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Question

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(

Vedclass · Accepted Answer

$0.02$

Vedclass · Answer

Let $r_0$ and $r_T$ be densities of glycerin at ${0^ \circ }C$ and  ${T^ \circ }C$ respectively. Then,

${ho _T} = {ho _0}\left( {1 - \gamma \Delta T} ight)$ 

Where $\gamma $ is the cofficient of volume expansion of glycerine and $\Delta T$ is rise in temperature.

$\frac{{{ho _T}}}{{{ho _0}}} = 1 - \gamma \Delta Y\,\,or\,\,\gamma \Delta T = 1 - \frac{{{ho _T}}}{{{ho _0}}}$

$Thus,\frac{{{ho _0} - {ho _T}}}{{{ho _0}}} = \gamma \Delta T$

$Here,\gamma  = 5 	imes {10^{ - 4}}{K^{ - 1}}\,and\,\Delta T = {40^ \circ }C = 40\,K$

The fractional change in the density of glycerin

$ = \frac{{{ho _0} - {ho _T}}}{{{ho _0}}} = \gamma \Delta T = \left( {5 	imes {{10}^{ - 4}}{K^{ - 1}}} ight)\left( {40\,K} ight) = 0.020$

અ.નં.	$l(m)$	$\Delta T{(^o}C)$	$\Delta l(m)$
$(1)$	$2$	$10$	$4\times 10^{-4}$
$(2)$	$1$	$10$	$4\times 10^{-4}$
$(3)$	$2$	$20$	$2\times 10^{-4}$
$(4)$	$3$	$10$	$6\times 10^{-4}$

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