{"id":59,"date":"2025-12-06T16:33:51","date_gmt":"2025-12-06T16:33:51","guid":{"rendered":"https:\/\/engineersinstitute.com\/blog\/?p=59"},"modified":"2025-12-06T16:36:31","modified_gmt":"2025-12-06T16:36:31","slug":"gate2026-sample-questions-mcq-for-chemical-engineering-gate-ch-basics-level-concepts-practice-session-12","status":"publish","type":"post","link":"https:\/\/engineersinstitute.com\/blog\/gate2026-sample-questions-mcq-for-chemical-engineering-gate-ch-basics-level-concepts-practice-session-12\/","title":{"rendered":"GATE2026 Sample Questions MCQ for Chemical Engineering-GATE CH , Basics level Concepts Practice-Session 12"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Use the discount code <a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\"><strong>FREEMOCK10<\/strong> to get <strong>10 free mock tests<\/strong> <\/a>on our portal. Hurry, offer valid for a limited time!<\/p>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-8f761849 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-8f761849 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"683\" src=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-07_48_53-PM-1-1024x683.png\" alt=\"\" class=\"wp-image-54\" srcset=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-07_48_53-PM-1-1024x683.png 1024w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-07_48_53-PM-1-300x200.png 300w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-07_48_53-PM-1-768x512.png 768w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-07_48_53-PM-1.png 1536w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n<\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"683\" src=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-08_21_27-PM-1-1024x683.png\" alt=\"\" class=\"wp-image-53\" srcset=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-08_21_27-PM-1-1024x683.png 1024w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-08_21_27-PM-1-300x200.png 300w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-08_21_27-PM-1-768x512.png 768w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/ChatGPT-Image-Dec-6-2025-08_21_27-PM-1.png 1536w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\">These questions are provided <strong>only for review purposes<\/strong>, to give you an idea of the type of <strong>theoretical MCQs<\/strong> you may encounter. Since this is a web format, <strong>formula-based, diagram-based, or chart-based questions are not included<\/strong>, and numerical questions are also omitted as formulas cannot be supported in blog links.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In the <strong>actual test or online mock tests<\/strong>, you will get the <strong>complete set of MCQs<\/strong>, including theory and numericals, in the real exam format. You can try a <strong>full-length demo test for free<\/strong>. We are confident that this portal will become your <strong>first choice for GATE practice with high-quality, GATE-level questions<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">Access complete test papers with 6200+ questions on our ONLINE TEST PORTAL for GATE EXAM.<\/a><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Below are <strong>110 MCQs (10 per subject)<\/strong> \u2014 Theoretical, slightly difficult GATE style questions based on Basics fundamentals. Each question is followed by options (a\u2013d), the correct answer, and a short explanation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/drive.google.com\/file\/d\/12pYM6tyHroMQROM04ftlL70TLGx_gvDW\/view\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" width=\"739\" height=\"1024\" src=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/PYQ-1-739x1024.png\" alt=\"\" class=\"wp-image-57\" srcset=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/PYQ-1-739x1024.png 739w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/PYQ-1-217x300.png 217w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/PYQ-1-768x1064.png 768w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/PYQ-1-1109x1536.png 1109w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/PYQ-1.png 1220w\" sizes=\"auto, (max-width: 739px) 100vw, 739px\" \/><\/a><\/figure>\n\n\n\n<p class=\"has-light-green-cyan-background-color has-background wp-block-paragraph\"><strong><a href=\"https:\/\/drive.google.com\/file\/d\/12pYM6tyHroMQROM04ftlL70TLGx_gvDW\/view\" target=\"_blank\" rel=\"noopener\">Download GATE SOLVED PAPERS 2000-2025 Chemical Engineering-CH  ( BEST-SELLERS 11000+Copies SOLD out)<\/a><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">ENGINEERING MATHEMATICS \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>A 2\u00d72 matrix A has eigenvalues 2 and 5. For vector b \u2260 0, the system A x = b has a unique solution for every b. Which statement is correct?<br>a) A is singular.<br>b) det(A) = 0.<br>c) A is invertible.<br>d) A has a zero eigenvalue.<br>Correct Answer: c<br>Explanation: Distinct nonzero eigenvalues imply det(A)=product=10 \u22600 so A is invertible; hence unique solution exists.<\/li>\n\n\n\n<li>For f(x)=e^{2x} approximated by Taylor series about x=0 up to second order, the remainder term R_2(x) behaves as:<br>a) O(x^2)<br>b) O(x^3)<br>c) O(x)<br>d) O(1)<br>Correct Answer: b<br>Explanation: Truncating after second-order (terms up to x^2) leaves next term proportional to x^3 (third-order), so remainder is O(x^3).<\/li>\n\n\n\n<li>The ODE y&#8221; + y = \u03b4(t \u2212 t0) (impulse at t0) with zero initial conditions yields solution expressible by:<br>a) Convolution with sine kernel (Green\u2019s function)<br>b) Polynomial solution only<br>c) No causal solution<br>d) Exponential blow-up<br>Correct Answer: a<br>Explanation: Linear constant-coefficient second-order ODE solution is convolution of impulse with system\u2019s impulse response (sin kernel).<\/li>\n\n\n\n<li>For function \u03c6(x,y)=x^2 y + sin(xy), directional derivative at (1,0) in direction (1,1)\/\u221a2 equals:<br>a) 0<br>b) 2\/\u221a2<br>c) 1\/\u221a2<br>d) 3\/\u221a2<br>Correct Answer: c<br>Explanation: \u2207\u03c6 = (2xy + y cos(xy), x^2 + x cos(xy)). At (1,0): \u2207\u03c6=(0,1). Directional derivative = \u2207\u03c6 \u00b7 (1,1)\/\u221a2 = 1\/\u221a2.<\/li>\n\n\n\n<li>A square system Ax=b where A is symmetric positive definite. Conjugate Gradient (CG) method will:<br>a) Diverge for SPD matrices<br>b) Converge in at most n iterations in exact arithmetic<br>c) Require preconditioning to converge at all<br>d) Solve eigenvalue problem instead of linear system<br>Correct Answer: b<br>Explanation: CG for SPD matrices converges in at most n steps (n=dimension) in exact arithmetic.<\/li>\n\n\n\n<li>The principal value integral \u222b_{\u2212\u03c0}^{\u03c0} f(x) dx with f periodic can be expanded in Fourier series. The Gibbs phenomenon appears near:<br>a) Smooth maxima\/minima only<br>b) Discontinuities of function or derivative<br>c) Inflection points only<br>d) Where f is analytic<br>Correct Answer: b<br>Explanation: Gibbs overshoot occurs at jump discontinuities (or discontinuous derivatives) in periodic functions\u2019 Fourier approximations.<\/li>\n\n\n\n<li>For the eigenvalue problem of Laplacian \u2207^2 \u03c6 + \u03bb \u03c6 = 0 on 0&lt;x&lt;L with \u03c6(0)=\u03c6(L)=0, eigenvalues \u03bb_n scale as:<br>a) n^2 (\u03c0^2\/L^2)<br>b) n (\u03c0\/L)<br>c) constant independent of n<br>d) exponential in n<br>Correct Answer: a<br>Explanation: 1-D Dirichlet Laplacian eigenvalues \u03bb_n = (n\u03c0\/L)^2 scale as n^2.