Correct Answer - B::C
To increases the range of ammeter a parallel resistance (called shunt) is requried which is given by
`S=((i_(g))/(i-i_(g)))G` for option (c)
`S=((50xx10^(-6))/(5xx10^(-3)-50xx10^(-6)))(100)=1Omega`
To change it into voltmeter, a high resistance `R` is put in series where R is given by `R=(V)/(i_(g))-G`
For option (b) `R=(10)/(50xx10^(-6))-100=200kOmega`