A cubical block of wood $10 c m$ on a side floats at the interface between oil and water, as in Fig, with its lower face $2 c m$ below the interface. The density of the oil is $0.6 g c m^{- 3}$ . The mass of the block is

340 g

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680 g

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80 g

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10 g

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Solution

The correct option is C: $680 g$

As the equilibrium is maintained,
weight of wood $=$ total upthrust experienced by wood= total weight of fluid displaced by block
$\Rightarrow$ weight of wood $= (100 \times 2 \times 1) g + (8 \times 100 \times 0.6) g$
$\Rightarrow$ Mass of wood $= (200 + 800 \times 0.6) g r a m$
$\Rightarrow$ Mass of wood $= (480 + 200) g r a m$
$\Rightarrow$ Mass of wood $= 680 g r a m$

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A cubical block of wood 10 cm on a side floats at the interface between oil and water, as in Fig, with its lower face 2 cm below the interface. The density of the oil is 0.6 gcm−3. The mass of the block is

The correct option is C: 680gAs the equilibrium is maintained,weight of wood = total upthrust experienced by wood= total weight of fluid displaced by block⇒ weight of wood =(100×2×1)g+(8×100×0.6)g⇒ Mass of wood =(200+800×0.6)gram⇒ Mass of wood =(480+200) gram⇒ Mass of wood =680 gram

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