GATE (TF) Textile 2008 Question Paper Solution | GATE/2008/TF/73

Common Data Questions

A cotton fabric has 25 ends per cm, 28 picks per cm, warp count 30 tex, weft count 15 tex, warp crimp 12%. The diameter of the yarn is given by; 4.44 x 10^-3 (yarn tex/fibre density)^1/2

Question 73 (Textile Engineering & Fibre Science)

The fabric thickness (mm) will be

(A)	0.04
(B)	0.36
(C)	3.62
(D)	13.62

[Show Answer]

Option B is correct
Given in the question-

Ends per cm(EPCM)=25

Picks per cm(PPCM)=28

Warp count=30 tex

Weft count=15 tex

Warp crimp=12%

By formula-

$Fabric thickness(t)=h_1+d_1=h_2+d_2$

$Fabric thickness(t)=h_1+d_1$

$Fabric thickness(t)=\frac{136 \times \sqrt Warp crimp percentage}{ppcm\times 2.54} \times 1000 \times 2.54+19.58 \times 10^-3$

d1 as calculated in question no. 71.

$Fabric thickness(t)=\frac{136 \times \sqrt 12}{28\times 2.54 \times 1000} \times 2.54+19.58 \times 10^-3$

Fabric thickness(t)=0.03640+0.01958

Fabric thickness(t)=0.036 cm

Fabric thickness(t)=0.36 mm (Ans)

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