Question

The value of coefficient of volume expansion of glycerin is \(5 \times 10^{-4} K^{-1}\). The fractional change in the density of glycerin for a rise of \(40 ^{\circ} C\) in its temperature, is

• A. 0.025
• B. 0.01
• C. 0.015
• D. 0.02

Answer 1

The Correct Option is D

 To calculate the fractional change in the density of glycerin, we can use the relation between the coefficient of volume expansion and density. The coefficient of volume expansion, denoted by \(\beta\), is defined as:

\[ \beta = \frac{\Delta V}{V_0 \Delta T} \]

where \(\Delta V\) is the change in volume, \(V_0\) is the initial volume, and \(\Delta T\) is the change in temperature.

The relation between volume and density is given by:

\[ \rho = \frac{m}{V} \]

where \(\rho\) is the density, \(m\) is the mass, and \(V\) is the volume. Since mass \((m)\) remains constant, any change in volume results in a change in density.

The fractional change in density \(\left(\frac{\Delta \rho}{\rho}\right)\) can be related to the coefficient of volume expansion as:

\[ \frac{\Delta \rho}{\rho} = -\beta \Delta T \]

Given:

• The coefficient of volume expansion of glycerin, \(\beta = 5 \times 10^{-4} \, K^{-1}\).
• The rise in temperature, \(\Delta T = 40^{\circ} C\).

Substituting these values into the equation:

\[ \frac{\Delta \rho}{\rho} = - (5 \times 10^{-4}) \times 40 \]

\[ \frac{\Delta \rho}{\rho} = -0.02 \]

The negative sign indicates a decrease in density as temperature increases, which is typical because as a liquid expands, its density decreases. Therefore, the magnitude of the fractional change in density is \(0.02\).

Hence, the correct answer is

0.02

.