Solveeit Logo
Login

Question

Physics Question on Thermal Expansion

The value of coefficient of volume expansion of glycerin is 5×10−4K−15 \times 10^{-4} K^{-1}. The fractional change in the density of glycerin for a rise of 40∘C40 ^{\circ} C in its temperature, is

A

0.025

B

0.01

C

0.015

D

0.02

Answer

0.02

Explanation

Solution

Let p0  and  pTp_0 \, \, and \, \, p_T be densities of glycerin at 0∘C0^{\circ}C
and T∘^{\circ}C respectively. Then
pT=p0(1−γΔ)p_T = p_0 (1- \gamma \Delta)
where γ\gamma is the coefficient of volume expansion of
glycerin and ΔT\Delta T is rise in temperature.
pTp0=1−γΔT  or  γΔT=1−pTp0\frac{p _T}{ p_0} = 1- \gamma \Delta T \, \, or \, \, \gamma \Delta T = 1-\frac{p _T}{ p_0}
Thus , p0−pTp0=γΔT\frac{p_0 - p_T}{p_0} = \gamma \Delta T
Here, γ=5×10−4K−1  and  ΔT=40∘C=40K\gamma =5 \times 10^{-4} K^{-1} \, \, and \, \, \Delta T = 40^\circ C = 40 K
∴  \therefore \, \, The fractional change in the density of glycerin
=p0−pTp0=γΔT=(5×10−4K−1)(40K)=0.020=\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.