Work, Power & Energy

Molecules hitting the forward and rear surfaces will bounce back with speeds given above. Let mass of one molecule be m0. Then,
ΔP0(forward)=2m0(u+v)
ΔP0(Rear)=2m0(u−v)
Let the rates of collision with front and rear surfaces be R1 and R2 respectively
So, R1α(u+v)
R2α(u−v)
Force ΔP.R.
So, F1R1.2m0(u+v),F2=R2.2m0(u−v)
So, F1−F2α2m0(u+v2)−2m0(u−v)2
α‌‌2m0[4uv]
α‌‌‌uv so, (A) is correct
Clearly, the net force due to gas is proportion to v, i.e. it is variable hence acceleration of plate is variable. Finally the plate will start moving with terminal velocity
Hence (C) is correct
Resistive force =ΔP.A∝V
⇒ Hence (D) is correct