A uniform metal wire of length $$l$$ has $$10 \Omega$$ resistance. Now this wire is stretched to a length $$2l$$ and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is

The given circuit shows a uniform straight wire $$A B$$ of $$40 \mathrm{~cm}$$ length fixed at both ends. In order to get zero reading in the galvanometer $$G$$, the free end of $$J$$ is to be placed from $$B$$ at:

The terminal voltage of the battery, whose emf is $$10 \mathrm{~V}$$ and internal resistance $$1 \Omega$$, when connected through an external resistance of $$4 \Omega$$ as shown in the figure is:

A wire of length '$$l$$' and resistance $$100 \Omega$$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is: