Question

If the density of a block is $981kg/{m^3}$ , it shall:
(A) Sink
(B) Float
(C) Float completely immersed in water
(D) Float completely out of water

Answer 1

Hint To solve this question, we should first find the buoyant force acting on the block, when it is completely immersed in the water. If this comes out to be more than the weight of the block then it will float otherwise it will sink.

Formula Used: The formula used to solve this question is given as,
$\Rightarrow {F_B} = \rho g{V_i}$
Here, ${F_B}$ is the buoyant force acting on a body, $\rho$ is the density of the fluid in which the body is immersed, $g$ is the acceleration due to gravity, and ${V_i}$ is the part of the volume of the body immersed in the fluid.

Complete step by step answer
In the question, we are given the density of the block to be $981kg/{m^3}$ , i.e.
$\Rightarrow {\rho _{block}} = 981kg/{m^3}$
Now, we know that if the Buoyant Force acting on the Block is greater than the weight of the block, then it shall float on water, whereas, if the Buoyant Force acting on the block is less than weight of the block, the block shall sink. However, if the Buoyant Force acting on the block equals the weight of the block, then it will stay in equilibrium at that position.
Now, from our previous knowledge, we know two things.
Firstly, density of water is $1000{\text{ }}kg/{m^3}$ i.e.
${\rho _{water}} = 1000{\text{ }}kg/{m^3}$ And, secondly, we know that density of an object is the ratio of its mass to volume. This means that density of the block is,
$\Rightarrow {\rho _{block}} = \dfrac{{Mass}}{{Volume}} \\\ \Rightarrow {\rho _{block}} = \dfrac{M}{V} \\\$
Where, $M$ is the mass of the block and $V$ is the volume of the block. Now, we can rearrange the terms to get,
$\Rightarrow V = \dfrac{M}{{{\rho _{block}}}}$
Putting the value of ${\rho _{block}}$ in the above equation we get,
$\Rightarrow V = \dfrac{M}{{981}}$
Now, we know from the Archimedes Principle that the buoyant force acting on the block can be given as,
$\Rightarrow {F_B} = {\rho _{water}}g{V_i}$
Now we assume that it is completely immersed in water. So that ${V_i} = V$ . So the buoyant force becomes
$\Rightarrow {F_B} = {\rho _{water}}gV$
So, putting the value of ${\rho _{water}}$ and $V$ we get,
$\Rightarrow {F_B} = (1000)g\left( {\dfrac{M}{{981}}} \right)$
Now putting the value of $g = 9.81m/{s^2}$
$\Rightarrow {F_B} = (1000)(9.81)\left( {\dfrac{M}{{981}}} \right) \\\ \Rightarrow {F_B} = \left( {\dfrac{{9810}}{{981}}} \right)(M) \\\$
This gives,
$\Rightarrow {F_B} = 10M$ ........................(1)
The weight of the block is given as,
$weigh{t_{block}} = Mg$
Putting $g = 9.81m/{s^2}$
$\Rightarrow weigh{t_{block}} = 9.81M$ …………………………..(2)
Thus, we can easily see that,
$\Rightarrow 10M > 9.81M$
From (1) and (2)
$\Rightarrow {F_B} > weigh{t_{block}}$
Here ${F_B}$ is the value of the buoyant force when it is completely immersed in water.
Since it is greater than the weight, it will push the block out of water and thus the block cannot remain completely immersed.
Hence, the option C is incorrect.
Also, it cannot throw the block completely out of the water, as some upward force is needed for the equilibrium of the block.
Hence, the option D is also incorrect.
So, the Block will float on the surface of water, with some fraction of its volume immersed.
Hence, the correct answer is option B.

Note
We could also have attempted this question very easily by simply taking the ratio of the density of the block to that of the water. We know from the law of floatation that the fraction of volume of a body immersed in a fluid is equal to the ratio of the density of the body to that of the fluid. Using this method we can conclude the answer to this question in seconds.