GATE (TF) Textile 2014 Question Paper Solution | GATE/2014/TF/40

Question 40 (Textile Engineering & Fibre Science)

The surface helix angle of a fibre in a yarn of 0.3 mm diameter is 30^o. Assuming ideal helix geometry, the helix angle (in degrees), of a fibre situated at 0.1 mm from the yarn axis, accurate to one decimal place, will be …………………

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Given in the question–
Fibre dia(D)=0.3 mm

Helix angle of a fibre in a yarn( $\alpha$ )=30^o

Fibre situation(d)=0.1 mm

Helix angle( $\theta$ ) =?

Formula:
By helix geometry

$tan\alpha=2\times \pi D$
&
$tan\theta=2\times \pi d$

Where,
$\alpha$ is helix angle and D is diameter

By deviding both equetions.

We get,

$\frac{tan \alpha}{tan \theta}=\frac{D}{d}$

$\frac{tan 30^o}{tan \theta}=\frac{0.3}{0.1}$

$\frac{tan 30^o}{tan \theta}=\frac{0.3}{0.1}$

$\frac{0.577}{tan \theta}=\frac{0.3}{0.1}$

$\frac{0.577\times 0.1}{0.3}=tan \theta$

$0.192=tan \theta$

$tan \theta=0.192$

$\theta=tan^-1(0.192)$

$\theta=10.87^o$ (Answer)

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