GATE
MECHANICAL
ENGINEERING
Multiple Choice Questions GATE Mechanical Engineering Vol 2, 1e Copyright © By Publishers ISBN 9-788192-27629-8
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ENGINEERING MATHEMATICS
Linear Algebra: Matrix algebra, Systems of linear equations, Eigen values and eigen vectors. Calculus: Functions of single variable, Limit, continuity and differentiability, Mean value theorems, Evaluation of definite and improper integrals, Partial derivatives, Total derivative, Maxima and minima, Gradient, Divergence and Curl, Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems.
Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Cauchy’s and Euler’s equations, Initial and boundary value problems, Laplace transforms, Solutions of one dimensional heat and wave equations and Laplace equation.
Complex variables: Analytic functions, Cauchy’s integral theorem, Taylor and Laurent series. Probability and Statistics: Definitions of probability and sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Poisson,Normal and Binomial distributions.
Numerical Methods: Numerical solutions of linear and non-linear algebraic equations Integration by trapezoidal and Simpson’s rule, single and multi-step methods for differential equations.
APPLIED MECHANICS AND DESIGN
Engineering Mechanics: Free body diagrams and equilibrium; trusses and frames; virtual work; kinematics and dynamics of particles and of rigid bodies in plane motion, including impulse and momentum (linear and angular) and energy formulations; impact.
Strength of Materials: Stress and strain, stress-strain relationship and elastic constants, Mohr’s circle for plane stress and plane strain, thin cylinders; shear force and bending moment diagrams; bending and shear stresses; deflection of beams; torsion of circular shafts; Euler’s theory of columns; strain energy methods; thermal stresses.
Theory of Machines: Displacement, velocity and acceleration analysis of plane mechanisms; dynamic analysis of slider-crank mechanism; gear trains; flywheels.
Vibrations: Free and forced vibration of single degree of freedom systems; effect of damping; vibration isolation; resonance, critical speeds of shafts.
Design: Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of the design of machine elements such as bolted, riveted and welded joints, shafts, spur gears, rolling and sliding contact bearings, brakes and clutches.
FLUID MECHANICS AND THERMAL SCIENCES
Fluid Mechanics: Fluid properties; fluid statics, manometry, buoyancy; control-volume analysis of mass, momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoulli’s equation; viscous flow of incompressible fluids; boundary layer; elementary turbulent flow; flow through pipes, head losses in pipes, bends etc.
Thermodynamics: Zeroth, First and Second laws of thermodynamics; thermodynamic system and processes; Carnot cycle. irreversibility and availability; behaviour of ideal and real gases, properties of pure substances, calculation of work and heat in ideal processes; analysis of thermodynamic cycles related to energy conversion.
Applications: Power Engineering: Steam Tables, Rankine, Brayton cycles with regeneration and reheat. I.C. Engines: air-standard Otto, Diesel cycles. Refrigeration and air-conditioning: Vapour refrigeration cycle, heat pumps, gas refrigeration, Reverse Brayton cycle; moist air: psychrometric chart, basic psychrometric processes. Turbomachinery: Pelton-wheel, Francis and Kaplan turbines — impulse and reaction principles, velocity diagrams.
MANUFACTURING AND INDUSTRIAL ENGINEERING
Engineering Materials: Structure and properties of engineering materials, heat treatment, stress-strain diagrams for engineering materials.
Metal Casting: Design of patterns, moulds and cores; solidification and cooling; riser and gating design, design considerations.
Forming: Plastic deformation and yield criteria; fundamentals of hot and cold working processes; load estimation for bulk (forging, rolling, extrusion, drawing) and sheet (shearing, deep drawing, bending) metal forming processes; principles of powder metallurgy.
Joining: Physics of welding, brazing and soldering; adhesive bonding; design considerations in welding.
Machining and Machine Tool Operations: Mechanics of machining, single and multi-point cutting tools, tool geometry and materials, tool life and wear; economics of machining; principles of non-traditional machining processes; principles of work holding, principles of design of jigs and fixtures
Metrology and Inspection: Limits, fits and tolerances; linear and angular measurements; comparators; gauge design; interferometry; form and finish measurement; alignment and testing methods; tolerance analysis in manufacturing and assembly.
Computer Integrated Manufacturing: Basic concepts of CAD/CAM and their integration tools. Production Planning and Control: Forecasting models, aggregate production planning, scheduling, materials requirement planning.
Inventory Control: Deterministic and probabilistic models; safety stock inventory control systems.
Operations Research: Linear programming, simplex and duplex method, transportation, assignment, network flow models, simple queuing models, PERT and CPM.
GENERAL APTITUDE
Verbal Ability: English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction.
Numerical Ability: Numerical computation, numerical estimation, numerical reasoning and data interpretation.
FLUID MECHANICS
FM 1 Basic Concepts and Properties of Fluids FM 3 FM 2 Pressure and Fluid Statics FM 33 FM 3 Fluid Kinematics & Bernouli Equation FM 80 FM 4 Flow Analysis Using Control Volumes FM 124 FM 5 Flow Analysis Using Differential Method FM 172
FM 6 Internal Flow FM 211
FM 7 External Flow FM 253
FM 8 Open Channel Flow FM 289
FM 9 Turbo Machinery FM 328
HEAT TRANSFER
HT 1 Basic Concepts & Modes of Heat-Transfer HT 3 HT 2 Fundamentals of Conduction HT 34 HT 3 Steady Heat Conduction HT 63 HT 4 Transient Heat Conduction HT 94 HT 5 Fundamentals of Convection HT 114 HT 6 Free and Force Convection HT 129 HT 7 Radiation Heat Transfer HT 155
HT 8 Heat Exchangers HT 181
THERMODYNAMICS
TD 1 Basic Concepts and Energy Analysis TD 3 TD 2 Properties of Pure Substances TD 28 TD 3 Energy Analysis of Closed System TD 52 TD 4 Mass and Energy Analysis of Control Volume TD 76 TD 5 Second Law of Thermodynamics TD 106
TD 6 Entropy TD 136
TD 7 Gas Power Cycles TD 166
TD 8 Vapor and Combined Power Cycles TD 199 TD 9 Refrigeration and Air Conditioning TD 226
BASIC CONCEPTS AND PROPERTIES OF FLUIDS
Common Data For Q. 1 and 2In an automobile tire the pressure is 245 kPa and the air temperature is 298 K.
The volume of tire is 0.050 m3 and gas constant of air is 0.287 kPa-m kgK3/ . FM 1.1 The pressure in the tire at air temperature of 32 K2 when volume of tire is
constant, will be
(A) 336 kPa (B) 26 kPa
(C) 310 kPa (D) 1854.02 kPa
FM 1.2 What amount of air should be come out to obtain pressure to its original value at same temperature ?
(A) 0.1812 kg (B) 0.1672 kg
(C) 0.014 kg0 (D) 0.3484 kg
FM 1.3 Consider Carbon dioxide at 1 atm2 and 400 Cc . What will be the density of
Carbon dioxide and cp at this state and the new pressure when the gas is cooled isentropically to 150 Cc ? (For Carbon dioxide k= . and R= m s2 2 )
(A) ρ =0.797kg m/ 3, 4 . kg cp − = , p2= kPa (B) ρ=1.3#10-4kg m/ 3, kg cp − = , p2= 5.5 kPa (C) ρ =7.97kg m/ 3, kg cp − = , p2= 5.5 kPa (D) ρ =7.97kg m/ 3, kg cp − = , p2= 5.5 Pa
FM 1.4 A Cane of beverage contains 455 ml of liquid. The mass of cane with liquid is
0.369 kg while an empty cane weighs 0.1 3 N9 . What will be the specific weight,
density and specific gravity of liquid respectively ? (A) 0.977kN m/ 3, 99.6kg m/ 3, 0 0996.
