First slide

Thermal expansion

Question

A solid body is in the form of a cube of side A cm at —10°C. When it is heated through a small temperature difference of t°C, its volume increases by B cc. When it is completely converted into liquid at 0°C, its coefficient of volume expansion becomes ‘n’ times that of the substance in solid form. The fractional change in density of the liquid when heated through a temperature difference of t°C from 0°C is

Difficult

A

$\frac{1}{1 - \frac{A^{3}}{n B}}$
B

$\frac{1}{\left( 1+\frac{{{A}^{3}}}{nB} \right)}$
C

$\left( 1+\frac{{{A}^{3}}}{nB} \right)$
D

$\left( 1-\frac{{{A}^{3}}}{nB} \right)$

Solution

V₁=A³ ; V₂–V₁=B ; ${{\gamma }_{s}}=\frac{\Delta V}{V\Delta t}=\frac{B}{{{A}^{3}}t}$
$\frac{{{d}_{0}}}{{{d}_{t}}}=1+{{\gamma }_{R}}t\Rightarrow \frac{{{d}_{0}}-{{d}_{t}}}{{{d}_{0}}}=1-\frac{1}{1+{{\gamma }_{R}}t}=\frac{{{\gamma }_{R}}t}{1+{{\gamma }_{R}}t}$
$\frac{\Delta d}{{{d}_{0}}}=\frac{1}{1+\frac{1}{{{\gamma }_{R}}t}}$
Given that ${{\gamma }_{R}}=n{{\gamma }_{s}}=\frac{nB}{{{A}^{3}}t};\frac{\Delta d}{d}=\frac{1}{1+\frac{{{A}^{3}}}{nb}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App