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Applications of floatation - class-IX
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A cube of size 10 cm is floating in equilibrium in a tank of water. When a mass of 10 gm is placed on the cube, the depth of cube inside water increases by $\mathrm { g } = 10 \mathrm { m } / \mathrm { s } ^ { 2 }$ density of water $= 1000 \mathrm { kg } / \mathrm { m } ^ { 3 } )$
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A
1 mm
💡 Explanation:
Let $x$ be the initial depth upto which the cube is sinked in water.
Let $d$ be the density of the cube
Then
$x\times 10 \times 10\times 1 \times g$
$=10\times 10 \times 10\times 1 \times g\times d$........(1)
$\Rightarrow x=10d$
Let $x^1$ be the new depth, then
$x^1\times 100 \times g=1000\times d \times g+10g$........(2)
subtracting (1) and (2) we get
$\Rightarrow 100x^1=10x+10$
$x^2-x= \frac{1}{10cm}=1mm$
Hence,
option $B$ is correct answer.