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Question

A block of density 2000kg/m3 and mass 10kg is suspended by a spring of stiffness 100N/m. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density 1000kg/m3 . If the block is in equilibrium position.

A
the elongation of the spring is 1cm.
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B
the magnitude of buoyant force acting on the block is 50N.
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C
the spring potential energy is 12.5J.
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D
magnitude of spring force on the block is greater than the weight of the block.
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Solution

The correct options are
A the magnitude of buoyant force acting on the block is 50N.
B the spring potential energy is 12.5J.
There are three forces when the block is in equilibrium.
1. weight of the block downward.
2. buoyancy force due to the liquid upward
3. force due to the spring upward

Weight of block: mbg=Vbρbg=Vb×2000×10
Buoyancy force upward: Fbuoyancy=Vbρlg=Vb×1000×10
Upward force due to spring: Fspring=kx=100x
Equating the upward force with the downward force,
Vb×1000×10+100x=Vb×2000×10
100x=Vb×10000=102000×10000
x=12m
Potential Energy stored in the spring: PE=12kx2=12×100×(12)2=12.5Nm=12.5J

134054_71878_ans.JPG

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