GATE Electronics and Communications (EC) 2019 Solved Paper

at $\mathit{F}=10\mathit{H}\mathit{z}$ we have one pole
At $\mathit{F}={10}^2\mathit{H}\mathit{z}$ we can see two more poles are added as slope is decreased by $40\mathit{d}\mathit{B}∕$ decade
At $\mathit{F}={10}^3\mathit{H}\mathit{z}$ we have a zero
At $\mathit{F}={10}^4\mathit{H}\mathit{z}$ we have two zero's
At $\mathit{F}={10}^5\mathit{H}\mathit{z}$ we have two pole's
At $\mathit{F}={10}^6$ we have one pole
$\therefore$ Total poles ${\mathit{N}}_{\mathit{p}}=6$
And total zeros ${\mathit{N}}_{\mathit{z}}=3$