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## Machine Design MCQ Set 1

1. The Concentration of the two phases in a closed system at the Interphase is
a) Changes continuously
b) Never changes
c) Becomes zero
d) Increases till the driving force becomes zero

Answer: b [Reason:] The concentration changes only if the component of the two phases added or removed. Generally, Interphase occurs at equilibrium. Once the additional component is added to a system at equilibrium, the concentration changes till it become uniform but it will be differ from the previous.

2. Diffusion of components between the phases at equilibrium is
a) Zero
b) Infinity
c) Changes continuously
d) Diffusion never occurs

Answer: a [Reason:] At equilibrium, the concentration becomes uniform so the rate of diffusion stops.

3. Consider a steady-state condition; the concentration at any point in the equipment never changes with time.
a) True
b) False

Answer: a [Reason:] Generally at a steady state, the net transfer remains same so the concentration will be same at every point.

4. Find the x & y co-ordinate in the figure below at steady state. Here, the gas phase and liquid phase mole fraction representation is given below a) x- distance & y- concentration
b) x-time & y- concentration
c) x-concentration & y- time
d) x-concentration & y-distance

Answer: a [Reason:] Generally, the concentration changes with distance but never changes with time at steady state. From the concept of driving force, the y- coordinate should be concentration.

5. The real driving force of the mass transfer is
a) Chemical potential
b) Physical potential

Answer: a [Reason:] Chemical potential represents the dynamic equilibrium of the mass transfer.

6. According to Lewis and Whitman theory, the departure from concentration equilibrium at the Interphase is due to
a) Low mass transfer rates
b) High mass transfer rates
c) Moderate mass transfer rate
d) None of the mentioned

Answer: b [Reason:] Theoretically proved by Lewis and Whitman, that if the mass transfer rates are higher the concentration deviates from equilibrium.

7. At the Interphase, the Interphase concentration of the component in both phase phases have equal chemical potential is due to differential concentration.
a) True
b) False

Answer: b [Reason:] The concentration rise at the Interphase is not a barrier to the diffusion. They are equilibrium concentration which is similar to equal chemical potential.

8. The equilibrium concentrations in the gas and the liquid phases, in mole fraction, give rise to a curve known as
a) Equilibrium distribution curve
b) Equilibrium concentration curve
c) Differential distribution curve
d) Differential concentration curve

Answer: a [Reason:] The equilibrium distribution curve represents the phase-phase equilibrium curve with the co-ordinates of mole fractions in both the phases.

## Machine Design MCQ Set 2

1. Wood ash leaching for alkali is known as ________
a) Lixivation
b) Lixartion
c) Lixation
d) None of the mentioned

Answer: a [Reason:] lixivation is common leaching process used for removing alkali from wood ashes.

2. The boiling point of the solvent is ______
a) Boiling solvent
b) Decoction
c) Leaching
d) None of the mentioned

Answer: b [Reason:] To get the leached product the solvents are used at its boiling point is known as decoction.

3. The removal of soluble materials from the solid is known as ________
a) Elution
b) Decoction
c) Extraction
d) None of the mentioned

Answer: a [Reason:] Generally elution is the process of removing soluble materials from the solids.

4. Leaching is generally used for cement industries
a) True
b) False

Answer: b [Reason:] Leaching generally used for metallurgical industries for ore removal purposes.

5. For getting leached product from copper ore it is mandatory to add hydrochloric acid to the ore.
a) True
b) False

Answer: b [Reason:] For getting leached product from the ore we have to add sulphuric acid to the copper ore.

6. Find the process a) Leaching
b) Extraction
c) Evaporation
d) None of the mentioned

Answer: a [Reason:] Here the sugar gets leached from the sugar beet with help of solvent water.

7. Which of the following process can accelerate leaching?
a) Heating
b) Drying
c) Crushing
d) Crushing or grinding

Answer: d [Reason:] Since crushing or grinding can result in removal of more solute from the solids.

