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.