Question

A test tube loaded with lead shots, weighs $150\,gf$ and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that $27\,gf$ of lead shots have to be removed, so as to float it to level X. Find R.D of alcohol.

Answer 1

Hint:The relative density of alcohol is calculated with reference to the density of water. The relative density of a fluid is directly proportional to the ratio of weight of the fluid to the weight of the reference fluid. Buoyant force acts on the test tube, it is directly proportional to the weight of the object immersed in fluid.

Formulae used:
$W=mg$
The weight of the fluid displaced= weight of the test tube
Weight of alcohol displaced= weight of the test tube
$\text{R}\text{.D=}\dfrac{\text{W}\,\text{of}\,\text{alcohol}\,\text{displaced}}{\text{W}\,\text{of}\,\text{water}\,\text{displaced}}$

Complete step-by-step solution:
The relative density of a substance is the ratio of its density with a reference material. Here, the reference substance is water; this means that relative density of alcohol is to be calculated with reference to water.
Weight is the gravitational force acting on an object. It is given by-
$W=mg$
Here,
$W$ is the weight of the object
$m$ is the mass of the object
$g$ is acceleration due to gravity
Given, lead shots of $150\,gf$ are added to water and water level rises upto a mark X.
By the Archimedes principle,
The weight of the fluid displaced= weight of the test tube $=\,150\,gf$
Then the test tube is made to float in alcohol,
Again, by the Archimedes principle,
Weight of alcohol displaced= weight of the test tube= $150-27=123\,gf$
 In both cases fluid rises upto level X, therefore,
Volume of alcohol displaced= volume of water displaced
Therefore, R.D of alcohol is given by-
$\begin{align}
  & \text{R}\text{.D=}\dfrac{\text{W}\,\text{of}\,\text{alcohol}\,\text{displaced}}{\text{W}\,\text{of}\,\text{water}\,\text{displaced}} \\
 & \Rightarrow \text{R}\text{.D}=\dfrac{123}{150} \\
 & \therefore \text{R}\text{.D}=0.82 \\
\end{align}$

Therefore, the relative density of alcohol is $0.82$.

Note:
The Archimedes principle states that the buoyant force acting on an object is proportional to the fluid displaced by it. Density is mass divided by volume. The $gf$ or the gram-force unit is a unit of force which is defined as the mass of one gram multiplied by acceleration due to gravity.