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

Let \(p_0 \, \, and \, \, p_T\) be densities of glycerin at \(0^{\circ}C\)
and T\(^{\circ}\)C respectively. Then 
\(p_T = p_0 (1- \gamma \Delta)\) 
where \(\gamma\) is the coefficient of volume expansion of 
glycerin and \(\Delta T\) is rise in temperature. 
\(\frac{p _T}{ p_0} = 1- \gamma \Delta T \, \, or \, \, \gamma \Delta T = 1-\frac{p _T}{ p_0}\) 
Thus , \(\frac{p_0 - p_T}{p_0} = \gamma \Delta T\) 
Here, \(\gamma =5 \times 10^{-4} K^{-1} \, \, and \, \, \Delta T = 40^\circ C = 40 K\)
\(\therefore \, \,\) The fractional change in the density of glycerin 
\(=\frac{p_0 - p_T}{p_0} = \gamma \Delta T = (5 \times 10^{-4} K^{-1}) (40 K) =0.020\)
So, the correct answer is (D): 0.020.

Answer 2

The correct option is (D): 0.020.
\(d_{f}=\frac{d_{1}}{(1+γΔT)}\)
Fractional change \(= \frac{d_{1}−d_{f}}{d_{1}} = 1− \frac{d_{f}}{d_{1}}\)
\(= 1−(1+γΔT)^{−1} = 1−(1−γΔT)\)
\(∵(1+x)^{n}≈1+nx=γΔT\)
\(= 5×10^{−4}×40 = 0.020\).

Answer 3

The Correct option is (D)

Given:

The value of the coefficient of volume expansion of \(5\times10^{−4}K^{−1}\)

The original density of glycerine is

\(ρ=ρ_{0}​(1+YΔT)\)

\(⇒ρ−ρ_{0}​=ρ_{0}​YΔT​\)

Thus, the fractional change in the density of glycerin for a rise of \(40\degree C\) in its temperature.

\(\frac{ρ−ρ_{0}}{ρ_{0}}​​=YΔT=5\times10^{−4}\times40\)

\(200\times10^{−4}=0.020​\)

Answer 4

The density of a body is the mass per unit volume while volume is the 3D space occupied by an object. As volume changes with temperature, density also changes by an inverse relation. The change \((\Delta)\) in volume for every unit change \((\Delta)\) in temperature is given by the coefficient of volume expansion. The energy of the system rises with a rise in temperature.

Formulae used:
\(ρ=\frac{M}{V}\)
\(ρ=ρ_{0}(1+γΔT)\)

Complete step-by-step solution:
The density of a body is the measure of its mass per unit volume. Its SI unit is \(kgm^{−2}\). It is given as-
\(ρ=\frac{M}{V}\)
Where,
ρ = density
M = mass
V = volume
Let the initial volume of glycerin = ρ₀.
We know that the new density on volume expansion is given as-
\(ρ=ρ_{0}(1+γΔT)\)
Here,
\(γ\) = the coefficient of volume expansion
\(ΔT\) = the change in temperature
Given temperature changes by 40 units, the value of γ is \(5\times10^{−4}K^{−1}\)
In order to calculate new density, substitute the given values in the above equation:
\(ρ=ρ_{0}(1+γΔT)\)

\(⇒ρ=ρ_{0}(1+5\times10^{−4}\times40)\)

\(⇒ρ=ρ_{0}(1+0.02)\)

\(∴ρ=1.02ρ_{0}\)

\(\text{Fractional change in density} = \frac{\text{Change in density}}{\text{Initial density}}\)

\(\frac{Δρ}{ρ}=\frac{1.02ρ_{0}−ρ_{0}}{ρ_{0}}\)

\(∴\frac{Δρ}{ρ}=0.02\)

Hence, the fractional change in the density is 0.02.