GATE (TF) Textile 2008 Question Paper Solution | GATE/2008/TF/47

Question 47 (Textile Engineering & Fibre Science)

In a flat yarn, the number of filaments in the yarn cross section is 271. The yarn is divide into 5 segment of equal radial increments which are numbered as 1, 2, 3, 4 and 5, from yarn core to surface respectively. The approximately number of fibres in the 4th segment is

(A)50
(B)80
(C)90
(D)110
[Show Answer]

Option B is correct.

Total number of filament in the yarn cross section=271

Let us assume,

The radius of 5th segment=5x

Then, area of 5th segment will be= \pi \times (5x)^2

Area up to 5th segment=25\times x^2 \times \pi

Now, Area of 4 the segment=Area up to 4 th segment-Area up to 3 rd segment.

Area of 4 the segment=\pi \times (4x)^2-\pi \times (3x)^2

Area of 4 the segment=\pi \times 16 \times x^2-\pi \times 9 \times x^2

Area of 4 the segment=7 \times x^2\times \pi

By unitary rule-

Up to 5th segment area has………………………………………………=271 total no. of filament

Area 25\times x^2 \times \pi has……………………………….=271 filament

4th segment area(7 \times x^2\times \pi ) has................=\frac{271 \times 7 \times x^2\times \pi }{ 25\times x^2 \times \pi} filament

Total number of filaments in 4 th segment=75.88

Total number of filaments in 4 th segment(Aprroximately)=80 (Ans)

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