ISRO Mechanical Engineering Past Paper PDF 2020 - MADE EASY

Detailed Solutions MECHANICAL ISRO ENGINEERING Written Test of Scientist/Engineer Examination Date of Test : 12-01-2020...

Detailed Solutions MECHANICAL ISRO ENGINEERING Written Test of Scientist/Engineer Examination Date of Test : 12-01-2020 Set-A MADE EASY has taken due care in making solutions. If you find any discrepency/ typo/technical error, kindly mail us at: [email protected] Students are requested to share their expected marks. www.madeeasy.in Corporate Office: 44-A/1, Kalu Sarai, New Delhi - 110016 | Ph: 011-45124612, 9958995830 Delhi | Hyderabad | Noida | Bhopal | Jaipur | Lucknow | Indore | Pune | Bhubaneswar | Kolkata | Patna  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 ISRO (Scientist/Engineer) Examination Mechanical Engineering : Paper Analysis Exam held on 12.01.2020 Sl. Subjects No. of Qs. Level of Difficulty 1 Fluid Mechanics 4 Easy 2 Heat and Mass Transfer 8 Moderate 3 Theory of Machine 12 Moderate 4 Material Science 11 Easy 5 Production Engineering 8 Moderate 6 Strength of Material 9 Easy 7 Thermodynamics 6 Moderate 8 Engineering Mechanics 2 Moderate 9 Power Plant Engineering 6 Moderate 10 Fluid Machinery 2 Moderate 11 Machine Design 2 Easy 12 IC Engines 1 Easy 13 Industrial Engineering 1 Easy 14 Refrigeration and Air-conditioning 1 Moderate 15 Mathematics 7 Moderate Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 2  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.1 A centrifugal pump having an impeller diameter of 127 mm, delivers a power of 12 hp. If the impeller diameter is changed to 254 mm, what is the power, if other parameters are kept constant? (a) 48 hp (b) 192 hp (c) 24 hp (d) 96 hp Ans. (a) The main output of centrifugal pump is manometric head. Take same manometric head.  Poutput = P = ηo × mgH P = ηo × ρQgH ⎡ Aflow ∝ D 2 ⎤ Here, Q = Aflow × Vflow ⎢ ⎥ ⎢⎣Vflow ∝ H ⎥⎦ Q ∝ D2 H P ∝ D2 H × H P ∝ D2(H)3/2 P = Constant (which is specific quantity) D (H )3/2 2 P1 P2 = (Here H1 = H2; similar output) D1 (H1)3/2 2 D2 (H 2 )3/2 2 2 2 P2 ⎛D ⎞ ⎛ 254 ⎞ = ⎜ 2⎟ = ⎜ ⎝ 127 ⎟⎠ = 4 P1 ⎝ D1 ⎠ P2 = 4P1 = 4 × 12 = 48 hp End of Solution Q.2 In a circular tube of diameter 100 mm and length 13 m with laminar low, the friction factor is estimated to be 0.05. Calculate the Reynolds number? (a) 950 (b) 2300 (c) 1280 (d) None of the above Ans. (c) 64 For laminar flow, f= Re 64 64 Re = = = 1280 f 0.05 End of Solution Q.3 An open tank is filled with water to a height of 20 m. What is the velocity of the water flow at the outlet, if the outlet is at the base of the tank? (a) 40 m/s (approx.) (b) 20 m/s (approx.) (c) 10 m/s (approx.) (d) 5 m/s (approx.) Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 3  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Ans. (b) 20 m V= 2gh = 2 × 10 × 20 = 20 m/s End of Solution Q.4 For a non-dimensional specific speed value of 1, for maximum efficiency, which of the following turbines is preferred? (a) Pelton wheel (b) Francis turbine (c) Kaplan turbine (d) Tyson wheel Ans. (c) Non-dimensional specific speed, N P KS = ρ ⋅ (gH )5/ 4 ⎛N P ⎞ 1 KS = ⎜ 5/ 4 ⎟ ⋅ ⎝ (H ) ⎠ ρ ⋅ (g)5/ 4 NS KS = ρ × (g)5/ 4 Here, KS = 1 NS = 1× ρ × (g)5/4 = 1000 × (9.81)5/ 4 = 31.6227 × 17.3614 = 549.01 units ⇒ For Kaplan turbine specific speed value ranges from 300 to 1000, hence option (c) is correct. End of Solution Q.5 An experiment is conducted with a fluid of density 1 kg/m3 at 10 m/s velocity. The free stream static pressure is 100 kPa and the local static pressure is 101 kPa. What is the pressure coefficient at the location? (a) 70 (b) 80 (c) 20 (d) 50 Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 4  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Ans. (c) PLocal − PFree Stream (101 − 100) × 103 Cp = = = 2 1 2 1 2 ρU × 1 × 10 2 ∞ 2 End of Solution Q.6 In a P-V diagram of a steady flow compressor, the intercooling can be represented as (a) Vertical line (b) Horizontal line (c) Parabolic line (d) None of the above Ans. (b) P P3 intercooling P2 P1 V End of Solution Q.7 Which of the following is a non-Newtonian fluid? (a) Air (b) Water (c) Gasoline (d) None of the above Ans. (d) End of Solution Q.8 A hot body at 1000K transfers 2000 kJ of heat to a body at 500 K. Determine the