The coefficient of volume expansion of glycerine is 49 × 10 ^ - 5...

Let ${V}_{0}$ be the initial volume of glycerine, i.e, at $0 ° C$ (dry). If ${V}_{t}$ be its volume at $30 ° C$

Then ${V}_{t} = {V}_{0} \left(1 + v ∆ t\right)<br /><br />$= {V}_{0} \left(1 + 49 \times {10}^{- 5} \times30\right)$<br /><br />${V}_{t} = {V}_{0}\left(1 + 0.01470\right) = 1.0147070{V}_{0}$<br /><br />$⇒ \dfrac{{V}_{0}}{{V}_{t}} = \dfrac{1}{1.01470}$<br /><br />Let${ρ}_{o}and&{ρ}_{t}$be the initial and final densities of glycertine then initial density,${ρ}_{0} = \dfrac{m}{{v}_{0}} and final density, ${ρ}_{t} = \dfrac{m}{{V}_{t}}$

Where, $m =$mass of glycertine

$\dfrac{∆ ρ}{{ρ}_{0}} =$ fractional change in the density

$\dfrac{{ρ}_{t} - {ρ}_{0}}{{ρ}_{0}} = \dfrac{m\left(\dfrac{1}{{V}_{t}} - \dfrac{1}{{V}_{0}}\right)}{\dfrac{m}{{V}_{0}}} = \left(\dfrac{{V}_{0}}{{V}_{t}} -1\right)$

$\dfrac{∆ ρ}{{ρ}_{0}} = \left(\dfrac{1}{1.01470} -1\right) = - 0.0145$

Here, negative sign shows that density decreases with rise in temperature

$\dfrac{∆ ρ}{{ρ}_{0}} =0.0145 = 1.45 \times {10}^{- 2}$

$\dfrac{∆ ρ}{{ρ}_{0}} =1.45 \times {10}^{- 2}$