GATE (TF) Textile 2006 Question Paper Solution | GATE/2006/TF/67

Question 67 (Textile Engineering & Fibre Science)

The corresponding weight of the fabric in GSM works out to be

(A)200
(B)240
(C)270
(D)290
[Show Answer]

Option C is correct.

Given in the question-

Fabric construction: 2/2 Twill

i.e., Ends(E)=4

Interlacement(I)=2

Count of yarn(C): 60 tex

g/m=60/1000

Crimp in yarn(c): 5%

m=0.39

k=200,

GSM of the fabric=?

By formula-

GSM=2 \times \frac{threads}{cm} \times (1+crimp percentage) \times \frac{gm}{meter}

We have calculated earlier in question no. 66, Threads/m=2284

GSM=2 \times 2284 \times (1+5 percentage) \times \frac{60}{1000}

GSM=2 \times 2284 \times (1+\frac{5}{100}) \times \frac{60}{1000}

GSM=2 \times 2284 \times (1+0.05) \times 0.06

GSM=2 \times 2284 \times 1.05 \times 0.06

GSM=287.78

GSM=290

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