A uniform conducting wire of length is $24a$, and resistance $R$ is wound up as a current carrying coil in the shape of an equilateral triangle of side '$a$' and then in the form of a square of side'$a$'. The coil is connected to a voltage source $V_0$. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is$1:\sqrt{y}$, where $y$ is

For triangle shape :
Number of turns is :
$N_t=\frac{\text{Total length of the wire}}{\text{Perimeter of the triangle}}$
$N_t=\frac{24a}{3a}=8$
Magnetic moment is :
$M_t=N_{t}I{A}_t=8\times I\times \frac{a^2\sqrt{3}}{4}$

For square shape :
Number of turns is :
$N_s=\frac{\text{Total length of the wire}}{\text{Perimeter of the triangle}}$
$N_s=\frac{24a}{4a}=6$
Magnetic moment is :
$M_s=N_{s}I{A}_s=6\times I\times a^2$
$I$ will be same in both the cases.
Therefore,
$\frac{M_t}{M_s}=\frac{8\times I\times \frac{a^2\sqrt{3}}{4}}{6\times I\times a^2}$
$\frac{M_t}{M_s}=\frac{1}{\sqrt{3}}$
$\Rightarrow y=3$