+2 votes
in JEE by kratos

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.

(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.

1 Answer

+6 votes
by kratos
 
Best answer

Correct option: ( B, C)

Explanation:

Kx = V(2000) (10) – V (1000) (10)

Kx = 50 N ... (b)

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