SSE JE Electrical Engineering 10 Dec 2020 Papers

Deflecting torque is directly proportional to current, then the scale of the ammeter is linear and the meter is PMMC type.
Given data
${\mathrm{I}}_1=5\mathrm{A},{\mathrm{\theta }}_1=80°$
For ${\mathrm{I}}_1$ current let the deflection be ${\mathrm{\theta }}_{\mathrm{1}}$
${\mathrm{I}}_2=2.5\mathrm{A},{\mathrm{\theta }}_2=?$
For ${\mathrm{I}}_2$ current let the deflection be ${\mathrm{\theta }}_2$
The current passing through the ammeter is directly proportional to the deflection of the pointer
$\mathrm{I}\propto \mathrm{\theta }$
$\Rightarrow \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=\frac{{\mathrm{\theta }}_1}{{\mathrm{\theta }}_2}$
$\Rightarrow \frac{5}{2.5}=\frac{80°}{{\mathrm{\theta }}_2}$
$\Rightarrow {{\mathrm{\theta }}_{}^{}}_2^{}=40°$