GATE (TF) Textile 2011 Question Paper Solution | GATE/2011/TF/49

Question 49 (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.

Density of the above fibre increased by 2.2 % when drawn. The corresponding change (%) in crystallinity is approximately

(A)50
(B)100
(C)150
(D)200
[Show Answer]

Given 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

Formula-

Crystallinity percentage=(\frac {1.45}{1.36} \times \frac{1.36-1.33}{1.45-1.33}) \times 100

Crystallinity percentage=(1.066 \times \frac{0.03}{0.12}) \times 100

Crystallinity percentage=(1.066 \times 0.25) \times 100

Crystallinity percentage=0.2665 \times 100

Crystallinity percentage=26.65

Again given in the question-

Density of fibre increased by 2.2%

The corresponding change in crystallinity %=?

\rho_f1=\rho_f + \rho_f \times 2.2%

\rho_f1=1.36 + 1.36 \times 2.2%

\rho_f1=1.39

Now by formula-

Crystallinity_1 percentage=\frac {\rho_c}{\rho_f1} \times \frac{\rho_f1-\rho_a}{\rho_c-\rho_a} \times 100

Crystallinity_1 percentage=(\frac {1.45}{1.39} \times \frac{1.39-1.33}{1.45-1.33}) \times 100

Crystallinity_1 percentage=1.043 \times \frac{0.06}{0.12}) \times 100

Crystallinity_1 percentage=1.043 \times 0.5 \times 100

Crystallinity_1 percentage=0.5215 \times 100

Crystallinity_1 percentage=52.15

Now,

The corresponding change in crystallinity %=(\frac{52.15-26.65}{26.65}) \times 100

The corresponding change in crystallinity %=(\frac{25.5}{26.65}) \times 100

The corresponding change in crystallinity %=0.9566 \times 100

The corresponding change in crystallinity %=95.67
=100 (Ans)

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