<\/li>\n\n\n\n<li>Using trapezoidal rule to integrate a smooth periodic function over one period, error typically:<br>a) O(h)<br>b) O(h^2) but can be superconvergent for periodic analytic functions<br>c) O(1) independent of h<br>d) Exponential growth with decreasing h<br>Correct Answer: b<br>Explanation: Trapezoidal rule is O(h^2) generally; for periodic analytic integrands it can show spectral accuracy (fast convergence).<\/li>\n\n\n\n<li>For random variable X with mean \u03bc and variance \u03c3^2, Chebyshev\u2019s inequality states P(|X\u2212\u03bc| \u2265 k\u03c3) \u2264:<br>a) 1\/k<br>b) 1\/k^2<br>c) e^{\u2212k}<br>d) 1\/(k\u221a2\u03c0)<br>Correct Answer: b<br>Explanation: Chebyshev gives \u2264 1\/k^2.<\/li>\n\n\n\n<li>A boundary-value ODE solved by finite-difference gives a linear system with dominant diagonal. Best iterative solver choice is:<br>a) Jacobi always<br>b) Gauss\u2013Seidel or SOR for faster convergence<br>c) Direct LU never recommended<br>d) Randomized methods only<br>Correct Answer: b<br>Explanation: For diagonally dominant systems, Gauss\u2013Seidel or SOR typically converges faster than Jacobi; direct LU acceptable but iterative preferred for large sparse problems.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">PROCESS CALCULATIONS &amp; THERMODYNAMICS \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"11\" class=\"wp-block-list\">\n<li>A continuous steady-state process has a recycle stream. If fresh feed doubled and product composition remains unchanged, what must be true about recycle ratio R?<br>a) R decreases to maintain composition.<br>b) R remains unchanged.<br>c) R increases proportionally.<br>d) R must be zero.<br>Correct Answer: a<br>Explanation: To maintain same outlet composition with higher fresh feed, recycle must reduce fraction of recycled material so that overall mix stays same.<\/li>\n\n\n\n<li>A binary ideal solution A\u2013B at T has vapor-phase composition yA related to liquid xA by Raoult\u2019s law. If vapor pressure PA\u00b0(T) >> PB\u00b0(T), at given xA the dew point:<br>a) Approaches pure A boiling point<br>b) Approaches pure B boiling point<br>c) Independent of PA\u00b0 and PB\u00b0<br>d) Equals arithmetic mean of component bps<br>Correct Answer: a<br>Explanation: If PA\u00b0 much larger, A more volatile; dew point dominated by A; behaves close to pure A.<\/li>\n\n\n\n<li>For an open steady-flow device, first law gives enthalpy balance. If kinetic and potential energy changes negligible, then \u0394H = Q \u2212 W. For an adiabatic turbine, which term \u22480?<br>a) Q only<br>b) W only<br>c) \u0394H only<br>d) All nonzero<br>Correct Answer: a<br>Explanation: Adiabatic implies Q\u22480; turbine does work so \u0394H = \u2212W.<\/li>\n\n\n\n<li>Fugacity f of a real gas approaches which expression in low-pressure limit?<br>a) f \u2192 P (ideal gas)<br>b) f \u2192 constant independent of P<br>c) f \u2192 P^2<br>d) f undefined at low P<br>Correct Answer: a<br>Explanation: At low P gases approach ideal behavior so fugacity \u2248 pressure.<\/li>\n\n\n\n<li>For a closed system undergoing reversible isothermal compression of ideal gas, \u0394S =?<br>a) 0<br>b) nR ln(V2\/V1)<br>c) nC_v ln(T2\/T1)<br>d) negative irrespective of path<br>Correct Answer: b<br>Explanation: Isothermal reversible: \u0394S = \u222b dQ_rev\/T = nR ln(V2\/V1).<\/li>\n\n\n\n<li>In a VLE flash of binary mixture, overall mole fraction z fixed. If distillate composition xD equals 0.9 and bottoms xB 0.1, then feed z=0.5\u2014what is approximate reflux ratio effect?<br>a) Higher reflux yields purer distillate (approaching xD)<br>b) Reflux has no impact on compositions at steady-state<br>c) Lower reflux increases number of stages but not purity<br>d) Reflux only affects energy consumption but not separation<br>Correct Answer: a<br>Explanation: Reflux increases internal liquid flow and sharpens separation, improving purity for given stages.<\/li>\n\n\n\n<li>Residual properties (e.g., H^R) at given T,P measure deviation from ideal gas. Residual enthalpy H^R is zero when:<br>a) Real gas equals ideal gas behavior (low pressure limit)<br>b) At critical point always<br>c) For incompressible liquids only<br>d) Never zero<br>Correct Answer: a<br>Explanation: Residual properties vanish when gas behaves ideally (P\u21920).<\/li>\n\n\n\n<li>For an endothermic reversible reaction at equilibrium, increasing temperature shifts equilibrium toward:<br>a) Products (Le Chatelier)<br>b) Reactants<br>c) No change<br>d) Depends on activation energy only<br>Correct Answer: a<br>Explanation: For endothermic, heat acts as reactant; raising T favors product formation.<\/li>\n\n\n\n<li>A steady-state evaporator removes water from solution. If molar flow of feed is F with concentration xF, vapor flow V, and bottoms B, material balance for solute gives:<br>a) F xF = B xB<br>b) F = B + V must hold but not solute balance<br>c) Mass balance only for solvent<br>d) Energy balance replaces solute balance<br>Correct Answer: a<br>Explanation: Solute (non-volatile) conserved: F xF = B xB.<\/li>\n\n\n\n<li>The Joule\u2013Thomson coefficient \u03bcJT = (\u2202T\/\u2202P)_H for real gases can be positive or negative. For ideal gas \u03bcJT =:<br>a) 0<br>b) positive<br>c) negative<br>d) undefined<br>Correct Answer: a<br>Explanation: Ideal gas enthalpy depends only on T, so no temperature change with pressure at constant enthalpy \u2192 \u03bcJT=0.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">FLUID MECHANICS \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"21\" class=\"wp-block-list\">\n<li>Flow through long circular pipe is laminar at Re=500. For fully developed laminar flow, velocity profile is parabolic; doubling pipe radius with same pressure gradient yields centreline velocity change by factor:<br>a) 2<br>b) 4<br>c) 8<br>d) 16<br>Correct Answer: d<br>Explanation: For laminar Hagen\u2013Poiseuille, volumetric flow \u221d R^4 for same pressure gradient; centreline velocity scales similarly \u2192 factor 16.<\/li>\n\n\n\n<li>A sudden contraction in duct causes vena contracta. For incompressible flow, which phenomenon increases downstream loss?<br>a) Flow separation and eddies \u2192 energy loss<br>b) Viscosity drop \u2192 energy gain<br>c) Static pressure increase only<br>d) No loss if flow is steady<br>Correct Answer: a<br>Explanation: Contraction causes separation and vortices causing head loss.<\/li>\n\n\n\n<li>Darcy friction factor for fully rough turbulent flow depends mainly on:<br>a) Reynolds number only<br>b) Relative roughness only (independent of Re sufficiently high)<br>c) Fluid compressibility only<br>d) Flow acceleration only<br>Correct Answer: b<br>Explanation: In fully rough regime, friction factor independent of Re and controlled by relative roughness.<\/li>\n\n\n\n<li>For two-phase bubbly flow in a vertical column at low gas holdup, the slip velocity between phases is dominated by:<br>a) Bubble rise due to buoyancy and drag balance<br>b) Compressibility of liquid phase<br>c) Heat transfer only<br>d) Chemical reaction rates<br>Correct Answer: a<br>Explanation: Bubble rise determined by buoyancy versus drag; at low holdup bubbles rise relative to liquid.<\/li>\n\n\n\n<li>For flow in packed bed, Ergun equation combines viscous and inertial terms. If particle size decreases keeping bed porosity constant, pressure drop per unit length:<br>a) Increases (inversely with d_p and d_p^2 terms)<br>b) Decreases<br>c) Unchanged<br>d) Oscillatory with d_p<br>Correct Answer: a<br>Explanation: Ergun includes terms proportional to 1\/d_p and 1\/d_p^2 so smaller particles \u2192 higher \u0394P.