GATE (TF) Textile 2020 Question Paper Solution | GATE/2020/TF/55

Question 55 (Textile Engineering & Fibre Science)

The work of adhesion (W_SL) depends on the surface tension $(\gamma_{LV})$ of the liquid and the contact angle $(\theta)$ formed on a surface and is expressed as $\gamma _{LV}(1+cos\theta))$ . The W_SL for a given fabric and a liquid is reduced to 1/3^rd of the original value after oil repellent treatment. If the measured contact angle of the untreated fabric is 60^o, the percentage change in the contact angle after the treatment is __100__.

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Given in the question:

Work of adhesion(W_SL)= $\gamma _{LV}(1+cos\theta))$

Where, $(\gamma_{LV})$ =Surface tension of the liquid
W_SL=Work of adhesion
and
Let ,W_SL1=1 then, W_SL2=(1/3)*W_SL1 (After oil treatment)

Let,measured contact angle of untreated fabric is $(\theta_1)$ & Measured contact angle of treated fabric is $(\theta_2)$
So , $(\theta_1)$ =60^o

Percentage change in the contact angle after treatment( $\Delta\theta$ )=?

Calculation:

Work of adhesion(W_SL)= $\gamma _{LV}(1+cos\theta))$

i.e $\frac{WSL_1}{WSL_2}=\frac{(1+cos\theta_1)}{(1+cos\theta_2)}$

$\frac{WSL_1}{(1/3)*WSL_1}=\frac{(1+cos60^o)}{(1+cos\theta_2)}$

$3=\frac{(1+1/2)}{(1+cos\theta_2)}$

$3=\frac{(3/2)}{(1+cos\theta_2)}$

${(1+cos\theta _2)}= 3 \div (3/2)$

${(1+cos\theta_2)}=3 \times \frac{2}{3}$

${(1+cos\theta_2)}=2$

${cos\theta_2}=2-1$

$cos\theta_2=1$

$\theta_2=Cos^{-1}1$

$\theta_2=0^o$

Now, $\Delta \theta = \frac{\theta_1 - \theta_2 }{\theta_1} \times 100$

$\Delta \theta = \frac{60-0}{60} \times 100$

$\Delta \theta = \frac{60}{60} \times 100$

$\Delta\theta=100$ (Answer)