EDUCRUCIAL

Important questions of JB Gupta book Model Test Paper 2 : (We are working to provide detailed solutions of each question. We are updating this page.) Question 1 : In a magentic material, hysteresis loss takes place primarily due to  (a) rapid reversals of its magnetisation  (b) flux density lagging behind magnetis ing force  (c) molecular friction  (d) its high retentivity Show Answer Ans: D   Question 2 : On which of the following factors does the resolution of a potentiometer depend? (a) Size of wire  (b) Type of contact (c) Composition of wire material  (d) Shape of wire cross-section Show Answer Ans: A   Question 3 : The braking retardation is usually in the range (a) 0.15 to 0.30 km phps  (b) 0.30 to 0.6 km phphs  (c) 0.6 to 2.4 km phps  (d) 3 to 5 km phps  (e) 10 to 15 km phphs Show Answer Ans: D Question 4 : Which of the following methods is used for reduction/elimination of harmonic torques? (a) Chording (b) Integral slot windings (c) Skewing (d) Incre

  Question : The voltages at V1 and V2 of the arrangement shown in the figure will be respectively A. 6 V and 5.4 V B. 5.4 V and 6 V C. 3 V and 5.4 V D. 6 V and 3 V Solution : For diode D1 it is clear that anode voltage is higher. So it is forward bias. Voltage V2 = 6 - Vt Vt is cut in voltage of diode. V2 = 6 - 0.6 V2 = 5.4 V For diode D2, cathode voltage is higher and hence it is reverse biased.  So V1 = 6 V So option A is correct.  For any further queries please comment below. 

  Question : When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3m A. Determine the current flowing in the diode when 0.15v forward bias is applied at room temperature. Solution : Given Io = 0.3 * 10-6A and VF = 0.15v The current flowing through the PN diode under forward bias is I = Io (e^(40VF) -1) for germanium diode = 0.3 * 10^(-6) (e^(40*0.15) -1) = 120.73mA. For any further queries please comment below.  

Post Name: VS JE Electrical Exam Date: 22-Feb-2020 Exam Time: 12:30-14:30 Sections in this paper : 10% - General knowledge  10% - English 60% - Electrical Engineering topics (Syllabus same as GATE)  10% - Computer knowledge 10% - Gujarati grammar (We are working on preparing solutions for each question. )  All  answers are as per asnwer key provided by GSECL.  GENERAL KNOWLEDGE QUESTIONS (10 marks) :  Question No.1 Who was sworn in as the 47th Chief Justice of India recently? (A) Justice Sharad Arvind Bobde (B) Justice Kurian Joseph (C) Justice Madan Lokur (D) Justice Dipak Misra Show Answer Ans: A As on Feb 2022, Chief justice of India - N. V. Raman RBI Governor - Shaktikant das Deputy governor RBI - T Rabi Shankar Gujarat energy, petrochemical, money minister - Kanubhai Desai Question No.2 What is the full form of the pesticide DDT? (A) Dichloro Diphenyl Tri-chloromethane (B) Dichloro Diphenyl Tri-chloroethane (C) Diamino Diphenyl Tri-chloroethane (

Question : [GSECL JE]  A Uniformly loaded DC distributor is fed at both ends with equal voltages, The voltage drop at the midpoint is...... that of the DC distributor fed at one end. એકસમાન રૂપે ભારિત કરવામાં આવેલા DC વિતરકને બંને છેડેથી સમાન વૉલ્ટેજ પૂરો પાડવામાં આવે છે. મધ્યબિંદુ પરનો વોલ્ટેજ ઘટાડો એક છેડે પૂરા પાડવામાં આવેલા DC વિતરકથી...................છે. A. Twice  B. One-forth C. One - half D. One - third Answer : For distributor fed at one end : Voltage drop for uniformly loaded distributor fed at one end = iR(lx - (x²/2) (Derivation of above equation is given in DC distribution chapter of V. K. Mehta book)  For voltage drop at mid point x = l/2 Substituting x = l/2 Voltage drop at mid point = 3iRl²/8 ...........................(1) For distributor fed at both ends : Voltage drop for uniformly loaded distributor fed at both ends with equal voltages = iR(lx - x²)/2 For voltage drop at mid point x = l/2 Substituting x = l/2 Voltage drop at mid point = iRl²/8 .......................