<\/li>\n\n\n\n<li>A fluid with viscosity \u03bc and density \u03c1 flows over a flat plate. Boundary layer thickness \u03b4 scales as:<br>a) (x\/\u221aRe_x) where Re_x = \u03c1 U x \/ \u03bc<br>b) x * Re_x<br>c) independent of x<br>d) inverse of x<br>Correct Answer: a<br>Explanation: For laminar boundary layer \u03b4 ~ x \/ sqrt(Re_x) ~ (\u03bc x \/ (\u03c1 U))^{1\/2}.<\/li>\n\n\n\n<li>Cavitation in pumps occurs when local pressure drops below:<br>a) Liquid vapour pressure<br>b) Ambient atmospheric pressure always<br>c) Hydraulic grade line only<br>d) Pump suction pressure higher than discharge<br>Correct Answer: a<br>Explanation: Cavitation when local pressure &lt; vapour pressure causing vapor pockets.<\/li>\n\n\n\n<li>For compressible flow through a converging nozzle, choking occurs when Mach number at throat:<br>a) Reaches 1<br>b) Exceeds 2 only<br>c) Always subsonic<br>d) Equals 0.5<br>Correct Answer: a<br>Explanation: Choked flow when M=1 at throat; mass flow becomes independent of downstream pressure.<\/li>\n\n\n\n<li>In laminar flow between parallel plates separated by 2h, driven by constant pressure gradient, the velocity profile is:<br>a) Parabolic in y with zero at walls<br>b) Uniform plug profile<br>c) Sinusoidal<br>d) Exponential decay<br>Correct Answer: a<br>Explanation: Poiseuille flow between plates yields parabolic profile.<\/li>\n\n\n\n<li>For flow measurement by orifice meter, Reynolds number effect on discharge coefficient Cd is significant when:<br>a) Re is low (transition from laminar influences flow separation)<br>b) Re is always irrelevant<br>c) Cd increases unbounded with Re<br>d) Cd is only function of fluid temperature<br>Correct Answer: a<br>Explanation: At low Re separation\/viscous effects alter Cd; at high Re Cd asymptotes.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">MECHANICAL OPERATIONS \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"31\" class=\"wp-block-list\">\n<li>In size reduction by crushing, energy required per unit mass often follows Rittinger\u2019s and Kick\u2019s laws. For fine grinding dominant mechanism is:<br>a) Rittinger\u2019s law (energy \u221d new surface area)<br>b) Kick\u2019s law (energy \u221d log size reduction ratio)<br>c) Bond\u2019s law only for coarse crushing<br>d) Energy independent of particle size<br>Correct Answer: a<br>Explanation: Rittinger applies for fine grinding where surface generation dominates.<\/li>\n\n\n\n<li>A settling tank treating slurry shows particle hindered settling at high solids concentration. Compared to free settling, hindered settling velocity:<br>a) Decreases due to hydrodynamic interference<br>b) Increases due to crowding<br>c) Remains equal<br>d) Becomes negative<br>Correct Answer: a<br>Explanation: In hindered regime, interactions reduce settling velocity.<\/li>\n\n\n\n<li>For filtration with constant pressure, cake builds. If cake resistance increases linearly with time, filtrate flow rate:<br>a) Decreases with time (inverse relation)<br>b) Increases linearly<br>c) Unchanged<br>d) Oscillatory<br>Correct Answer: a<br>Explanation: Increasing resistance lowers flow rate for constant pressure per Darcy\u2019s law.<\/li>\n\n\n\n<li>In centrifugation for solids removal, G-factor (centrifugal acceleration) scales with:<br>a) \u03c9^2 r \/ g (i.e., rotational speed squared)<br>b) \u03c9 r only<br>c) independent of \u03c9<br>d) inversely with radius<br>Correct Answer: a<br>Explanation: Centrifugal acceleration \u221d \u03c9^2 r; normalized by g gives G-factor.<\/li>\n\n\n\n<li>Cyclone separation efficiency for fine particles improves by:<br>a) Increasing inlet tangential velocity and decreasing diameter<br>b) Increasing cyclone diameter only<br>c) Removing vortex finder<br>d) Increasing gas viscosity<br>Correct Answer: a<br>Explanation: Higher tangential velocity and smaller cyclone give higher centrifugal forces and better capture.<\/li>\n\n\n\n<li>In a fluidized bed, minimum fluidization velocity Umf depends primarily on:<br>a) Particle diameter, density difference, and fluid viscosity<br>b) Reaction kinetics only<br>c) Bed height only<br>d) Gravity inverted<br>Correct Answer: a<br>Explanation: Umf from balance of drag and weight\u2014function of particle size\/density and fluid properties.<\/li>\n\n\n\n<li>For crushing in jaw vs hammer mill, which produces narrower PSD (particle size distribution) for same top size?<br>a) Jaw crusher gives broader, hammer mill may produce more fines \u2014 depends on mechanism<br>b) Both identical always<br>c) Hammer mill produces no fines<br>d) Jaw generates monodisperse output<br>Correct Answer: a<br>Explanation: Compression crushers (jaw) give broader distribution; impact mills produce more fines \u2014 depends on operating parameters.<\/li>\n\n\n\n<li>For mixing viscous fluids in stirred tank, power number correlates with Reynolds number; at low Re (laminar), mixing time:<br>a) Is large and independent of impeller speed squared scaling<br>b) Decreases strongly with speed even in laminar<br>c) Is minimal and constant<br>d) Negative correlation with viscosity only<br>Correct Answer: a<br>Explanation: In laminar regime, mixing dominated by viscous diffusion and scales differently; power number varies as Re^{-1}.<\/li>\n\n\n\n<li>In filtration cake formation, specific cake resistance \u03b1 depends on:<br>a) Particle size, porosity and compressibility of cake<br>b) Only filter medium area<br>c) Temperature only<br>d) Stirring speed<br>Correct Answer: a<br>Explanation: \u03b1 relates to cake microstructure \u2014 particle size, packing\/porosity and compressibility under pressure.<\/li>\n\n\n\n<li>In pneumatic conveying of solids, dilute-phase conveying needs:<br>a) Higher gas velocity to keep particles suspended compared to dense-phase<br>b) Lower gas velocity than dense-phase<br>c) No gas, just gravity<br>d) Fixed velocity regardless of system<br>Correct Answer: a<br>Explanation: Dilute-phase keeps particles entrained and needs higher velocities.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">HEAT TRANSFER \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"41\" class=\"wp-block-list\">\n<li>For steady one-dimensional conduction through composite slab with two layers having thermal conductivities k1 and k2 and thickness L1 and L2, overall thermal resistance is:<br>a) L1\/k1 + L2\/k2<br>b) (L1 + L2)\/(k1 + k2)<br>c) (k1 L2 + k2 L1)<br>d) Product of Ls only<br>Correct Answer: a<br>Explanation: Resistances in series add: R = L1\/k1 + L2\/k2.<\/li>\n\n\n\n<li>For external forced convection over a cylinder at moderate Re, heat transfer coefficient h correlates with Nu_D which depends on:<br>a) Reynolds number and Prandtl number (Nu = f(Re,Pr))<br>b) Only Prandtl number<br>c) Only material conductivity<br>d) Independent of flow conditions<br>Correct Answer: a<br>Explanation: Convective Nu correlations for external flow are functions of Re and Pr.<\/li>\n\n\n\n<li>In a finned surface, when fin tip convective loss is negligible, adding more fin length beyond certain value:<br>a) Produces diminishing returns due to axial conduction along fin<br>b) Linearly increases heat removal forever<br>c) Reduces base temperature always<br>d) Makes finoretical invalid<br>Correct Answer: a<br>Explanation: Fin effectiveness saturates due to increased conduction resistance along fin.<\/li>\n\n\n\n<li>For a plate heat exchanger using LMTD method, LMTD becomes invalid if:<br>a) Heat capacity rates vary significantly along exchanger (phase change or variable properties)<br>b) Flow is purely countercurrent with constant properties<br>c) For steady counterflow without phase change<br>d) For identical inlet temperatures<br>Correct Answer: a<br>Explanation: LMTD assumes constant capacity rates; with phase change or varying properties, correction factors needed.<\/li>\n\n\n\n<li>Radiative heat transfer between two large parallel grey plates (\u03b51, \u03b52) separated by vacuum: net heat flux depends on:<br>a) (1\/\u03b51 + 1\/\u03b52 \u2212 1) factor and difference of T^4<br>b) Only emissivity of plate1<br>c) Only temperature difference linear term<br>d) Independent of emissivity if one is blackbody<br>Correct Answer: a<br>Explanation: Two-surface radiation exchange factor includes (1\/\u03b51 + 1\/\u03b52 \u22121) in denominator times \u03c3(T1^4 \u2212 T2^4).<\/li>\n\n\n\n<li>Boiling heat transfer in pool boiling: Transition from nucleate to film boiling at critical heat flux leads to:<br>a) Sudden decrease in heat transfer coefficient (dangerous)<br>b) Continuous increase in heat transfer<br>c) No change at critical point<br>d) Only change in surface roughness<br>Correct Answer: a<br>Explanation: At CHF, vapor blankets surface leading to film boiling with much lower HTC \u2192 temperature runaway risk.<\/li>\n\n\n\n<li>For a counterflow heat exchanger with equal heat capacity rates (C_min=C_max), the effectiveness \u03b5 is:<br>a) \u03b5 = 1 \u2212 exp(\u2212NTU) \/ (1+exp(\u2212NTU)) simplifying to tanh(NTU\/2) relationship<br>b) \u03b5 independent of NTU<br>c) \u03b5 = NTU always<br>d) \u03b5 = 0 always<br>Correct Answer: a<br>Explanation: For counterflow with equal C, \u03b5 = tanh(NTU).<\/li>\n\n\n\n<li>For transient conduction in semi-infinite solid subjected to a sudden surface temperature change, temperature at depth x and time t depends on error function erf( x \/ (2\u221a\u03b1t) ). For large t, penetration depth:<br>a) ~ \u221a(\u03b1 t)<br>b) ~ \u03b1 t<br>c) ~ t^2<br>d) independent of t<br>Correct Answer: a<br>Explanation: Thermal diffusion length scales as \u221a(\u03b1 t).<\/li>\n\n\n\n<li>In convection correlations, Prandtl number Pr = \u03bd\/\u03b1 (momentum to thermal diffusivity). For Pr >> 1, thermal boundary layer relative to velocity boundary layer is:<br>a) Thinner (thermal diffuses slower \u2192 thinner)<br>b) Thicker<br>c) Equal thickness always<br>d) Inverse of Reynolds number<br>Correct Answer: a<br>Explanation: High Pr \u2192 thermal diffusivity low \u2192 thermal boundary layer thinner than velocity BL.<\/li>\n\n\n\n<li>For design of shell-and-tube heat exchanger, baffle spacing too large causes:<br>a) Poor shell-side heat transfer due to bypassing and reduced turbulence<br>b) Increased pressure drop always beneficial<br>c) Enhanced heat transfer irrespective of spacing<br>d) No impact on flow pattern<br>Correct Answer: a<br>Explanation: Large spacing reduces crossflow and turbulence, permitting bypass and lowering heat transfer.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">MASS TRANSFER \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"51\" class=\"wp-block-list\">\n<li>In a single-stage countercurrent gas absorption, if gas enters lean and leaves rich, the driving force for transfer depends on:<br>a) Difference between gas phase partial pressure and equilibrium partial pressure at liquid composition<br>b) Absolute partial pressure only<br>c) Liquid viscosity only<br>d) Gas molecular weight only<br>Correct Answer: a<br>Explanation: Mass transfer driving force is difference between bulk and equilibrium concentrations\/partial pressures.<\/li>\n\n\n\n<li>For distillation with ideal stages, McCabe-Thiele analysis assumes:<br>a) Equilibrium stage operation and constant molar overflow<br>b) Non-equilibrium mass transfer only<br>c) Variable latent heats always<br>d) No reflux allowed<br>Correct Answer: a<br>Explanation: McCabe\u2013Thiele uses equilibrium stages and constant molar overflow to draw operating lines.<\/li>\n\n\n\n<li>In film theory for gas\u2013liquid transfer, the overall mass transfer coefficient K_G relates to individual film coefficients by:<br>a) 1\/K_G = 1\/k_G + H\/ (k_L) where H is Henry\u2019s law constant (gas\u2013liquid interface relation)<br>b) K_G = k_G + k_L always<br>c) Independent of Henry\u2019s constant<br>d) Inversely proportional to Schmidt number only<br>Correct Answer: a<br>Explanation: Resistances in series in gas and liquid films combine; Henry\u2019s law links concentrations.<\/li>\n\n\n\n<li>For plate evaporator operation, if boiling point elevation increases with concentration, the number of effects required for specified economy:<br>a) Increases (reduces possible vapor reuse effectiveness)<br>b) Decreases<br>c) Unchanged<br>d) Dependent only on feed temperature<br>Correct Answer: a<br>Explanation: BPE reduces driving force per effect; need more effects to achieve same evaporation.<\/li>\n\n\n\n<li>In adsorption using Langmuir isotherm, at high concentrations surface coverage \u03b8 approaches:<br>a) 1 (monolayer saturation)<br>b) 0<br>c) Proportional to pressure linearly forever<br>d) Infinite<br>Correct Answer: a<br>Explanation: Langmuir limits to monolayer capacity.<\/li>\n\n\n\n<li>For drying a porous solid under constant-rate period, surface remains at wet-bulb temperature. Transition to falling-rate occurs when:<br>a) Surface becomes dry and internal diffusion controls rate<br>b) Air temperature drops below wet bulb always<br>c) Constant-rate continues indefinitely<br>d) Only at zero humidity<br>Correct Answer: a<br>Explanation: Constant-rate while free moisture at surface; falling-rate when surface dries and internal transport limits.<\/li>\n\n\n\n<li>In packed column distillation, flooding occurs due to:<br>a) Excessive vapor velocity causing liquid hold-up and pressure rise<br>b) Too low liquid flow leading to dry packing only<br>c) Perfect countercurrent with no limits<br>d) Packing chemical incompatibility only<br>Correct Answer: a<br>Explanation: Flooding when vapor velocity high enough to prevent liquid flow down \u2014 causes high pressure drop.<\/li>\n\n\n\n<li>For binary gas diffusion through stagnant gas, Fick\u2019s law in multicomponent systems requires correction; effective diffusivity scales approximately inversely with:<br>a) Total pressure for gases (D \u221d 1\/P)<br>b) Temperature only<br>c) Molecular size independent<br>d) Viscosity only<br>Correct Answer: a<br>Explanation: Gas-phase binary diffusivity D_AB \u221d T^{3\/2}\/(P \u221aM), so inversely proportional to pressure.<\/li>\n\n\n\n<li>In countercurrent liquid\u2013liquid extraction, the number of theoretical stages required decreases if distribution coefficient:<br>a) Increases (favors solute transfer to extractant)<br>b) Decreases<br>c) Equals 1 always<br>d) Independent of stage number<br>Correct Answer: a<br>Explanation: Higher distribution coefficient shifts solute more to extractant per stage \u2192 fewer stages required.<\/li>\n\n\n\n<li>For crystallization by cooling, supersaturation control is critical. High nucleation vs growth rates produce:<br>a) Fine crystals (many nuclei) vs coarse crystals (dominated by growth)<br>b) Coarse always<br>c) No crystals if rate too high<br>d) Crystal size independent of kinetics<br>Correct Answer: a<br>Explanation: High nucleation yields many small crystals; controlled nucleation\/growth yields larger crystals.