GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Stiffness of helical spring
$\mathit{K}=\frac{\mathit{G}{\mathit{d}}^4}{8{\mathit{D}}^3\mathit{n}}$
Where,
d=Spring Wire diameter
D= Mean Coil diameter
n= Number of coils
G= Shear Modulus
d increased by $2\%$ i.e. ${\mathit{d}}^{\prime }=1.02\mathit{d}$
$\frac{{\mathit{k}}^{\prime }}{\mathit{k}}={\left(\frac{{\mathit{d}}^{\prime }}{\mathit{d}}\right)}^4$
$\frac{{\mathit{k}}^{\prime }}{\mathit{k}}={\left(\frac{1.02\mathit{d}}{\mathit{d}}\right)}^4={1.02}^4=1.08242$
Percentage change in spring Stiffness $=\frac{{\mathit{k}}^{\prime }-\mathit{k}}{\mathit{k}}\times 100=8.242\%$