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Standard XII

Physics

Atmospheric Pressure

A block of wo...

Question

A block of wood of side $40 c m$ floats in water in such a way that it lower face is $5 c m$ below the free surface of water. What is the weight of the blocks?

64 k g

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16 k g

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8 k g

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Cannot be determined as density of wood is not given

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Solution

The correct option is D $8 k g$
Volume of part of wooden block immersed in water $V_{i n} = (40 \times 40 \times 5) c m^{2} = (40 \times 40 \times 5) \times 10^{- 6} m^{3}$
Density of water $σ_{w} = 10^{3} k g / m^{3}$
Weight of wooden block $=$ Weight of water displaced
$M g = V_{i n} σ_{w} g$
$M g = (40 \times 40 \times 5) \times 10^{- 6} \times 10^{3} g$
$\Rightarrow M = 8 k g$ .

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A block of wood of side 40 cm floats in water in such a way that it lower face is 5 cm below the free surface of water. What is the weight of the blocks?

The correct option is D 8 kgVolume of part of wooden block immersed in water Vin=(40×40×5) cm2=(40×40×5) ×10−6 m3Density of water σw=103 kg/m3Weight of wooden block = Weight of water displacedMg=VinσwgMg=(40×40×5)×10−6×103g⇒M=8 kg.

A block of wood of side $40 c m$ floats in water in such a way that it lower face is $5 c m$ below the free surface of water. What is the weight of the blocks?