###### Question 48 (Textile Engineering & Fibre Science)

A 60.5 tex yarn with an unevenness CV of 10% is produced from 0.5 tex polyester fibre. The index of irregularity of the yarn, accurate to one decimal place, will be ………………..

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**Given in the question**–

Tex of yarn=60.5

CV_{eff}(%)=10

Tex of polyester fibre=0.5

Index of irregularity of yarn=?**Calculation**–**Index of irregularity=CV _{eff}/CV _{lim}**

CV

_{lim}=100/

Where, n is the number of fibres in the yarn cross section

n=Tex of yarn/tex of fibre

n=60.5/0.5

**n=121**

Now,

CV

_{lim}=100/

CV

_{lim}=100/11

**CV**

_{lim}(%)=9.09**Index of irregularity=10/**9.09

**Index of irregularity=1.10**(Answer)