net entropy change? (a) +4 kJ/kg (b) – 2 kJ/kg (c) +2 kJ/kg (d) – 4 kJ/kg Ans. (c) A B 1000 K 500 K 2000 KJ −2000 KJ ΔSA = = –2 1000 K 2000 KJ ΔSB = = 4 500 K ∴ ΔSnet = –2 + 4 = +2 End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 5  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.9 Highest ratio of specific heat is possible for? (a) Argon (b) Helium (c) Hydrogen (d) Air Ans. (a, b) Since for both γ = 1.66 End of Solution Q.10 Which of the following can be used to determine the variation of saturation pressure with temperature along phase boundaries? (a) Joule-Thomson relation (b) Carnot equation (c) Rankine-Hougoniot’s relation (d) Clausius-Clapeyron relation Ans. (d) End of Solution Q.11 Air at 27°C and 100 kPa enters in a steady flow to a nozzle at a velocity of 100 m/s. If the inlet area of the nozzle is 0.5 m2, what is the mass flow rate through the system? (a) 116 kg/s (b) 232 kg/s (c) 58 kg/s (d) 143 kg/s Ans. (c) 100 kPa 300 K 100 m/s 2 0.5 m (2) (1) Mass flow rate, m = ρ1 × A1 × V1 ∴ P1 = ρ1R1T1 P1 100 ⇒ ρ1 = = = 1.1614 kg/m3 RT1 0.287 × 300 m = 1.1614 × 0.5 × 100 = 58.07 kg/s End of Solution Q.12 Critical point of water is (a) 22.06 kPa (b) 22.06 MPa (c) 22.06 atm (d) 22.06 mbar Ans. (b) End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 6  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.13 In a shower mixer, cold water at 27 deg C, flowing at 5 kg/min is mixed with hot water at 77 deg C flowing at 15 kg/min. The exit temperature of the mixture is (a) 45.4 deg C (b) 64.5 deg C (c) 34.5 deg C (d) 68.4 deg C Ans. (b) For steady state, the rate of enthalpy entering the mixer = The rate of enthalpy leaving mixture (5 × Cp × 27) + (15 × Cp + 77) = (5 + 15) × Cp × (Texit)mixture w w w ⇒ Texit = 64.5°C. End of Solution Q.14 Which of the following uses a regenerator? (a) Brayton cycle (b) Ericsson cycle (c) Stirling cycle (d) Both (b) and (c) Ans. (d) End of Solution Q.15 Ratio of convective mass transfer to the mass diffusion rate is called? (a) Sherwood number (b) Schmidt number (c) Rayleigh number (d) Strouhal number Ans. (a) End of Solution Q.16 A gas is contained in a cylinder with a moveable piston of 100 kg mass. When 2500 J of heat flows into the gas, the internal energy of the gas increases by 1500 J. What is the distance through which the piston rises? (a) 2 m (b) 1 m (c) 2.5 m (d) 0.5 m Ans. (b) PAtm MPiston = 100 kg A Gas Q = 2500 J (ΔU)Gas = 1500 J Applying 1st law Q = (ΔU)Gas + W 2500 J = 1500 J + W Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 7  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 WGas = 1000 J WGas = (FAtm + FPiston) × ΔxPiston Rises As area of piston is not given, assuming W.D. on atmosphere to be zero. WGas = FPiston × Δx = mgΔx 1000 J = 100 × 9.81 × Δx Δx = 1.01 m End of Solution Q.17 A refrigerator with COP of 5 is used in a room at 300 K. What will be the heat intake through a section of refrigerator wall of area 100 cm × 100 cm with a thickness of 10 cm, assuming only conduction? Value of thermal conductivity of the wall can be taken as 1 W/cmK. (a) 5000 W (b) 1000 W (c) 7500 W (d) 3000 W Ans. (*) Assuming the refrigerator working on reversed Carnot cycle (Ideal Refrigerator) 300 k A = (1 × 1) m2 250 k 10 cm TL TL COP = 5 = = TH − TL 300 − TL ⇒ TL = 250 K KA ΔT q= Watt b 100 × (1 × 1)(350 − 250) = Watt 0.1 = 50,000 watt (No answer matching) End of Solution Q.18 A mass of 1 kg of air at 27°C and 0.98 atm is taken through a diesel cycle. If the compression ratio of the engine is 16, calculate the temperature of the air after compression? (For calculation, take the ratio of specific heats of air as 1.5) (a) 1200 deg C (b) 1473 deg C (c) 927 deg C (d) 768 deg C Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 8  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Ans. (c) m = 1 kg, T1 = 27°C = 300 K, rc = 16 T2 (γ–1) = 160.5 = 4 T1 = rc T2 = 4 × 300 = 1200 K ∴ T1 = 927° C End of Solution Q.19 An aircraft is in its take off roll at sea level with ambient temperature of 18°C. What is the approximate speed of the aircraft if the temperature measured by a probe at the exit of the engine diffuser is 36°C? (Assuming air stagnation at diffuser outlet and Cp of air as 1.0 kJ/kgK) (a) 12 m/s (b) 3 m/s (c) 6 m/s (d) 18 