A cube of iron of edge 5 cm floats on the surface of mercury, contained in a tank. Water is then poured on top, so that the cube just gets immersed. The depth of water layer in cm is (20+x). (Specific gravities of iron and mercury are 7.8 and 13.6, respectively.) Find the value of x.

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Solution

When in equilibrium it is in completely submerged state,
$W e i g h t o f i r o n c u b e + W e i g h t o f w a t e r l a y e r = B u o y a n t f o r c e d u e t o m e r c u r y d i s p l a c e m e n t$
$⟹ (5^{3}) (7.8) g + (h \times 5^{2}) (1) g = (5^{3}) (13.6) g$
$⟹ h = 29 c m$

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A cube of iron of edge 5 cm floats on the surface of mercury, contained in a tank. Water is then poured on top, so that the cube just gets immersed. The depth of water layer in cm is (20+x). (Specific gravities of iron and mercury are 7.8 and 13.6, respectively.) Find the value of x.

When in equilibrium it is in completely submerged state,Weight of iron cube+Weight of water layer=Buoyant force due to mercury displacement⟹(53)(7.8)g+(h×52)(1)g=(53)(13.6)g⟹h=29cm

When in equilibrium it is in completely submerged state,
$W e i g h t o f i r o n c u b e + W e i g h t o f w a t e r l a y e r = B u o y a n t f o r c e d u e t o m e r c u r y d i s p l a c e m e n t$
$⟹ (5^{3}) (7.8) g + (h \times 5^{2}) (1) g = (5^{3}) (13.6) g$
$⟹ h = 29 c m$