(B) 9.77kN m/ 3, 996kg m/ 3, 0 996.
(C) 9.77N m/ 3, 996kg m/ 3, 9 96.
(D) 97.7kN m/ 3, 996kg m/ 3, 0 996.
FM 1.5 The specific gravity of a gas contained in a tank at the temperature of 25 Cc is 2#10−3. If the atmospheric pressure is 10.1 kPa, the gage pressure is
(A) 70 kPa (B) 7 kPa
(C) 0.7 kPa (D) 70 kPa
FM 1.6 Consider steam at state near the saturation line : ( ,p T1 1)=(1.31MPa, 250cC), 4 m s and . )
Rsteam k 2 2
−
= = . If the steam expands isentropically to a new pressure of 414 kPa, what will be the density ρ1 and the density ρ2 ?
(A) ρ1=5.44kg m/ 3,ρ2=5.04kg m/ 3 (B) ρ1=2.28kg m/ 3,ρ2=5.44kg m/ 3
(C) ρ1=5.44kg m/ 3,ρ2=2.28kg m/ 3 (D) ρ1=5.04kg m/ 3,ρ2=5.44kg m/ 3 FM 1.7 A 30 m3 cylinder contains Hydrogen at 25 Cc and 200 kPa What amount of
Hydrogen must be bled off to maintain the Hydrogen in cylinder at 20 Cc and 600 kPa ? (R=0.2968kPa m kg K. 3/ . )
(A) 271.35 kg (B) 206.99 kg
(C) 478.34 kg (D) 64.36 kg
FM 1.8 Wet air with 100% relative humidity, is at 30 Cc and 1 atm. If Rair = m s − , Rwater=461m s K2/ 2− and vapor pressure of saturated water at 30 Cc is 4242 Pa, what will be the density of this wet air using Dalton’s law of partial Pressures ? (A) 1.12kg m/ 3 (B) 1.09kg m/ 3
(C) 0.03kg m/ 3 (D) 1.147kg m/ 3
FM 1.9 In a formula one race, at the start of the race the absolute pressure of a car tire is
362.5 kPa and at the end of the race the absolute pressure of car tire is measured
to be 387.5 kPa. If the volume of the tire remains constant at 0.022 m3 then
percentage increase in the absolute temperature of the air in the tire is
(A) 6.9% (B) 69%
(C) 0.69% (D) Not increased
FM 1.10 A compressed air tank contains 24 kg of air at a temperature of 80 Cc . If the
reading of gage mounted on the tank is 300 kPa, what will be the volume of tank
in m3 ?
(A) 404 (B) 4.04
(C) 0.404 (D) 40.4
FM 1.11 A small submersible moves in 30 Cc water (pv =4.242 kPa) at 2-m depth, where ambient pressure is 133 kPa. Its critical cavitation number is Ca.0 2. . At what
velocity will cavitation bubbles form ?
(A) 22.72 m/s (B) 32.66 m/s (C) Zero (D) 32.13 m/s
FM 1.12 What will be the speed of sound of steam at 150 Cc and 400 kpa? (k = 1.33, R= 461m s K2/ 2− )
(A) 50.9 m/s (B) 509 m/s (C) 30.3 m/s (D) 303 m/s
FM 1.13 A liquid has a weight density of 9268N m/ 3 and dynamic viscosity of 131 5. N s m− / 2
. What will be the kinematic viscosity of the liquid in m2/sec ?
(A) 0.0139 (B) 1.39 (C) 0.139 (D) 13.9
FM 1.14 A 72 m long and 30 m diameter blimp is approximated by a prolate spheroid
whose volume is given by v= 32pLR2. The weight of 20 Cc gas within the blimp
for (a) helium at 1.1 atm and (b) air at 1.0 atm, is (RHe 2077m s/
2 2 −
= , Rair=287m s K2/ 2− )
(A) WHe=60.97kN, Wair=401.1kN (B) WHe=401.1kN, Wair=6.97kN
(C) WHe=6.2kN, Wair =40.9kN (D) WHe=40.9kN, Wair=6.2kN
FM 1.15 The oil having viscosity of 4.56#10−2N s m− / 2, is contained between two parallel
plates. The bottom plate is fixed and upper plate moves when a force F is
applied. If the distance between the stationary and moving plates is 2.54 mm and
the plate with velocity of 1m/ sec ?
(A) 2 32. N (B) 23.2 N
(C) 232 N (D) 0.232 N
FM 1.16 A thin moving plate is separated from two fixed plates by two fluids of different viscosity as shown in figure below. If the contact area is A, the force required for
the flow to be steady laminar viscous flow, is
(A) F h h VA 2 2 m m =; + E (B) F h2 h VA 2 m m =: − D (C) F h h VA 2 2 m m =; − E (D) F h h2 VA 2 m m =: + D
FM 1.17 A large movable plate is located between two large fixed plates. Two fluids having the different viscosities are contained between the plates. If the moving plate has a velocity of 6m/sec, what will be the magnitude of the shearing stresses on plate
1 and plate 2 respectively, that act on the fixed plates ?
(A) 10N m/ 2, 15N m/ 2 (B) 20N m/ 2, 15N m/ 2
(C) 15N m/ 2, 15N m/ 2 (D) 15N m/ 2, 20N m/ 2
FM 1.18 A thin flat plate of area A is moved horizontally between two plates, one stationary
and one moving with a constant velocity Vm as shown in figure below. If velocity of flat plate is Vp and dynamic viscosity of oil is μ, the force must be applied on the plate to manage this motion is
(A) A h V h V V p p m 1 2 μ ; + - E (B) μA V( p-Vm) h2 (C) h AVp 1 μ (D) ( ) A h V h V V p p m 1 2 μ ; - - E
FM 1.19 A Newtonian fluid having the specific gravity of 0.91 and Kinematic viscosity of
is given by the relation: U u sin y 2d p = a k
What will be the magnitude of the shearing stress developed on the plate in term of U and δ ?
(A) 0.571U N m/ 2
δ (B) 5.71Uδ N m/ 2
(C) 5.71Uδ N m/ 2 (D) 0.571Uδ N m/ 2
FM 1.20 A 50cm#30cm#20cm block of 15 kg mass is to be moving at a constant
velocity of 0.8m s/ on an inclined plane. If a 0.8 mm thick oil film with a dynamic
viscosity of 0.006 Pa s− is there between the block and inclined plane, what amount
of force is required in x-direction ? (g=10m s/ )2
(A) 55 N (B) 55.55 N
(C) 6.42 N (D) 414.75 N
FM 1.21 A closed rectangular container is half filled with water at 45 Cc . If the air in
remaining half section of container is completely escaped. The absolute pressure in the escaped space at same temperature (saturation pressure of water at 45 Cc
is9.593 kPa) is
(A) P>Psaturation (B) P<Psaturation (C) P=Psaturation (D) Not determined
FM 1.22 Consider two parallel plates as shown in figure below. If the fluid is glycerin (ρ 1264kg m/ 3
= , μ =1.5N s m− / 2) and the distance between plates is 9 mm. What
will be the shear stress required to move the upper plate at V= m s and the
Reynolds number respectively ?