8. The process elution is also known as elutriation.
a) True
b) False

Answer: a [Reason:] The process of removing the soluble materials from the solid.

9. Temperature of the solvent be low.
a) True
b) False

Answer: b [Reason:] The temperature of the solvent should be high in order to get more solubility of solute.

10. If the solvent viscosity is low and temperature of the solvent is high then the leaching rate decreases.
a) True
b) False

Answer: b [Reason:] The leaching rate increases with diffusivity since the viscosity is low and the temperature is high.

11. In order to reduce the time required for the removing solute the sugar beets are cut into slices known as
a) Cossettes
b) Coset
c) Consot
d) All the mentioned

Answer: a [Reason:] The cut portions can improve the solute removal. The cut sets are named cossettes.

## Machine Design MCQ Set 3

1. What is the melting point of chromium?
a) 2118 K
b) 2218 K
c) 2318 K
d) 2418 K

Answer: a [Reason:] It is a chemical element with atomic number 24. It is a steely-gray, lustrous, hard and brittle metal and has a high melting point.

2. What is the melting point of cobalt?
a) 1449 K
b) 1559 K
c) 1669 K
d) 1769 K

Answer: d [Reason:] It is found only in chemically combined form. It occurs as two crystallographic structures i.e. hcp and fcc.

3. Which one is having highest melting point?
a) Bismuth
b) Beryllium
d) Copper pure

Answer: b [Reason:] Melting point of beryllium is 1550 k while that of bismuth, cadmium and pure copper are 545 K, 594 K and 1358 K respectively.

4. What is the melting point of germanium?
a) 1211 K
b) 1311 K
c) 1411 K
d) 1511 K

Answer: a [Reason:] It is a lustrous, hard, grayish-white metalloid in the carbon group. It is a semiconductor which reacts and form complexes with oxygen in nature.

5. Which one is having minimum melting point?
a) Tin
b) Iron pure
c) Germanium
d) Gold

Answer: c [Reason:] Melting point of germanium is 1211 K while that of tin, pure iron and gold are 1505 K, 1810 K and 1336 K respectively.

6. What is the melting point of nichrome?
a) 1572 K
b) 1672 K
c) 1772 K
d) 1872 K

Answer: b [Reason:] Nichrome generally refers to an alloy of nickel, chromium and iron. Nichrome alloys are used in resistance wire.

7. What is the melting point of palladium?
a) 1527 K
b) 1627 K
c) 1727 K
d) 1827 K

Answer: d [Reason:] It is a rare and lustrous silvery-white metal. Palladium has the lowest melting point and is the least dense.

8. Which one is having maximum melting point?
b) Magnesium
c) Platinum pure

Answer: c [Reason:] Melting point of pure platinum is 2045 K while that of palladium, magnesium and lead are 1827 K, 923 K and 601 K respectively.

9. What is the melting point of tin?
a) 505 K
b) 605 K
c) 705 K
d) 805 K

Answer: a [Reason:] It is a main group metal in group 14 of the periodic table. It has 10 stable isotopes, the largest number in the periodic table.

10. Which one is having minimum melting point?
a) Titanium
b) Zinc
c) Uranium
d) Rhodium

Answer: b [Reason:] Melting point of zinc is 693 K while that of titanium, uranium and rhodium are 3269 K, 1406 K and 2236 K respectively.

## Machine Design MCQ Set 4

1. __________ energy helps for molecular diffusion.
a) Thermal
b) Electrical
c) Internal
d) All of the mentioned

Answer: a [Reason:] Thermal energy is heat. Heat can cause diffusion among molecules.

2. For two particles at steady state collision of A and B, J flux of A is 2 mol/sq.m sec find the J flux of B.
a) 1
b) 2
c) 3
d) None of the mentioned

Answer: d [Reason:] At steady state collision, J flux of A = – J flux of B = -2.

