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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

A cubical blo...

Question

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 $0.6 g c m^{- 3}$ . The mass of block is

706 g

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607 g

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760 g

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670 g

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Solution

The correct option is C
760 g

Weight of block = Weight of displaced oil + Weight of displaced water

$\Rightarrow m g V_{1} p_{0} g + V_{2} p w g$

$\Rightarrow m = (10 \times 10 \times 6) \times 0.6 + (10 \times 10 \times 4) \times 1 = 760 g .$

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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 is0.6 g cm−3. The mass of block is

The correct option is C 760 g Weight of block = Weight of displaced oil + Weight of displaced water ⇒ mgV1p0g+V2pwg ⇒ m=(10×10×6)×0.6+(10×10×4)×1=760g.

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 $0.6 g c m^{- 3}$ . The mass of block is

The correct option is C
760 g

Weight of block = Weight of displaced oil + Weight of displaced water

$\Rightarrow m g V_{1} p_{0} g + V_{2} p w g$

$\Rightarrow m = (10 \times 10 \times 6) \times 0.6 + (10 \times 10 \times 4) \times 1 = 760 g .$