$A$ piece of wire of resistance $6 \, \Omega$ is connected to a battery of $12 \, V$. Find the amount of current flowing through it. Now,the same wire is redrawn by stretching it to double its length. Find the resistance of the new (redrawn) wire.

(D) Given: Resistance $(R) = 6 \, \Omega$,Potential difference $(V) = 12 \, V$.
Using Ohm's law: $V = IR$.
Therefore,current $(I) = \frac{V}{R} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.
For the second part,the original resistance is $R = \rho \frac{l}{A}$.
When the wire is stretched to double its length,the new length $l' = 2l$. Since the volume of the wire remains constant,$V = A \times l = A' \times l'$.
Thus,$A' = \frac{A \times l}{l'} = \frac{A \times l}{2l} = \frac{A}{2}$.
The new resistance $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right) = 4R$.
Substituting the original resistance value: $R' = 4 \times 6 \, \Omega = 24 \, \Omega$.