The correct option is A 300 N
Mass of the chain, m=20 kg
and length, l=10 m
Total force on the table = Thrust force + weight of chain lying on table.
Mass per unit length of chain λ=ml
Taking dx element of the hanging part of the chain situated at distance x from ground, whose mass is dm
dm=λdx
Let velocity of this element when it hits the surface be v :
From W−E theorem
Wgravity=△K.E
(dm g)x=12dmv2
or v=√2gx ...(1)
As we know, thrust FT=v⋅dmdt
=v⋅λ⋅(dxdt)
⇒FT=v2⋅λ
Putting the value of v from (1)
FT=(√2gx)2×ml=2gxml=2×10×5×2010
FT=200 N
Weight of chain lying on the table Wl=ML×lg=2010×5×10=100 N
Therefore, total force on the table =200+100=300 N