GATE (TF) Textile 2011 Question Paper Solution | GATE/2011/TF/45

Question 45 (Textile Engineering & Fibre Science)

Warp and weft yarns with diameter of 0.4 mm and 0.6 mm, respectively, are used to produce plain woven fabric with end spacing of 0.8 mm and pick spacing of 1.2 mm. Assuming the degree of flattening to be 0.8 in both warp and weft yarns, the approximate fabric cover would be

(A)	0.56
(B)	0.66
(C)	0.76
(D)	0.86

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Given in the question-

Warp yarns diameter( $d_1$ )=0.4mm

Weft yarns diameter( $d_2$ )=0.6mm

End spacing( $s_1$ )=0.8mm

Pick spacing( $s_2$ )=1.2mm

Degree of flattering(In both warp and weft,f)=0.8

Fabric cover=?

By formula-

Fabric cover(C)=C1+C1-C1.C2

Where,

C1 is the warp cover factor= $\frac{d_1}{f \times s_1}$

C1= $\frac{0.4}{0.8 \times 0.8}$

C1=0.625

C2 is the warp cover factor= $\frac{d_2}{f \times s_2}$

C2= $\frac{0.6}{0.8 \times 1.2}$

C2=0.625

Now,

Fabric cover(C)=C1+C1-C1.C2

Fabric cover(C)=0.625+0.625-0.625 x 0.625

Fabric cover(C)=1.25-0.3906

Fabric cover(C)=0.859

Fabric cover(C)=0.86 (Ans)-Option D is correct

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