m/s Ans. (*) V2 To = T1 + 2cp V2 ⇒ 36 = 18 + 2 × 1 × 103 ⇒ V2 = 2 × 103 (36 – 18) V= 2 × 103 × 18 = 60 10 m/s End of Solution Q.20 The external surface of a wall of 3 m height, 5 m width and 0.5 m thickness is at a temperature of 2°C. If a heat loss of 150 W from the room is measured across the wall, find the inner wall temperature? The thermal conductivity of wall material can be taken as 1W/mK (a) 280 K (b) 285 K (c) 268 K (d) 282 K Ans. (a) T1 A = (3 × 5) m2 T2 q = 150 watt 0.5 m Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 9  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 KA ΔT q= Watt b 1 × (3 × 5)(T1 − 2) 150 Watt = 0.5 ⇒ T1 = 7 °C = 280 K End of Solution Q.21 The typical range of Prandtl number for water is (a) 0.004–0.300 (b) 1.7–13.7 (c) 50–500 (d) 2000–1000 Ans. (b) End of Solution Q.22 Analogy between momentum and heat transfer is known as (a) Stanton-Prandtl analogy (b) Grassoff-Meyer analogy (c) Chilton-Colburn analogy (d) None of the above Ans. (c) End of Solution Q.23 Which of the following statement is true for a Rayleigh flow at M = 1? (a) Enthalpy is maximum (b) Entropy is maximum (c) Enthalpy is minimum (d) Both (b) and (c) Ans. (b) Mab = 1/√k Tmax T ba Ma < 1 Maa = 1 Cooling (Ma → 0) Heating (Ma → 1) smax Ma > 1 Ma = Mach number Cooling (Ma → ∞ ) s Entropy increases with heat gain, and thus we proceed to the right on the Rayleigh line as heat is transferred to the liquid. The Mach number is Ma = 1 at point a, which is the point of maximum entropy. End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 10  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.24 A fluid passes through a well insulated tube of 4.7 cm2 cross section area and 5 m length. If the pipe friction coefficient is 0.07 and the flow velocity is 3 m/s, then the flow can be represented using? (a) Rayleigh flow model (b) Isentropic flow model (c) Gino flow model (d) Fanno flow model Ans. (d) Fanno flow is the adiabatic flow through a constant area duct where the effect of friction is considered. End of Solution Q.25 A fluid having a density of 1 g/cc is in a state with Grashof number 2 × 106 and prandtl number 0.7. Assuming acceleration due to gravity as 10 m/s2, calculate the Rayleigh number? (a) 1.4 × 106 (b) 2.86 × 106 (c) 3.7 × 10 6 (d) 8.4 × 106 Ans. (a) Ra = Gr × Pr = 2 × 106 × 0.7 = 1.4 × 106 End of Solution Q.26 A cold liquid enters a counter flow heat exchanger at 15 deg at a rate of 8 kg/s. A hot stream of the same liquid enters the heat exchanger at 75 deg at 2 kg/s. Assuming the specific heat of the fluid as 4 kJ/kg°C, determine the maximum heat transfer rate. (a) 960 kW (b) 240 kW (c) 1920 kW (d) 480 kW Ans. (d) Maximum heat trasnfer rate in heat exchanger  p )min × (Th − Tc ) = (mC i i = (2 × 4)(75 – 15) = 480 kW End of Solution Q.27 The temperature of a surface with 0.2 m2 area is 17 deg C. Calculate the wavelength corresponding to maximum monochromatic emissive power (a) 20 micrometers (b) 30 micrometers (c) 10 micrometers (d) 40 micrometers Ans. (c) W ien’s Displacement Law : ien’s Assuming body as black body, λmT = 2898 μmK Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 11  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 2898 2898 λm = = = 9.993 μm T (17 + 273)  10 μm End of Solution Q.28 For a surface, the direct and diffuse components of the solar radiation are 400 and 300 W/m2 and the direct radiation makes 60 deg angle with the normal. If the surface temperature is 300 K and effective sky temperature is 200 K, calculate net rate of radiation heat transfer. (Assume solar absorptivity and emissivity as 0.1 each; For calculation. Take Stefan Boltzmanns constant as 6 × 10–8 W/m2 K4) (a) 11 W/m2 (b) 45 W/m2 (c) 33 W/m2 (d) 64 W/m2 Ans. (a) Sun Atmosphere GD 60° GSolar GSky EEmitted αsGSolar ∈GSky EAbsorbed The total solar energy incident on the surface GSolar = (GDirect × cos θ) + GDiffuse = (400 × cos 60) + 300 w/m2 = 500 w/m2 Net rate of radiation H.T.