Question

A body weighs \[25gf\] in air and \[20gf\] in water. What would be its weight in a liquid of density $0.8g/c{m^3}$ ?

Answer 1

Hint: A body changes its weight in different mediums due to change in relative density. When we know about the density of both the mediums and actual weight of the body in one of those mediums, we can easily predict the body’s weight in the other medium with the help of a basic formula.

Formulae used :
$\dfrac{{{\rho _m}}}{{{\rho _o}}} = \dfrac{{loss{\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} weight{\kern 1pt} in{\kern 1pt} {\kern 1pt} medium{\kern 1pt} {\kern 1pt} m}}{{actual{\kern 1pt} weight{\kern 1pt} of{\kern 1pt} the{\kern 1pt} {\kern 1pt} object}}$ where ${\rho _m}$ is the density of the medium and ${\rho _o}$ is the density of the object.

Complete step by step solution:
According to the question,
Weight of the body in air ${W_A} = 25kgf$
Weight of the body in water ${W_w} = 20kgf$
Density of water ${\rho _w} = 1g/c{m^3}$
Density of the liquid ${\rho _l} = 0.8g/c{m^3}$
If we use the formula which helps us find the loss in weight in different mediums, we could easily solve this question.
Now, using the formulae for relative density for air and water, we get
$
   \Rightarrow \dfrac{{{\rho _w}}}{{{\rho _o}}} = \dfrac{{loss{\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} weight}}{{actual{\kern 1pt} {\kern 1pt} weight}} \\
   \Rightarrow \dfrac{1}{{{\rho _o}}} = \dfrac{5}{{25}} \\
   \Rightarrow {\rho _o} = 5g/c{m^3} \\
 $
Therefore, density of the object is $5g/c{m^3}$
Now, using the above results and applying the same formulae for finding the loss of weight in the liquid with density $0.8g/c{m^3}$, we get
$
   \Rightarrow \dfrac{{{\rho _l}}}{{{\rho _o}}} = \dfrac{{loss{\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} weight}}{{actual{\kern 1pt} {\kern 1pt} weight}} \\
   \Rightarrow \dfrac{{0.8}}{5} = \dfrac{{loss{\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} weight}}{{25}} \\
   \Rightarrow loss{\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} weight = 4kgf \\
    \\
 $
Therefore, the weight of the liquid can be calculated by subtracting this loss of weight in the liquid medium from the actual weight of the body
${W_L} = {W_A} - loss{\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} weight$
$
  {W_L} = (25 - 4)kgf \\
  {W_L} = 21kgf \\
 $
Therefore, the body weighs $21kgf$ in a liquid of density $0.8g/c{m^3}$

Note:
Sometimes, you might encounter a question where the density of the liquid will be high. In such cases, the object will become extremely light in that medium and may even not penetrate the medium if the liquid’s density is more than the body itself.