Quiz: Mechanical Engineering 25th June |_00.1
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Quiz: Mechanical Engineering 25th June

Quiz: Mechanical Engineering
Exam: AE/JE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. When a fluid flows in concentric circles, it is known as __ .
(a) free circular motion
(b) free rotational motion
(c) free spiral vortex flow
(d) free cylindrical vortex flow

Q2. The magnitude of water hammer does not depend upon
(a) temperature of fluid
(b) length of pipe
(c) elastic properties of pipe material
(d) time of valve closure

Q3. If the particles of a fluid attain such velocities that vary from point to point in magnitude and direction as well as from instant, the flow is __.
(a) Uniform flow
(b) Steady flow
(c) Turbulent flow
(d) Laminar flow

Q4. One dimensional flow is _.
(a) restricted to flow in a straight line
(b) uniform flow
(c) one which neglects changes in a transverse direction
(d) the most general flow

Q5. The pitch of lead screw is 5 mm and pitch of thread to be cut is 1mm, find the change gears.
(a) 0.5
(b) 1
(c) 1/5
(d) 5

Q6. What is the addendum of a cycloidal gear tooth?
(a) cycloid
(b) involute
(c) epicycloid
(d) hypocycloid

Q7. For such element only under normal stresses, the radius of Mohr circle is:

Quiz: Mechanical Engineering 25th June |_40.1

(a) σ
(b) σ /2
(c) 2σ
(d) 0.6 σ

Q8. In a Mohr’s circle of σ-τ plane (σ = normal stress, τ = shear stress), the vertical diameter represents
(a) Maximum shear stress
(b) Maximum normal stress
(c) Principal stress
(d) Minimum normal stress

Q9. A rotating mass having moment of inertia of 30 kgm² rotates at 800 rpm and is travelling in a curve of 170-meter radius at a speed of 240 km/hr. It will experience a gyroscopic reaction of………
(a) 10 kgf-m
(b) 100 kgf-m
(c) 1000 kgf-m
(d) 10000 kgf-m

Q10. In order to double the period of a simple pendulum________________.
(a) the mass of its bob should be doubled
(b) the mass of its bob should be quadrupled
(c) its length should be doubled
(d) its length should be quadrupled

Solutions
S1. Ans.(d)
Sol.

Quiz: Mechanical Engineering 25th June |_50.1

Vortex motion is defined of the motion of rotating fluid mass. If motion takes place without expenditure of external energy and pattern of motion can be described with the help of concentric stream lines is called free cylindrical vortex motion.

S2. Ans.(a)
Sol.

S3. Ans.(c)
Sol. Turbulent flow is defined as the flow in which energy are in disorganized form occurs at higher velocities, which is not in same in magnitude and in directions.

S4. Ans.(c)
Sol. One dimensional flow is the flow which flow only in longitudinal direction and negligible flow in transverse direction.

S5. Ans.(c)
Sol. Change of gear = (Thread pitch)/(Pitch of lead screw)
▭(change of gear=1/5)

S6. Ans.(c)
Sol. Cycloid profile has addendum of gear at epicycloid and dedendum of gear at hypocycloid.

S7. Ans.(a)
Sol.

Quiz: Mechanical Engineering 25th June |_60.1
σ_x= -σ
σ_y= σ
Radius of Mohr circle = |(σ_x-σ_y)/2|
= |(-σ-σ)/2|
= σ

S8. Ans.(a)
Sol. In Mohr’s circle vertical axis represent shear stress & Horizontal axis represent normal stress so the vertical diameter of circle represents maximum shear stress.

S9. Ans.(b)
Sol.
Given, moment of inertia of rotating mass (I) =30 k〖g-m〗^2

As We know Angular-velocity,
ω=2πN/60
=(2π×800)/60
=83.77 rad/sec
Angular velocity of precession, ω_p = V/r
=(240×1000)/(3600×170) =0.392 rad/sec
Gyroscopic reaction = I〖ωω〗_p
=30×83.77×0.392
=985.3 N-m
Or
▭(Gyroscopic reactiojn =100.42 kgf-m)

S10. Ans.(d)
Sol. we know, time period of simple pendulum T=2π√(l/g)
If l^’=4l
Then T^’= (2π√(l^’/g))
=2π√(4l/g)
=2(2π√(l/g))
▭(T^’=2T)

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