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

(A)	1
(B)	2
(C)	3
(D)	4

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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

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

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

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

$Denier=188324640 \times 10^-8$

Denier=1.88
=2 (Ans)

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