(A) 100Pa, 460 (B) 10Pa, 4600
(C) 10000Pa, 4.6 (D) 1000Pa, 46
FM 1.23 The velocity profile in a pipe flow is given by u=u ( −rn Rn), where r is the
radial distance from the centre. If the viscosity of the fluid is μ then the drag force applied by the fluid on the pipe wall in the direction of flow across length L
of the pipe is (R=radius of circular pipe).
(A) π μn u L (B) n uμ R
(C) 2nπμu L0 (D) 2nπμu0
FM 1.24 Consider air at 20 Cc with μ=1.8#10-5Pa-s. Its viscosity at 400°C by (a) The
Power-law (n=0.7) (b) the sutherland law (S = 110 K) respectively, are (A) μp= . 1#10 -s,μs=1. #10 -s - -(B) μp= . 1#10 -s,μs= . #10 -s - -(C) μp= . #10 -s,μs= . 1#10 -s - -(D) μp=1. #10 -s,μs= . 1#10 -s -
-FM 1.25 Consider a block of mass m slides down on an inclined plane of a thin oil film
as shown in figure below. The film contact area is A and its thickness is h. The
terminal velocity V of the block is
(A) V sin A mgh m q = (B) V cos A mgh m q = (C) V sin h mgA m q = (D) V cos h mgA m q =
FM 1.26 A thin layer of glycerin flows down on an inclined plate of unit width with the velocity distribution: U u h y h y = −
If the plate is inclined at an angle α with the horizontal, the expression for the surface velocity U will be
(A) U =h singma (B) U= mgsinh a
(C) U = gh sinm a (D)U= msingha
FM 1.27 A shaft of 8.0 cm diameter and 30 cm length is pulled steadily at V= m s
through a sleeve of 8.02 cm diameter. The clearance is filled with oil of ν =0.003m s2/
and S G. .= . , the force required to pull the shaft is (ρ =w 998kg m/ 3)
(A) 793 N (B) 795 N (C) 79.3 N (D) 7.95 N
FM 1.28 Match List I (Properties of fluids) with List II (Definition/ Result) and select the correct answer using the codes given below :
List-I List-II
a. Ideal fluid 1. Viscosity does not vary with rate of deformation
b. Newtonian fluid 2. Fluid of zero viscosity
c. μ ρ/ 3. Dynamic viscosity
d. Mercury in glass 4. Capillary depression
5. Kinematic viscosity 6. Capillary rise Codes a b c d (A) 1 2 4 6 (B) 1 2 3 4 (C) 2 1 3 6 (D) 2 1 5 4
FM 1.29 Match List I (Fluid properties) with List II (Related terms) and select the correct answer using the codes given below :
List-I List-II
a. Capillarity 1. Cavitation
b. Vapour pressure 2. Density of water
c. Viscosity 3. Shear forces
d. Specific gravity 4. Surfaces Tension
Codes a b c d (A) 1 4 2 3 (B) 1 4 3 2 (C) 4 1 2 3 (D) 4 1 3 2
FM 1.30 The hydrogen bubbles have diameter D - . 1 mm. Assume an ‘‘air-water”
interface at 30 Cc and surface tension σ =0.0712N m/ . What will be the excess
pressure within the bubble ?
(A) 1.42 kPa (B) 2.85 kPa (C) 28.5 kPa (D) 14.2 kPa
excess pressure inside the rain drop is
(A) 973.3 Pa (B) 97.33 Pa
(C) 9.73 Pa (D) 97.33 kPa
FM 1.32 A shower head emits a cylindrical water jet of diameter 0.73 mm into air. The
pressure inside the jet is approximately 300 Pa greater than the air pressure. What will be the surface tension of water ?
(A) 0.0365 N/m (B) 0.73 N/m (C) 0.365 N/m (D) 0.073 N/m
FM 1.33 A thin wire ring of 6 cm diameter is lifted from a 20 Cc water surface. How much
lift force is required if σ =0.0728N m/ ?
(A) 0.274 N (B) 0.0274 N (C) 0.137 N (D) 0.0137 N
FM 1.34 A 4 mm diameter glass tube is immersed in water and mercury. The temperature of the liquid is 20 Cc and the values of the surface tension of water and mercury
at 20 Cc in contact with air are 0.0734N m/ and 0.51N m/ , respectively. The
angle of contact for water is zero and that for mercury is 128c. What will be the
capillary effect for water and mercury in millimeters, respectively ? (A) 4.60, 3.82 (B) 2.35, 7.48
(C) 3.82, 4.60 (D) 7.48, 2.35
FM 1.35 The system shown in figure below is used to estimate the pressure inside the tank by measuring the height of liquid in the 1 mm diameter tube. The fluid is
at 60 Cc . What will be the capillary rise if the fluid is (a) water (σ =0.0662N m/
, ρ =983kg m/ 2, 0 c , θ ) and (b) Mercury (σ =0.47N m/ , ρ =13500kg m/ 3, 130c , θ ) ? (A) hw=0.0275m, hm=−0.0456 m (B) hw =−0.0275 m, hm=0.00 1m (C) hw=0.0275 m, hm=−0.00 1 m (D) hw =0.0137 m, hm=−0.00456 m
FM 1.36 A glass tube of 4.6 mm diameter is inserted into milk and milk rises upto 3.5 mm in the tube. If the density of milk is 960kg m/ 3 and contact angle is 15c, the
surface tension of milk is
(A) 0.2315N m/ (B) 0.025N m/
(C) 0.0236N m/ (D) 0.02315N m/
FM 1.37 A liquid film suspended on a rectangle wire frame of one movable side of 12 cm.
What amount of surface tension is required if the movable side of frame is to be moved with 0.018 N ?
(A) 0.075N m/ (B) 0.00432N m/
FM 1.38 In figure shown, a vertical concentric annulus with outer radius ro and inner radius ri is lowered into the fluid of surface tension σ and contact angle θ <45c . If the gap is very narrow, what will be the expression for the capillary rise h in
the annulus gap ?
(A) ( ) cos h g ro ri r s q = − (B) (cos ) h g ro ri r s q = − (C) (cos ) h g ro ri rs q = − (D) ( cos ) h g ro ri r s q = −
FM 1.39 A solid cylindrical needle of diameter 1.6 mm and density 7824kg m/ 3 may float
on a liquid surface. Neglect buoyancy and assume a contact angle of 0c. What
will be the surface tension σ ?
(A) 0.0772 N/m (B) 0.154 N/m (C) 0.772 N/m (D) 0.0154 N/m
Common Data For Linked Answer Q. 40 and 41
A Frustum-shaped body is rotating at a uniform angular velocity ω =200rad s/
in a container. The gap of 1.2 mm on all sides between body and container is
filled with oil of viscosity 0.1 Pa s− at 20cC.
