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A cube of wood supporting $200\,g$ mass just floats in water, when the mass is removed, the cube rises by $1\,cm$ , the linear dimension of cube is
A. $10\,cm$
B. $20\,cm$
C. $10\sqrt 2 \,cm$
D. $5\sqrt 2 \,cm$

Answer
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Hint: This problem is based on Archimedes’ principle; it states that when a body is immersed fully or partially, the amount of fluid displaced is equal to the upward force known as buoyant force. In this problem when 200 g of mass is removed, the cube rises by $1cm$ hence we can use Archimedes' principle that the weight of $200g$ is equal to the weight of the water displaced by immersing $1cm$ of the wood.

Complete step by step answer:
Given: Mass of wood supports = $200\,g$
When mass is removed, the cube of wood is raised by $1cm$. Hence according to the Archimedes’ principle, weight of the $200\,g$ is equal to the weight of the water displaced by immersing 1 cm of the cube of wood.Therefore, According to Archimedes’ principle the weight of the liquid displaced is equal to buoyant force. Hence, Mathematically
$\Delta {F_b} = W$ ……….. $\left( 1 \right)$

We known that, density = $\dfrac{\text{mass}}{\text{volume}}$
Therefore, mass = density \[ \times \] volume
According to Newton’s second law F = ma
Therefore equation $\left( 1 \right)$ becomes,
$\Delta V \times \rho \times g = mg$
$\Rightarrow \Delta V \times \rho = m$
Substituting the given data
${x^2} \times 1 \times 1gc{m^{ - 3}} = 200\,g$
On simplifying the above equation, we get
${x^2} = 200$
Therefore, $x = 10\sqrt 2 \,cm$

Hence, option C is correct.

Note: The upthrust or the buoyant force applied by the fluid on the body or object when an object is fully or partially immersed and buoyancy is the force that makes the body or object to float. Buoyancy is caused by the difference in pressure that is acting on opposite sides of a body or object immersed in a static fluid. The S.I unit for buoyant force is Newton $\left( N \right)$.