UPPSC Lec'21 ME: Daily Practices Quiz. 22-Nov-2021 |_00.1
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UPPSC Lec’21 ME: Daily Practices Quiz. 22-Nov-2021

Quiz: Mechanical Engineering
Exam: UPPSC AE/Lecturer
Topic: Fluid Mechanics

Each question carries 3 marks
Negative marking: 1 mark
Time: 8 Minutes

Q1. Newton’s Law of viscosity states that “Shear stress” is directly proportional to
(a) Velocity
(b) Velocity gradient
(c) Shear strain
(d) Viscosity

Q2. If σ be the surface tension of the water, ρ be the mass density g be the gravitational accelerations and d be the diameter of the glass tube, then the capillary rise of water in the glass tube h will be
(a) 2σ/ρgd
(b) σ/2ρgd
(c) 4σ/ρgd
(d) 8σ/ρgd

Q3. The pressure difference (in Pa) in a droplet of a fluid of 0.002m diameter with surface tension of 0.01 N/m is
(a) 10
(b) 20
(c) 4 π
(d) 0.0004 π

Q4. A force of 400 N is required to open a process control valve. What is the area of diaphragm needed for a diaphragm actuator to open the valve with a control gauge pressure of 70 kPa?
(a) 0.0095 m^2
(b) 0.0086 m^2
(c) 0.0057 m^2
(d) 0.0048 m^2

Q5. An oil which has kinematic viscosity 0.2 stokes flows through a circular pipe of 1cm radius. The velocity at which the flow will be critical is about
(a) 2.0 m/s
(b) 1.5 m/s
(c) 0.5 m/s
(d) 4.0 m/s
Q6. ‘A fluid is at rest’ it means that
(a) The fluid has zero normal stress and non-zero shear stress.
(b) The fluid has non-zero normal stress and zero shear stress.
(c) The fluid has non-zero normal stress and shear stress.
(d) The fluid has zero-normal stress and zero shear stress.

Solutions

S1. Ans. (b)
Sol. Newton’s Law of viscosity states that shear stress is directly proportional to rate of shear strain or velocity gradient.
τ∝du/dy
τ=μ du/dy
S2. Ans. (c)
Sol. The capillary rise of water in the glass tube h will be
h=4σcosθ/ρgd
But, for water the value of θ is zero.
So,
h=4σcos0/ρgd=4σ/ρgd

S3. Ans. (b)
Sol. ∆P=4σ/d=4×0.01/0.002=20 Pa

S4. Ans. (c)
Sol. Given that Force (F) = 400 N, Gauge Pressure (P) = 70 kPa
Area of diaphragm(A)
A=Force/Pressure=F/P=400/(70×1000)=0.0057 m^2

S5. Ans. (a)
Sol. Critical Reynold number,
Re=2000
Given that radius of circular pipe R=1cm
So, Diameter D=2R=2cm=0.02m
Also given that Kinematic viscosity, ϑ=0.2stokes=0.2×10^(-4) m^2/sec
Re=(V×D)/ϑ
2000=(V×0.02)/(0.2×10^(-4) )
V=2m/sec
S6. Ans. (b)
Sol. A fluid is at rest means that the fluid has non-zero normal stress and zero shear stress.

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Published by VINAYAK KUMAR

Graduated from BIT Sindri, Postgraduated from IIT BHU Varanasi

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