The current sensitivity of a galvanometer increases by 20%. If its . . .

Question : The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 25%, the voltage sensitivity will

a) decreased by 1%
b) increased by 5%
c) increased by 10%
d) decreased by 4%

Doubt by Yashika

Solution : 
Is' = Is+20% of Is
Is' = Is+(20Is/100)
Is' = Is +0.2Is
Is' =1.2Is

Similarly 
R' = R+25% of R
R' = R+(25R/100)
R' = R+0.25R
R' = R+0.25R
R' = 1.25R

We know,
Voltage sensitivity = Current Sensitivity / Resistance

Initial
Vs = Is/R

Final 
Vs'=Is'/R'
Vs'=1.2Is/1.25R
Vs'=(1.2/1.25)Vs [ ∵Vs = Is/R]
Vs'=(120/125)Vs

Percentage change in Voltage Sensitivity 
=[(Final Vs - Initial Vs)/(Initial Vs)]×100%

The -ve sign shows that voltage sensitivity has decreased. So, Voltage sensitivity decreased by 4%.

Hence, d) would be the correct option.