GATE (TF) Textile 2005 Question Paper Solution | GATE/2005/TF/45

The permissible minimum angle of lead is 30^o. For a ring of 40mm diameter, the minimum bobbin diameter is

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Option C is correct.

Given-

Angle of lead( $\theta$ )= $30^o$

Dia of ring( $d_r$ )=40mm

Bobbin dia( $d_b$ )=?

By formula-

$Sin(\theta)=\frac{d_b}{d_r}$

$Sin(30)=\frac{d_b}{40}$

$0.5=\frac{d_b}{40}$

$d_b=0.5 \times 40$

$d_b=20 mm$ (Ans)