Question

If density of milk is $ 1.02{\text{ }}gm/cc $ , weight of one liter of milk is
 $
  (A) \;\;\;\;\;1{\text{ }}kg \\
  (B) \;\;\;\;\;1.02{\text{ }}kg \\
  (C) \;\;\;\;\;9.6{\text{ }}N \\
  (D) \;\;\;\;\;10{\text{ }}N \\
  $

Answer 1

Hint: Density is given by the formula $ \rho = \dfrac{m}{v} $ where $ \rho $ is density, $ m $ is mass and $ v $ is the volume. The formula can be interchanged according to the values given in a particular question. This question also just interchanges the values given. Read the question properly to solve such questions.

Complete answer:
In this particular question we are given that the density of milk is $ 1.02{\text{ }}gm/cc $ . And we have to find the weight of $ 1{\text{ }}L $ milk. Here the volume of milk $ 1{\text{ }}L $ and the density of milk are given so by using the formula of density we can find out the mass and then substitute the value of mass in another formula to find out the weight
Density is denoted by sign known as Rho
So formula of density is
 $ \begin{array}{*{20}{l}}
  {\rho = \dfrac{m}{v}} \\
  {m = \rho \times v}
\end{array} $
Given that volume $ = {\text{ }}1{\text{ }}L $ now this $ 1{\text{ }}L $ can also be written as $ 1{\text{ c}}{{\text{m}}^3} \times 1000 $
 So $ v = 1{\text{ }}L{\text{ = 1c}}{{\text{m}}^3} \times 1000 $ and $ \rho = {\text{ }}1.02{\text{ }}gm/cc $
Substituting in the formula, we get
  $ m = 1.02{\text{ }}\dfrac{{gm}}{{cc}}{\text{ }} \times {\text{ }}1{\text{ L}} $
 $ {\text{ = 1}}{\text{.02 }}\dfrac{{gm}}{{cc}} \times 1{\text{ c}}{{\text{m}}^3} \times 1000 $
 $ \; = {\text{ }}1.{\text{ }}02{\text{ }}gm \times 1000 $
 $ = 1020gm $
 We get the value here in gm. We need to convert this value in kg to get to the final result. To convert it into kg we will just divide the answer by $ 1000 $
So
 $
  \dfrac{{1020}}{{1000}} \\
   = 1.02kg \\
  $
Therefore the correct option is $ B.\;\;\;\;\;1.02{\text{ }}kg $.

Note:
This question is quite simple to solve. You just need to first interchange the formula of density and then substitute the values to get the final answer. There is no alternate way for this particular question. Here in the above question the unit of density is given as $ \dfrac{{gm}}{{cc}} $, this means grams per centimeter cube. It is a CGS unit of density.