Question

Using a block that is $12.0\;cm$ wide, $7.0\;cm$ long and $9.0\;cm$ tall answer the following questions;
(a) If the block weights $500.0\;grams$ , how much of the block will be below the surface of the water?
(b) How much would the block have to weigh so that it floats half way under the water?
(c) If the entire block was under water how much water would it displace?

Answer 1

As a block is immersed in water, we can use the Archimedes’ principle which states that the buoyant force applied by the liquid on the object is equal to the weight of the liquid displaced by the object. The volume of the liquid displaced is certainly the volume of the block immersed in water. Hence, for the first question from the mass, the volume can be found. For the second and third question, volume is known and the weight can be found out.

Formula used:
Archimedes principle ${{F}_{b}}=\rho gV$
where, ${{F}_{b}}$ = Buoyant force, $\rho$ = Density of the fluid, $g$ = Acceleration due to gravity and $V$ = fluid volume.

Complete step by step answer:
Let us note down the given data;
Length of the block $l=12cm$
Width or Breadth of the block $b=7cm$
Height of the block $h=9cm$
Density of water (in $g\;cm^{-3} )$ $\rho =1g\;c{{m}^{-3}}$
Now, let us consider each question separately,

(a) Mass of the block $m=500g$. Let us suppose the block will be immersed in water up to height $x\;cm$.The volume of the water displaced is equal to the volume of the block immersed.Hence, for equilibrium the buoyancy force is balanced by the given weight of the block which can be expressed as,
${{F}_{b}}=mg$
The buoyant force, from the Archimedes principle is expressed as,
$\rho gV=mg$
$\Rightarrow \rho V=m$
Here, the meaning of the symbols in the above equation is written as
$\text{Mass of block=Density of water}\times \text{Volume of water displaced}$
We can substitute here the given value of mass and the dimensions of block
$500g=1gc{{m}^{-3}}\times 12cm\times 7cm\times xcm$
To simplify, let us not consider the units as they are getting balanced
$500=84x$
$\therefore x=5.95cm$
Hence, the block will be $x=5.95cm$ below the surface of water.

(b) As the block is immersed up to half height,
The height of block that is immersed in water is = $\left( \dfrac{9}{2} \right)cm$
Thus, the volume of the block immersed in water = $12cm\times 7cm\times \left( \dfrac{9}{2} \right)cm$
As per the Archimedes principle, the volume of the water displaced will be equal to the volume of the block immersed in water. The weight of the block is balanced by the buoyancy force, and the buoyancy force is expressed as
${{F}_{b}}=\rho gV$
Substituting the given values,
${{F}_{b}}=1gc{{m}^{-3}}\times 9.8m{{s}^{-2}}\times 12cm\times 7cm\times \left( \dfrac{9}{2} \right)cm$
Converting the metric value of acceleration due to gravity without mentioning the units,
${{F}_{b}}=1\times 9.8\times {{10}^{2}}\times 12\times 7\times \left( \dfrac{9}{2} \right)$
$\therefore {{F}_{b}}=370818\,dyne$
This is the weight of the block if the block is half immersed in water.

(c) From the Archimedes principle we know that the volume of the block immersed in water is equal to the volume of the water displaced.
The volume of water displaced = $12\,cm\times 7\,cm\times 9\,cm$
$\therefore$ The volume of water displaced = $756\,cm^3$

Note: From the data given, we must be able to understand that we have to use the Archimedes principle here. In the second question, we can further find the required mass of the block if the block is half immersed by dividing the obtained weight by acceleration due to gravity, as the weight of block is the product of the mass of block and acceleration due to gravity as shown
$\dfrac{370818}{9.81\times {{10}^{2}}}=m$
$\therefore m=378\,g$