Question

A spring balance read $10\,kg$ when a pail of water is suspended from it. What be its reading when (a)an ice cube of mass $1.5\,kg$ is put into the pail.,(b) an iron piece of mass $7.8\,kg$ suspended by another string immersed with half its volume inside the water in the pail Relative density of iron is 7.8 .

Answer 1

Hint-Relative density of iron is the ratio of density of iron ${\rho _{iron}}$ to density of water ${\rho _{water}}$,
$Relative\,density = \dfrac{{{\rho _{iron}}}}{{{\rho _{water}}}}$
${\rho _{water}} = {10^3}\,kg/{m^3}$
Density is mass divided by volume .$\rho = \dfrac{m}{V}$
Upthrust acting on a body immersed in water is given by the equation
$U = V{\rho _{water}}g$

Step by step solution:
Given ,
Mass of pail of water,${m_{water}} = 10\,kg$
Mass of ice cube ${m_{ice = }}1.5\,kg$
Mass of iron piece ${m_{iron}} = 7.8\,kg$
Relative density of iron = 7.8
Part (a)
When ice cube is put into the pail, total reading= ${m_{water}} + {m_{iron}} = 10\,kg + 1.5\,kg = 11.5\,kg$
Part (b)
Relative density of iron is the ratio of density of iron ${\rho _{iron}}$ to density of water ${\rho _{water}}$,
$Relative\,density = \dfrac{{{\rho _{iron}}}}{{{\rho _{water}}}}$
. Its value is given as 7.8 and we know that ${\rho _{water}} = {10^3}\,kg/{m^3}$. Substitute these values in the equation. We get
$
  7.8 = \dfrac{{{\rho _{iron}}}}{{{{10}^3}\,kg/{m^3}}} \\
  {\rho _{iron}} = {10^3}\,kg/{m^3} \times 7.8 \\
   = 7.8 \times {10^3}\,kg/{m^3} \\
 $
Density is mass divided by volume .$\rho = \dfrac{m}{V}$
Therefore, volume of iron
$
  V = \dfrac{m}{\rho } \\
   = \dfrac{{7.8\,kg}}{{7.8 \times {{10}^3}\,kg/{m^3}}} \\
   = {10^{ - 3}}\,{m^3}\, \\
 $
Upthrust acting on a body immersed in water is given by the equation
$U = V{\rho _{water}}g$
Given only half of the iron is immersed in water .therefore, volume inside water =$\dfrac{V}{2}$
Therefor, $U = \dfrac{V}{2}\rho g$
Substitute all the given values
$U = \dfrac{{{{10}^{ - 3}}}}{2} \times {10^3} \times 9.8 = 4.9\,N$
$ \Rightarrow U = \dfrac{{4.9}}{{9.8}}\,kgf = 0.5\,kgf$
 The weight of the iron piece is balanced by the string to which it is attached. Still iron pieces will exert an equal and opposite force of \[0.5\,kgf\]. Hence the total reading shown by the spring balance is $10\,kg + 0.5\,kg = 10.5\,kg$

Note:Upthrust acting on a body in water is calculated as
$U = V{\rho _{water}}g$ . Here the density taken is the density of water and not of the body immersed. The volume shown in the equation is the volume of the water displaced. So it will be equal to the volume of the immersed part of the body. Be careful not to take the volume of the entire body. We need only the volume of the immersed part