$A$ wire of resistance $12 \, \Omega$ is bent to form an equilateral triangle. What is the equivalent resistance between any two corners of the triangle?

$A$ wire of length $25 \ m$ and cross-sectional area $5 \ mm^2$ having resistivity of $2 \times 10^{-6} \ \Omega \ m$ is bent into a complete circle. The resistance between diametrically opposite points will be (in $\Omega$)

$A$ wire of resistance $20 \Omega$ is divided into $10$ equal parts. $A$ combination of two parts are connected in parallel and so on. Now,the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is . . . . . . $\Omega$.

$A$ and $B$ are two points on a uniform ring of radius $r$. The resistance of the ring is $R$. $\angle AOB = \theta$ as shown in the figure. The equivalent resistance between points $A$ and $B$ is . . . . . . .

When two resistances are connected in parallel,the equivalent resistance is $\frac{6}{5} \, \Omega$. When one of the resistances is removed,the effective resistance is $2 \, \Omega$. The resistance of the wire removed will be ................ $\Omega$.

In the following star circuit diagram,the equivalent resistance between the points $A$ and $H$ will be ..............$r$.