Four wires of equal length and of resistances $10 \ \Omega$ each are connected in the form of a square. The equivalent resistance between two opposite corners of the square is ............. $\Omega$. (in $Omega$)

$A$ uniform ring of conducting wire of radius $r$ has resistance $R$. $A$ and $B$ are two points on the ring which subtend an angle $\theta$ at the centre of the ring. The equivalent resistance between points $A$ and $B$ is:

$10$ resistors,each of resistance $R$,are connected in series to a battery of $emf$ $E$ and negligible internal resistance. When those are connected in parallel to the same battery,the current is increased $n$ times. The value of $n$ is:

$A$ ring is made of a wire having a resistance $R_0 = 12 \,\Omega$. Find the points $A$ and $B$,as shown in the figure,at which a current-carrying conductor should be connected so that the resistance $R$ of the sub-circuit between these points is equal to $\frac{8}{3} \,\Omega$.

If you are provided a set of resistances $2\, \Omega, 4\, \Omega, 6\, \Omega$ and $8\, \Omega$. Connect these resistances so as to obtain an equivalent resistance of $\frac{46}{3}\, \Omega$.

The resistors of resistances $2 \ \Omega$,$4 \ \Omega$,and $8 \ \Omega$ are connected in parallel. The equivalent resistance of the combination will be: