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 `=800kg//m^(3)` and spring constant of the spring `=50N//m`. Take `g=10m//s^(2)`

Explanation

The specific gravity of the block `= 0.8`. Hence the height inside water `= 3 cm xx 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 xx 10^(-6) m^(3)`.
Hence, the force of buoyancy
`= (27 xx 10^(-6)m^(3)) xx 1(1000 kg//m^(3)) xx (10 m//s^(2)) = 0.27 N`.
The spring is compressed by `0.6 cm` and hence the upward force exerted lby the spring
`= 100 N//m xx 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 blok is
`W' = (27 xx 10^(-6)m) xx (800 kg//m^(3)) xx (10 m//s^(2)) = 0.22 N`.
Thus, `W = 0.27 N + 0.6 N - 0.22 N = 0.65 N`.