Two finishing routes must leave the same dry add-on. One uses 80% wet pickup and the other 50%. If no finish is lost, by what factor must the lower-pickup bath concentration increase?
A 0.625
B 1.30
C 2.30
D 1.60
Answer: D. 1.60. Dry add-on scales with pickup times solids fraction. Maintaining it requires a concentration ratio of 0.80/0.50.
At steady operation, fabric enters a compactor at 22 m/min and leaves at 20 m/min with no slip. Find length reduction relative to entering cloth.
A 9.09%
B 10.00%
C 2.00%
D 90.91%
Answer: A. 9.09%. Each minute, 22 m enters and 20 m leaves. Relative reduction is 2/22.
Water penetrates a finished fabric at 9.81 kPa. Convert to hydrostatic head using 1,000 kg/m³ for density and 9.81 m/s² for g.
A 0.10 m
B 1.00 m
C 0.01 m
D 9.81 m
Answer: B. 1.00 m. From p = ρgh, head is 9,810 Pa divided by 1,000 × 9.81.
From a nominal 4% solids add-on, 20% of the deposited solids is lost during rinsing. What dry add-on remains on the dry fabric basis?
A 2.0%
B 3.2%
C 4.8%
D 0.8%
Answer: B. 3.2%. The retained share is 80% of 4%, yielding 3.2%.
Line speed increases from 18 to 24 m/min, and heated length rises from 15 to 18 m. How does residence time change?
A It decreases by 10%
B It decreases by 25%
C It increases by 20%
D It stays constant
Answer: A. It decreases by 10%. Original residence is 15/18 minute. New residence is 18/24 minute, or 90% of the original.
Measurements show equal overall finish mass but declining repellency after washing. What additional comparison helps distinguish finish redistribution from gross wash-off?
A Change base fabric composition
B Record dye-bath pH alone
C Compare dry yarn counts only
D Assess surface finish distribution
Answer: D. Assess surface finish distribution. Equal total mass does not show where finish remains. Surface distribution measurements can detect changes in functional coverage.
Over six hours, 45 g of vapour passes through 0.075 m² of membrane. At a constant rate, what is daily transmission per square metre?
A 7.5 g/(m²·day)
B 180 g/(m²·day)
C 2,400 g/(m²·day)
D 600 g/(m²·day)
Answer: C. 2,400 g/(m²·day). The daily equivalent is 45 × 4 = 180 g over the test area. Divide by 0.075 m².
Dry finish adds 3% to mass while fabric area contracts by 5%. What is the final-to-initial areal density ratio?
A 1.030
B 1.050
C 0.978
D 1.084
Answer: D. 1.084. Areal density changes by the mass ratio divided by the area ratio: 1.03/0.95 ≈ 1.084.
The acceptable cure time is at least 72 seconds in a 21.6 m hot zone. What is the highest allowable steady line speed by residence time alone?
A 21.6 m/min
B 18 m/min
C 15 m/min
D 30 m/min
Answer: B. 18 m/min. Seventy-two seconds is 1.2 minutes. Divide 21.6 m by 1.2 min.
For a finish that must repel external water and carry sweat, which measurement pair shows a tradeoff?
A Repellency rises, wicking falls
B Both desired properties improve
C The dye recipe stays fixed
D Only yarn length is reduced
Answer: A. Repellency rises, wicking falls. Improved resistance to wetting can coincide with slower movement of liquid sweat. Both properties must be measured for the application.
During finishing, 180 g/m² fabric contracts 6% in length and grows 2% in width. What is the new areal density if dry mass is unchanged?
A 194.4 g/m²
B 180.0 g/m²
C 187.7 g/m²
D 172.6 g/m²
Answer: C. 187.7 g/m². The area factor is 0.94 × 1.02 = 0.9588. Divide 180 g/m² by that factor.
Surface resistance drops from 10¹² Ω to 10⁹ Ω after treatment. By what factor did resistance decrease?
A 1,000
B 10,000
C 100
D 10
Answer: A. 1,000. Divide 10¹² by 10⁹ to obtain 10³, or 1,000.
Dry fabric runs at 400 kg/h and retains liquor at 65% wet pickup. The liquid is 12% solids by mass. Find the drying water load if all water evaporates.