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">CHEMICAL REACTION ENGINEERING (CRE) \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"61\" class=\"wp-block-list\">\n<li>A first-order irreversible reaction A \u2192 products in a PFR with constant density: doubling residence time \u03c4 doubles conversion from 0.5 to:<br>a) 0.75<br>b) 1.0<br>c) 0.25<br>d) 0.66<br>Correct Answer: a<br>Explanation: For first-order, X = 1 \u2212 exp(\u2212k\u03c4). If at \u03c4 gives X=0.5 \u2192 exp(\u2212k\u03c4)=0.5 \u2192 k\u03c4=ln2. At 2\u03c4, exp(\u22122k\u03c4)=exp(\u22122ln2)=1\/4 \u2192 X=1\u22121\/4=0.75.<\/li>\n\n\n\n<li>For irreversible second-order reaction A + B \u2192 products with equal initial concentrations C0 in batch reactor, concentration-time relation follows:<br>a) 1\/C = 1\/C0 + kt<br>b) ln(C\/C0) = \u2212kt<br>c) C = C0 exp(\u2212kt)<br>d) 1\/C^2 = &#8230;<br>Correct Answer: a<br>Explanation: Second-order with equal stoichiometry gives 1\/C = 1\/C0 + kt.<\/li>\n\n\n\n<li>For an isothermal CSTR carrying a first-order reaction, space time \u03c4 required to achieve conversion X relates to plug flow space time \u03c4_PFR for same X as:<br>a) \u03c4_CSTR > \u03c4_PFR for same X (except X\u21920 or X\u21921?)<br>b) \u03c4_CSTR = \u03c4_PFR always<br>c) \u03c4_CSTR &lt; \u03c4_PFR always<br>d) No relation<br>Correct Answer: a<br>Explanation: For same conversion, PFR is more efficient (less \u03c4) than CSTR for first-order; so \u03c4_CSTR > \u03c4_PFR.<\/li>\n\n\n\n<li>Residence Time Distribution (RTD) measured by step tracer in a reactor shows non-ideal behavior. A broad E(t) indicates:<br>a) Large spread in residence times and potential back-mixing<br>b) Perfect plug flow<br>c) Single residence time only<br>d) No effect on conversion<br>Correct Answer: a<br>Explanation: Broad distribution indicates varying particle times and mixing\/backflow reducing ideality.<\/li>\n\n\n\n<li>For heterogeneous catalytic reactions, effectiveness factor \u03b7 accounts for:<br>a) Internal diffusion limitations inside porous catalyst decreasing observed rate<br>b) Temperature gradients only<br>c) Reaction mechanism changes only<br>d) Catalyst color change<br>Correct Answer: a<br>Explanation: \u03b7 = observed rate\/ rate if entire catalyst at surface concentration; accounts for pore diffusion limitations.<\/li>\n\n\n\n<li>For a reversible reaction A \u21cc B with equilibrium constant K_eq, infinite residence time in closed batch will yield conversion determined by:<br>a) Equilibrium composition given by K_eq and stoichiometry<br>b) Zero conversion always<br>c) Complete conversion for any K_eq<br>d) Depends only on catalyst presence<br>Correct Answer: a<br>Explanation: At infinite time, reaction reaches equilibrium\u2014extent determined by K_eq.<\/li>\n\n\n\n<li>In autocatalytic reactions, reaction rate increases with extent initially. A sign of autocatalysis in concentration\u2013time profile is:<br>a) Sigmoidal (S-shaped) curve indicating induction then acceleration<br>b) Pure exponential decay<br>c) Linear decrease always<br>d) No dependence on concentration<br>Correct Answer: a<br>Explanation: Autocatalytic systems show induction (slow), then rapid increase then leveling \u2192 S-curve.<\/li>\n\n\n\n<li>For non-isothermal reactor, hot-spot formation is hazardous if reaction is:<br>a) Highly exothermic with low thermal conductivity and insufficient heat removal<br>b) Endothermic only<br>c) Zero-order at all temperatures<br>d) Only homogeneous gas-phase<br>Correct Answer: a<br>Explanation: Exothermicity + poor heat removal can cause runaway and hot spots.<\/li>\n\n\n\n<li>Turnover frequency (TOF) for heterogeneous catalyst is defined as:<br>a) Moles of reactant converted per active site per unit time<br>b) Mole fraction of catalyst active sites only<br>c) Pressure normalized rate only<br>d) Independent of active site density<br>Correct Answer: a<br>Explanation: TOF quantifies intrinsic catalytic activity per site per time.<\/li>\n\n\n\n<li>For Langmuir\u2013Hinshelwood mechanism (both reactants adsorb), reaction rate typically depends on:<br>a) Surface coverages of both species and adsorption constants<br>b) Gas-phase concentrations only always linear<br>c) Catalyst only as inert support<br>d) Only temperature, not adsorption<br>Correct Answer: a<br>Explanation: LH mechanism rate expression includes adsorption equilibria and surface coverages.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">INSTRUMENTATION &amp; PROCESS CONTROL \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"71\" class=\"wp-block-list\">\n<li>A process variable exhibits unstable oscillatory behavior. Root locus shows poles crossing into right-half plane as gain increases. Addition of derivative action (D) in PID typically:<br>a) Improves damping and reduces overshoot (adds phase lead)<br>b) Always destabilizes system<br>c) Removes steady-state error only<br>d) Converts system to first-order only<br>Correct Answer: a<br>Explanation: Derivative provides predictive action adding phase lead, improving damping\/overshoot.<\/li>\n\n\n\n<li>A sensor with time constant \u03c4_s is in series with process of time constant \u03c4_p. For accurate faster control, sensor dynamics ideally:<br>a) \u03c4_s &lt;&lt; \u03c4_p (sensor much faster than process)<br>b) \u03c4_s >> \u03c4_p<br>c) \u03c4_s = \u221e<br>d) \u03c4_s = 0 recommended but impossible<br>Correct Answer: a<br>Explanation: Faster sensor captures changes quickly; slow sensor lags control action and degrades performance.<\/li>\n\n\n\n<li>Bode plot of open-loop shows phase margin negative at crossover frequency. Which conclusion?<br>a) Closed loop likely unstable (insufficient phase margin)<br>b) System has good robustness<br>c) No relation to stability<br>d) Nyquist not applicable<br>Correct Answer: a<br>Explanation: Negative phase margin indicates phase lag exceeds 180\u00b0 at gain crossover \u2192 likely instability.<\/li>\n\n\n\n<li>For first-order plus dead-time process, Smith predictor control structure primarily compensates for:<br>a) Time delay (dead time), improving loop response<br>b) Process gain only<br>c) Sensor noise only<br>d) Nonlinearity in actuator<br>Correct Answer: a<br>Explanation: Smith predictor uses model to predict process output over dead time and compensate delay.<\/li>\n\n\n\n<li>A thermocouple measures temperature by Seebeck effect. Its accuracy is reduced by:<br>a) Contact resistance and thermal lag at measurement junction<br>b) Increasing ambient pressure only<br>c) Magnetic fields only<br>d) Catalyst poisoning<br>Correct Answer: a<br>Explanation: Poor contact and junction lag degrade reading and response time.<\/li>\n\n\n\n<li>In frequency response, Nyquist plot encirclement of \u22121 point determines:<br>a) Closed-loop stability when combined with open-loop pole count (Nyquist criterion)<br>b) Mass balance in reactor<br>c) Steady-state gain only<br>d) Bode plot invalidation<br>Correct Answer: a<br>Explanation: Nyquist criterion relates encirclements of \u22121 to closed-loop stability.<\/li>\n\n\n\n<li>A pressure transmitter calibrated linearly produces 4\u201320 mA proportional to 0\u2013100 kPa. If loop reading is 12 mA, estimated pressure:<br>a) 50 kPa<br>b) 100 kPa<br>c) 0 kPa<br>d) 25 kPa<br>Correct Answer: a<br>Explanation: 4 mA \u21920 kPa, 20 mA \u2192100 kPa; 12 mA is midpoint \u219250 kPa.