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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. Density of wood =`800 kgm^-3` and spring constant of the spring `=50Nm^-1 Take g=10ms^-2`. , |
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Answer» The specific gravity of the block `=0.8` Hence the height inside water `=3cmxx0.8=2.4cm` the height outside water `=3cm-2.4=0.6cm` 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 `=27xx10^(-6)m^(3)` Hence the force buoyancy `=(27xx10^(-6)m^(3))xx(1000kg//m^(3))xx(10m//s^(2))=0.27N` The spring is compressed by 0.6 cm and hence the upward force exerted by the spring `=50N//mxx0.6cm=0.3N` 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=(27xx10^(-6)m)xx(800kg//m^(3))xx(10m//s^(2))=0.22N` Thus, `W=0.27N+0.3N-0.22N=0.35N` |
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