Question 66 (Textile Engineering & Fibre Science)
A yarn specimen of 200 mm extends by 10% when loaded with 500 cN force. The length of the specimen after removal of load was found to be 202 mm. Percentage elastic recovery yarn is
| (A) | 30 |
| (B) | 50 |
| (C) | 70 |
| (D) | 90 |
[Show Answer]
Option D is correct
Given in the question-
Length of yarn specimen(l1)=200 mm
Extended length of yarn specimen(l2)=200+200 x 10%
l2= 220 mm
Load(f)=500 cN
Length of specimen after removal of load(l3)=202 mm
Percentage elastic recovery of yarn=?
Extension in length(l22)=Extended length of yarn specimen(l2)-Length of yarn specimen(l1)
l22=220-200
l22=20 mm
Recover length(l33)=Extended length of yarn specimen(l2)-Length of specimen after removal of load(l3)
l33=220-202
l33=18 mm
Now,
Percentage elastic recovery of yarn=Recover length(l33) x 100/Extension in length(l22)
Percentage elastic recovery of yarn=18 x 100/20
Percentage elastic recovery of yarn=90 (Ans)