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Question

A cubical block of wood of edge 3 cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it.
(Take density of wood =800 kg/m3, spring constant of the spring =50 N/m and g=10 m/s2.)

A
0.35 N
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B
0.6 N
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C
0.27 N
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D
0.78 N
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Solution

The correct option is A 0.35 N

The specific gravity of the block =0.8. Hence the height of cubical block inside water =3 cm×0.8=2.4 cm. The height outside water =3 cm2.4=0.6 cm.
Suppose, the maximum weight that can be put without wetting it, is W. The cubical block in this case is completely immersed in the water.
So, the volume of the displaced water
= volume of the cubical block =27×106 m3

Hence, the force of buoyancy
=Vρwg=(27×106 m3)×(1000 kg/m3)×10 m/s2
=0.27 N
The spring is compressed by 0.6 cm and hence the upward force exerted by the spring
=50 N/m×0.006 m=0.3 N
The force of buoyancy and the spring force taken together balance the weight of the block plus the weight W put on the block.
The weight of the block is
W=(27×106 m)×(800 kg/m3)×10 m/s2
=0.216 N
Thus, weight W=0.27 N+0.3 N0.216 N
=0.35 N

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