2014:
The resistance of a wire is given by $R = \rho \frac{l}{A}$, where $\rho$ is the resistivity of the material, $l$ is the length of the wire and $A$ is the area of cross-section of the wire.
When the length of the wire is doubled, its resistance becomes $R’ = \rho \frac{2l}{A} = 2R$.
When the diameter of the wire is halved, its area of cross-section becomes $\frac{A}{2}$.
The new resistance of the wire is $R’’ = \rho \frac{l}{\frac{A}{2}} = 4R$.
Therefore, the new resistance is 4 times the original resistance, which is 10 ohm. So, the new resistance is 40 ohm.
2015:
The current in a circuit is given by $I = \frac{V}{R}$, where $V$ is the voltage and $R$ is the resistance.
In this case, the voltage is 2 V, the resistance of the cell is 1
