Question

A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just float in water. [Specific gravity of wood is 0.8 and that of wood is 11.3.]
(A) 54.8g
(B) 11.11g
(C) 19.3g
(D) 24.8g

Answer 1

Hint : The specific gravity of an object is equal to the ratio of density of an object to the density of water. Taking density of water as unity, this formula reduces to mass per unit volume of the object. The volume of the water displaced is equal to the volume of the block plus the volume of the lead piece.

Formula used: In this solution we will be using the following formula,
$Volume = \dfrac{{{\text{Mass }}}}{{{\text{Specific gravity}}}}$

Complete step by step answer:
We are given a cubical block of wood whose mass is given as 200 g. A lead piece fastened underneath it and let its mass be $m$ .
When the block is put into water, the mass of water displaced by it is equal to the sum of the mass of wood and that of lead.
Therefore, the mass of water displaced is given by the following equation. $\therefore {\text{mass of water displaced = m + 200g}}$
The specific gravity of an object is defined as the mass of the object divided by its volume. We are given that the specific gravity of wood is 0.8 and that of lead is 11.3.
For the wooden block and the lead piece, we can write,
Volume of wooden block is V, and $V = \dfrac{{{\text{Mass of wooden block}}}}{{{\text{Specific gravity of wood}}}} = \dfrac{{200}}{{0.8}} = 250{m^3}$
Volume of lead given by, $v = \dfrac{{{\text{Mass of lead piece}}}}{{{\text{Specific gravity of lead}}}} = \dfrac{m}{{11.3}}$
Since, the specific gravity of water is 1, we can write,
${\text{Volume of water displaced = }}\dfrac{{{\text{Mass of water displaced}}}}{{{\text{Specific gravity of water}}}} = \dfrac{{m + 200}}{1}$
So, the total volume of water that will be displaced is equal to the volume of the wooden block plus the volume of the lead piece. Therefore, we can write
$m + 200 = \dfrac{m}{{11.3}} + 250$
Simplifying the equation,
$\Rightarrow m - \dfrac{m}{{11.3}} = 250 - 200 = 50$
$\Rightarrow m \times \dfrac{{10.3}}{{11.3}} = 50$
Thus, the value of $m$ is calculated from here.
$\Rightarrow m = 54.85g$
This is the required mass of the lead piece which will just allow the block to float in water.
Hence the correct answer is Option A.

Note:
To avoid confusion, we must always remember that, in these problems, the unit of force is taken as gram-weight or kg-weight, not dyne or Newton. This is because while equating, the conversion factor g, the acceleration due to gravity, cancel out.
The specific gravity of an object has no dimensions as it is the ratio of two same quantities. The specific gravity can also be called the relative density because it is defined with reference to a certain material, in this case, water.