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Question

A cubical block of plastic 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. Density of plastic = 800 kg / m3 and spring constant of the spring = 100 N/m. Take g = 10 m/s2.
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A
0.65N
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B
0.59N
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C
6.5N
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D
5.5N
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Solution

The correct option is A 0.65N
The specific gravity of the block = 0.8. Hence the height inside waterA=3 cm × 0.8 = 2.4 cm. The height outside water =3 cm - 2.4= 0.6 cm. Suppose the maximum weight that can be put without wetting it is W. The block in this case is completely immersed in the water. The volume of the displaced water
= volume of the block = 27 ×106m3.
Hence, the force of buoyancy
= (27×106m3)×1(1000kg/m3)×(10m/s2)=0.27N.
The spring is compressed by 0.6 cm and hence the upward force exerted by the spring= 100 N/m × 0.6 cm = 0.6 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=(27×106m)×(800kg/m3)×(10m/s2) = 0.22 N
Thus, W = 0.27 N + 0.6 N - 0.22 N= 0.65 N

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