To increase the current sensitivity of a moving coil galvanometer by `50%` its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

Explanation

Voltage sensitivity is given by `(phi)/(V)=((NAB)/(k))(1)/(R)`
Current sensitivity is given by `(phi)/(1) = ((NAB)/(k))`
According to the question, `(N'A'B)/(k) = (150)/(100) ((NAB)/(k))`
`N'A'=(3)/(2)NA`,
`(phi')/(V)=((N'A'B)/(k))=(1)/(R')=((3)/(2)NAB)/(k(2R))=(3)/(4)((NAB)/(k))(1)/(R)=(3)/(4) (phi)/(V)`
Percentage decrease `=(1-(3)/(4))/(1)xx100=(1)/(4)xx100=25%`