The rake angle of a cutting tool is 15°, shear angle 45° and cutting velocity 35 m/min. The velocity of chip along the tool face will be_______.
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Solution
Velocity of chip along tool face (V) =\(\frac{Vsin\phi }{cos(\phi -\alpha )}\)
where, v =cutting velocity = 35 m/min.
α= rake angle= 15°
φ= shear angle = 45°
V=\(\frac{35\times sin45^{\circ}}{cos(45-15)}=\frac{35\times sin45^{\circ}}{cos30}=\frac{35\times 0.707}{0.866}\)
= 28.57 m/min.
The three times estimates of a PERT activity are optimistic time = 8 min; most likely time = 10 min. and pessimistic time =14 min; Then the expected time of activity will be _____.
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Solution
The PERT equation for expected time is
E=\(\frac{(O+4M+P)}{6}\)
where, O = Optimistic time = 8 min
M =Most likely time =10 min.
P = Pessimistic time =14 min.
E=\(\frac{8+4\times 10+14}{6}=\frac{8+40+6}{6}=10.33 min\)
Weekly production requirements of a product are 1000 items.The cycle time of producing one product on a machine is 10 min.The factory works on two shift basis in which total available time is 16 h. Out of available time, about 25% is expected to be wasted on break downs, material unavailability and quality related problems. The factory works for 5 days a week. Then umber of machines required to fulfill the production requirements will be ______
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Solution
Given: Weekly production requirement = 1000 products Cycle Time (CT) for producing one product/machine = 10 min Total available time (Ttotal) = 16 h
Time loss = 25%Effective time= 16 –\(\frac{25}{100}\)×16=16-4= 12 h
Total effective hours/week = 5 × 12 = 60 h
Number of products/machine =\(\frac{Total\, effective\, hours}{Cycle\, Time}\)
=\(\frac{60\times 60}{10}=360\)
Now,Number of machines required to fulfill the production requirements =\(\frac{Number\, of\, products/week}{Number\, of\, products/machines}\)
=\(\frac{Number\, of\, products/week}{Number\, of\, products/machines}\)
For a rhombohedral space lattice, which one of the following is correct?
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Solution
For a rhombohedral space lattice,
a =b = c, α = β= γ≠90°
where a, b, c are lattice parameters and α, β, γ are included angles
The stiffness of the beam shown in figure will be _______.
where, I = 275 × 10-6 m4
l= 0.9m
E = 200 GPa
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Solution
Using the formula for strain energy (u) =\(\int_{0}^{L}\frac{p^{2}}{2EA}dx\)
where , E→Elasticity
A→Area
P→Load
L→Length of beamStiffness (k) =\(\frac{AE}{L}\)
using both of the above formula,
K=\(\frac{2}{3}\times \frac{200\times 10^{9}\times 275\times 10^{-6}}{(0.9)^{3}}= 5.02\times 10^{7}N/m\)
In case of multiple disc clutch, if S1 is the number of discs on driving shaft and S2 is the number of discs on the driven shaft then, the number of pairs of contact surfaces is :
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Solution
Number of pairs of contact surfaces =S1+ S2– 1
The figure is showing a rigid body undergoing planar motion.The tangential accelerations of points P and Q on the body are \(\frac{300}{150}\)mm/s2 and \(\frac{600}{300}\)mm/s2 respectively in the direction shown. The angular acceleration of rigid body will be ______.
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Solution
Angular acceleration (w) =\(\frac{W_{p}+W_{q}}{80}\)
Here, wp= tangential accelerationof ‘P’ = 300 mm/s2
wq = tangential acceleration of Q = 600 mm/s2
w =\(\frac{300+600}{80}\)= 11.25 rad /s2
Steam pressure at inlet and exit of a nozzle are 16 bar and 5.2 bar respectively and discharge is 0.28m3/s. Critical pressure ratio is 0.5475. If the exit pressure is reduced to 3.2 bar, then the flow rate will be _______.
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Solution
Flow rate will be 0.28 m3/s. Because it will have the same valuei.e. constant.
Match List I with List II.
List I List II
A.Hydrometer 1.Vapour pressure
B.Bombcalorimeter 2.Composition of products combustion.
C.Reid bomb 3.Specific gravity
D.Orsat appratus 4.Heating value.
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Solution
Hydrometer= Specific gravity
Bomb calorimeter = Heating value
Reid bomb = Vapour pressure
Orsat appratus= Composition of products combustion.
Stream and velocity potential functions for a two-dimensional flow field given by u= 2x and v = –2y are:
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Solution
As,\(\frac{\partial \Psi }{\partial y}=2x,\frac{\partial \phi }{\partial y}=0-2y\)