A galvanometer having a resistance of $$8 \Omega$$ is shunted by a wire of resistance $$2 \Omega$$. If the total current is $$1 \mathrm{~A}$$, the part of it passing through the shunt will be

Two cells with the same emf $$E$$ and different internal resistances $$r_1$$ and $$r_2$$ are connected in series to an external resistance $$R$$. If the potential difference across the first cell is zero then value of $$R$$.

The current sensitivity of a galvanometer having 20 divisions is $$10 \mu \mathrm{A} /$$ div. If the resistance of the galvanometer is $$100 \Omega$$ then the value of the resistance to be used to convert this galvanometer in to an voltmeter to read up to $$1 \mathrm{~V}$$ is :

The electric flux from cube of side $$1 \mathrm{~m}$$ is '$$\Phi$$' When the side of the cube is made $$3 \mathrm{~m}$$ and the charge enclosed by the cube is made one third of the original value, then the flux from the bigger cube will be :