In ah NC control machine, if the specification of motor is 1% pulse, and the pitch of lad screw is 3.9 mm, the expected positioning accuracy will be ………………..
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Solution
For each revolution of lead screw, 360 pulses will
be developed. So,
For 1°/pulse, the positioning accuracy will be
=\(\frac{3.9}{360}=0.0108 mm\)
A shell of 200 mm diameter and 100 mm height is being produced by applying deep drawing of a 9 mm steel metal. The diameter of blank used to produce the shell_________
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Solution
The diameter of blank (Db) =\(\sqrt{(D)^{2}+4Dh}\)
where, D= diameter, h= height
Given : D= 200 mm, h= 100 mm
Now,Db =\(\sqrt{(200)^{2}+4(200)(100)}\)
=\(\sqrt{40,000+80,000}\)
\(\sqrt{120,000}= 346.41 mm\)
Under uniform cooling, the volumetric solidification shrink age and volumetric solid contraction for a casting are given as 5% and 7.2% respectively. The diameter of casting (spherical casting of 27 mm diameter) after solidification will be _______
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Solution
The diameter of casting after solidification (D)
\(=\sqrt[3]{\left ( 1-\frac{5}{100} \right )\left ( 1-\frac{7.5}{100} \right )}\times 27=\sqrt[3]{0.95\times 0.928}\)×27
\(=\sqrt[3]{0.8816}\times \times 27= 25.89 mm\)
A uniform ladder of length 9 m and weight 30 N is placed as shown in figure. If the ladder is just to slip, then the frictional force acting on the ladder at the point of contact between ladder and the floor will be
………….
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Solution
FBD of ladder,
Now, Given:PQ= 9m, OP = 5 m
then, OQ=\(\sqrt{(9)^{2}-(5)^{2}}=\sqrt{81-25}=\sqrt{56}= 7.48 m\)
sin α=\(\frac{OQ}{PQ}=\frac{7.48}{9}\),cos α=\(\frac{OP}{PQ}=\frac{5}{9}\)
Now,∑Fx = 0, RQ= μRP
∑Fx = 0, RP= ω = 30 N
Taking moments about ‘P’,
RQ× OQ= 30 × PS cos α
RQ×7.48=30×9⁄2×5⁄9
⇒RQ=\(R_{Q}=\frac{30\times 9\times 5}{2\times 9\times 7.48}=\frac{1350}{134.64}= 10.03 N\)
Now,μ = \(\frac{R_{Q}}{R_{P}}=\frac{10.03}{30}=0.33\)
Now, Frictional force (F) = μRP= 0.33 × 30 = 10.03 N
In the figure given below, if the speed of the input of shaft of spur gear train is 2400 rpm and the speed of the output shaft is 100 rpm, then the module of gear 4 will be_____
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Solution
Using the following formula,Centre distance between two gears (A& B)
\(\frac{m(T_{A}+T_{B})}{2}\)............ (1)
Now, for gears 1 and 2, using abve equations,
35=\(\frac{1(60+T_{2})}{2}\)
⇒70 = 60 + T2 ⇒ T2 = 10
Now,\(\frac{N_{4}}{N_{2}}=\frac{T_{2}}{T_{1}}\times \frac{T_{3}}{T_{4}}\)
⇒\(\frac{1000}{2400}=\frac{10}{60}\times \frac{10}{T_{4}}\)
T4=\(\frac{2400}{60}=40\)
For gears 3 and 4, using equations (1),
35=\(\frac{m(T_{3}+T_{4})}{2}=\frac{m(10+40)}{2}\)
m=\(\frac{35\times 2}{50}=\frac{35}{25}=\frac{7}{5}=1.4\)
It is given that the actual demand is 59 unit, previous forecast 64 unit and smoothing factor 0.3. The forecast for next period will be ________
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Solution
Forecasting formula :
Ft+1= α Dt+ (1 – α) Ft
Given : α= 0.3, Dt= 59, Ft= 64
Ft+1= 0.3 × 59 + (1 – 0.3) × 64
= 17.7 + 44.8 = 62.5
Process A has fixed cost of ₹40,000 and variable cost of ₹9/unit and the process B has fixed cost of ₹16,000 and variable cost of ₹24/unit. The production quantity at which total costs of A and B becomes equal will be––––––––
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Solution
Total cost of process A = Fixed cost + cost/piece × N
= 40,000 +9N............ (1)
Total cost of process B = 16,000 + 24 N............ (2)
Now, when total costs of A and B becomes equal, then equating (1) and (2),
40,000 + 9N= 16,000 + 24N
24,000 = 15N
N=\(\frac{24000}{15}=1600\)
In a machine tool gear box, the smallest and largest spindles are 100rpm and 1120 rpm respectively. If the reare 8 speeds in all, the fourth speed will be ––––––––
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Solution
The fourth speed will be =\(\frac{1120}{8}\times 2=280\, rpm\)
A MS plate 400 × 800 × 30 mm3 is to be shaped along its wider face. The ratio of return time to cutting time is 2 :3 and the feed/cycle is 2 mm. Tool approach and over travel are 50 mm each and cutting speed of tool is 24m/min. The machining time required for machining will be––––––––
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Solution
Length of stroke = l = 50 + 50 + 800 = 900 mm
Time taken for cutting stroke =\(\frac{0.96}{24}\)= 0.04 m/min.
Return time : cutting time = 2 : 3 (given)
So, return time = 23× cutting time
=2⁄3× 0.04 = 0.0267 min.Cycle time= Time taken for cutting stroke + Return time
= 0.04 + 0.0267 = 0.067 min.Width for shaping = 410 mm
Number of cycles =\(\frac{410}{2}\)= 205
Now,total time taken for machine
= 205 × 0.067 = 13.74 min
Match List-I with List-II.
List-I List-II
A.Fluorspar 1.BCC structure
B.Alpha-iron 2.HCP structure
C.Silver 3.SC structure
D.Zinc 4.FCC structure
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Solution
Fluorspar ⇒ SC
α-iron ⇒ BCC
Silver ⇒ FCC
Zinc ⇒ HCP