The correct option is B Antinode occurs at x=0.25 m
For the stationary wave represented by, y=0.02 cos(10πx)cos(50πt+π2)
Considering a snapshot of string at some given instant of time, i.e., t= fixed
Nodes will be formed where displacement of particle is zero.
It will happen when,
cos(10πx)=0
⇒10πx=π2,3π2,5π2,......
⇒x=0.05,0.15,0.25,....
So, nodes occurs at x=0.15 m and x=0.25 m.
On comparing given equation with standard equation of standing wave,
y=2Acos(kx)cos(ωt+ϕ),
Wavenumber, k=10π m−1
Angular frequency, ω=50π s−1
So, Velocity of wave,
v=ωk=50π10π=5 m/s
Also, wavelength,
λ=2πk=2π10π=0.2 m