FM 1.40 The power required at the top surface to maintain this motion is (A) hD 24 2 3 πμω (B) hD 32 2 4 πμω (C) hD 4 2 4 πμω (D) hD 16 2 2 πμω
FM 1.41 The reduction in power required at the top surface when oil viscosity is 0.0078 Pa s−
at 80cC, will be
(A) 5.29 W (B) 67.824 W
FM 1.42 A fluid of surface tension σ = 0.0728N m/ and contact angle θ =0c is filled
between 0.75 mm apart two parallel plates as shown in figure. If the density of fluid is ρ =998kg m/ 3, the capillary height h will be
(A) 2 mm (B) 10 mm
(C) 20 mm (D) 1 mm
FM 1.43 A 56 kg block slides down on a smooth inclined plate. A gap of 0.1 mm between
the block and plate contains oil having viscosity 0.4N s m− / 2. If the velocity
distribution in the gap is linear and the area of the block in contact with the oil is 0.4 m2, the terminal velocity of the block is
(A) 0.03125m s/ (B) 0.3125m s/
(C) 3.125m s/ (D) 0.03125mm s/
FM 1.44 Two 50 cm long concentric cylinders are mounted on a shaft. The inner cylinder is completely submerged in fluid and is rotating at 200 rpm and the outer cylinder is fixed. The fluid film thickness between two cylinders is 0.1 cm2 and outer
diameter of the inner cylinder is 20 cm. If the torque transmitted by the shaft to rotate inner cylinder is 0.8 N, the viscosity of the fluid is
(A) 0.0173N s m− / 2 (B) 0.0231N s m− / 2
(C) 0.173N s m− / 2 (D) 0.0346N s m− / 2
FM 1.45 A layer of water having the viscosity of 1.2#10−3N s m− / 2 flows down on inclined
U u h y h y = −
If the velocity of water U= m sec and h= m, what will be the magnitude
of the shearing stress that the water exerts on the fixed surface in N m/ 2 ?
(A) 7.20 (B) 0.720
(C) 7 2. #10−3 (D) 0 072.
FM 1.46 A 2.5 mm diameter aluminum sphere (ρ =2700kg m/ 3) falls into an oil of density 875kg m/ 3. If the time to fall 75 cm is 48 s then the oil viscosity is
(A) 0.0589kg m s/ −
(B) 0.589kg m s/ −
(C) 0.397kg m s/ −
(D) 0.0397kg m s/ −
FM 1.47 Consider a concentric shaft fixed axially and rotates inside the sleeve. If the shaft of radius ri rotates at ωrad s/ inside the sleeve of radius r0 and length L and
the applied Torque is T, what will be the relation for the viscosity μ of the fluid between shaft and sleeve ?
(A) ( ) r L T r r 2 i i 0 μ= πω - (B) T( ) 2 03 0 μ= πω -(C) T( ) 2 3 0 μ πω = - (D) T( ) 2 3 0 μ πω = +
FM 1.48 The velocity profile for laminar one-dimensional flow through a circular pipe is given as u r( )=umax( −r R2 2), where R is the radius of the pipe and r is the
radial distance from the centre of the pipe. If an oil at 40 Cc flows through a 15 m
long pipe with R=0.0 m and maximum velocity of umax= m s, what will
be the friction drag force applied by the fluid on inner surface of the pipe when
0.0010kg m s/
μ= - ?
(A) 0.0942 N (B) 0.942 N
FM 1.49 A 1 m diameter cylindrical tank has a length of 5 m long and weight 125 N. If it
is filled with a liquid having a specific weight of 10.9kN m/ 3, the vertical force
required to give the tank an upward acceleration of 2.75m/ sec2 is
(A) 550 kN (B) 55N
(C) 5. N5 (D) 55 kN
FM 1.50 A cylindrical rod of diameter D, length L and density ρs falls due to gravity inside a tube of diameter Do. The clearance, (Do−D)<<D is filled with a film of viscous fluid ( , )ρ μ .The expression for terminal fall velocity would be
(A) V=rsgD D( om−D) (B)
( )
V = rsgD Dmo+D
(C) V= rsg D(mo−D) (D) V = rsgD D( m−Do)
FM 1.51 The belt as shown in figure below moves at steady velocity of 2.5m s/ and skims
the top of a tank of oil SAE 30 W (μ=0.29kg m s/ - ) at 20 Cc with L = m, 0 cm
b = and h= cm. What power P in watts is required to remain belt in
motion ?
(A) 11 Watts (B) 44 Watts (C) 109 Watts (D) 1.1 Watts
FM 1.52 Two balls of Steel and Aluminum can float on water due to surface tension effect. The density of steel and aluminium balls are to be 7800kg m/ 3 and 2700kg m/ 3,
respectively. Which metal ball would have maximum diameter to float on water at 20cC and what will be the diameter of that ball when surface tension of water
at 20 Cc is 0.073N m/ ?
(A) steel, 4. mm1 (B) Aluminium, 2.4 mm
(C) Aluminium, 4.1 mm (D) Steel, 2. mm4
FM 1.53 For a cone-plate viscometer of radius R= cm, the angle θ = 3c and the gap is
filled with liquid as shown in figure. If the viscous torque T= . and rotation
rate is 94.2rad s/ , the liquid viscosity will be
(A) 0.0116kg m s/ −
(B) 0.116kg m s/ −
(C) 0.193kg m s/ −
(D) 0.0193kg m s/ −
FM 1.54 A solid cone of base r0 and initial angular velocity ω0 is rotating inside a conical
neglected, the cone’s angular velocity ω is
(A) exp sin r t mh 5 3 0 03 ω=ω ;- μ θE (B) exp sin mhr t 3 5 0 0 3 ω=ω ;- μ θE (C) exp sin mhr t 3 5 0 04 ω=ω ;- μ θE (D) exp sin mhr t 3 5 0 02 ω=ω ;- πμ θE
FM 1.55 The rotating-cylinder viscometer as shown in figure below shears the fluid in a narrow clearance Δ = -R (r R) with a linear velocity distribution in the gap. If
the driving torque measured is T and the bottom friction is included then the
expression for μ is (A) ( ) ( ) R L R T r R μ= πω + (B) ( / ) ( ) 2 3 4 μ= πω + -(C) ( / ) ( ) 2 2 4 μ= πω + - (D) ( / ) ( ) 2 3 4 μ= πω
-FM 1.56 For a 300 mm long sliding lubricated bearing, the viscosity of oil is 0.008kg m s/ −
during steady operation at 80 Cc . The average oil film thickness between the shaft
and journal is 1.2 mm. If shaft of 80 mm diameter is rotated at 750 rpm, the
amount of torque needed to overcome bearing friction would be
(A) 0.0063 N m− (B) 0.063 N m−
(C) 0.63 N m− (D) 6.3 N m−
FM 1.57 =0.063 N m− A disk of radius R= cm, rotates at 1200 . .r p m inside an oil
container of viscosity μ=0.29kg m s/ - as shown in figure below. The oil film
thickness is h= mm. If the velocity profile is linear and neglecting shear on the
(A) 0.716 N m− (B) 6.83 N m−
(C) 0.0716 N m− (D) 14.3 N m−
FM 1.58 A soap bubble of diameter D coalesces with another bubble of diameter D to
form a single bubble D with the same amount of air. For an isothermal process, D as a function of D D, , patm and surface tension σ is
(A) p Da + sD =(p Da + sD )+(p Da + sD )
(B) p Da + sD =(p Da + sD )−(p Da + sD )
(C) p Da − sD =(p Da − sD )+(p Da − sD )
(D) p Da + sD =(p Da + sD )+(p Da − sD )
FM 1.59 A skater of mass m moving at constant speed Vo, suddenly stands stiff with skates pointed directly forward and allows herself to coast to a stop. If blade length is L
, water film thickness h, water viscosity μ and blade width is b then how far will
she travel (on two blades) before she stops ? (A) x Lb V mho m = (B) x V mh Lb o m = (C) x Lb V mho m = (D)x V mh Lb o m =
FM 1.60 Two thin flat plates are tilted at an angle φ and placed in a tank of surface tension σ and contact angle θ as shown in figure below. At the free surface of the liquid in the tank, the distance between two plates are L and width is b into the
paper. What will be the expression for σ in terms of other variables ?