= ( α Solar × GSolar ) + ∈σ TSky − T5 4 4 ( ) = (0.1 × 500) + 0.1 × 6 × 10–8(2004 – 3004) = 11 w/m2 Here, σ = 6 × 10–8 w/m2-k4 instead of (5.67 × 10–8) w/m2-k4 End of Solution Q.29 Which of the following element is added to High Speed Tool Steels as a scavenger to remove slag impurities during melting and also for increasing cutting efficiency of tools. (a) Chromium (b) Vanadium (c) Molybdenum (d) Managanese Ans. (b) End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 12  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.30 Region of disorder created by movement of dislocations in superlattice is called: (a) Twin (b) Stacking fault (c) Anti-phase boundary (d) Orowon loop Ans. (c) End of Solution Q.31 The state of stress during deep drawing forming operation of a cup is (a) In the flange of blank, uni-axial compression and in wall of cup, bi-axial tension and compression (b) In the flange of blank, uni-axial tension and in wall of cup, bi-axial tension and compression (c) In the flange of blank, uni-axial tension and compression and in the wall of cup, simple uni-axial tension (d) Both flange and wall of the cup will have bi-axial compression and tension Ans. (c) End of Solution Q.32 During turning of a metallic rod at a given condition, the tool life was found to decrease from 100 min. to 25 min. When cutting speed was increased from 50 m/min. to 100 m/min. How much will be life of tool if machined at 80 m/min? (a) 38.06 min. (b) 39.06 min. (c) 40.06 min. (d) 41.06 min. Ans. (b) n n VT 1 1 = V2T2 or 50 × 100n = 100 × 25n n ⎛ 100⎞ 100 ⎜⎝ 25 ⎟⎠ or = 50 or 4n = 2 or n = 0.5 n n VT 1 1 = V3T3 50 × 1000.5 = 80 × T30.5 or 50 × 10 = 80 × T30.5 25 or = T30.5 4 1/ 0.5 2 ⎛ 25⎞ ⎛ 25⎞ 625 or T3 = ⎜ ⎟ =⎜ ⎟ = ⎝ 4⎠ ⎝ 4⎠ 16 = 39.0625 min End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 13  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.33 In a single pass rolling operation, the thickness of a 100 mm wide plate is reduced from 20 mm to 15 mm. The roller radius is 125 mm and rotational speed is 8 rpm. The average flow stress for the plate material is 400 MPa. The power required for rolling operation in kW is? 12.5π (a) (b) 10π 3 40π 20π (c) (d) 3 3 Ans. (d) Force, F = σ0 RΔh b = 400 N/mm2 × 125 mm × 5 mm × 100 mm = 1000 kN Arm length, a = 0.5 125 × 5 = 12.5 mm Torque, T = F × a = 1000 kN × 12.5 mm Power = 2Tw 2π × 8 = 2 × 1000 × 12.5 × W 60 20π = kW 3 End of Solution Q.34 A dummy activity is used in PERT network to describe (a) Precedence relationship (b) Necessary time delay (c) Resource restriction (d) Resource idleness Ans. (a) End of Solution Q.35 In small castings, which of the following allowances can be ignored. (a) Shrinkage Allowance (b) Rapping Allowance (c) Draft Allowance (d) Machining Allowance Ans. (c) Shrinkage and machining allowance will not depend on size of casting. Rapping allowance is required for all castings. For small casings draft allowance can be ignored. End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 14  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.36 In an arc welding process, two weld coupons were made using two different welding processes. For the first coupon, the voltage, current and welding speed used are 15 V, 300 A and 30 mm/min respectively. Whereas the second coupon is welded with 60 kV, 200 mA and 25 mm/s. If the heat transfer efficiency for welding the first coupon is half of that of second coupon, the ratio of heat input per unit length is? (a) 50 : 2 (b) 75 : 8 (c) 5 : 8 (d) 5 : 2 Ans. (b) Given: V1 = 15 V V2 = 60 kW = 60 × 103 V I1 = 300 A I2 = 200 mA = 200 × 10–3 A 30 1 v1 = 30 mm/min = = mm/sec 60 2 v2 = 25 mm/sec ηh2 ηh = 1 2 VI ηh J/s Hs = [Heat supplied per unit length] v Ratio of heat supplied from first coupon to second. V1I1 η 15 × 300 ηh2 Hs1 v 1 h1 1/ 2 2 = = H s2 V2 I 2 60 × 103 × 200 × 10−3 ηh2 ηh2 v2 25 150 75 = = 16 8 ⇒ Hs : Hs = 75 : 8 1 2 End of Solution Q.37 In Gas Tungsten Arc Welding (GTAW) which of the following polarity is generally used for getting higher penetration (a) Direct Current Straight Polarity (DCSP) (b) Direct Current Reverse Polarity (DCRP) (c) Alternating Current High Frequency (ACHF) (d) All of the above Ans. (a) In GTAW process, non consumable electrode is used, depth of penetration is more if heat generation is more on the workpiece. In Direct Current Straight Polarity (DCSP) heat generation is more on the workpiece hence higher penetration. End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 15  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.38 The laws of adhesive wear, commonly