Correct Answer  Option 3 : 40
Concept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady state error for different inputs is given by
Input

Type 0

Type  1

Type 2

Unit step

\(\frac{1}{{1 + {K_p}}}\)

0

0

Unit ramp

∞

\(\frac{1}{{{K_v}}}\)

0

Unit parabolic

∞

∞

\(\frac{1}{{{K_a}}}\)

From the above table, it is clear that for type – 1 system, a system shows zero steadystate error for stepinput, finite steadystate error for Rampinput and \(\infty \) steadystate error for parabolicinput.
Calculation:
The given forward path transfer function is \(G\left( s \right) = \frac{K}{{s\left( {s + 2} \right)}}\)
Velocity error constant, \({K_v} = \mathop {\lim }\limits_{s \to 0} s \times \frac{K}{{s\left( {s + 2} \right)}} = \frac{K}{2}\)
Steadystate error \( = \frac{2}{K} = 0.05\)
⇒ K = 40