GATE (TF) Textile 2010 Question Paper Solution | GATE/2010/TF/31

Question 31 (Textile Engineering & Fibre Science)

30 bales of 3.5 Mi and 40 bales of 4.0 Mi cotton fibres (Mi is Microair value of cotton fibres) are already available in the raw material inventory of the mill. What is the approximate Mi value required of another 30 bales so that the resultant mean Mi of the total bales will be 4.0 ?

(A)	3.4
(B)	4.7
(C)	5.2
(D)	6.0

[Show Answer]

Given in the question-
No. of bales of 3.5 Mi (N1)=30

Mircronaire value(Mi 1)=3.5

No. of bales of 4.0 Mi (N2)=40

Mircronaire value(Mi 2)=4.0

No. of bales of Mi (N3)=30

Resultant mean micronaire (Mi R)=4.0

Total no. of bales(N)=N1+N2+N3

Total no. of bales(N)=N1+N2+N3

Total no. of bales(N)=30+40+30

Total no. of bales(N)=100

Here,

Length of 100 bales=length of 30 bales ,3.5 mi value+length of 40 bales ,4.0 Mi value+ length of 30 bales ,Mi value

$\frac{N}{Mi R}=\frac{N1}{Mi 1}+\frac{N2}{Mi 2}+\frac{N3}{Mi 3}$

It is whole by unitery rule-

3.5 Mi , means
3.5 micro gram weight have=1 inch length

then, 30 micro gm weight have=30/3.5 length

Here, we have considered no. of bales as weight for calculations.

Similarily done all,

$\frac{100}{4}=\frac{30}{3.5}+\frac{40}{4}+\frac{30}{Mi}$

$25=8.57+10+\frac{30}{Mi}$

$25=18.57+\frac{30}{Mi}$

$25-18.57=\frac{30}{Mi}$

$\frac{30}{Mi}=6.43$

$\frac{30}{6.43}=Mi$

Mi=4.67

i.e Mi value required of another 30 bales =4.67 (Ans)

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