GATE MT 2016 Q.24 Aluminium powder having an apparent density of 810 kg.m^-3 is compacted in a cylindrical die at 600 MPa. The density of the as-pressed aluminium compact is 1755 kg.m^-3. If the height of the as-pressed compact is 12 mm, the fill height (in mm) required is \_\_\_\_\_\_.
Correct answer is 26 mm. Mass of powder is constant. mi=mf  ⇒ ρiVi = ρfVf ρihi = ρf hf ⇒ hi = (ρf hf)/ρi On substitution, (1755×12)/ 810= 26 mm  


Correct answer is 26 mm.


Mass of powder is constant.


mi=mf ⇒ ρiVi= ρfVf


ρihi = ρf hf⇒ hi = (ρf hf)/ρi


On substitution, (1755×12)/ 810= 26 mm


 


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radius = 1.75 m
height (h) = 12 m
volume = 
= ×  × 1.75 × 1.75 × 12
= 38.5 m3
since, 1 m3 = 1 kiloliter
capacity of the pit = (38.5 × 1)
= 38.5 kilolitres Correct option is A) 9.378 m2


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result must have significant figures equal to least figures being multiplid do area of metal sheet 9.378 m2  (four significant digits) given radius of cone (r) = 3.5/2 m height of cone (h) = 12 m find out we have to find the capacity in kiloleters solution volume of cone = (1/3) × πr2h on taking π=22/7 = (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 = (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2) = 38.5 m3 capacity of conical pit in kilo liters: 1000 m3 = 1 liter 38.5 m3 = 1000 × 38.5 liters = 38,500 liters = 38.5 kilo liters answer the capacity of the container is 38.5 kilo liters v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> given: volume of cylinder = 1617 cm3let r be the radius of base and h be the height of coner:h = 2:3∴ r/h = 2/33r = 2hh = 3r/2 …(i)formula: volume of cylinder = πr2h∴ 1617 = (22/7) r2hr2h = 514.5using (i) we have∴ r2 × (3r/2) = 514.53r3 = 1029r3 = 343r = 7 cmh = 21/2 cmtotal surface area of cylinder = 2πr2 + 2πrh= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)= 308 + 462= 770 cm2therefore total surface area of cylinder = 770 cm2 we are asked to determine the minimum mass of the ore containing calcium phosphate (ca3(po4)2) that must be processed to obtain 1.00 kg of phosphorus (p). to solve this problem, we shall follow these steps:step 1: convert grams to moles of phosphorus using the molar mass of phosphorus.step 2: perform a mole-to-mole comparison between phosphorus(p) and calcium phosphate (ca3(po4)2), 2 mol p: 1 mol ca3(po4)2.step 3: compute for the mass of calcium phosphate (ca3(po4)2) using the obtained moles of calcium phosphate and the molar mass of calcium phosphate (ca3(po4)2). step 4: convert the mass from g to kg. take into account the percentage given in the problem to determine the number of kilograms of the ore that is required.[readmore]step 1: convert grams to moles of phosphorus using the molar mass of phosphorus. we are given:mass of phosphorus = 1.00 kgrecall:1 kg = 103 g molar mass of p (from the periodic table) = 30.97 g/molmole of phosphorus=1.00 kg p(103 g p1 kg p)(1 mol p30.97 g p)mole of phosphorus = 32.289 molstep 2: perform a mole-to-mole comparison between phosphorus(p) and calcium phosphate (ca3(po4)2), 2 mol p: 1 mol ca3(po4)2.mole of ca3(po4)2 =32.289 mol p(1 mol ca3(po4)22 mol p)mole of ca3(po4)2 = 16.145 molstep 3: compute for the mass of calcium phosphate (ca3(po4)2) using the obtained moles of calcium phosphate and the molar mass of calcium phosphate (ca3(po4)2). the molar mass of calcium phosphate (ca3(po4)2) is:ca3(po4)2          3 × 40.078 g/mol ca = 120.234 g/mol                          2 × 30.97 g/mol p = 61.94 g/mol                          8 × 16.00 g/mol o = 128 g/mol                         sum = 310.174 g/molmass of ca3(po4)2=16.145 mol ca3(po4)2(310.174 g ca3(po4)21 mol ca3(po4)2)mass of ca3(po4)2 = 5,007.759 g step 4: convert the mass from g to kg. take into account the percentage given in the problem to determine the number of kilograms of the ore that is required.recall:1 kg = 103 g percent (%) = per 100 (i.e. 57.1 kg calcium phosphate per 100 kg ore)mass of ca3(po4)2=5,007.759 g ca3(po4)2(1 kg ca3(po4)2103 g ca3(po4)2)(100 kg ore 57.1kg ca3(po4)2)mass of ore = 8.77 kgthe mass of ore, containing calcium phosphate, needed to obtain 1.00 kg of phosphorus is 8.77 kg. Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm ?solution:
step 1 of 2:
the intermediate pressure in a compressor is given by . the relative uncertainty of and the relative uncertainty of is 2% . we have to find the relative uncertainty in .


\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_




step 2 of 2:


the intermediate pressure in a compressor is = ((


it is given that the relative uncertainty in , = 0.05
 
 relative uncertainty in , = 0.02


so the relative uncertainty in the intermediate presser 




 = 


 = 
 
 = 0.0269


 therefore the relative uncertainty in the intermediate presser is 0.0269 mpa.
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ mass (molar mass) → mol (avogadro's number) → molecules[readmore]3.51×104 g ×1 mol163.94 g×6.022×1023 molecules1 mol = 1.29x1026 moleculesthere are 1.29x1026 molecules in the sample given,diameter of pencil = 7 mmradius of pencil = diameter of graphite = 1 mmradius of graphite = volume of graphite = πr2h = weight of graphite = volume × specific gravity of graphite= volume of wood = volume of pencil – volume of graphite= = ∴ total weight of the pencil = weight of wood + weight of graphite= 0.165 + 2.64 = 2.805 g.