GATE (TF) Textile 2011 Question Paper Solution | GATE/2011/TF/48

Question 48 (Textile Engineering & Fibre Science)

Consider the following data for a synthetic fibre.
Density of amorphous region (\rho _a) as 1.33 g/cm3, density of crystalline region (\rho_c) as 1.45 g/cm3, density of fibre (\rho_f) as 1.36 g/cm3 and diameter of the fibre as 14 micron.

Denier of the fibre is approximately

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Given in the question-

Density of amorphous region(\rho_a)=1.33 g/cm3

Density of crystalline region(\rho_c)=1.45 g/cm3

Density of fibre(\rho_f)=1.36 g/cm3

Diameter of the fibre(d)=14 micron
d=14 \times 10^-6 m

d=14 \times 10^-4 cm
As we know the definition of denier-

Weight of 9000 meter length .

so that denier is the mass.

Mass=Volume x density

Mass=(\frac{\pi \times d^2}{4}) \times 100 \times 9000 \ density of fibre

Where d in cm

Mass=(\frac{\pi \times (14 \times 10^-4 )^2}{4}) \times 100 \times 9000 \ density of fibre

Denier=(\frac{3.14 \times (14 \times 10^-4 )^2}{4}) \times 100 \times 9000 \ 1.36

Denier=\frac{3.14 \times (14 \times 10^-4 )^2}{4} \times 100 \times 9000 \ 1.36

Denier=188324640 \times 10^-8

=2 (Ans)

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