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Chapter 1: Problem 13

If the density of a substance is \(2 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), then the mass of \(5 \mathrm{~m}^{3}\) of this substance is _______. (1) \(1000 \mathrm{~kg}\) (2) \(10000 \mathrm{~g}\) (3) \(10000 \mathrm{~kg}\) (4) Both (1) and (2)

Short Answer

Expert verified
Answer: (3) \(10000~kg\).

Step by step solution

01

Write down the given information

We are given: - Density of the substance, \(\rho = 2\times 10^3~kg/m^3\) - Volume of the substance, \(V = 5~m^3\) We need to find the mass of the substance.
02

Recall the formula for density

The formula for density is \(\rho = \frac{m}{V}\), where \(\rho\) is density, \(m\) is mass, and \(V\) is volume.
03

Rearrange the formula to solve for mass

We need to solve for mass, so we rearrange the density formula to isolate \(m\): \(m = \rho \times V\)
04

Plug in the given values

Substitute the given values for density and volume into the formula: \(m = (2\times 10^3~kg/m^3) \times (5~m^3)\)
05

Calculate the mass

Calculate the mass by multiplying the density and volume: \(m = 10^4~kg\) So, the mass of the substance is \(10000~kg\).
06

Match the answer with given options

The calculated mass matches with option (3) \(10000~kg\). Therefore, the correct answer is: (3) \(10000~kg\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Calculation
To determine the mass of a substance when its density and volume are known, you need to use a specific calculation. This involves multiplying the density by the volume. The mass is a measure of how much matter is contained in the substance, often expressed in kilograms or grams.

It's crucial to match the units. If the density is in kilograms per cubic meter \(kg/m^3\), and the volume is in cubic meters \(m^3\), the mass will be calculated in kilograms. Always ensure that units align to avoid errors in calculations. In our example, with a density of \(2 \times 10^3 \, kg/m^3\) and a volume of \(5 \, m^3\), multiplying the two gives us a mass of \(10000 \, kg\).

  • Check units for consistency
  • Use the formula: \(m = \rho \times V\)
  • Remember, mass tells you how much there is of the substance!
Volume
Volume refers to the amount of space that a substance occupies. It's a three-dimensional measure, and its units include cubic meters \(m^3\), cubic centimeters \(cm^3\), and liters for liquids.

Understanding the concept of volume is essential when dealing with physical properties like density. A larger volume means more space is filled by the substance, and combined with density, it helps determine the mass. When solving problems involving volume, typical measurements might include objects or containers, and it often requires a good grasp of geometric computations.

  • Volume is typically measured in \(m^3\) or \(cm^3\)
  • It's crucial for calculating mass when the density is known
  • Helps in understanding how much space a substance covers
Density Formula
The density formula is a fundamental concept in physics and chemistry. It is expressed as \(\rho = \frac{m}{V}\), where \(\rho\) is the density, \(m\) is the mass, and \(V\) is the volume.

Density helps describe how compact or concentrated a substance is. Higher density implies that more mass is packed into a given volume. This concept is vital in identifying substances, quality control, and understanding buoyancy. By rearranging the density formula to solve for mass, \(m = \rho \times V\), we can effectively calculate the mass when density and volume are known.

  • Key for understanding material properties
  • Used in many scientific and engineering applications
  • Starts with \(\rho = \frac{m}{V}\) and rearranges to find mass

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Most popular questions from this chapter

The volume of mercury and water is \(50 \mathrm{ml}\) each. What is the ratio of their mass, if their densities are in the ratio \(68: 5\) ?

Match the entries given in column A with the appropriate ones in column \(B\). $$ \begin{array}{lllll} \hline & {\text { Column A }} & {\text { Column B }} \\ \hline \text { A. } & 1 \mathrm{~g} \mathrm{~cm}^{-3} & (\quad) & \text { a. } 1000 \mathrm{~kg} \mathrm{~m}^{-3} \\ \text { B. } & \text { Convection current } & ( \quad ) & \text { b. } \mathrm{cm}^{3} \\ \text { C. } & \text { Volume } & ( \quad ) & \text { c. } \text { Measurement of large distances } \\ \text { D. } & \text { Triangulation method } & (\quad ) & \text { d. } \text { Mass/volume } \\ \text { E. } & \text { Mass } & (\quad ) & \text { e. } \text { Change in density } \\ \text { F. } & \text { Density } & \text { ( ) } & \text { f. } \text { kg } \\\ \hline \end{array} $$

Density of the material of a paper is given as \(0.5 \mathrm{~g} \mathrm{~cm}^{-3}\). The mass of the paper is \(1 \mathrm{~g}\) and its length and breadth are \(10 \mathrm{~cm}\) and \(5 \mathrm{~cm}\) respectively. Arrange the following steps in a sequence to find the thickness of the paper. (a) The thickness of the paper is \(=\frac{\text { volume }(\mathrm{V}) \text { of the paper }}{\text { length } \times \text { breadth }}\) (b) The density (d) of the material of the paper is \(=\frac{\text { mass of the paper(m) }}{\text { volume of the paper(V) }}\) (c) Then the volume \((\mathrm{V})\) of the paper \(=\frac{\text { mass of the paper }(\mathrm{m})}{\text { density }(\mathrm{d}) \text { of the paper }}\) (d) The volume \((\mathrm{V})\) of the paper is \(=\) length \(\times\) breadth \(\times\) thickness of the paper. (1) abcd (2) badc (3) \(\mathrm{abdc}\) (4) \(\mathrm{bcda}\)

A density bottle weighs \(100 \mathrm{~g}\) when filled with liquid and \(80 \mathrm{~g}\) when filled with water. If the weight of an empty density bottle is \(20 \mathrm{~g}\), find the density of the liquid.

5 litre of a liquid weighs \(5 \mathrm{kgf}\). The density of the liquid is ______. (1) \(1 \mathrm{~kg} \mathrm{~m}^{-3}\) (2) \(1 \mathrm{~g} \mathrm{~cm}^{-3}\) (3) \(100 \mathrm{~kg} \mathrm{~m}^{-3}\) (4) \(100 \mathrm{~g} \mathrm{~m}^{-3}\)
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