage is , Q.78, , An ac circuit resonates at a frequency of 10, kHz. If its frequency is increased to 11 kHz,, then :, , Q.85, , (A) Impedance will increase by 1.1 times, (B) Impedance will remain remain unchanged, , Q.86, , (C) Impedance will increase and become, inductive, (D) Impedance will increase and become, capacitive, Q.79, , n an ac circuit 6 ohm resistor, an inductor of, 4 ohm and a capacitor of 12 ohm are, connected n series with an ac source of 100, volt (rms). The average power dissipated in, the circuit will be –, (A) 600 W, , Q.80, , Q.81, , (B) 500 W, , (C) 400 W, (D) 200 W, n an ac circuit emf and current are E = 5, cos t volt and = 2 sin t ampere, respectively. The average power dissipated in, this circuit will be –, (A) 10 W, (B) 2.5 W, (C) 5 W, (D) Zero, The equations of alternating e.m.f. and current, in an A.C. circuit are E = 5 cos t volt and, = 2 sin t ampere respectively. The average, power loss in this circuit will be –, (A) 1 watt, , Q.82, , Q.87, , 3, , (B), (D), , 1, 2, 1, 5, , n an L–C–R series circuit the loss of power, is in –, (A) Only R, (B) Only L, (C) Only C, (D) both L and C, n an ac circuit the readings of an ammeter, and a voltmeter are 10 A and 25 volt, respectively, the power in the circuit will be –, (A) More than 250 W, (B) Always less than 250 W, (C) 250 W, (D) Less than 250 W or 250 W, A choke coil of 100 ohm and 1 H is connected, to a generator of E = 200 sin (100t) volt. The, average power dissipated will be –, (A) Zero, (B) 200 W, (C) 141 W, (D) 100 W, A choke coil of negligible resistance carries, 5 mA current when it is operated at 220 V., The loss of power in the choke coil is –, (A) Zero, (B) 11 W, (C) 44 × 103 W, (D) 1.1 W, The ratio of apparent power and average, power in an A.C. circuit is equal to –, (A) Reciprocal of power factor, (B) Efficiency, (C) Power factor, (D) Form factor, , Q.88, , n an A.C. circuit, a resistance of 3, an, inductance coil of 4and a condenser of 8, are connected in series with an A.C. source, of 50 volt (R.M.S.). The average power loss in, the circuit will be, (A) 600 watt, (C) 400 watt, , Q.89, , (B) 2.5 watt, , (C) 3 watt, (D) Zero, The series combination of resistance R and, inductance L is connected to an alternating, source of e.m.f. e = 311 sin (100 t). If the, value of wattless current is 0.5A and the, impedance of the circuit is 311, the power, factor will be –, , 1, 2, 1, , Q.90, , (B) 500 watt, (D) 300 watt, , , ), 2, ampere an A,V, V = 200 sin (100 t) volt. The, power loss in the circuit will be –, In an A.C. circuit, i = 5 sin (100t–, , (A) 20 volt, , (B) 40 volt, , (C) 1000 watt, , (D) 0 watt, , When N identical bulbs are connected in, parallel, total power consuption is P, what, would be the power consuption when they, connected in series(A) P, (C) P/N, , (B) PN, (D) P/N2, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

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Physics by Shivendra Sir, Q.91, , Q.92, , Two bulbs of 500 watt and 300 watt work on, 200 volt r.m.s. the ratio of their resistances, will be –, (A) 25 : 9, , (B) 3 : 5, , (C) 9 : 25, , (D) 5 : 9, , An air core coil and an electric bulb are, connected in series with an A.C. source. If, an iron rod is put in the coil, then the, intensity of bulb’s will –, , Q.96, , Q.97, , (A) Be same, (B) Increase, (C) Decrease, (D) Decrease, increase, Q.93, , If a bulb and a coil are connected in series, with D.C. source and a iron core put in the, coil then the glowing of bulb –, (A) Decreases, (C) No change, , Q.94, , Q.95, , (A) 300 watt, (B) 50 watt, (C) 150 watt, (D) 25 watt, The A.C. meters are based on the principle, of –, (A) Heating effect, (B) magnetic effect, (C) Chemical effect, (D) Electromagnetic effect, The correctly marked ammeter for A.C. current, is shown in (1), , (B) Increases, (D) Zero, , (3), , Three bulbs of 40, 60 and 100 watt are, connected in series with the source of 200, volt. Then which of the bult will be glowing, the most –, , Q.99, , (2), , 0 1 2 3 4 5 6 7 8, 0, , 01, , 3, , 2, , (4) None of these, , 1 2 3 4, , Alternating current can not be measured by, direct current meters, because –, (A) alternating current can not pass through, an ammeter, , (A) 100 watt, (B) 60 watt, , (B) the average value of current for complete, cycle is zero, , (C) 40 watt, (D) All are glowing equally, , (C) some amount of alternating current is, destroyed in the ammeter., , If two bulbs each of 220V, 30 watt are, connected in series, then we get electric, power as –, , (D) None of these, , (A) 60 watt, (C) 6 watt, , Q. No., , Q.98, , Two electric bulbs of 100 watt (220 volt) are, connected in series and these are connected, with other bulb of 100W (220V) in parallel then, total power in watt will be –, , 1, , 2, , Q.100 The A.C. meters measure its –, (A) root mean square value, (B) peak value, , (B) 15 watt, (D) 30 watt, , 3, , 4, , 5, , 6, , 7, , (C) square mean value, (D) None of the above, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , Ans., , D, , A, , B, , B, , B, , C, , A, , D, , D, , C, , D, , D, , A, , A, , B, , C, , B, , A, , A, , A, , Q. No., , 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , Ans., , B, , D, , C, , C, , C, , C, , A, , B, , C, , B, , C, , D, , B, , C, , A, , C, , A, , D, , D, , C, , Q. No., , 41, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 50, , 51, , 52, , 53, , 54, , 55, , 56, , 57, , 58, , 59, , 60, , Ans., , A, , D, , A, , B, , B, , C, , B, , C, , B, , C, , B, , C, , B, , B, , A, , C, , D, , B, , A, , D, , Q. No., , 61, , 62, , 63, , 64, , 65, , 66, , 67, , 68, , 69, , 70, , 71, , 72, , 73, , 74, , 75, , 76, , 77, , 78, , 79, , 80, , Ans., , B, , D, , B, , A, , A, , D, , B, , B, , C, , C, , D, , D, , C, , C, , B, , A, , A, , C, , A, , D, , Q. No., , 81, , 82, , 83, , 84, , 85, , Ans., , D, , B, , A, , D, , D, , 86, A, , 87, A, , 88, D, , 89, D, , 90, D, , 91, B, , 92, C, , 93, C, , 94, C, , 95, B, , 96, C, , 97, A, , 98, B, , 99, B, , 100, A, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

