Question

A block B of specific gravity 2 and another block C of specific gravity 0.5 Both are joined together and they are floating inside water such that they are completely dipped inside water the ratio of the masses of the blocks B and C is
(A) 2:1
(B) 3:2
(C) 5:3
(D) 2:3

Answer 1

Hint: We know that a body at rest in a fluid is acted upon by a force pushing upward called the buoyant force, which is equal to the weight of the fluid that the body displaces. If the body is completely submerged, the volume of fluid displaced is equal to the volume of the body. If the body is only partially submerged, the volume of the fluid displaced is equal to the volume of the part of the body that is submerged. Based on this concept we have to answer this question.

Complete step by step answer
The two blocks joined together are B and C. It is said that even though the blocks are completely dipped in the water they are floating in the water. Therefore, there are two forces acting on the body; the weight of the body and buoyant force. Let us consider the weight of the body. The weight of the full body is the sum of the weight of the blocks B and C.
We know that weight of a body is the product of the mass of that body and earth's gravity, i.e.$w=m g, w^{\prime}$ is the weight of the body, 'm' is the mass of the body and 'g' is the gravitational acceleration. Let ' $w_{B}$ ' be the weight of the block B. Therefore the weight of block B will become,$w_{B}=m_{B} \times g,$ where $^{\prime} m_{B}^{\prime}$ is the mass of the block B and 'g' is gravity. Now, let ${ }^{\prime} w_{C}{ }^{\prime}$ be the weight of the block C, then we have, $w_{C}=m_{C} \times g,$ were ${ }^{\prime} m_{C}^{\prime}$ is the mass of the block C. Now we have total weight of the body,
$\boldsymbol{w}=\left(\boldsymbol{m}_{B} \times \boldsymbol{g}\right)+\left(\boldsymbol{m}_{C} \times \boldsymbol{g}\right)$
$w=\left(m_{B}+m_{C}\right) g$
The other force acting on the body is the buoyant force, $F_{B}$
We know that the buoyant force is the weight of the liquid displaced by the body.
Therefore, in this case the buoyant force of the total body is the sum of the buoyant force of the block B and C.
Therefore, we have,
$F_{B}=\rho_{w}\left(V_{B}+V_{C}\right) g,$ where ${ }^{\prime} F_{B}^{\prime}$ is buoyant force, ${ }^{\prime} \rho_{w}^{\prime}$ is the density of water, ${ }^{\prime} V_{B}^{\prime}$ is the volume of block B, $^{\prime }V_{C}^{\prime }$ is the volume of block C and ${ }^{\prime} g^{\prime}$ is gravity.
We know that density is mass per unit volume, i.e. $\rho=\dfrac{M}{V}$ where $^{\prime} \rho^{\prime}$ is density, 'M' is mass and $v$ ' is volume.
From the above equation, we can find the volume of block B as,
$V_{B}=\dfrac{m_{B}}{\rho_{B}}$
In the question, we are given the specific gravity of block B.
We know that specific gravity is the ratio of the density of a substance and density of water. We are given the specific gravity of block B to be 2. Therefore, we can write,
$\dfrac{\rho_{B}}{\rho_{w}}=2$
Therefore, we get the density of block B as,
$\rho_{B}=2 \rho_{w}$
Substituting this in the equation of volume of block B, we get
$V_{B}=\dfrac{m_{B}}{2 \rho_{w}}$
We are also given the specific gravity of block C to be 0.5
Similarly, we get volume of block C as, $V_{C}=\dfrac{m_{C}}{\rho_{C}}$
$V_{C}=\dfrac{m_{C}}{0.5 \rho_{w}}$
Now we can write buoyant force of the body as, $F_{B}=\rho_{w}\left(V_{B}+V_{C}\right) g$
$F_{B}=\rho_{w}\left(\dfrac{m_{B}}{2 \rho_{w}}+\dfrac{m_{C}}{0.5 \rho_{w}}\right) g$
$F_{B}=\dfrac{\rho_{w}}{\rho_{w}}\left(\dfrac{m_{B}}{2}+\dfrac{m_{C}}{0.5}\right) g$
$F_{B}=\left(0.5 m_{B}+2 m_{C}\right) g$
In the question it is said that the body is floating in water. We know that when a body floats in water, the weight of the body will be equal to the buoyant force.
Therefore,
$w=F_{B}$
We know the values of ' $w$ and ${ }^{\prime} F_{B}$ ". Substituting this in the above equation,
$\left(m_{B}+m_{C}\right) g=\left(0.5 m_{B}+2 m_{C}\right) g$
$\left(m_{B}+m_{C}\right)=\left(0.5 m_{B}+2 m_{C}\right)$
Let us take ${ }^{\prime} m_{C}{ }^{\prime}$ common on both sides of the equation,
we get $m_{C}\left(\dfrac{m_{B}}{m_{C}}+1\right)=m_{C}\left(\dfrac{0.5 m_{B}}{m_{C}}+2\right)$
$\Rightarrow \left(\dfrac{m_{B}}{m_{C}}+1\right)=\left(\dfrac{0.5 m_{B}}{m_{C}}+2\right)$
$\Rightarrow \dfrac{m_{B}}{m_{C}}-\dfrac{0.5 m_{B}}{m_{C}}=2-1$
$\Rightarrow \dfrac{0.5 m_{B}}{m_{C}}=1$
$\Rightarrow \dfrac{m_{B}}{m_{C}}=\dfrac{1}{0.5}$
Multiplying denominator and numerator with 2 on both
sides we get, $\dfrac{2 m_{B}}{2 m_{C}}=\dfrac{1}{0.5} \times \dfrac{2}{2}$
$\dfrac{m_{B}}{m_{C}}=\dfrac{2}{1}$
Therefore, the ratio of the masses of blocks B and C is 2: 1

Hence the correct answer is option A.

Note: To answer such a question, it should be known to us that Archimedes' principle is very useful for calculating the volume of an object that does not have a regular shape. The oddly shaped object can be submerged, and the volume of the fluid displaced is equal to the volume of the object. It can also be used in calculating the density or specific gravity of an object.
Let us explain with the help of an example, for an object denser than water, the object can be weighed in air and then weighed when submerged in water. When the object is submerged, it weighs less because of the buoyant force pushing upward. The object's specific gravity is then the object's weight in air divided by how much weight the object loses when placed in water. But most importantly, the principle describes the behaviour of anybody in any fluid, whether it is a ship in water or a balloon in air.