<\/li>\n\n\n\n<li>For a PI controller in cascade control where inner loop is fast, outer loop benefits by:<br>a) Inner loop rejecting disturbances quickly; outer loop handles setpoint changes giving overall tighter control<br>b) Slowing down outer loop only<br>c) Removing need for sensors entirely<br>d) Always destabilizing the plant<br>Correct Answer: a<br>Explanation: Cascade control uses inner fast loop to stabilize and reject disturbances, improving outer loop performance.<\/li>\n\n\n\n<li>A transmitter-sensor pair exhibits Gaussian white noise. To improve signal-to-noise ratio for slow process, recommended action:<br>a) Apply low-pass filtering or increase measurement averaging time<br>b) Increase controller derivative gain<br>c) Reduce measurement resolution<br>d) Increase sampling rate to infinity<br>Correct Answer: a<br>Explanation: Low-pass filtering\/averaging reduces high-frequency noise; derivative amplifies noise.<\/li>\n\n\n\n<li>For model predictive control (MPC), primary advantage over PID is:<br>a) Ability to handle multi-variable constraints and anticipate future changes via model optimization<br>b) Always simpler to tune than PID<br>c) No requirement for a process model<br>d) Only suited for linear steady-state systems<br>Correct Answer: a<br>Explanation: MPC optimizes future control moves considering constraints and multivariable interactions using a model.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">PLANT DESIGN &amp; ECONOMICS \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"81\" class=\"wp-block-list\">\n<li>When sizing a storage tank for liquid feed with variable demand, safety factor accounts for:<br>a) Surge, inventory for downtime, and emergency reserves beyond nominal capacity<br>b) Only corrosion allowance<br>c) Only color of tank coating<br>d) Tank shape solely<br>Correct Answer: a<br>Explanation: Design includes operational surges, downtime inventory and contingencies.<\/li>\n\n\n\n<li>Discounted cash flow uses net present value (NPV). A project with positive NPV at chosen discount rate means:<br>a) Project adds value and is financially acceptable<br>b) Must be rejected regardless<br>c) Equal to payback period zero<br>d) Independent of discount rate choice<br>Correct Answer: a<br>Explanation: Positive NPV \u21d2 present value of inflows > outflows at discount rate \u21d2 acceptable.<\/li>\n\n\n\n<li>For piping layout, pressure drop minimization vs capital cost trade-off often leads to:<br>a) Selecting larger diameter pipe increasing capex but reducing operating cost (energy)<br>b) Always smallest diameter to minimize capex<br>c) Material has no effect on pressure drop<br>d) Only pump selection matters, not diameter<br>Correct Answer: a<br>Explanation: Larger diameter lowers velocity and \u0394P leading to energy savings but higher material cost\u2014trade-off optimization required.<\/li>\n\n\n\n<li>In equipment scale-up from lab to pilot, Reynolds and other similarity numbers must be matched to preserve:<br>a) Hydrodynamic and heat\/mass transfer similarity (Re, Pr, Fr etc. as appropriate)<br>b) Only geometric similarity, nothing else<br>c) Scaling independent of physics<br>d) Chemical composition only<br>Correct Answer: a<br>Explanation: Dynamic similarity requires dimensionless numbers matching for comparable behavior.<\/li>\n\n\n\n<li>Depreciation by straight-line method implies:<br>a) Equal annual depreciation expense over useful life<br>b) Depreciation accelerating over time<br>c) No depreciation at all<br>d) Depreciation equals salvage value always<br>Correct Answer: a<br>Explanation: Straight-line spreads cost equally across life.<\/li>\n\n\n\n<li>For cost estimation, Lang factor approach multiplies purchased equipment cost by factor to estimate total installed cost; Lang factor is crude because:<br>a) It ignores plant-specific details like material, complexity and site conditions<br>b) It is exact for all processes<br>c) Suitable only for academic problems<br>d) Only valid in metric units<br>Correct Answer: a<br>Explanation: Lang factors are rough multipliers and not detailed estimates.<\/li>\n\n\n\n<li>In facility layout, minimizing total material handling cost often leads to:<br>a) Locating high-flow units closer to each other (minimize transport)<br>b) Random placement of equipment<br>c) Maximizing distances always preferred<br>d) Ignoring safety and utility lines<br>Correct Answer: a<br>Explanation: Minimizing logistics cost groups frequently interacting units nearby.<\/li>\n\n\n\n<li>Safety relief valve set pressure must be:<br>a) At or below maximum allowable working pressure with consideration to accumulation and codes (e.g., API\/ASME)<br>b) Above ultimate tensile strength always<br>c) Equal to atmospheric pressure always<br>d) Set arbitrarily by operator<br>Correct Answer: a<br>Explanation: Relief devices adhere to design codes with set pressure below MAWP and account for transient accumulation.<\/li>\n\n\n\n<li>For process optimization, objective function maximizing profit subject to material balances and constraints typically solved by:<br>a) Nonlinear programming (NLP) methods when relations nonlinear<br>b) Only trial-and-error ever<br>c) Linear methods always suffice<br>d) Ignoring constraints gives optimum<br>Correct Answer: a<br>Explanation: Complex plant economics with nonlinearities require NLP; LP only for linear problems.<\/li>\n\n\n\n<li>In estimating heat exchanger area, U (overall heat transfer coefficient) uncertainty leads to area design margin by safety factor because:<br>a) Small change in U can cause significant change in required area since A = Q\/(U \u0394T_lm)<br>b) U variations negligible always<br>c) Area independent of U<br>d) U only matters for radiation<br>Correct Answer: a<br>Explanation: Area inversely proportional to U; uncertainty in U handled via margin or conservative design.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\"><strong>CHEMICAL TECHNOLOGY \u2014 10 QUESTIONS<\/strong><\/a><\/p>\n\n\n\n<ol start=\"91\" class=\"wp-block-list\">\n<li>In steam reforming of methane to syngas, catalyst deactivation is often due to:<br>a) Carbon deposition (coking) and sintering at high temperatures<br>b) Growth of catalyst improving activity<br>c) Only heat transfer limitations<br>d) Pressure alone causing deactivation<br>Correct Answer: a<br>Explanation: Coking and sintering reduce active surface and block pores, deactivating catalysts.<\/li>\n\n\n\n<li>For ammonia synthesis (Haber process), equilibrium favors ammonia at:<br>a) High pressure and moderate temperature (Le Chatelier: reduce moles)<br>b) Low pressure always<br>c) Extremely high temperature only<br>d) No dependence on T or P<br>Correct Answer: a<br>Explanation: N2 + 3H2 \u2192 2NH3 reduces moles; high P shifts equilibrium to product; high T disfavors exothermic synthesis.<\/li>\n\n\n\n<li>In petroleum refining, catalytic cracking primarily increases yield of:<br>a) Light distillates (gasoline-range hydrocarbons) from heavy fractions<br>b) Heavy fuel oil only<br>c) Produces only coke without distillate<br>d) Converts to pure methane only<br>Correct Answer: a<br>Explanation: Catalytic cracking breaks long chains to shorter gasoline-range molecules.