GATE (TF) Textile 2008 Question Paper Solution | GATE/2008/TF/71

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 71 (Textile Engineering & Fibre Science)

The ratio of warp way modular length to sum of thread diameter is

(A)	0.119
(B)	0.19
(C)	1.19
(D)	11.9

[Show Answer]

Option C 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%

The ratio of warp way modular length to sum of thread diameter=?

Let the warp way modular length=l1

Sum of diameter=D

Sum of diameter(D)=Warp diameter(d1)+Weft diameter(d2)

$d_1=4.44 \times 10^-3 \times \sqrt \frac{Warp count}{Density of cotton fibre}$ cm

$d1=4.44 \times 10^-3 \times \sqrt \frac{30}{1.54}$

$d1=4.44 \times 10^-3 \times \sqrt 19.48$

$d1=4.44 \times 10^-3 \times 4.41$

$d1=19.58 \times 10^-3$ cm

Similarily-

$d_2=4.44 \times 10^-3 \times \sqrt \frac{Weft count}{Density of cotton fibre}$ cm

$d_2=4.44 \times 10^-3 \times \sqrt \frac{15}{1.54}$ cm

$d_2=4.44 \times 10^-3 \times \sqrt 9.74$ cm

$d_2=4.44 \times 10^-3 \times 3.12$ cm

$d_2=13.94\times 10^-3$ cm

Sum of diameter(D)=Warp diameter(d1)+Weft diameter(d2)

$D=19.58 \times 10^-3+13.94\times 10^-3$ cm

$D=33.52 \times 10^-3$ cm

By formula-

Warp crimp%=(l1-P2) x 100/P2

Where,
P2 is the pick spacing=1/PPCM

P2=1/28

$12=\frac {(l_1-\frac{1}{28}) \times 100}{\frac{1}{28}}$

$12 \times \frac{1}{28} =(l_1-\frac{1}{28}) \times 100$

$\frac{12}{28} =(l_1-\frac{1}{28}) \times 100$

$\frac{0.12}{28} =(l_1-\frac{1}{28})$

$\frac{0.12}{28}+\frac{1}{28} =l_1$

$l_1=0.04$ cm

Now,

The ratio of warp way modular length to sum of thread diameter= $\frac{l_1}{D}$

The ratio of warp way modular length to sum of thread diameter= $\frac{0.04}{33.52 \times 10^-3 }$

The ratio of warp way modular length to sum of thread diameter=1.19 (Ans)

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