(A) ( ) ( ) cos tan gbh L h 2 σ= ρ -θ φ- φ (B) ( ) cos tan gh L h 2 σ = ρ - φ φ (C) ( ) ( ) cos tan gh L h 2 σ= ρ -θ φ- φ (D) ( ) ( ) cos tan gh L h 2 σ = ρ +θ φ+ φ ***********
SOLUTIONS
FM 1.1 Option (A) is correct.
We have p =310kPa, v = . m R = . a-m kgK3/ T =298 K and T =
Treating air as an Ideal gas, the final pressure in the tire from the ideal gas law,
T p v T p v = p T T p #
= = 298323#310 =336 kPa v =v (constant) FM 1.2 Option (C) is correct.
Amount of air needs to be bled off to restore pressure (p2=310kPa) is
m m m Δ = -m RT p v = . . 0.1812 kg 0 287 298 310 0 050 # # = = and m RT p v = . . 0.1672 kg 0 287 323 310 0 050 # # = = Hence Δm =0.1812−0.1672=0.014 kg FM 1.3 Option (C) is correct. We have p =10atm=1013250Pa T =400cC=400+273=673K
From ideal gas law
ρ ( ) ( ) . m g RT p # = = = cp . . ( ) g k kR # − = − = − =
For gas cooled isentropically to T = cC= , the formula is p p T T k k =b l − p p a T T k k .. # # = b l − = b l − =135.5 kPa FM 1.4 Option (B) is correct.
Specific weight γ = Volume of fluidWeight of fluid g v mg v W γ=ρ = = Volume of fluid
Total weight weight of Cane
= − . . . . mg 355 10 0 153 355 10 0 369 9 81 0 153 6 6 # # # = − − = −− . 9.77kN m/ 355 10 3 47 6 3 # = − = Density ρ . . g m g m g -g = = =
Specific gravity S G. water rr = = = FM 1.5 Option (D) is correct. We have S G. .= # − , T= c =( + )= , patm.= a
Density of gas ρ =S G #. . Density of ater 2#10 3#1000 2kg m/ 3
= − =
From gas equation p =rRT
2#287#298 171 kPa
= = (absolute pressure) Also pabsolute =patmospheric+pgage
pgage =171−101 =70 kPa
FM 1.6 Option (C) is correct. For ideal gas ρ1
RT p = . ( ) . 5.44kg m/ 461 273 250 1 31 10 461 543 1 31 10 6 6 3 # # # # = + = =
For isentropic expansion
T T p p k k =b l− T . T p p kk .. # # # # = − = − b l c m =393 K Now ρ2 RT p = 2.28kg m/ 461 393 414 103 3 # # = = FM 1.7 Option (D) is correct. We have v= m, p = a, T = cC= + = , a p = T = cC= + =
The initial mass of Hydrogen in cylinder
m RT p v = . 271.35 kg 0 2968 298 800 30 # # = =
Final mass of Hydrogen in cylinder
m RT p v = . 206.99 kg 0 2968 293 600 30 # # = =
Thus the amount of Hydrogen that must be bled off is
m
Δ =271 35. −206 99. =64.36 kg FM 1.8 Option (D) is correct.
Dalton’s law of Partial Pressure is
ptotal =pair+pwater= mvaR Ta +mvwR Tw or mtotal m m R T p v R T p v a w a a w w
= + = + For an ideal gas Since ptotal =pair+pwater= atm= Pa
pair =101325−pwater=101325−4242=97083 Pa Now ρ ma v mw R Tp R Tp a a w w = + = + ( ) 287 30 273 97083 461 303 4242 # # = + + = 28797083#303+4614242#303 1.147kg m/ 3 =
FM 1.9 Option (A) is correct.
We have p = . a p = . a v =v = =v . m
From ideal gas law,
T p v T p v = T T p p = v =v T p p T # = .. T . T 362 5 387 5 1 069 1 1 # = = Increase in temperature =T −T 1.069T1 T1 T1(1.069 1) = − = − . T 0 069 1 = or 6.9% of T1 FM 1.10 Option (B) is correct
We have m= g, T= c =( + ) = , pgage =300 kPa From gas equation ρ
RT pabsolute = ( ) RT patm. pgage # # = + = + =3.96kg m/ 3
Thus, Volume of tank v =mr
. 4.04 m 3 96 16 3 = = FM 1.11 Option (D) is correct By definition Cacritical 0.25 ( ) V p p 2 a v 2 r = = − . 0 25 ( ) V 998 2 133000 4242 2 # = − V . ( ) 998 0 25 2 133000 4242 # = − =32.13m s/ FM 1.12 Option (B) is correct
The ideal gas formula Predicts:
Speed of sound a , kRT = . # #( + ) . 509m s/ 1 33#461#423 = = FM 1.13 Option (C) is correct. We have γ = 9268N m/ 3, μ =131 5. Ns m/ 2 Weight density γ =rg ρ . . kg m/ 9 81 9268 944 75 3 = = Kinematic viscosity ν = =mr 944 75131 5.. 0.139Nsec. /m kg = =0.139m2/sec. FM 1.14 Option (A) is correct.
The volume of blimp is
v R L ( ) 3 2 3 2 15 72 2 2 # # # p p = = =33929 m3
(a) ρHe ( ) R T p He He # # = = = (b) ρair R T p air air # # = = =
Then the respective gas weights are
WHe =rHegv= # # =
Wair =rairgv= # # =
FM 1.15 Option (A) is correct.
We have μ=4 6.5 #10-2Ns m/ 2, y= . # − , V= sec., A= .
When force F is applied on the plate, shear force comes in the action. F =t#A y V A # # m = τ=μVy 4.56 10 . 0.129 2 54 10 1 2 3 # # # # = − − =2 32. N FM 1.16 Option (D) is correct.
Assuming a linear velocity distribution on each side of the plate.
F =t1A+t2A 1V Ah V Ah 1 2 2 m m =b l +b l h V h V A h h VA m m m m =; + E =: + D FM 1.17 Option (C) is correct.
From Newton’s law of viscosity
τ =mdudy =mUy 1 τ . . 0 02 0 008 6 # = =15N m/ 2 and τ2 =0 01. #0 004.6 =15N m/ 2 FM 1.18 Option (A) is correct.
The magnitudes of shear forces acting on the upper and lower surfaces of the plate are Fshear upper, A A dy du , w upper t m = = A h V h AV p p m m = − = Fshear lower, A A dy du , w lower t m = = ( ) A h Vp Vm m = −
therefore from force balancing
F =Fshear upper+Fshear lower
h AV h A V V p p m m m = + − A h V h V V p p m m = ; + − E
FM 1.19 Option (A) is correct.
We have S G. .= . , ν=4#10-4m2/sec.
From Newton’s law of viscosity (at the surface of plate)
(y 0) τ = dy du y m = = c m ...(i) dy du y= c m U cos y 2 2 y 0 p d pd = = a k : D = p2Ud
From equation (i) τ =nr#p2Ud μ=νρ ( . .S G 1000) U 2 n# # #p # d = ( . ) . U 4#10 4# 0 91#1000 #1 57# d = − 0.571Ud N m/ 2 = FM 1.20 Option (B) is correct. We have m= g,V=0.8m s/ , μ=0.006 Pa s -y =0.8mm=8#10−4m
The force balance from figure gives
Fx
Σ = F−Fshearcos c−FNsin c= ...(i) Fy
Σ = FNcos c−Fshearsin c−W= ...(ii)
Weight W =m#g= # = N and Fshear =tsAs ( . ) ( . . ) . A y V s # m # # # = = − Fshear =0.9 N Equation (ii) gives FN
20 ( 20 ) cos sin Fshear W c c = + . 159.95 cos sin N 20 0 9# 20 150 c c = + =
By substituting the value of Fshear and FN in equation (i),we get
F =Fshearcos c+FNsin c
. cos . sin
0 9# 20c 159 95# 20c
FM 1.21 Option (C) is correct. We have
The saturation pressure of water at 45cC=9.593kPa
When air is fully escaped, the space is filled with vapor and the container have a two-phase mixture of saturated water vapor.
