Quiz: Mechanical Engineering 31 July 2020

Quiz: Mechanical Engineering
Exam: ISRO (TECHNICAL ASST.)
Topic: Miscellaneous

Each question carries 4 mark.
Negative marking: 1 mark
Time: 15 Minute

Q1. Annealing of steel takes place at a temperature
(a) Just above the lower critical temperature
(b) Just below the lower critical temperature
(c) Just above the upper critical temperature
(d) Just below the upper critical temperature

Q2. Weber number is the ratio of
(a) Square root of ratio of inertia force to surface tension force
(b) Square root of ratio of inertia force to viscous force
(c) Square root of ratio of inertia force to gravity force
(d) None of these

Q3. Kelvin-Plank statement is a corollary of
(a) First law of thermodynamics
(b) Second law of thermodynamics
(c) Zeroth law of thermodynamics
(d) Third law of thermodynamics

Q4. Diffuser is of converging cross section at outlet for
(a) Supersonic flow
(b) Sonic flow
(c) Subsonic flow
(d) None of the above

Q5. If the center distance between the involute gears is increased then which of the following will be the effects of it?
1. Pitch circle diameter is increased.
2. Pressure angle is decreased.
3. Velocity ratio is increased.
4. Pressure angle is increased.
(a) Only 1
(b) 1 and 3
(c) 1 and 4
(d) 2 and 3

Q6. Which of the following factors contributes in increase of material removal rate in electro chemical machining?
(a) Increase in valency of metal ion
(b) Increase in density of metal
(c) Increase in voltage between workpiece and tool
(d) None of these

Q7. The locus of point on a circle which rolls without slipping along a straight line is known as
(a) An involute
(b) A cycloid
(c) A parabola
(d) An ellipse

Q8. If for a turning job, the length of the work is 400 mm and the taper is 1:20, the set-over required for the tail stock to achieve the taper is
(a) 15 mm
(b) 20 mm
(c) 4 mm
(d) 5 mm

Q9. A solid metal brick weighs 90 N in air and 50 N, when submerged in water then what is the weight of the water displaced by body?
(a) 30 N
(b) 60 N
(c) 40 N
(d) 50 N

Q10. If a boat covers a 50 km in 10 min when it is running downstream and covers the same distance in 20 min when running upstream then find out the speed of flowing river water
(a) 37
(b) 41.6
(c) 20.8
(d) 16

Solutions

S1. Ans (b)
Sol. This process is mainly suited for low-carbon steel. The material is heated up to a temperature just below the lower critical temperature of steel.

S2. Ans (a)
Sol. Weber Number is a dimensionless characteristic number used to analyse fluid flows where there is an interface between two different fluids. As a dimensionless quantity, it describes as the square root of the ratio between the inertial forces and the surface tension forces for liquids flowing through a fluid medium.

S3. Ans (b)
Sol. The Kelvin–Planck statement of the second law of thermodynamics states that it is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work.

S4. Ans (a)
Sol.
1. If Mach Number (Ma<1) then flow is subsonic.
Nozzle is converging and diffuser is diverging at outlet.
2. If Mach Number (Ma>1) then flow is supersonic.
Nozzle is diverging and diffuser is converging at outlet.

S5. Ans (c)
Sol. For the involute gear profile increase in the center to center distance leads to increase in pressure angle so the pitch circle radius of the gears gets changed and it is increased. But velocity ratio between gears will remain constant.

S6. Ans (c)
Sol. We know that
Volume MRR in ECM = (A.V)/(F.ρ.R.v)
Here,
A = Atomic weight of metal ion of workpiece
V = Voltage between work-piece and tool
F = Faraday’s constant
ρ = Density of metal
R = Resistance between workpiece and tool
v = Valency of metal ion

S7. Ans (b)
Sol. A cycloid is the curve traced by a point on a circle as it rolls along a straight line without slipping.

S8. Ans.(b)
Sol. Set-over or offset required on tail-stock for taper turning=taper X length of workpiece
Set-over or offset length= 1/20 X 400=20 mm

S9. Ans.(c)
Sol. We know,
Weight of the water displaced=Buoyancy on body= (weight of body in air – weight of body in water)
=90-50
=40 N

S10. Ans.(c)
Sol.
Let speed of river is v and speed of boat in still river is u and total distance covered either side is 50 km.
Given,
For Downstream flow-
(50×1000)/(u+v)=10×60

u+v = 83.33……….(1)

For up-stream flow-
(50×1000)/(u-v)=20×60

u-v = 41.66………..(2)

Solving equation (1) and (2)..

Speed of river (v) = 20.83 m/s