A cubical block of wood $10 \,cm$ on a side floats at the interface between oil and water with its lower surface horizontal and $4\, cm$ below the interface. The density of oil is $0.6gc{m^{ - 3}}$. The mass of block is ...... $gm$
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(c) Weight of block
= Weight of displaced oil + Weight of displaced water
==> $mg = {V_1}{\rho _0}g + {V_2}{\rho _W}g$
==> $m = (10 \times 10 \times 6) \times 0.6 + (10 \times 10 \times 4) \times 1$ $= 760\, gm.$
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