The correct option is D 0.9
Let,
u1= Speed of ball before collision with surface
v1= Speed of ball after collision with surface
u2= Speed of plate before collision=0
v2= Speed of plate after collision =0
The coefficient of restitution, e=speed of separationspeed of approach
Speed of separation = v1−v2=v1 ...i
Speed of approach=u1−u2=u1 ...ii
∴e=v1u1 ....iii
Since ball is dropped from height 1 m, it's speed while just colliding with plate is :
u1=√2gh=√2×10×1=√20
using, v2=u2+2as
v=u2−2gh
Taking upwards=+ve, v=0, u=v1, a=−g, s=+h=0.81 m
⇒0=v21−2(10×0.81)
∴v1=√2×10×0.81
From Eq. (iii)
⇒e=√2×10×0.81√20=√0.81
∴e=0.9