referred to as Archard’s law can be expressed as _____. (If ‘Q’ is the total volume of wear debris produced, ‘W’ is the normal load, ‘L’ is the sliding distance, ‘H’ is the hardness of the softest contacting surfaces being worn away, and ‘k’ is a non-dimensional wear coefficient dependent on the materials in contact and their exact degree of cleanliness) (a) Q = kWL/H (b) Q = kW/LH (c) Q = kHW/L (d) Q = k/LWH Ans. (a) kWL Archald’s law, Q = H End of Solution Q.39 In a four-bar linkage if ‘S’ is the length of the shortest link, ‘L’ is the length of longest link and ‘P’ and ‘Q’ are length of other links, then the criteria for getting a triple rocker mechanism in which no links will fully rotate is : (a) S + Q > P + L (b) S + L > P + Q (c) S + L = P + Q (d) S + L > P + Q Ans. (d) All three links i.e., Input, Output and Coupler will be Rocker, when Grashof’s law will not be satisfied, i.e., we will not be getting continuous relative motion in 4-bar mechanism. i.e., (S + L) > (P + Q) End of Solution Q.40 The piston of an engine moves with simple harmonic motion. The crank rotates at 120 r.p.m. with a stroke of 2 meters. The velocity of the piston, when it is at a distance of 0.8 metre from the center is : (a) 4.8 π (b) 2.4 π (c) 1.2 π (d) 0.6 π Ans. (b) A l r ODC IDC O B IDC ODC Centre 2r = 2 m x 0.8 m 1m Here N = 120 r.p.m. 2π × 120 ω= = 4π rad/s 60 Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 16  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Stroke = 2 m ⇒ 2r ⇒ r = 1 m ⇒ Crank Radius Here, x = (1 – 0.8) = 0.2 meter Here S.H.M. L Therefore, n= = Large r ⎡ ⎛ 0 ⎤ ⎞ x = r ⎢(1 − cos θ) + ⎜ n − n 2 − sin2 θ ⎟ ⎥ (n2 ⇒ very large) ⎢⎣ ⎝ ⎠ ⎥⎦ 0.2 = 1[1 – cos θ] 8 4 cos θ = 1 – 0.2 = 0.8 = = 10 5 16 ⎛ 3⎞ sin θ = 1− = ⎜ ⎟ 25 ⎝ 5⎠ 120 3 Velocity v = rω sin θ = 1 × 2π × × 60 5 12π v= = (2.4)π m/s 5 End of Solution Q.41 If a block slides outward on a link at a uniform rate of 30 m/s, while the link is rotating at a constant angular velocity of 50 rad/s counter clockwise, the Coriolis component of acceleration is _____ m/s2. (a) 1000 (b) 1500 (c) 3000 (d) 4500 Ans. (c) v = 30 m/s ω = 50 rad/s Coriolis component of acceleration: ac = 2vω = 2 × 30 × 50 = 3000 m/s2 End of Solution Q.42 In a screw jack of lead angle α, the effort required to lift the load W is given by ____ (φ = Friction Angle) (a) P = W tan(α – φ) (b) P = W tan(α + φ) (c) P = W cos(α – φ) (d) P = W cos(α + φ) Ans. (b) Effort required to lift the load (W), i.e., P = W tan(α + φ) End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 17  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.43 When the axes of first and last gear are co-axial, then gear train is known as (a) simple gear train (b) compound gear train (c) reverted gear train (d) epicyclic gear train Ans. (c) Reverted gear train is used to connect co-axial shafts. End of Solution Q.44 A disc spinning on its axis at 20 rad/s will undergo precession when a torque 100 N-m is applied about an axis normal to it. If the mass moment of inertia is 1 kg-m2, then the angular velocity of precession is? (a) 0.2 rad/s (b) 5 rad/s (c) 10 rad/s (d) 200 rad/s Ans. (b) ω = 20 rad/s Gyroscopic couple, C =100 N-m I = 1 kg-m2 ωp = ? C = I⋅ω⋅ωp 100 = 1 × 20 × ωp ωp = 5 rad/s End of Solution Q.45 Which of the following is an absorption type dynamometer? (a) Prony brake dynamometer (b) Epicyclic-train dynamometer (c) Torsion dynamometer (d) Belt transmission dynamometer Ans. (a) Absorption Dynamometers 1. Prony brake Dynamometer 2. Rope brake Dynamometer 3. Hydraulic Dynamometer Transmission Dynamometers 1. Belt Transmission Dynamometer 2. Epi-cyclic Dynamometer 3. Torsion Dynamometer End of Solution Q.46 Which of the following mechanism provides intermittent rotary motion? (a) Chebyschev Linkage (b) Geneva Mechanism (c) Peaucellier Mechanism (d) Roberts Mechanism Ans. (b) There are two intermittent motion mechanisms 1. Geneva mechanism 2. Ratchet mechanism End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 18  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 π Q.47 If m, n are integers and m + n is odd then the value of ∫ sinmx ⋅ cosnx ⋅ d x is 0 π (a) 0 (b) 2 (c) π (d) 1 Ans. (a) π π 1 I= ∫ sin(mx) ⋅ cos(nx)d x = 2 ∫0 [sin(m + n)x + sin(m − n)x ]d x 0 π/2 2 = 2 ∫ [sin(m + n)x + sin(m − n)x]d x 0 π /2 ⎡ − cos(m + n)x − cos(m − n)x ⎤ = ⎢ + ⎥ ⎣ m+n m−n ⎦ ⎧ π ⎫ ⎧ π ⎫ ⎪ cos(m + n) 2 − 1⎪ ⎪ cos(m − n) 2 − 1⎪ = −⎨ ⎬−⎨ ⎬ ⎪ m+n ⎪ ⎪ m−n ⎪ ⎩ ⎭ ⎩ ⎭ =0 – 0 = 0 (∵ m + n is integer ⇒ m – n is also an integer) End of Solution x2 x4 x5 Q.48 1+ x + − − +..... = 2 8 15 (a) etan x (b) ecos x (c) esin x (d) ex sin x Ans. (c) Let us take option (c). i.e., f (x) = esinx ⇒ f (0) = 1 f ′(x) = esinx ⋅ cosx ⇒ f ′(0) = 1 f ′′(x) = esinx ⋅ cos2x – esinx sinx ⇒ f ′′(0) = 1 Similarly f ′′′(0) = 0 f iv′(0) = –3..... Now Maclaurin series expansion of f (x) is x2 x3 x4 f (x) = f (0) + xf ′(0) + f ′′(0) + f ′′′(0) + f iv′ (0) +...... 2! 3! 4! Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 19  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 x2 x3 x4 esinx = 1+ x(1) + (1) + (0) + (−3) +........ 2! 3! 4! x2 x4 = 1+ x + − −...... 2 8 End of Solution Q.49 The Laplace transform of eat cos ωt (s − a) ω (a) (b) (s − a) + ω 2 2 (s − a)2 + ω 2 a s (c) (d) (s − a) + ω 2 2 (s − a)2 + ω 2 Ans. (a) s We know that L{cosωt} = s + ω2 2 By first shifting property, (s − a) L{e at cosωt} = (s − a)2 + ω 2 End of Solution Q.50 If v = yzi + 3zxj + zk, then curl v is (a) –3xi + yj + 2zk (b) 3xi – yj + 2zk (c) –3xi – yj – 2zk (d) 3xi + yj – 2zk Ans. (a)  ˆ ˆ ˆ V = yzi + 3z x j + zk then iˆ jˆ kˆ    ∂ ∂ ∂ curl V = ∇ × V = ∂x ∂y ∂z yz 3z x z = iˆ[0 − 3x] − jˆ[0 − y ] + kˆ[3z − z] = −3xiˆ + yjˆ + 2zkˆ End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 20  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.51 cos(z) can be expressed as 1 iz 1 iz (a) (e + e −iz ) (b) (e − e −iz ) 2 2 1 iz 1 iz (c) (e + e −iz ) (d) (e − e −iz ) 2i 2i Ans. (a) 1 1 cosz = [2 cos z ] = [cos z + i sin z + cos z − i sin z ] 2 2 1 iz = (e + e − iz ) 2 End of Solution Q.52 In a vector field, Divergence of the gradient is (a) curl (b) unity (c) zero (d) Laplacian Ans. (d)     Div[g rad ϕ ] = ∇ ⋅ (∇ϕ) = (∇ ⋅ ∇)ϕ = ∇2 ϕ which is Laplacian operator. End of Solution Q.53 If a continuously differentiable vector function is the gradient of a scalar function, then its curl is (a) infinite (b) indeterminate (c) unity (d) zero Ans. (d) Let φ(x, y, z) = c be a scalar function then ∂φ ˆ ∂φ ˆ ∂φ gradφ = iˆ +j +k ∂x ∂y ∂z iˆ jˆ kˆ  ∂ ∂ ∂ So, curl grad φ = ∇ × gradφ = ∂x ∂y ∂z ∂φ ∂φ ∂φ ∂x ∂y ∂z = 0iˆ + 0jˆ + 0kˆ = 0 End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 21  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.54 What does ‘a’ and ‘e’ indicate in the surface texture symbol shown below a e (a) Roughness value, machining allowance (b) Production method, roughness value (c) Machining allowance, sampling length (d) Heat treatment method, sample length Ans. (a) End of Solution Q.55 Two steel trusses, XY and YZ of identical size supports a load of 200 N as shown in figure. The length of the truss is 1 m. The force in the truss XY in N is X Z 30° 30° 1m 1m Y 200 N (a) 100 N (b) 200 N (c) 150 N (d) 50 N Ans. (b) According to Lami’s theorem FXY FYZ 200 = = sin120° sin120° sin120° So, FXY = FYZ = 200 N End of Solution Q.56 A block weighing 100 N is resting on a plane inclined with horizontal as shown in figure. What horizontal force P is necessary to hold the body from sliding down the plane? (Coefficient of friction can be taken as 0.25) 100 N P 45° (a) 30 N (b) 120 N (c) 60 N (d) 15 N Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 22  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Ans. (c) µN N P cos θ θ P P θ W μ = 0.25 W sin θ Psinθ + wcosθ θ W ∵ ΣFy = 0 N= W cos θ + P sin θ and ΣFx = 0 W sin θ = P cos θ + μN ⇒ W sin θ = P cos θ + μ(W cos θ + P sin θ) ⇒ W sin θ – μW cos θ = P cos θ + μP sin θ W (sinθ − μ cos θ) 100 × (sin45° − 0.25 × cos 45°) ⇒ P= = cos θ + μ sinθ cos 45° + 0.25 × sin45° = 60 N End of Solution Q.57 The ratio of maximum shear stress to average shear stress in a beam of rectangular section is (a) 5.1 (b) 2/3 (c) 3/2 (d) 1.0 Ans. (c) Maximum shear stress in a rectangular cross-section beam, 3W τmax =...(i) 2 bd Average shear stress in a rectangular cross-section beam, W W τavg = =...