Then vapor pressure Pv=Psaturation( cC)= . a FM 1.22 Option (D) is correct.
Shear stress is given by
τ .. a
L
V #
m
= = =
and the Reynolds Number is
Re VL # .# . b m r = = FM 1.23 Option (C) is correct. Velocity profile u =u ( −rn Rn)
We know that the wall shear stress in pipe flow w τ drdu r R m =− = u dr d R r n n r R m =− − = : D u R nr n n r R m =− − − = ; E = mu nR
Then the drag force applied by the fluid on the pipe wall becomes
F =twAw R ( ) n u R L n u L # # m p pm = = FM 1.24 Option (B) is correct.
(a) From the Power-law for air p μ 0 TT n 0 m = b l 1.8 10 3.221 10 kg m s/ 293 673 . 5 0 7 5 # # # − = − b l = −
(b) From the sutherland law s μ 0 ( /T TT) (. ST S) 0 1 5 0 m = ; + + E 1.8 10 ( ) ( / ) ( ) 673 110 673 293 . 293 110 5 1 5 # # = − = + + G 3.225#10 5kg m s/ − = −
FM 1.25 Option (A) is correct.
Assume a linear viscous velocity distribution in the film below the block. Then a force balance in x - direction gives:
Fx Σ =Wsinq−tA sin W h V A maX q m = −: D = = or Wsinθ h V A m = V sin sin A hW A mgh m q m q = = W=mg FM 1.26 Option (C) is correct.
flow. The FBD is shown below. In equilibrium condition Fx Σ = Wsinα =t# #l b = sin mg α =t#l sin vg ρ α =t#l m=rV sin g vg γ α l # t = ρ= gγ sin l h γ# # # α =t#l τ =ghsina ...(i)
From the Newton’s law of viscosity, shear stress at the plate ( =0) τ dudy y m = = c m hU hUy y m = − = ; E = mhU ...(ii) From equation (i) and (ii), we get
h U μ =ghsina U sin 2 2 m g a = FM 1.27 Option (A) is correct.
Assuming a linear velocity distribution in the clearance, the force is balanced by resisting shear stress in the oil.
F A ( ) R V D L wall i t mD # p = =b l F R R V D L i i m p = − ...(i)
For the given oil
μ =rn=( #r )#n
0.87 #998#0.003 2.63kg m s/ −
= =
Then by substituting in equation (i), we get
F ( . . ) . . . . 0 0401 0 0400 2 63#0 4#p#0 08#0 3 = − =79 .2 79 b793 N FM 1.28 Option (D) is correct List-I List-II
a. Ideal fluid 2. Fluid of zero viscosity
b. Newtonian fluid 1. Viscosity does not vary with rate of deformation
c. μ ρ/ 5. Kinematic viscosity
d. Mercury in glass 4. Capillary depression. So , correct pairs are a-2, b-1, c-5, d-4.
FM 1.29 Option (D) is correct.
List-I List-II
a. Capillarity 4. Surface tension
b. Vapour pressure 1. Cavitation
c. Viscosity 3. Shear forces
d. Specific gravity 2. Density of water
So, correct pairs are a-4, b-2, c-3, d-2.
FM 1.30 Option (C) is correct
For a droplet or bubble with one spherical surface
p Δ = 2sR . . . 0 005 10 2 0 0712 5 10 2 0 0712 3 6 # # # # = − = − 28480Pa-28.5kPa = FM 1.31 Option (B) is correct. We have d =3 , σ 7.3 10 2N m/ # =
-We know that surface tension on liquid droplet is given by the relation,
p d 4s = . 3 10 4 7 3 10 3 2 # # # = − − 97.33 Pa = FM 1.32 Option (D) is correct
For a liquid cylinder, the internal excess pressure is
p Δ =Rs σ =D #p R 200 ( . ) 2 0 00073 # = . 200#0 000365 = =0.073N m/ FM 1.33 Option (B) is correct
There are two surface, inside and outside the ring. So the total force measured is
F =2(spD)=2psD
2#p#0.0728#(0.06)
= =0.0274 N
FM 1.34 Option (D) is correct.
We have d =4mm=4# −3m
The capillary effect is given by the equation,
h cos g d 4 # # r s q =
where σ = Surface tension in N m/ θ = Angle of contact
Capillary effect for water
σ =0.0734N m/ , θ = 0c, ρ =1000kg m/ 3 h . . cos 1000 9 81 4 10 4 0 0734 0 3 # # # # c = − 7.48#10 3m 7.48mm = − =
σ =0.51N m/ , θ =128c ρ =S G # =13.6#1000=13600kg m/ 3 h . . cos 13600 9 81 4 10 4 0 51 128 3 # # # # # c = − 2.35#10 3m =− − =−2.35 mm
Here the negative sign indicates the capillary depression. In magnitude h =2.35 mm
FM 1.35 Option (C) is correct. (a) For water, capillary rise
hw cosgD . . ( .cos ) # # # # c r s q = = =0.0275 m (b) For Mercury hm cosgD . . cos . # # # # c r s q = = =−0.00912 m
Here negative sign shows the capillary depression.
FM 1.36 Option (D) is correct. We have ρ = 960kg m/ 3, D= . mm= . # − m R= D . 1.9 10 m 2 3 8#10 3 3 # = − = − . mm h= =0.0025m, contact angleφ =15c
The surface tension of milk milk σ 2cosgRh 960 9 81. 21 9. cos10153 2 5. 10 3 # # # # # # c f r = = − − 0.02315N m/ =
FM 1.37 Option (A) is correct.
We have b= cm, =0.12m, F= .
From the surface tension force relation, s
σ 2Fb 2 0 018.( .0 12) #
FM 1.38 Option (B) is correct.
From the figure above, the force balance on the annular fluid is
Force in vertical direction =Weight of fluid film
( ) cos # 2 ro 2 ri σ θ π + π =rg#p(ro −ri)h h (cos ) g ro ri r s q = −
FM 1.39 Option (A) is correct.
The needle “dents” the surface downward and the surface tension forces are upward as shown in figure. Then a vertical force balance gives:
Vertical forces =Weight of needle cos L 2σ θ# =r pg D #L cos 2σ θ =r pg D 2σ =r pg D θ =0c"cos0c=1 σ g D # r p = = 7824#9 81. #83 14. #( .0 0016)2 0.0772N m/ = FM 1.40 Option (B) is correct.
The wall shear stress anywhere on the surface of the frustum at a distance r from
the axis of rotation is w
τ =mdudr =mVh =m whr
The shear force on the area dA,
dF dA h rdA w t m w = = Torque dT rdF h r dA m w = = T h r dA A mw =
#
The shaft power required at top surface is
Pshaft top, T h r dA A # w w mw = =
#
h Ar dA mw =#
...(i)For the top surface dA =2prdr
Hence Pshaft top, ( )
h r r r dr D mw p = =
#
h r dr h r r D r D pmw pmw # = = = : D=
#
= pmw32h2D4 FM 1.41 Option (C) is correct.By putting the value in expression of shaft power at top (20cC), Pshaft top, ( . ) . ( . ) ( ) ( . ) hD 32 32 0 0012 3 14 0 1 200 0 12 2 4 2 4 # # # # pmw = = 67.824 W =
The power is proportional to viscosity. Thus the power required at 80 Cc is Pshaft top, , cC P , , C C C shaft top 20 80 20 m m # = c c c . . 67.824 5.29 W 0 1 0 0078 # = =
Therefore, the reduction in the required power input at 80 Cc is Pshaft top, , cC−Pshaft top, , cC =67 824. −5 29.