  Question : [GSECL JE ELECTRICAL]  A 1 kVA, single phase, 50 Hz, 100/250 V trarsformer gives the following results on short circuit test conducted at high voltage Side:  Voltmeter reading: 12.5 V, Ammeter reading: 4 A, Wattmeter reading: 80 W.  The equivalent resistance referred to high voltage side will be____.  ઉચ્ચ વોલ્ટેજ બાજુએ હાથ ધરાયેલ લઘુ પરિપથ પરીક્ષણ માટે 1 KVA, એકલચરણ, 50 Hz, 100/250 નું એક પરિવર્તક નીચે મુજબના પરિણામો આપે છે વોલ્ટમીટર વાંચન: 12.5 V, એમીટર વાંચન: 4 A, વૉટમીટર વાંચન: 80 W.  ઉચ્ચ વોલ્ટેજ બાજુએ સંદર્ભિત સમતુલ્ય અવરોધ _______ રહેશે. (a) 20 ohm (b) 5 ohm (c) 40 ohm (d) 10 ohm Answer : (b) 5 ohm To find resistance referred to high voltage side, we need short circuit test data only which is given in question.  High voltage side resistance = Short circuit power/(current)² Resistance referred to HV side = 80 W/ (4 A)² Resistance referred to HV side = 5 ohm.  Option (b) - correct. 

  Question : A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to transmission line. At the receiving end of the line a load is supplied through a step-down transformer of 10 MVA, 138/69 kV rating. A 0.72 pu load, evaluated on load side transformer ratings as base values, is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the value of the load (in per unit) in generator circuit will be  (a) 36 (b) 1.44 (c) 0.72 (d) 0.18  [GATE E.E. 2006] Answer : Load = 0.72 pu at load side base of 10 MVA and 69 kv. We have to find load in generating circuit. So it is required to convert 0.72 pu to new generating side base of 20 MVA, 13.8 kV. Z(new) = Z(old) * (KVA(new)/KVA(old)) * (kV(old)/kV(new))² Z(new) = 0.72 * (20/10) * (69/13.8)² Z(new) = 36 pu For any further queries, please comment below. 

  Question : A 220 Ω resistor is connected in series with the gate of an SCR as shown in fig. The gate current required to fire the SCR is 7 mA. What is the input voltage (Vin) required to fire the SCR? Answer : Vin will not be equal to only IG * R but we have to add junction voltage 0.7 volt also.  So correct Vin = 0.7 + IG*R = 0.7 + (7mA)(220 ohm) Vin = 2.24 V

 When charge Q is given to a spherical conductor of radius R , then potential at the surface of sphere is V = KQ/R  C = Q/ V  C = R/K 

QUESTION:- if A:B = 2:3 and B:C = 4:5, then C:A IS equal to: A) 15:8 B) 12:10) C) 8:5 D) 8:15 SOLUTION :- Here, option A) 15:8 is correct answer.  IF any queries in solution please comment below. 

  Question : The Average weight of A, B, C is 45 kg.if the average weight of A and B is 40 kg and that of B And C is 43 kg, then the weight of B is: Solution : Sum of average weight = A+B+C   = 45                                                    3                                               A+B+C = 45×3                                               A+B+C = 135......1) Sum of average weight of A and B A+B = 40    2 A+B = 40×2 A+B = 80......2) Sum of average weight of B and C B+C = 43    2 B+C = 43×2 B+C= 86.......3) equation 2) + equation 3) A+B+B+C =80 +86 = 166 A+2B + C = 166 =A+2B+C  - A+B+C =166 - 135 = 31 B= 31 Kg Any queries about Solution please comment below. 

  Question : The ratio of the incomes of  A  and  B  is  5 : 4  and the ratio of their expenditures is  3 : 2 . If at the end of the year, each saves  R s . 1 6 0 0 , then the income of  A  is: A. 3400 B. 3600 C. 4000 D. 4400 Solution : Let their income be x and expenditure be y A's income = 5X B's income = 4X A's expenditure = 3y B's expenditure = 2y Save money = 1600 RS. Income - expenditure = sabes So, 5X - 3Y = 1600...... 1)        4X - 2y = 1600.......2) Equation (1) is multiply by 2 and equation (2) by 3, 10X - 6y = 3200.....3) 12X - 6y = 4800.....4) Here, X = 800 A's income = 5X                     = 5 × 800                     = 4000 RS. Option (c) is answer. Any queries in Solution please comment below.