- Home
- Standard 11
- Physics
ग्लिसरीन का आयतन प्रसार गुणांक $5 \times 10^{-4} K ^{-1}$ है। ग्लिसरीन के तापक्रम में $40^{\circ} C$ वृद्धि करने पर उसके घनत्व में आंशिक परिवर्तन होगा
$0.01$
$0.015$
$0.02$
$0.025$
Solution
Let $r_0$ and $r_T$ be densities of glycerin at ${0^ \circ }C$ and ${T^ \circ }C$ respectively. Then,
${\rho _T} = {\rho _0}\left( {1 – \gamma \Delta T} \right)$
Where $\gamma $ is the cofficient of volume expansion of glycerine and $\Delta T$ is rise in temperature.
$\frac{{{\rho _T}}}{{{\rho _0}}} = 1 – \gamma \Delta Y\,\,or\,\,\gamma \Delta T = 1 – \frac{{{\rho _T}}}{{{\rho _0}}}$
$Thus,\frac{{{\rho _0} – {\rho _T}}}{{{\rho _0}}} = \gamma \Delta T$
$Here,\gamma = 5 \times {10^{ – 4}}{K^{ – 1}}\,and\,\Delta T = {40^ \circ }C = 40\,K$
The fractional change in the density of glycerin
$ = \frac{{{\rho _0} – {\rho _T}}}{{{\rho _0}}} = \gamma \Delta T = \left( {5 \times {{10}^{ – 4}}{K^{ – 1}}} \right)\left( {40\,K} \right) = 0.020$