A 31.2 kg/h
B 352.0 kg/h
C 228.8 kg/h
D 260.0 kg/h
Answer: C. 228.8 kg/h. Retained liquor is 260 kg/h, of which 88% is water. Multiply 260 by 0.88.
Breaking force is 1,200 cN after resin finishing and 1,500 cN before it. What percentage of the original strength is retained?
A 80%
B 20%
C 125%
D 25%
Answer: A. 80%. Retention is the treated value divided by the starting value, 1,200/1,500.
Two chemical softeners initially lower bending rigidity by equal amounts, but only one retains the reduction after five washes. Which inference is supported by the data?
A One is more wash-durable
B Both are permanently identical
C One causes no moisture uptake
D One contains no polymer
Answer: A. One is more wash-durable. Retention of the property after the same care cycles shows greater durability under that test.
At 2.4 MJ per kilogram evaporated, how much heat per hour is saved if the dryer water load falls by 80 kg/h?
A 80 MJ/h
B 240 MJ/h
C 192 MJ/h
D 33.3 MJ/h
Answer: C. 192 MJ/h. At unchanged specific energy, multiply 80 kg/h by 2.4 MJ/kg.
Find the final length of 100 m of fabric after 4% compaction followed by 2% residual shrinkage during washing.
A 94.00 m
B 96.00 m
C 98.00 m
D 94.08 m
Answer: D. 94.08 m. The remaining length is 100 × 0.96 × 0.98, which equals 94.08 m.
Four cure trials use two temperatures and two times. Which design distinguishes effects of both factors and a possible interaction?
A Only hottest and longest
B Change both together only
C One treatment without repeats
D All temperature–time combinations
Answer: D. All temperature–time combinations. Testing each temperature at each time makes it possible to assess how the factors operate separately and together.
For 180 kg of dry textile and a 3% dry add-on, how much 40%-solids stock supplies all required finish solids?
A 18.0 kg
B 5.4 kg
C 13.5 kg
D 7.2 kg
Answer: C. 13.5 kg. The cloth needs 5.4 kg dry solids. Divide by the 0.40 solids fraction.
Overfeed is defined here as (feed speed − chain speed)/chain speed. Chain speed is 24 m/min and overfeed is 5%. Find feed speed.
A 24.5 m/min
B 22.8 m/min
C 25.2 m/min
D 29.0 m/min
Answer: C. 25.2 m/min. Feed speed is 1.05 times the chain speed, or 25.2 m/min.
During finishing, adding a selective enzyme keeps mechanical conditions fixed but increases mass loss markedly while fuzz is reduced. Which control best confirms that the enzyme caused the extra loss?
A Larger sample with more enzyme
B Matched bath without enzyme
C Different fibre and bath
D Higher speed and pressure
Answer: B. Matched bath without enzyme. Comparing a matched treatment that omits only the enzyme isolates its contribution to mass loss and fuzz reduction.
A finish improves crease recovery by 80° and retains 52° of that improvement after five washes. What fraction of the first gain remains?
A 35%
B 52%
C 80%
D 65%
Answer: D. 65%. Divide the retained improvement by the initial improvement: 52/80.
Suppose coating thickness doubles while intrinsic vapour permeability stays the same. At fixed vapour pressure difference, what happens to steady diffusion flux?
A It stays unchanged
B It doubles
C It quadruples
D It halves
Answer: D. It halves. Under that model, flux is inversely proportional to transport path thickness for fixed permeability and driving force.
During pad application, left pickup is 60% and right pickup is 75%, with uniform bath solids. What is the right-to-left ratio of dry add-on before losses?
A 1.15
B 1.25
C 0.80
D 1.35
Answer: B. 1.25. Dry add-on is proportional to pickup at the same solids concentration. The ratio is 75/60.
Doubling a finish bath’s volume leaves the active chemical mass unchanged. With unchanged pickup and liquor density, what happens to expected dry add-on?
A It doubles
B It halves
C It stays fixed
D It becomes zero
Answer: B. It halves. At constant pickup, add-on scales with bath solids concentration. Doubling volume halves concentration in the stated model.