Question

A piece of iron weighs $155gf$ in air and $133gf$ in water. What is its volume and the relative density of the iron? What would be the same piece of iron appear to weigh in alcohol of density $0.8g/c{m}^3$?

Answer 1

Step 1: Given data,
Weight of iron in air = $155gf$
Weight of iron in water = $133gf$
The density of iron = $0.8g/c{m}^3$
We know that the density of water at $4^{\circ }C$is = $1g/c{m}^3$

Step 2: Finding the volume of iron,
The loss in weight = $155gf-133gf=22gf$
So, the weight of water displaced by the iron when fully immersed will also be = $22gf$ (Mass will be $22g$)
$Volumeofwater=\frac{mass}{density}=\frac{22g}{1g/c{m}^3}=22cc$
Hence, the volume of iron will be equal to the volume of water displaced = $22cc$

Step 3: Finding the relative density of iron,
$RelativeDensity=\frac{Weightofthebody}{Lossofweightofthebodyinwater}=\frac{155gf}{22gf}=7.05$

Hence, the relative density of iron is $7.05$

Step 4: Finding the density of iron,
$Density=\frac{mass}{volume}=\frac{155g}{22c{m}^3}=7.05g/c{m}^3$

Step 5: Finding the weight of iron in alcohol,
$RelativeDensity=\frac{Densityofiron}{Densityofalcohol}=\frac{7.05g/c{m}^3}{0.8g/c{m}^3}=8.8$

$RelativeDensityisalsoequalto==\frac{Weightofthebody}{Lossofweightofthebodyinalcohol}=\frac{155gf}{155gf-weightofironinalcohol}=8.8(frompreviousequation)$

So, by cross-multiply we get, $Weightofironinalcohol=155gf-\frac{155gf}{8.8}=137.4g$

Hence, the weight of iron in alcohol is $137.4g$