JEE AdvancedFluid MechanicsHard
Question
A block of density 2000 kg/m3 and mass 10 kg is suspended by a spring stiffness 100 N/m. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density 1000 kg/m3 If the block is in equilibrium position.
Options
A.the elongation of the spring is 1 cm.
B.the magnitude of buoyant force acting on the block is 50 N.
C.the spring potential energy is 12.5 J.
D.magnitude of spring force on the block is greater than the weight of the block.
Solution
Kx = 10 × 10 − 
× 10 × 10
Kx = 50 N
Ustored =
× (100) 
= 
= 12.5 J
× 10 × 10Kx = 50 N
Ustored =
× (100) = = 12.5 JCreate a free account to view solution
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