GATE (TF) Textile 2003 Question Paper Solution | GATE/2003/TF/60

Question 60 (Textile Engineering & Fibre Science)

A ring frame running at 14000 rpm is producing yarn at 15 m/min. The difference in yarn twist (turns / cm) in the yarn when the package diameter changes from 20 mm to 36 mm will be

(A)	8.0
(B)	9.2
(C)	7.1
(D)	10.1

[Show Answer]

Option C is correct.

Ring spindle speed=14000 rpm

FRD=15 m/min

Dia of packade(d1)=20mm

Dia of packade(d1)=36mm

The difference in yarn twist (turns / cm) in the yarn when the package diameter changes from 20 mm to 36 mm =?

By formula-

$Yarn twist(per m)=\frac{spindle speed-\frac{FRD}{\pi \times Dia}}{FRD}$

At 20mm dia

$Yarn twist_1(per m)=\frac{14000-\frac{15}{3.14 \times 20 \times 10^-3}}{15}$

At 36 mm dia

$Yarn twist_2(per m)=\frac{14000-\frac{15}{3.14 \times 36 \times 10^-3}}{15}$

$Yarn twist_1-Yarn twist_2=\frac{14000-\frac{15}{3.14 \times 20 \times 10^-3}}{15}-\frac{14000-\frac{15}{3.14 \times 36 \times 10^-3}}{15}$

$Yarn twist_1-Yarn twist_2=\frac{15}{3.14 \times 20 \times 10^-3\times 15}-\frac{15}{3.14 \times 36 \times 10^-3 \times 15}$

$Yarn twist_1-Yarn twist_2=15.92-8.85$

$Yarn twist_1-Yarn twist_2=7.07$ ans

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