At a 200 mm tensile gauge length, a specimen breaks at a grip separation of 232 mm. Find elongation at break.
A 13.8%
B 116%
C 16%
D 32%
Answer: C. 16%. The extension is 32 mm, and the initial length is 200 mm. The ratio is 0.16.
Why does a hydraulic burst result require a diaphragm pressure correction?
A The warp changes linear density
B The specimen gains water mass
C The pressure gauge measures colour
D The diaphragm resists distension
Answer: D. The diaphragm resists distension. Part of the indicated pressure is needed to distend the diaphragm alone. That contribution is removed to isolate specimen resistance.
Find the moisture regain if a specimen weighs 54 g before oven drying and 50 g once dry.
A 8%
B 7.41%
C 108%
D 4%
Answer: A. 8%. The moisture mass is 4 g and the dry mass is 50 g. The regain is 4/50 × 100%.
In ring spinning trials, thin-place counts were recorded over different lengths. What reporting basis allows a fair comparison?
A Defects per fixed length
B Total defects only
C Mass of all packages
D Difference in package diameter
Answer: A. Defects per fixed length. Normalising counts to a common length compares defect frequency across runs of different lengths.
If 50 m of yarn has a mass of 1.20 g, determine its linear density.
A 12 tex
B 24 tex
C 60 tex
D 2.4 tex
Answer: B. 24 tex. A length of 50 m is 0.05 km. Dividing 1.20 g by 0.05 km gives 24 tex.
For ideal CIELAB Euclidean colour difference, ΔL* = 3, Δa* = 4 and Δb* = 0. What is ΔE*ab?
A 25
B 5
C 12
D 7
Answer: B. 5. The Euclidean distance is the square root of 3² + 4² + 0².
Repeated tensile forces are 42 N, 45 N, 47 N, 48 N and 53 N. What is their range?
A 47 N
B 8 N
C 95 N
D 11 N
Answer: D. 11 N. The range is the maximum minus the minimum: 53 − 42 = 11 N.
In a fabric bow measurement, the weft line deviates 18 mm across a 1.2 m width. What is bow as a percentage of width?
A 15%
B 6.67%
C 0.15%
D 1.5%
Answer: D. 1.5%. Divide the maximum deviation by the width in matching units: 18/1,200 × 100%.
For a yarn of 25 tex that breaks at 4 N, calculate breaking tenacity.
A 16 cN/tex
B 0.16 cN/tex
C 6.25 cN/tex
D 100 cN/tex
Answer: A. 16 cN/tex. Convert 4 N to 400 cN and divide by 25 tex.
Mass transfer is 0.11 g/h through 0.010 m² of exposed fabric. Find the hourly water vapour transmission rate per area.
A 0.11 g/(m²·h)
B 11 g/(m²·h)
C 1.1 g/(m²·h)
D 110 g/(m²·h)
Answer: B. 11 g/(m²·h). Divide the measured hourly mass transfer by the exposed area, 0.010 m².
Length between wash marks falls from 500 mm to 480 mm. Report the percentage shrinkage.
A 20%
B 0.4%
C 4%
D 4.17%
Answer: C. 4%. Length decreases by 20 mm relative to the 500 mm original span.
To compare results from two yarn evenness instruments, which experimental setting should remain consistent?
A Evaluated test length
B Number of unused bobbins
C Machine case colour
D Instrument desk height
Answer: A. Evaluated test length. A reported unevenness value depends on sampling and test conditions, including the evaluated yarn length.
Tested fabric areas are unequal. Before comparing abrasion mass losses as percentages, what should each loss be divided by?
A Its initial specimen mass
B The average cycle count
C The largest specimen area
D Its final specimen mass
Answer: A. Its initial specimen mass. Percentage mass loss is computed relative to the original mass of each specimen.
Test readings from five specimens are 30, 32, 36, 38 and 39 N. The method requires reporting the median. What value goes in the report?
A 38 N
B 32 N
C 35 N
D 36 N
Answer: D. 36 N. The median is the middle of the five ordered readings, which is 36 N.
Under identical measurement conditions, which CIELAB coordinate change indicates a lighter result?
A Lower b*
B Higher a*
C Lower L*
D Higher L*
Answer: D. Higher L*. L* is the lightness coordinate. Higher L* corresponds to greater lightness in the specified measurement.
Counting 85 yarn turns in a 250 mm test length gives what twist level?
A 425 turns/m
B 340 turns/m
C 85 turns/m
D 21.25 turns/m
Answer: B. 340 turns/m. The test length is 0.25 m. The count divided by length is 85/0.25.
During a pilling assessment, why should specimens be judged under specified lighting conditions?
A Remove surface fuzz
B Reduce fabric mass
C Increase pilling while rating
D Make visual grades comparable
Answer: D. Make visual grades comparable. Controlled illumination improves consistency when comparing specimens with visual reference standards.
Under fixed pressure difference, 40 cm³/s flows through 20 cm² of fabric. What is volume flow per unit area?
A 0.5 cm³/(cm²·s)
B 800 cm³/(cm²·s)
C 2 cm³/(cm²·s)
D 20 cm³/(cm²·s)
Answer: C. 2 cm³/(cm²·s). The flow normalised by area is 40/20 in the given units.
Unlike regain, moisture content is based on total moist mass. Calculate content for a 40 g moist specimen containing 2 g of water.
A 5.26%
B 10%
C 5%
D 2%
Answer: C. 5%. Water accounts for 2/40 of the total moist mass, or 5%.
Two thickness tests use a 200 mm² foot. The first applies 0.10 N and the second 0.20 N. What pressure does the second apply?
A 10 kPa
B 1.0 kPa
C 0.5 kPa
D 0.1 kPa
Answer: B. 1.0 kPa. Pressure is force divided by area. The second test applies 0.20 N to 0.0002 m², or 1,000 Pa.
Imagine sampling only the first metre of every fabric roll. Which defect in the sampling plan is most likely?
A Poor lengthwise coverage
B Excessive warp density
C Too many colour readings
D Excessive operator bias correction
Answer: A. Poor lengthwise coverage. Measurements from the same end location may miss changes along the roll.
Using a 100 cm² cutter, a circular swatch weighs 1.85 g. What is the fabric areal density?
A 1.85 g/m²
B 18.5 g/m²
C 1,850 g/m²
D 185 g/m²
Answer: D. 185 g/m². A specimen of 100 cm² covers 0.01 m². Divide 1.85 g by that area.
Repeated grip slippage causes specimens to break beside the jaws. What is the best initial response?
A Shorten the report
B Lower the result manually
C Check gripping and retest
D Average those results
Answer: C. Check gripping and retest. Slippage and grip damage can invalidate measured extension and force. Correct the grip conditions and repeat according to the method.
Length and width changes differ after the same laundering treatment. How should the result be reported?
A Only the larger dimension
B The average specimen mass
C Separate values by direction
D One shared shrinkage value
Answer: C. Separate values by direction. Textile dimensions can change differently lengthwise and widthwise. Reporting both values preserves that distinction.
When comparing abrasion to a defined visual endpoint, what response variable is commonly recorded?
A Tear angle
B Cycles to endpoint
C Initial fabric width
D Fibre length
Answer: B. Cycles to endpoint. The result counts the rubbing cycles required before the specified appearance or damage endpoint occurs.