The correct option is C 40 J
Let, EA = Energy stored in spring ′A′
EB = Energy stored in spring ′B′
K1 = spring constant for spring ′A′
K2 = spring constant for spring ′B′
We know that spring potential energy E=12kx2=Fx2
Hence, EB=2Fx2=Fx ...(i) As, for the same extension, xA=xB=x , (FB=2F)
Given that energy stored in the spring A is 20 J
EA=Fx2=20 J ...(ii) (FA=F)
Dividing (ii) by (i), we get
EAEB=F.x2F.x
⇒EAEB=12
⇒20EB=12
⇒EB=40 J
Hence option C is the correct answer