Question

A hollow sphere of external radius \[R\] and thickness \[t(\ll R)\] is made of a metal of density \[\rho \]. The sphere will float in water if:
A. \[t\le \dfrac{R}{\rho }\]
B. \[t\le \dfrac{R}{3\rho }\]
C. \[t\le \dfrac{R}{2\rho }\]
D. \[t\ge \dfrac{R}{3\rho }\]

Answer 1

Hint: Any material can float in the water if it can displace a large amount of water compared to its weight. Here we are using a hollow sphere. So, we can find out the area of the hollow sphere from the given data. From this area, thickness and density we can calculate the mass of the hollow sphere. We can compare this mass with the mass of the displaced water to make the float as possible.

Complete Step by Step Answer:
For the floating of a hollow sphere, the mass of the displaced water has to be larger than that of the actual mass of the hollow sphere.
As we know the density of the material is \[\rho \].
Area of the hollow sphere can be written as,
\[A=4\pi {{R}^{2}}\], where R is the radius of the sphere.
Mass of the hollow sphere can be written as,
\[{{m}_{s}}=A\times t\times \rho \], where t is the thickness of the hollow sphere.
\[{{m}_{s}}=4\pi {{R}^{2}}\times t\times \rho \]
Now we can calculate the volume of the water displaced.
Volume of the water displaced will be equal to the volume of the hollow sphere.
\[V=\dfrac{4}{3}\pi {{R}^{3}}\]
To find the mass of the water displaced, we can multiply the volume with density of water. As we know the density of water can be written as 1. So, the mass of the water displaced will be equal to,
\[{{m}_{w}}=\dfrac{4}{3}\pi {{R}^{3}}\]
For the floating of hollow sphere, \[{{m}_{s}}\le {{m}_{w}}\]
We can assign the values of mass of water displaced and mass of sphere.
\[\Rightarrow 4\pi {{R}^{2}}\times t\times \rho \le \dfrac{4}{3}\pi {{R}^{3}}\]
Cancel the quantities on both sides.
\[\Rightarrow t\times \rho \le \dfrac{1}{3}R\]
So, the thickness of the hollow sphere has to be \[t\le \dfrac{R}{3\rho }\], for the floating purpose.
Therefore, the correct option is C.

Additional Information: According to the Archimedes principle, the weight of an object in the air is known as its actual weight. If the object is immersed in a fluid and the measured weight is known as its apparent weight. Buoyant force is the reason behind this apparent weight loss. This apparent weight loss will be the same of the weight of the fluid displaced by the immersed body.

Note: The condition for the floating is that the weight of the material has to be less than the weight of the displaced water. Here we don’t have to consider the weight, mass is enough to calculate. The acceleration due to gravity will be the same for both things. Don’t forget to consider the density of the water as 1.