A wire of resistanc e`9Omega` is broken in two parts. The length ratio being `1:2`. The two pieces are connected in parallel. The net resistance will be

A uniform metal wire of resistance R is stretched to twice its length. Now this wire is halved and the two part are connected in parallel. The equivalent resistance is-

A wire of 2 Omega is halved and two pieces are joined in parallel. Find its resistance?

A wire of resistance R_(1) is cur into five equal pieces. These five pieces of wire are then connected in parallel. If the resultant resistance of this combination be R_(2) , then the ratio (R_(1))/(R_(2)) is:

A uniform wire of resistance (50 Omega) is cut into 5 equal parts. These parts are now connected in parallel.the Equivalent resistance of the combination is

A uniform wire of resistance (50 Omega) is cut into 5 equal parts. These parts are now connected in parallel.the Equivalent resistance of the combination is

A uniform wire of resistance (50 Omega) is cut into 5 equal parts. These parts are now connected in parallel.the Equivalent resistance of the combination is

A wire has a resistance of 6 Omega . It is cut into two parts and both half values are connected in parallel. The new resistance is