<\/li>\n\n\n\n<li>In polymerization (free-radical), chain-transfer reactions affect:<br>a) Molecular weight distribution and average chain length<br>b) Catalyst color only<br>c) Polymer crystallinity independent of chemistry<br>d) Only melt index measured but not chain length<br>Correct Answer: a<br>Explanation: Chain-transfer terminates growing radical altering molecular weight distribution.<\/li>\n\n\n\n<li>For sulfuric acid production via contact process, vanadium pentoxide catalyst promotes:<br>a) Oxidation of SO2 to SO3 (heterogeneous catalysis)<br>b) Hydrogenation of SO3 to SO2<br>c) Serves as inert support only<br>d) Absorbs SO3 irreversibly<br>Correct Answer: a<br>Explanation: V2O5 catalyzes oxidation step in contact process.<\/li>\n\n\n\n<li>In fertilizer industry, urea is produced by reaction of ammonia and carbon dioxide; major downstream issue in urea prilling is:<br>a) Formation of fine dust and humidity leading to caking<br>b) Urea inertness making prilling unnecessary<br>c) No heat effects during prilling<br>d) Urea always liquid at ambient<br>Correct Answer: a<br>Explanation: Prills can cake under humidity and dust generation is concern.<\/li>\n\n\n\n<li>In petrochemical ethylene production by steam cracking, yield of ethylene increases by:<br>a) Increasing severity (higher temperature, shorter residence time) optimizing cracking conditions<br>b) Decreasing temperature always<br>c) Lowering steam to feed ratio only<br>d) Adding oxygen to feed<br>Correct Answer: a<br>Explanation: Higher T and short crack time favor monomers; steam dilutes and reduces coke but conditions tuned for yield.<\/li>\n\n\n\n<li>In paper\/pulp industry, Kraft process uses alkaline pulping to:<br>a) Remove lignin while preserving cellulose fibers<br>b) Convert cellulose to monomers only<br>c) Paper produced directly from raw wood without chemicals<br>d) Produce synthetic polymers only<br>Correct Answer: a<br>Explanation: Kraft chemically digests lignin enabling fiber separation.<\/li>\n\n\n\n<li>In LNG (liquefied natural gas) production, precooling and cryogenic refrigeration stages aim to:<br>a) Remove heavier hydrocarbons and cool gas to cryogenic temperatures before liquefaction<br>b) Heat gas to increase volume<br>c) Separate oxygen from methane only<br>d) Use only ambient cooling always<br>Correct Answer: a<br>Explanation: Pre-treatment removes heavy ends and dehydration; multi-stage refrigeration cools to about \u2212162\u00b0C.<\/li>\n\n\n\n<li>In electrodeposition (electroplating), current efficiency depends on:<br>a) Applied current density, bath composition and side reactions (e.g., hydrogen evolution)<br>b) Only electrode color<br>c) Temperature independent always<br>d) Electrodeposition has no competing reactions<br>Correct Answer: a<br>Explanation: Side reactions reduce current efficiency; optimal current density and bath chemistry maximize deposition.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><a href=\"https:\/\/engineersinstitute.com\/gate_test_series_chemical-engineering.php\">GENERAL APTITUDE (GA: Verbal + Quantitative) \u2014 10 QUESTIONS<\/a><\/strong><\/p>\n\n\n\n<ol start=\"101\" class=\"wp-block-list\">\n<li>(Verbal) Select the option that best completes sentence: \u201cDespite his _______ attempts to simplify the presentation, the audience found the topic abstruse.\u201d<br>a) ineffectual<br>b) successful<br>c) perfunctory<br>d) sporadic<br>Correct Answer: a<br>Explanation: \u201cIneffectual\u201d (unsuccessful) fits contrast with audience still finding topic abstruse.<\/li>\n\n\n\n<li>(Verbal) Choose the sentence with correct usage:<br>a) Neither of the candidates were available.<br>b) Each of the team members have submitted their report.<br>c) The committee is divided in its opinion.<br>d) Ten miles is a long distance for him to run.<br>Correct Answer: d<br>Explanation: \u201cTen miles\u201d as a measure uses singular verb \u201cis\u201d; c) could be acceptable but \u201cdivided in its opinion\u201d is less idiomatic than \u201cdivided in its opinions\u201d; d is clearly correct.<\/li>\n\n\n\n<li>(Verbal) Select the word most nearly opposite to \u201cephemeral\u201d:<br>a) perennial<br>b) fleeting<br>c) transient<br>d) evanescent<br>Correct Answer: a<br>Explanation: Ephemeral = short-lived; perennial = long-lasting.<\/li>\n\n\n\n<li>(Quant) If a rectangular tank 2 m by 3 m is filled from empty to half depth in 30 min at constant flow, time to fill remaining half depth (same cross-section) is:<br>a) 30 min<br>b) 60 min<br>c) 15 min<br>d) 45 min<br>Correct Answer: a<br>Explanation: Constant volumetric flow \u2192 equal volumes take equal times; half the depth = half the volume.<\/li>\n\n\n\n<li>(Quant) A and B together can finish job in 6 days. A alone takes 10 days more than B alone. Time for B alone is:<br>a) 12 days<br>b) 8 days<br>c) 15 days<br>d) 6 days<br>Correct Answer: a<br>Explanation: Let rates 1\/A=1\/(B+10). 1\/A+1\/B =1\/6. Solve: 1\/(B+10) + 1\/B = 1\/6 \u2192 (2B+10)\/B(B+10)=1\/6 \u219212B+60= B^2+10B \u2192 B^2 \u22122B \u221260=0 \u2192 (B\u2212? ) root positive B=12.<\/li>\n\n\n\n<li>(Quant) For events A and B with P(A)=0.4, P(B)=0.5, and P(A\u222aB)=0.8, P(A\u2229B) equals:<br>a) 0.1<br>b) 0.2<br>c) 0.3<br>d) 0.4<br>Correct Answer: b<br>Explanation: P(A\u222aB)=P(A)+P(B)\u2212P(A\u2229B) \u2192 0.8=0.9\u2212P_intersection \u2192 P_intersection=0.1? Wait compute: 0.4+0.5=0.9; 0.9 \u2212 x = 0.8 \u2192 x = 0.1. Corrected: 0.1.<br>(Adjusted Correct Answer: a)<br>Explanation corrected: P(A\u2229B)=0.4+0.5\u22120.8=0.1.<\/li>\n\n\n\n<li>(Quant) If sequence a_n defined by a_{n+1} = 0.5 a_n + 3 with a_0 = 0, steady-state limit as n\u2192\u221e:<br>a) 6<br>b) 3<br>c) \u221e<br>d) 0<br>Correct Answer: a<br>Explanation: Fixed point solves a = 0.5a + 3 \u2192 0.5a = 3 \u2192 a = 6.<\/li>\n\n\n\n<li>(Quant) If probability that a student passes Mathematics is 0.7 and passes Physics is 0.6, and events independent, probability student passes both =<br>a) 0.42<br>b) 0.30<br>c) 0.60<br>d) 0.70<br>Correct Answer: a<br>Explanation: Independent \u2192 multiply 0.7\u00d70.6=0.42.<\/li>\n\n\n\n<li>(Quant) For quadratic equation x^2 \u2212 (k+1)x + k = 0 to have equal roots, discriminant must be zero. Value of k:<br>a) 1<br>b) \u22121<br>c) 0<br>d) 2<br>Correct Answer: a<br>Explanation: Discriminant \u0394 = (k+1)^2 \u22124k = k^2 +2k +1 \u22124k = k^2 \u22122k +1 = (k\u22121)^2 \u21d2 \u0394=0 when k=1.<\/li>\n\n\n\n<li>(Verbal\/Reasoning) A sentence rearrangement: Which rearrangement yields coherent paragraph? (Given lines) \u2014 (For brevity present a conceptual style) Best practice for answering such GA questions is to:<br>a) Identify central theme and logical connectors, then order sentences accordingly<br>b) Randomly pick sentences until grammar fits<br>c) Always pick alphabetical ordering<br>d) None of the above<br>Correct Answer: a<br>Explanation: Coherence requires theme recognition and connectors to determine logical sequence.