3. Estimate the mole fraction if concentration of A is 2 mol/cu.m and the total concentration is 5 mol/cu.m.
a) 0.2
b) 0.25
c) 0.4
d) 1

Answer: c [Reason:] xA = cA/C = 2/5 = 0.4.

4. Find the partial pressure of A if the total pressure is 2 atm; Concentration of A is 2 mol/cu.m and total concentration is 5 mol/cu.m.
a) 0.2 atm
b) 0.4 atm
c) 0.6 atm
d) 0.8 atm

Answer: d [Reason:] yA= 2/5 = 0.4 pA = 0.4 x 2 = 0.8 atm.

5. From the reaction, Methane –> Carbon + 2(Hydrogen molecule) find the ratio of N Flux of A to the Total N flux( A+B); Given N Flux of B =-2 N flux of A.
a) 1
b) -1
c) 2
d) -2

Answer: b [Reason:] N flux of A/ Total flux = N flux of A/ N flux of A – 2 N flux of A = -1.

6. For a Steady state diffusion of A through Non- diffusing B. N flux of B= ______
a) 0
b) 1
c) 2
d) Infinity

Answer: a [Reason:] Non- diffusing B means N flux of B = 0.

7. The value of N flux of A under steady state non-diffusing B with diffusing A is _________
a) 0
b) Constant
c) Infinity
d) None of the mentioned

Answer: b [Reason:] Diffusing A with non diffusing B means Flux of A remains Constant while the Flux of B is 0.

8. If the value of N flux of A to the total N flux is 1 then it is a steady state diffusion of non-diffusing B.
a) True
b) False

Answer: a [Reason:] N Flux of A/ Total N flux For N flux of B = 0 then N flux of A/N flux of A = 1.

## Machine Design MCQ Set 5

1. Before collision, the distance between the two particles is mean free path.
a) True
b) False

Answer: b [Reason:] The mean free path can be found after collision.

2. Diffusion co-efficient in molecular diffusion is estimated by _________
a) Daltons law
b) Diffusion law
c) Ficks law
d) None of the mentioned

Answer: c [Reason:] Ficks law is the ratio of J-flux to the concentration gradient.

3. Find the total flux for a particle A and B in a steady state if flux of A and B is 2.44 and 4.44 mol/sq.m sec.
a) 2
b) 4.88
c) 6.88
d) 8

Answer: c [Reason:] Total flux= flux A+ flux B= 6.88.

4. Find the flux of A ( xA= 0.2) if Total flux is 5 mol/sq.m sec and J flux of A is 2 mol/sq.m sec.
a) 2
b) 2.5
c) 3
d) 3.5

Answer: c [Reason:] N flux of A = J flux of A + Total flux N * xA = 2+ 0.5*2 = 3.

5. Find J flux of A ( xA = 0.5) if the total N flux is 6 mol/sq.m sec and N flux of A is 3 mol/sq.m sec.
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] N flux of A = J flux of A + Total flux N * xA 3= J flux of A + 6 * 0.5 J flux = 0.

6. Diffusivity of the liquids can be determined by
a) Wilke- lee equation
b) Wilke- chan equation
c) Lee and chan equation
d) None of the mentioned

Answer: b [Reason:] Wilke- Chan derived an equation for diffusivity of liquids.

7. For the molecular diffusion of liquids, if the diffusing molecules having criteria of steady state diffusion of B over non diffusing A then the N flux of A is
a) 1
b) 0
c) Negative
d) Infinity

Answer: b [Reason:] For non-diffusing A then N flux of B is zero.

8. For liquid molecular diffusion of A and B, steady state equimolar counter diffusion the N flux of A is negative of N flux of B.
a) True
b) False

Answer: a [Reason:] For equimolar counter diffusion, the one component flux is negative of other.

9. Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec.
a) 0.67
b) 1.67
c) 2.67
d) None of the mentioned