GATE (TF) Textile 2015 Question Paper Solution | GATE/2015/TF/61

Question 61 (Textile Engineering & Fibre Science)

For a typical yarn tensile test, force ( $F$ ) in Newton and elongation ( $e$ )in cm are related as under $F=2+4e+3e^2$ .
If the yarn fails at an elongation of 3 cm, the work of rupture in N-m, accurate up to first decimal place is __0.5__.

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Given in the question–

Elongation at break(e)=3 cm

$F=2+4e+3e^2$

By integrating the equation both side-

$\int F de=\int_{0}^{e} (2+4e+3e^2)de$

$\int F de=\int_{0}^{3} (2+4e+3e^2)de$

$\int F de=\int_{0}^{3} 2 de +\int_{0}^{3} 4e de + \int_{0}^{3} 3e^2 de$

$\int F de=2\int_{0}^{3} 1 de +4\int_{0}^{3} e de + 3\int_{0}^{3} e^2 de$

$\int F de=2[e]_{0}^{3} +4[\frac{e^2}{2}]_{0}^{3} + 3[\frac{e^3}{3}]_{0}^{3}$

Work of rupture= $2[3-0]+\frac{4}{2}\times[3^2-0^0]+\frac{3}{3}\times [3^3-0^3]$

Work of rupture= $2[3]+2\times[9-0]+1\times [27-0]$

Work of rupture= $6+2\times 9+1\times 27$

Work of rupture=6+18+27

Work of rupture=51 N.cm

Work of rupture=51 x 10-² N.m

Work of rupture=0.51 N.m

Work of rupture=0.5 N.m (Answer)

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