Problem 22 - Master Resource Book in JEE Main Physics

Question

Density of substance at \(0^{\circ} \mathrm{C}\) is \(10 \mathrm{~g} / \mathrm{cc}\) and at \(100^{\circ} \mathrm{C}\), its density is \(9.7 \mathrm{~g} / \mathrm{cc}\). The coefficient of linear expansion of the substance is (a) \(1.03 \times 10^{-4} \mathrm{C}^{-1}\) (b) \(3 \times 10^{-4}{\underline{\phantom{xx}}}^{-4} \mathrm{C}^{-1}\) (c) \(19.7 \times 10^{-3 *} \mathrm{C}^{-1}\) (d) \(10^{-3}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}^{-1}\)

Step-by-Step Solution

Answer

Option (a): \(1.03 \times 10^{-4} \, \mathrm{C}^{-1}\).

Step 1: Understanding the Change in Density

First, identify the change in densities from the given problem. At \(0^{\circ} \mathrm{C}\), the density \(\rho_0\) is \(10 \mathrm{~g} / \mathrm{cc}\) and at \(100^{\circ} \mathrm{C}\), the density \(\rho_{100}\) is \(9.7 \mathrm{~g} / \mathrm{cc}\). Calculate the change in density, \(\Delta \rho = \rho_0 - \rho_{100} = 10 - 9.7 = 0.3 \mathrm{~g} / \mathrm{cc}\).

Step 2: Using the Formula Relating Density and Volume Expansion

We know that the density \(\rho\) is inversely proportional to the volume \(V\). Hence, when the volume increases with temperature, the density decreases. The formula relating density at two temperatures with the coefficient of volume expansion \(\beta\) is:\[ \frac{\rho_0}{\rho_t} = 1 + \beta \Delta T \]where \(t\) is the final temperature and \(\Delta T = t - 0\). Substituting the given values: \[ \frac{10}{9.7} = 1 + \beta \times 100 \]

Step 3: Solving for the Coefficient of Volume Expansion

From the equation \( \frac{10}{9.7} = 1 + 100\beta \), we rearrange to solve for \(\beta\):\[ \frac{10}{9.7} - 1 = 100\beta \]Calculate \( \frac{10}{9.7} \approx 1.03093 \), thus:\[ 100\beta = 1.03093 - 1 \approx 0.03093 \]\[ \beta \approx \frac{0.03093}{100} \approx 3.093 \times 10^{-4} \, \mathrm{C}^{-1} \]

Step 4: Converting Volume Expansion Coefficient to Linear Expansion Coefficient

The coefficient of linear expansion \(\alpha\) is related to the volume expansion coefficient \(\beta\) by \(\beta = 3\alpha\). Therefore, solve for \(\alpha\):\[ 3\alpha = 3.093 \times 10^{-4} \, \mathrm{C}^{-1} \]\[ \alpha = \frac{3.093 \times 10^{-4}}{3} \approx 1.031 \times 10^{-4} \, \mathrm{C}^{-1} \]

Step 5: Identifying the Correct Option

Compare the calculated value of \(\alpha\) with the given options. The closest match is (a) \(1.03 \times 10^{-4} \, \mathrm{C}^{-1}\).

Key Concepts

Coefficient of Linear Expansion

The concept of the coefficient of linear expansion is essential when discussing thermal expansion. It tells us how much a material will expand per unit length for every degree change in temperature. This expansion occurs along one dimension, such as length or height. Every material has its distinct coefficient of linear expansion denoted by the Greek letter \( \alpha \).Understanding \( \alpha \) is important in many practical applications. For instance, engineers need to consider it when constructing bridges or buildings to ensure that the components fit correctly at different temperatures. The formula for the linear expansion of an object is given by:\[\Delta L = \alpha \times L_0 \times \Delta T\]where \( \Delta L \) is the change in length, \( L_0 \) is the original length, and \( \Delta T \) is the temperature change. Thus, knowing \( \alpha \) helps in predicting the behavior of materials when temperatures fluctuate.

The relationship between linear and volume expansion is useful, especially in calculations involving changes in dimensions due to temperature, where often we use the relationship \( \beta = 3\alpha \) to find \( \alpha \) from \( \beta \). This relation comes from the assumption that expansion is uniform in all directions. Thus, \( \alpha = \frac{\beta}{3} \), as seen in our worked example where \( \alpha \) was determined to be \( 1.031 \times 10^{-4} \, \mathrm{C}^{-1} \).

Density Change with Temperature

As temperature increases, the volume of most substances also increases, leading to a decrease in density if the mass remains constant. This principle helps us understand phenomena like why hot air balloons rise – as the air inside the balloon heats up, its density decreases and it becomes buoyant.The relationship between volume and density is an inverse one. When a substance expands due to an increase in temperature, its volume rises but its mass does not change. Using the formula \[\rho = \frac{m}{V}\]where \( \rho \) is density, \( m \) is mass, and \( V \) is volume, you can see that as \( V \) increases, \( \rho \) decreases. This was demonstrated in our worked problem where the substance's density decreased from \( 10 \, \mathrm{g/cc} \) to \( 9.7 \, \mathrm{g/cc} \) as the temperature rose from \( 0^{\circ} \mathrm{C} \) to \( 100^{\circ} \mathrm{C} \).

By calculating the change in density and using the volume expansion coefficient, we can predict these density changes due to thermal effects, which impacts various fields from material science to meteorology.

Volume Expansion Coefficient

The volume expansion coefficient, denoted as \( \beta \), is a measure of how much a material's volume changes with a change in temperature. It is defined for scenarios where materials experience uniform expansion or contraction in all directions.This coefficient is crucial for understanding how a material behaves when subjected to temperature variations. The equation that describes the relationship between volume change and temperature is:\[\Delta V = \beta \times V_0 \times \Delta T\]In this formula, \( \Delta V \) is the change in volume, \( V_0 \) is the original volume, and \( \Delta T \) is the temperature change. When solving problems involving volume expansion, this formula allows you to calculate the expected change in volume based on known parameters.

In the problem we solved, we calculated \( \beta \) to be \( 3.093 \times 10^{-4} \, \mathrm{C}^{-1} \). This calculation used the relationship between density and volume expansion, showing that for the substance in question, this was the rate of volumetric increase per degree Celsius. Understanding \( \beta \) helps predict and explain the behavior of materials across different temperatures in both everyday applications like cooking and industrial applications like manufacturing.