<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1020\" height=\"627\" src=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/2025-07-09-21-32-31.png\" alt=\"\" class=\"wp-image-60\" srcset=\"https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/2025-07-09-21-32-31.png 1020w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/2025-07-09-21-32-31-300x184.png 300w, https:\/\/engineersinstitute.com\/blog\/wp-content\/uploads\/2025\/12\/2025-07-09-21-32-31-768x472.png 768w\" sizes=\"auto, (max-width: 1020px) 100vw, 1020px\" \/><\/figure>\n\n\n\n<p class=\"has-black-color has-luminous-vivid-amber-background-color has-text-color has-background has-link-color wp-elements-54f1a46b9b4f5ae8ccb7552a5abd696b wp-block-paragraph\"><strong>Join LIVE Class or access Recorded Lectures of GATE ( Access  with FREE Login for 2days full course ) , get Login at 9990357855<\/strong><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>GATE 2026 Chemical Engineering<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>GATE 2026 Chemical Engineering<\/li>\n\n\n\n<li>GATE CH 2026 syllabus<\/li>\n\n\n\n<li>GATE Chemical Engineering preparation<\/li>\n\n\n\n<li>GATE Chemical Engineering notes PDF<\/li>\n\n\n\n<li>GATE 2026 Chemical Engineering study material<\/li>\n\n\n\n<li>GATE Chemical previous year questions<\/li>\n\n\n\n<li>GATE Chemical PYQ solved<\/li>\n\n\n\n<li>GATE Chemical Engineering sample paper 2026<\/li>\n\n\n\n<li>GATE Chemical Engineering mock test<\/li>\n\n\n\n<li>GATE Chemical formula sheet<\/li>\n\n\n\n<li>GATE Chemical expected cutoff 2026<\/li>\n\n\n\n<li>GATE Chemical Engineering weightage topic wise<\/li>\n\n\n\n<li>GATE Mass Transfer questions<\/li>\n\n\n\n<li>GATE CRE questions<\/li>\n\n\n\n<li>GATE Fluid Mechanics PYQ<\/li>\n\n\n\n<li>GATE Heat Transfer MCQ<\/li>\n\n\n\n<li>GATE Process Calculations questions<\/li>\n\n\n\n<li>GATE Chemical short notes<\/li>\n\n\n\n<li>GATE 2026 Chemical chapter-wise test<\/li>\n\n\n\n<li>Best books for GATE Chemical Engineering<\/li>\n\n\n\n<li>GATE Chemical rank predictor<\/li>\n\n\n\n<li>GATE Chemical topper strategy<\/li>\n\n\n\n<li>GATE Chemical 70+ strategy<\/li>\n\n\n\n<li>GATE Chemical Engineering free coaching<\/li>\n\n\n\n<li>GATE Chemical online coaching<\/li>\n\n\n\n<li>GATE Chemical crash course<\/li>\n\n\n\n<li>GATE Chemical full course 2026<\/li>\n\n\n\n<li>GATE Chemical test series 2026<\/li>\n\n\n\n<li>GATE Chemical cut-off for PSU<\/li>\n\n\n\n<li>PSU through GATE Chemical Engineering<\/li>\n\n\n\n<li>M.Tech admissions through GATE Chemical<\/li>\n\n\n\n<li>IIT M.Tech Chemical cutoff 2026<\/li>\n\n\n\n<li>NIT M.Tech Chemical cutoff 2026<\/li>\n\n\n\n<li>GATE Chemical job opportunities<\/li>\n\n\n\n<li>GATE 2026 Chemical result date<\/li>\n\n\n\n<li>GATE 2026 Chemical exam pattern<\/li>\n\n\n\n<li>GATE Chemical 1-mark 2-mark questions<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">How to score 70+ in GATE Chemical Engineering<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Best online coaching for GATE 2026 Chemical Engineering<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Free study material for GATE Chemical Engineering PDF<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Weightage of Mass Transfer in GATE Chemical 2026<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Difficulty level of GATE Chemical Engineering<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Which is the toughest subject in GATE Chemical<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">GATE Chemical study plan 6 months \/ 3 months<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Practice questions for GATE 2026 Chemical<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Is GATE Chemical Engineering good for PSU jobs<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">GATE 2026 Chemical Engineering, GATE CH 2026 syllabus, GATE Chemical Engineering preparation, GATE Chemical Engineering notes PDF, GATE 2026 Chemical Engineering study material, GATE Chemical previous year questions, GATE Chemical PYQ solved, GATE Chemical Engineering sample paper 2026, GATE Chemical Engineering mock test, GATE Chemical formula sheet, GATE Chemical expected cutoff 2026, GATE Chemical Engineering weightage topic wise, GATE Mass Transfer questions, GATE CRE questions, GATE Fluid Mechanics PYQ, GATE Heat Transfer MCQ, GATE Process Calculations questions, GATE Chemical short notes, GATE 2026 Chemical chapter-wise test, Best books for GATE Chemical Engineering, GATE Chemical rank predictor, GATE Chemical topper strategy, GATE Chemical 70+ strategy, GATE Chemical Engineering free coaching, GATE Chemical online coaching, GATE Chemical crash course, GATE Chemical full course 2026, GATE Chemical test series 2026, GATE Chemical cut-off for PSU, PSU through GATE Chemical Engineering, M.Tech admissions through GATE Chemical, IIT M.Tech Chemical cutoff 2026, NIT M.Tech Chemical cutoff 2026, GATE Chemical job opportunities, GATE 2026 Chemical result date, GATE 2026 Chemical exam pattern, GATE Chemical 1-mark 2-mark questions, GATE Chemical practice questions, Sample GATE chemical questions, Sample Mock GATE chemical test<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Use the discount code FREEMOCK10 to get 10 free mock tests on our portal. Hurry, offer valid for a limited time! These questions are provided only for review purposes, to give you an idea of the type of theoretical MCQs you may encounter. Since this is a web format, formula-based, diagram-based, or chart-based questions are [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[121,125,106,105,140,135,102,136,122,128,127,126,109,104,103,108,114,112,111,124,134,129,138,107,110,123,117,137,118,115,119,116,113,120,132,130,131,133,139,141],"class_list":["post-59","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-best-books-for-gate-chemical-engineering","tag-gate-2026-chemical-chapter-wise-test","tag-gate-2026-chemical-engineering","tag-gate-2026-chemical-engineering-study-material","tag-gate-2026-chemical-exam-pattern","tag-gate-2026-chemical-result-date","tag-gate-ch-2026-syllabus","tag-gate-chemical-1-mark-2-mark-questions","tag-gate-chemical-70-strategy","tag-gate-chemical-crash-course","tag-gate-chemical-cut-off-for-psu","tag-gate-chemical-engineering-free-coaching","tag-gate-chemical-engineering-mock-test","tag-gate-chemical-engineering-notes-pdf","tag-gate-chemical-engineering-preparation","tag-gate-chemical-engineering-sample-paper-2026","tag-gate-chemical-engineering-weightage-topic-wise","tag-gate-chemical-expected-cutoff-2026","tag-gate-chemical-formula-sheet","tag-gate-chemical-full-course-2026","tag-gate-chemical-job-opportunities","tag-gate-chemical-online-coaching","tag-gate-chemical-practice-questions","tag-gate-chemical-previous-year-questions","tag-gate-chemical-pyq-solved","tag-gate-chemical-rank-predictor","tag-gate-chemical-short-notes","tag-gate-chemical-test-series-2026","tag-gate-chemical-topper-strategy","tag-gate-cre-questions","tag-gate-fluid-mechanics-pyq","tag-gate-heat-transfer-mcq","tag-gate-mass-transfer-questions","tag-gate-process-calculations-questions","tag-iit-m-tech-chemical-cutoff-2026","tag-m-tech-admissions-through-gate-chemical","tag-nit-m-tech-chemical-cutoff-2026","tag-psu-through-gate-chemical-engineering","tag-sample-gate-chemical-questions","tag-sample-mock-gate-chemical-test"],"_links":{"self":[{"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/posts\/59","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/comments?post=59"}],"version-history":[{"count":1,"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/posts\/59\/revisions"}],"predecessor-version":[{"id":68,"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/posts\/59\/revisions\/68"}],"wp:attachment":[{"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/media?parent=59"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/categories?post=59"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/engineersinstitute.com\/blog\/wp-json\/wp\/v2\/tags?post=59"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}