(i) A bd τmax 3 = τ avg 2 End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 23  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.58 A massless beam has a loading pattern shown in figure. Find the bending moment at mid span. 4 kN/m A B C 1m 1m (a) 1 kN-m (b) 3 kN-m (c) 2 kN-m (d) 0.0 kN-m Ans. (a) W = 4 kN 4 kN/m B C 0.5 m RA 1m 1m RC 4 × 1.5 RA = = 3 kN 2 B.MB = RA × 1 – W × 0.5 = 3 × 1 – 4 × 0.5 = 1 kN-m End of Solution Q.59 Steel machine element at the critical section is in biaxial stress state with two principal stress being 300 N/mm2 and 300 N/mm2 (equal magnitude). Find the von Mises stress (in N/mm2) in the member (a) 212.1 (b) 600 (c) 424.2 (d) 300 σ Ans. (d) σ = 300 MPa = σ1 = σ2 σ σ By using M.D.E.T, σper = σ12 + σ22 − σ1σ2 σ σper = σ = 300 MPa End of Solution Q.60 A machine component of 90 kg mass needs to be held in position using three springs as shown below. The spring constants K1, K2 and K3 are 4, 4 and 8 N/m respectively. Find the natural frequency of the system in rad/sec. K3 90 kg m K2 K1 Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 24  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 (a) 0.33 (b) 0.42 (c) 0.13 (d) 3.0 Ans. (a) k3 = 8 N/m M k2 = 4 N/m k= ? k1 = 4 N/m Here, k1 and k2 are in series whereas k3 is in parallel. 1 ⎛1 1⎞ 1 1 2 = ⎜ + ⎟ = + = k ⎝ k1 k2 ⎠ 4 4 4 1 1 = ⇒ k = 2 N/m k 2 Now, Total keq = (k + k3) = 2 + 8 = 10 N/m Natural frequency is keq 10 1 1 ωn = = = = rad/s m 90 9 3 ωn = 0.33 rad/s End of Solution Q.61 A cylindrical pressure vessel has diameter 200 mm and thickness 2 mm. Find the hoop and axial stress (N/mm2) in the cylindrical vessel, when it is subjected to an internal pressure of 5 MPa. (a) 125, 125 (b) 125, 250 (c) 250, 125 (d) 250, 125 Ans. (d) pd 5 × 200 σhoop = = = 250 MPa 2t 2×2 1 σLONG or σaxial = σhoop = 125 MPa 2 End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 25  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.62 For a long slender column of uniform cross-section, the ratio of critical buckling load for the case with both ends hinged to the case with both ends clamped is (a) 0.25 (b) 4.0 (c) 0.125 (d) 0.5 Ans. (a) ⎛ n ⋅ π2E Imin ⎞ ⎜ ⎟ (Pe )B.H ⎝ L2 ⎠ B.H n 1 (Pe )B.F = ⎛ = B.H = = 0.25 n ⋅ π E Imin ⎞ 2 nB.F 4 ⎜ 2 ⎟ ⎝ L ⎠ B.F End of Solution Q.63 If the length of a column subjected to compressive load is increased by three times its original length, the critical buckling load becomes (a) 1/3 of the original value (b) 3 times the original value (c) 1/9 of the original value (d) 1/27 of the original value Ans. (c) 1 Pe ∝ L2 Hence answer is option (c). End of Solution Q.64 A square bar of size 10 mm × 10 mm and length 1000 mm is subjected to 200 N axial tensile force. The bar is made of mild steel having modulus of elasticity of 200 GPa. Find the strain energy density stored in the bar under this state of loading? (a) 10 J/m3 (b) 20 J/m3 (c) 2 J/m 3 (d) 5 J/m3 Ans. (a) Strain energy per unit volume = modulus of resilience 2 ⎛ 200 ⎞ ⎜⎝ ⎟ σa2 (P/A)2 10 × 10 ⎠ = = = 2E 2E 2 × 200 × 103 1 = N-mm/mm3 = 10 N-m/m3 105 End of Solution Q.65 A solid circular shaft needs to be designed to transmit a torque of 55 Nm. If the allowable shear stress of the material is 280 N/mm2, find the diameter (in mm) of the shaft required to transmit the torque. (Assume, π = 22/7) (a) 5.62 (b) 10 (c) 31.62 (d) 25.0 Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 26  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Ans. (b) 16T 16 × 55 × 103 d3 = = πτper 22 × 280 7 d = 10 mm End of Solution Q.66 If S is the applied stress, c is the width of the crack and r the radius of curvature at the tip of the crack, Griffith’s crack theory gives the concentrated stress Sc as (a) 2S (c/r)1/2 (b) 2S (c/r)1/3 (c) S (c/r)1/2 (d) 2S (2c/r)1/2 Ans. (a) End of Solution Q.67 A metal alloy machine component is subjected to fluctuating tensile stress from 100 N/mm2 to 200 N/mm2. the material has yield and endurance strength of 450 and 200 N/mm2 respectively. Find the factor of safety of the machine component. (a) 5/6 (b) 12/7 (c) 1/3 (d) 6/5 Ans. (b) By using Soderberg’s equation, 1 σm kf σv = σ + σ N yt e σmax + σmin σm = = 150 MPa 2 σmax − σmin σv = = 50 MPa 2 k f = 1 (assumed) 1 150 (1) 50 = + N 450 200 12 N= 7 End of Solution Q.68 A diver of mass 100 kg is standing at the tip of a spring board of negligible mass. The natural frequency of the spring board with the diver is 1.6 Hz. What is the static deflection at the tip of the spring board