62.533 W =
FM 1.42 Option (C) is correct
With b the width of the plates into the paper, the capillary forces on each wall
together balance the weight of fluid held above the free surface. Weight of fluid =Surface tension force
( . ) g# #h#b ρ =2 # s( bcosq) or h ( .cos ) g # r s q = . ( . ) . cos 998 9 81 0 00075 2 0 0728 0 # # # # c = ,0.020m=20mm
FM 1.43 Option (A) is correct.
We have m= g, y= . mm, A = . m , α = 30c,μ =0.4Ns m/ 2
The FBD of the block shown below.
In equilibrium condition Fx Σ =0 Wsin c=tA sin mg c=m#Vy A y=film thickness V sin A mgy # c m = W=mg . . . 0 4 0 4 10 10 10 4 0 5 # # # # = − =0.03125m/ sec
FM 1.44 Option (A) is correct.
We have L= cm= m, N= rpm, h = . cm= . m ,
cm
D= R= = . cm= . m, T= .
T h R RL h R L # # mw p pmw = = ...(i) and ω N 20.94rad s/ 60 2 60 2# #200 p p = = =
From equation (i),we get
μ . . ( . ) . ( . ) R L T h 2 2 3 14 20 94 0 075 1 0 8 0 0012 3 3 # # # # # pw# = = 0.0173N s m− / 2 = FM 1.45 Option (D) is correct. We have μ=1.2#10-3Ns m/ 2, U= m sec, h= . m
From Newton’s law of viscosity
τ =mdudy ...(i)
At the fixed surface (at y= ) dy du h U h Uy y = − = ; E = hU
From equation (i) τ =m# hU =1 2. #10−3#20 1#. 3 =0.072N m/ 2 FM 1.46 Option (C) is correct.
According to stokes law
μ W3netDL#t p
= ...(i)
The net weight of the sphere in the fluid is
Wnet ( sphere fluid)g vfluid ( sphere fluid) g D6
3 # # # r r r r p = − = − (2700 875) 9 81. ( . ) 6 0 0025 3 # #p # = − =1.46#10−4N
Then from equation (i),we get
μ ( . ) ( . ) 1.46 10 3 0 0025 0 75 48 4 # # # # p # = ^ − h =0.397kg m s/ − FM 1.47 Option (C) is correct
Assuming a linear velocity distribution inside the annular clearance, the shear stress is τ Vr r rr i i mDD m w = = − ...(i)
This stress causes a force
dF =tdA=t(r di q)L ...(ii)
The torque of this force about the shaft axis is
dT =r dFi ...(iii)
Put equation (i), (ii) and (iii) together
T r dF r (r Ld ) r r r r r Ld i i i i i i i # # t q m w q =
#
=#
p =#
p − r r r L d r r r L d i i i i mw q mw q =#
p − = −#
p r r Lr i i pmw = − μ T r( r Lr) 2 i i 3 0 pw = −FM 1.48 Option (B) is correct.
The velocity profile is given by
( ) u r u R r max = c − m
The shear stress at pipe surface is expressed as s τ drdu u dr d R r max r R r R # m m =− =− − = ; E= R umax m =
Then the friction drag force
FD A R ( ) u RL max s s t m p = = As = pRL FD =4pmLumax ...(i)
By substituting the given values in equation (i), we get
FD =4#3 14. #( .0 0010)#(15)#5 =0.942 N
FM 1.49 Option (D) is correct.
In the figure WT =Weight oftank=125N
WL =Weight of liquid=mg=rvg=gv where γ =specific weight of liquid=10.9kN m/ 3
WL =10 9. #103#4p#(1)2#5 =42.8 kN From the Newton’s law of motion in vertical direction
Fy Σ =may FV−WT−WL =may FV− − =b125+9 81.42800l2.75 FV − =12033 FV =54958N-55kN
FM 1.50 Option (A) is correct.
At terminal velocity, the rod weight should equal the viscous drag.
W =Viscous Drag g v s # ρ (D D) / V DL 2 o # m p =; − E g D L s # # ρ π (DV DLD) /2 o m p = − V = rsgD D( mo−D)
FM 1.51 Option (C) is correct
The power is the viscous resisting force times the belt velocity.
P =Viscous resisting force#Velocity
A V
oil belt belt
t # # = h V b L V # # # m =b l ^ h V b h L m =
By substituting the values, we get
P . ( . ) ( . ) . 0 29 2 5 0 9 0 06 4 2 # # # = =108.75W,109W FM 1.52 Option (C) is correct.
We have ρsteel= g m, ρaluminum= g m , σwater= . m From surface tension force relation,
Fs=p sD s and W=mg=rvg=r pg D When the ball floats Fs =W
D s π σ =r pg D D g s rs =
For Steel Dsteel ( ) .
( . ) . m g steel s # # r s # = = = − 2.4 mm =
For Aluminum Daluminum .
( . ) . m g aluminium s # # # r s # = = = − 4.1 mm =
Hence Daluminum >Dsteel
So aluminum ball would be larger in size.
FM 1.53 Option (C) is correct
For any radius r#R, the liquid gap is h=rtanq. Then dT =tdA rw . tan cos r r r dr r m w q p q =a kb l L= cosdrq T sin sin 2 3 2 2 0 3 q pwm q pwm =
#
= μ sin 2 3 3 pw q =Substituting the numerical values, we get μ ( . ) ( . ) . sin 2 94 2 0 06 3 0 157 3 3 # # # # c p = =0.193kg m s/ − FM 1.54 Option (D) is correct.
At any radial position r<r on the cone surface and instantaneous rate ω dT =r dAt w sin r h r r dr # m w # p q = 9 C : D =hsinmwq# pr dr Torque T sin h r dr r w mw p =
#
2hsinr04 q pmw = ....(i)Since for cone T I dt dw =− mr ddt 10 3 02# w =− For cone I mr 10 3 0= 02
Then from equation (i),
mr dtd 10 3 02 w − 2hsinr04 q pm w =
Separating the variables and integrating both the sides,
d ωω ω ω
#
sin mr h r dt 3 2 10 t 02 04 0 # # q pm = −#
or ω exp sin mhr t 3 5 0 0 2 w pm q = ;− E FM 1.55 Option (B) is correct.For the fluid in the annular region
TA RdF R dA R R R RLd #t# # m wD q =
#
=#
=#
p b l R R L pmw D = Now Tbottom r dA r Rr rdr R t m wD p =#
=#
a k = DpwmR#
Rr dr R R 4 2pwm 4 D = Ttotal R R L R R pwm pwm D D = + μ ( / ) R LT R R 2pw 3 4 D = + μ ( / ) ( ) R L R T r R 2pw 3 4 = −+ FM 1.56 Option (B) is correct. We have L= mm= . m, μ =0.008kg m s/ . , tfilm = . mm= . m . , mm m D= = N= rpmTorque is given by T Area tfilm R mw # # = T t R RL t R L film # film mw p pmw = = As = pRL T ( ) t N R L film # pm p = ω= 260πN T NR Lt 60 4 film 2 3 # p m =
By substituting numerical values T . . ( . ) . 60 0 0012 4 2 0 008 750 0 04 3 0 3 # #p # # # # = FM 1.57 Option (A) is correct.