Physics by Shivendra Sir, Q.91, , Q.92, , Two bulbs of 500 watt and 300 watt work on, 200 volt r.m.s. the ratio of their resistances, will be –, (A) 25 : 9, , (B) 3 : 5, , (C) 9 : 25, , (D) 5 : 9, , An air core coil and an electric bulb are, connected in series with an A.C. source. If, an iron rod is put in the coil, then the, intensity of bulb’s will –, , Q.96, , Q.97, , (A) Be same, (B) Increase, (C) Decrease, (D) Decrease, increase, Q.93, , If a bulb and a coil are connected in series, with D.C. source and a iron core put in the, coil then the glowing of bulb –, (A) Decreases, (C) No change, , Q.94, , Q.95, , (A) 300 watt, (B) 50 watt, (C) 150 watt, (D) 25 watt, The A.C. meters are based on the principle, of –, (A) Heating effect, (B) magnetic effect, (C) Chemical effect, (D) Electromagnetic effect, The correctly marked ammeter for A.C. current, is shown in (1), , (B) Increases, (D) Zero, , (3), , Three bulbs of 40, 60 and 100 watt are, connected in series with the source of 200, volt. Then which of the bult will be glowing, the most –, , Q.99, , (2), , 0 1 2 3 4 5 6 7 8, 0, , 01, , 3, , 2, , (4) None of these, , 1 2 3 4, , Alternating current can not be measured by, direct current meters, because –, (A) alternating current can not pass through, an ammeter, , (A) 100 watt, (B) 60 watt, , (B) the average value of current for complete, cycle is zero, , (C) 40 watt, (D) All are glowing equally, , (C) some amount of alternating current is, destroyed in the ammeter., , If two bulbs each of 220V, 30 watt are, connected in series, then we get electric, power as –, , (D) None of these, , (A) 60 watt, (C) 6 watt, , Q. No., , Q.98, , Two electric bulbs of 100 watt (220 volt) are, connected in series and these are connected, with other bulb of 100W (220V) in parallel then, total power in watt will be –, , 1, , 2, , Q.100 The A.C. meters measure its –, (A) root mean square value, (B) peak value, , (B) 15 watt, (D) 30 watt, , 3, , 4, , 5, , 6, , 7, , (C) square mean value, (D) None of the above, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , Ans., , D, , A, , B, , B, , B, , C, , A, , D, , D, , C, , D, , D, , A, , A, , B, , C, , B, , A, , A, , A, , Q. No., , 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , Ans., , B, , D, , C, , C, , C, , C, , A, , B, , C, , B, , C, , D, , B, , C, , A, , C, , A, , D, , D, , C, , Q. No., , 41, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 50, , 51, , 52, , 53, , 54, , 55, , 56, , 57, , 58, , 59, , 60, , Ans., , A, , D, , A, , B, , B, , C, , B, , C, , B, , C, , B, , C, , B, , B, , A, , C, , D, , B, , A, , D, , Q. No., , 61, , 62, , 63, , 64, , 65, , 66, , 67, , 68, , 69, , 70, , 71, , 72, , 73, , 74, , 75, , 76, , 77, , 78, , 79, , 80, , Ans., , B, , D, , B, , A, , A, , D, , B, , B, , C, , C, , D, , D, , C, , C, , B, , A, , A, , C, , A, , D, , Q. No., , 81, , 82, , 83, , 84, , 85, , Ans., , D, , B, , A, , D, , D, , 86, A, , 87, A, , 88, D, , 89, D, , 90, D, , 91, B, , 92, C, , 93, C, , 94, C, , 95, B, , 96, C, , 97, A, , 98, B, , 99, B, , 100, A, , Method Guru (Physics by Shivendra Sir) - Shashtri Nagar, Sultanpur, 7014344748

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