when the diver is standing at the tip? (a) 0.1 mm (b) 981 mm (c) 98.1 mm (d) 9.81 mm Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 27  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Ans. (c) Here fn = 1.6 Hz ωn = 2πfn = 2π × 1.6 = 3.2π rad/s By Rayleigh’s method, if Δ = Static Deflection, then g ωn = Δ 9.81 3.2π = Δ 9.81 (3.2π)2 = Δ 9.81 Δ= = 0.09706 meter (3.2 π)2 Δ = 97.06 mm Closest answer is 98.1 mm. End of Solution Q.69 Which of the following is classified as a secondary bond in materials? (a) Ionic bonding (b) Covalent bonding (c) Metallic bonding (d) Hydrogen bonding Ans. (d) End of Solution Q.70 The number of atoms per unit cell for a FCC crystal structure is (a) 2 (b) 4 (c) 6 (d) 8 Ans. (b) End of Solution Q.71 Which of the following hardness tester uses Diamond cone type indenter? (a) Brinell (b) Vickers (c) Knoop (d) Rockwell Ans. (d) End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 28  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.72 In a two component system, if the non-compositional variable is only temperature, the number of degree of freedom in the case of a single phase field as per Gibbs Phase Rule is? (a) 0 (b) 1 (c) 2 (d) 3 Ans. (c) From Gibb’s phase rule with only temperature as variable, P + F= C + 1 1 + F=2 + 1 F=2 End of Solution Q.73 In an iron-carbon diagram, the percentage by weight of carbon at eutectoid compositon is (a) 1.12 (b) 0.76 (c) 0.24 (d) 0.03 Ans. (b) End of Solution Q.74 Of the various microstructures that may be produced for a given steel alloy, which among the following is the hardest form (a) Martensite (b) Pearlite (c) Bainite (d) Spheroidite Ans. (a) End of Solution Q.75 Which spring steel is widely used for aircraft engine valves? (a) Oil tempered wire (0.6 C to 0.7 C) (b) Chrome vanadium (c) Hard drawn wire (0.6 C to 0.7 C) (d) Phosphor bronze wire Ans. (b) End of Solution Q.76 If, Kf is the fatigue stress concentration factor and Kt is the theoretical stress concentration factor then, the notch sensitivity q is? (a) (Kf + 1)/(Kt + 1) (b) (Kf – 1)/(Kt – 1) (c) (Kt – 1)/(Kf – 1) (d) (Kf – 1)/(Kt + 1) Ans. (b) End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 29  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.77 Beach or clamshell marks on a failed metallic surface are typical of (a) Ductile fracture (b) Brittle fracture (c) Creep failure (d) Fatigue failure Ans. (d) Fast growth (rough region) Beach marks (slow growth) Smooth region Origin of crack End of Solution Q.78 In vibration isolation if ω/ωn is less than 2 then the transmissibility will be (a) Less than one (b) Equal to one (c) Greater than one (d) Zero Ans. (c) ε 1 0 1 2 ω/ωn ω By this curve it can be inferred, when < 2 ⇒ε >1 ωn End of Solution Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 30  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Q.79 The natural frequency of a simply supported beam of length l with mass M at its centre, flexural rigidity EI and negligible beam is 1 48E I 1 3E I (a) (b) 2π M l 3 2π M l 3 1 EI (c) (d) None of these 2π M l 3 Ans. (a) Simply supported Δ Mg = W Here Static deflection Wl 3 Mgl 3 Δ= = 48E I 48E I By Rayleigh’s method, Rayleigh’s g g ωn = = Δ Mgl 3 48E I 48E I ωn = Ml3 ωn 1 48E I fn = = 2π 2π M l 3 End of Solution Q.80 A machine component of natural frequency 20 rad/s is subjected to a base motion from the machine which is harmonic in nature with amplitude 3 m/s2 at 10 rad/s. What is the peak amplitude of relative displacement of the components if the damping is negligible? (a) 0.1 mm (b) 1.0 mm (c) 10.0 mm (d) 100.0 mm Ans. (c) ωn = 20 rad/s amax = 3 = (F0)/m (F0 = meω2 = m ⋅ amax) (amax = eω2) Here, ξ→ 0 ω = 10 rad/s ω 10 1 = = ωn 20 2 We know that: Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 31  ISRO : Mechanical Engineering Detailed Solutions : Exam held on 12.01.2020 Amplitude (F0 /s ) A= [ξ = 0] 0 ⎛ 2⎞2 2 ⎛ ω⎞ ⎡ 2ξω ⎤ ⎜1 − ⎜ ⎟ + ⎢ ⎥ ⎜⎝ ⎝ ωn ⎟⎠ ⎟⎠ ⎣ ωn ⎦ (F0 /s) (F0 /m) /(s/m ) = 2 = 2 ⎛ ω⎞ ⎛ ω⎞ 1− ⎜ 1− ⎜ ⎝ ωn ⎟⎠ ⎝ ωn ⎟⎠ (F0 / m ) ωn2 = 2 ⎛ ω⎞ 1− ⎜ ⎝ ωn ⎟⎠ amax / ωn2 3/(20)2 = 2 = 2 ⎛ ω⎞ ⎛ 1⎞ 1− ⎜ ⎟ 1− ⎜ ⎟ ⎝ ωn ⎠ ⎝ 2⎠ 3 / 400 4 = = = 0.01meter 3/4 400 = 10 mm End of Solution „„„„ Corporate Office: 44-A/1, Kalu Sarai, New Delhi-110016 | [email protected] | ß www.madeeasy.in Page 32