At any r#R, the viscous shear on both sides of the disk
total
τ =2#t=2#mwhr
and viscous force
dF =ttotal#dAw h ( ) r rdr mw p # =
Then viscous torque
dT =dF r. . h r dr r pmw = = pmwh r dr
Integrating both the sides
T . h r r dr R pmw = =
#
h R h R # pmw pmw = =Substituting the numerical values
T ( . ) . N ( . ) 0 001 60 0 29 2 0 05 4 # # # # # p p = ω=260πN ( . ) . ( . ) 0 001 60 2 2 0 29 1200 0 05 4 # #p # # # = =0.716 N m−
FM 1.58 Option (A) is correct
The masses remain the same for an isothermal process of an ideal gas
m +m =m v v 1 1 2 2 ρ +ρ =r3v3 or RT p D RT p D π π # + # 9 C 9 C 9 C 9 C =9RTp C#9pD C , RT p v D ρ= = π RT p r D RT p r D a s p a s p # # + + + ; E 9 C ; E 9 C=;pa+RTs rE#9pD C P P R a s − =
The temperature cancels out, and we may clean up and rearrange as follows
p Da + sD =(p Da 23+8sD22)+(p Da 13+8sD12) FM 1.59 Option (C) is correct
τ =mVh F A h VLb t m = =
Using F=ma to find the stopping distance Fx Σ F h VLb ma m dt dV x m
=− =− = = The ‘2’ is for two blades Separate and integrate once to find the velocity
V dV V V o
#
mhLb dt t m =−#
or log Ve V o : D =−2mmhLbt or V V eo mh Lb t 2 = − mIntegrate once again to find distance
x Vdt V eo mh dt Lb t 2 =
#
3 =#
3 − m Lb V mh 2 o m = FM 1.60 Option (C) is correctConsider the right side of the liquid column, the surface tension acts tangent to the local surface that is along the dashed line at right. This force has magnitude
F=sb as shown. Its vertical component is Fcos(θ φ- ) as shown. There are two
plates, therefore the total vertical force on the liquid column is
Fvertical =2sbcos(q−f)
Then the vertical force holds up the entire weight of liquid column between plates, which is
W =rgbh L( −htanf)
Set W equal to F, we get
( ) cos 2σ θ φ- =rgbh L( −htanf) or σ ( ) ( ) cos tan 2 q f r f = − − ***********
PRESSURE AND FLUID STATICS
FM 2.1 The barometric reading for a wall is given as 511 mmHg at the top and 5 8.5 mmHg8 at the bottom. For average air density of 1.18kg m/ 3, the height of
wall is (ρ =Hg 13600kg m/ 3)
(A) 205 m (B) 202 m
(C) 210 m (D) 200 m
FM 2.2 A vertical clean glass Piezometer tube has an inside diameter of 4 mm. When a pressure is applied, water at 26°C (γ =9790N m/ 3, σ =0.073N m/ , θ =0c)
rises into the tube to a height of 23.5 cm. After correcting for surface tension the applied pressure will be
(A) 147 Pa (B) 2448 Pa (C) 2300 Pa (D) 2154 Pa
FM 2.3 Consider a frictionless piston-cylinder of a gas car as shown in figure. The mass of piston is 4 kg and cross-sectional area is 35 cm2. During the compression stroke
of car engine a force of 70 N is exerted on the piston. If the atmospheric pressure is 105 kPa, the pressure inside the cylinder is
(A) 133.5 kPa (B) 13.35 kPa
(C) 60 kPa (D) None of these
FM 2.4 All fluids in figure shown below are at 20 Cc . What will be the Δp between points
A and B ?
Take the specific weights to be
Benzene : 8640N m/ 3 Mercury : 133100N m/ 3
air : 12N m/ 3
(A) 16 kPa (B) 13.35 kPa (C) 29.35 kPa (D) 26.17 kPa
FM 2.5 A one-tone load on the hydraulic lift shown in figure is to be raised by pouring oil into a thin tube. The density of oil is 780kg m/ 3 and diameter of hydraulic lift is 1.2 m. The height h, in order to begin to raise the weight should be
(A) 1.34 m
(B) 0.134 m
(C) 1.134 m
(D) 0.1134 m
FM 2.6 A closed cylindrical tank filled with water has a hemispherical dome and is connected to a piping system shown in figure below. The top part of the piping system has a liquid of specific gravity 0 7. and the remaining parts of the system
are filled with water. What will be the pressure at point A ?
(A) 33.35 kPa (B) 62.78 kPa
(C) 3.93 kPa (D) 6.278 kPa
FM 2.7 Water flows upward in a pipe inclined at 45c as shown in figure below and the
pressure difference between points (1) and (2) in the pipe is 34.4 kPa. What will be the mercury manometer reading h ?
(A) 20 cm (B) 44 cm (C) 36 cm (D) 12 cm
FM 2.8 The gage pressure of the air in the water tank shown in figure below is 59Kpa. The differential height hHg of the mercury column will be(S G. .mercury=13.6)
(A) 30 cm (B) 36 cm
(C) 13.6 cm (D) 51 cm
FM 2.9 The right leg of the manometer is open to the atmosphere as shown in figure. The gage pressure in the air gap in the tank is 25.68 kPa. What will be the specific weight of the oil in N m/ 3 ?
(A) 7831 (B) 10815
FM 2.10 An inverted U-tube manometer containing oil having specific gravity of 0.95, is located between two reservoirs as shown in the figure. The reservoir on the right, contains water and is open to the atmosphere and the reservoir on the left contains glycerin is closed and pressurized to 45kPa. What will be the depth of
water h in the figure ? (γglycerin= . m )
(A) 0.721 m (B) 6.21 m
(C) 7.21 m (D) 0 0721. m
FM 2.11 A water tank is divided into two compartments as shown in figure. An oil with density ρ =oil 5 .5 kg m is poured into one side and the water level rises a
certain amount on the other side to overcome this effect. The oil does not mix with water. What will be the final differential height of water shown in figure ?
(A) 33. 5 cm7 (B) 60 cm
(C) 45 cm (D) 0.3375 cm
FM 2.12 A tank contains water (γ =9790N m/ 3) and immiscible oil at 20 Cc as shown in
figure below. If the specific weight of oil is 8809N m/ 3, what will be the h ?
(A) 26 cm (B) 20 cm (C) 10 cm (D) 13 cm
FM 2.13 A tank is constructed of a series of cylinders as shown in figure. A mercury manometer is attached to the bottom of the tank. What will be the manometer reading h ?
(A) 37.6 m (B) 3.76 m
(C) 0.0376 m (D) 0.376 m
FM 2.14 The U-tube at right has a 1 cm internal diameter and contains a liquid (S.G. = 1.6) as shown in figure below. If 20 cm3 of water (γ =9790N m/ 3) is poured into
the right-hand leg, what will be the free surface height in each leg at equilibrium ?
(A) hR= . cm,hL= . cm (B) hR= . 6cm,hL= . cm (C) hR= . cm,hL= . 6cm (D) hR= . cm, hL= . cm
FM 2.15 Two compartments A and B of the tank are closed and filled with air and a liquid
shown in figure below. The liquid having the specific gravity of 0.6. If the pressure gage reads 3.5 kPa and weight of the air is negligible, the manometer reading h
will be
(A) 0.0424 m (B) 0.212 m
