
A log of wood of mass 20 kg floats on water. The weight that can be put on the raft to make it just sink should be
(density of wood=600 kgm$^{-3}$)
Answer
573.6k+ views
Hint:To solve this question, we will be using Archimedes principle which states that if an object is partially or fully submerged in fluid, it exerts an upward force called buoyant force. The apparent decrease of weight caused by the buoyant force is equal to the weight of the fluid, displaced by the object. From the relation between actual weight and apparent weight, we will define the principle of floatation.
Complete step by step answer:
According to the Principle of Floatation, a body floats in a liquid if the weight of the body is equal to the weight of the liquid displaced by it but the floating of a body occurs if and only if the density of the body is less than or equal to the density of fluid. It could be further explained as, when a solid body is dipped into a fluid, the fluid exerts an upward force of buoyancy on the solid and if the force of buoyancy equals to the weight of the solid, it will remain in the state of equilibrium, thus leading to the state of floatation.
The value of the buoyant force ${F_b}$ is given by:
${F_b} = \rho gV - W$
Here, $\rho ,g,V$ is the density of fluid, acceleration due to gravity and displaced volume respectively.
The volume V can be written as $\dfrac{m}{\rho }$.
Hence the force of buoyancy becomes,
${F_b} = \rho g \times \dfrac{m}{{{\rho _L}}} - W$
Here, ${\rho _L}$ is the density of wood. We will now solve the given equation,
${F_b} = \rho g \times \dfrac{m}{{{\rho _L}}} - mg$
Substitute m=20 kg, $\rho = 1000\,kg/{m^3}$, ${\rho _L} = 600\,kg/{m^3}$ in the above equation.
$\begin{array}{c}
{F_b} = 1000\,kg/{m^3} \times 9.8\,m/{s^2} \times \dfrac{{20\,kg}}{{600\,kg/{m^3}}} - 20\,kg \times 9.8m/{s^2}\\
= 133.33\,N
\end{array}$
Therefore, the weight that can be put on the raft is 133.33 N.
Note:It is advised to keep in mind that the principle of floatation is nothing but simply an application of Archimedes' principle as Archimedes made the first hypothesis about the relativity of displacement and density of the matter immersed and verified it before making it a principle.
Complete step by step answer:
According to the Principle of Floatation, a body floats in a liquid if the weight of the body is equal to the weight of the liquid displaced by it but the floating of a body occurs if and only if the density of the body is less than or equal to the density of fluid. It could be further explained as, when a solid body is dipped into a fluid, the fluid exerts an upward force of buoyancy on the solid and if the force of buoyancy equals to the weight of the solid, it will remain in the state of equilibrium, thus leading to the state of floatation.
The value of the buoyant force ${F_b}$ is given by:
${F_b} = \rho gV - W$
Here, $\rho ,g,V$ is the density of fluid, acceleration due to gravity and displaced volume respectively.
The volume V can be written as $\dfrac{m}{\rho }$.
Hence the force of buoyancy becomes,
${F_b} = \rho g \times \dfrac{m}{{{\rho _L}}} - W$
Here, ${\rho _L}$ is the density of wood. We will now solve the given equation,
${F_b} = \rho g \times \dfrac{m}{{{\rho _L}}} - mg$
Substitute m=20 kg, $\rho = 1000\,kg/{m^3}$, ${\rho _L} = 600\,kg/{m^3}$ in the above equation.
$\begin{array}{c}
{F_b} = 1000\,kg/{m^3} \times 9.8\,m/{s^2} \times \dfrac{{20\,kg}}{{600\,kg/{m^3}}} - 20\,kg \times 9.8m/{s^2}\\
= 133.33\,N
\end{array}$
Therefore, the weight that can be put on the raft is 133.33 N.
Note:It is advised to keep in mind that the principle of floatation is nothing but simply an application of Archimedes' principle as Archimedes made the first hypothesis about the relativity of displacement and density of the matter immersed and verified it before making it a principle.
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