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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Explanation

In case of floatation
`W = F_B`
`sigma Vg = rho_omega V'g`
`V` : volume of block
`V'` : immersed volume
`V' = (sigma)/(rho_omega) V = (800)/(1000) V = 0.8 V = 0.8 a^3`
Length of block inside water is `h`
`a^2 h = 0.8 a^3 rArr h = 0.8 a = 0.8 xx 3 = 2.4 cm`
Compressiond of spring `x = 3 -h= 0.6 cm`
In equilibrium
`mg + W = F'_B + kx`
`mg + sigma Vg = rho_omega Vg + kx`
`mg = (rho_omega - sigma)Vg + kx`
=`(1000 - 800) (3 xx 10^-2) xx 10 + 50 xx 0.6 xx 10^-2`
=`2 xx 10^3 xx 27 xx 10^-6 + 30 xx 10^-2`
=`0.054 + 0.3`
=`0.35 N`.

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