If a stunt of `1 Omega` is connected to a galvanometer of resistance `99 Omega`, what fraction of the main current will flow through the galvanometer?

Explanation

The galvanometer resistance G and the stunt resistance S are connected in parallel SO the galvanometer current
`I_G=I.S/(S+G)[I=mai n current]`
or, `I_G/I=S/(S+G)=1/(1+90)=1/100=1%`
i.e., 1% of the main current will flow through the galvanometer.