GATE (TF) Textile 2018 Question Paper Solution | GATE/2018/TF/49

Question 49 (Textile Technology & Fibre Science)

The diameter and length of torsion rod used in a projectile loom are increased by 5% and 20%, respectively. If the torque required to twist the rod increases by X%, then the value of X, accurate to two decimal places, is __1.27 to 1.31__.

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Given–
Let ,initial diameter of torsional rod=d₁

Then, final diameter of torsional rod(d₂)=d₁+d₁ x 5%
d₂=d₁+(d₁ x 5/100)
d₂=d₁+d₁ x 0.05

d₂=1.05 x d₁

Let, initial length of torsional rod=l₁

Then, final length of torsional rod(l₂)=l₁+l₁ x 20%
l₂=l₁+(l₁ x 20/100)
l₂=l₁+0.20 x l₁

l₂=1.2 xl₁

Let, initial torque required to twist the rod= $\tau_1$

Then, final torque required to twist the rod $(\tau_2)=\tau_1+(\tau_1\times X%)$
$(\tau_2)=\tau_1+(\tau_1\times\frac{X}{100})$

$(\tau_2)=\tau_1+(\tau_1\times0.01X)$
$(\tau_2)=(1+0.01X)\tau_1$

By formula:

$\tau\alpha\frac{d^4}{l}$

Where, $\tau$ is torque , d is diameter of rod and l is length of torsional rod

So,
$\frac{\tau_1}{\tau_2}=\frac{d_1^4}{d_2^4}\times\frac{l_2}{l_1}$

$\frac{\tau_1}{\tau_2}=\frac{d_1^4}{d_2^4}\times\frac{l_2}{l_1}$

$\frac{\tau_1}{(1+0.01X)\tau_1}=\frac{d_1^4}{(1.05)^4\times d_1^4}\times\frac{1.2\times l_1}{l_1}$

$\frac{\tau_1}{(1+0.01X)\tau_1}=\frac{d_1^4}{1.22\times d_1^4}\times\frac{1.2\times l_1}{l_1}$

$\frac{1}{1+0.01X}=\frac{1}{1.22}\times\ 1.2$

$\frac{1}{1+0.01X}=\frac{1}{1.22}\times\ 1.2$

$\frac{1}{1+0.01X}=\frac{1.22}{1.2}$

$\frac{1}{1+0.01X}=1.017$

$1=1.017\times(1+0.01X)$

$1.017\times(1+0.01X)=1$

$(1+0.01X)=\frac{1}{1.017}$

$(1+0.01X)=0.9832$

$(0.01X)=(1-0.9832)$

$(0.01X)=0.0167$

$X=\frac{0.017}{0.01}$

$X=\frac{0.017}{0.01}$

$X=1.70$ (Answer)