# A wooden box of cubical shape of side 40 cm and mass 24 kg floats in water with two horizontal faces. what is the depth of the immersion of box ?

$\phantom{\rule{0ex}{0ex}}Volumeofwaterrequiredtobalancethisweight=V\phantom{\rule{0ex}{0ex}}then\phantom{\rule{0ex}{0ex}}V\times {\rho}_{w}\times g=24\times g\phantom{\rule{0ex}{0ex}}V=\frac{24}{1000}=.024{m}^{3}\phantom{\rule{0ex}{0ex}}depthuptowhichitimmerse=\frac{.024}{40\times 40\times {10}^{-4}}=.15m=15cm$

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