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
(b) Velocity gradient
(c) Shear strain
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
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
(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.
S1. Ans. (b)
Sol. Newton’s Law of viscosity states that shear stress is directly proportional to rate of shear strain or velocity gradient.
S2. Ans. (c)
Sol. The capillary rise of water in the glass tube h will be
But, for water the value of θ is zero.
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)
S5. Ans. (a)
Sol. Critical Reynold number,
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
S6. Ans. (b)
Sol. A fluid is at rest means that the fluid has non-zero normal stress and zero shear stress.