Question 1.5: A test tube is containing 100 ml water at density 1000 kg/m³......

Given,

At 10 °C,      \text { Volume of water, } V_1=500  ml =0.51=0.5 \times 10^{-3}  m ^3

\text { Density of water, } \rho_1=1000  kg / m ^3

Height of water column = 20 cm = 0.2 m

At 90° C,        \rho_2=920  kg / m ^3

Assuming no water loss in evaporation, the mass of water will remain conserved

\begin{aligned} m & =\rho_1 V_1=\rho_2 V_2 \\ 1000 \times 5 \times 10^{-4} & =920 \times V_2 \\ V_2 & =5.4348 \times 10^{-4}  m ^3 \\ \text { Area of tube } & =\frac{5 \times 10^4}{0.2} \\ \pi r_{10}^2 & =2.5 \times 10^{-3}  m ^2 \\ r_{10} & =0.02821 \end{aligned}

Now, radius at 90°C will be

\begin{aligned} & r_{90}=r_{10}(1+\alpha \Delta T) \\ & r_{90}=0.02821\left(1+3.6 \times 10^{-6}(90-10)\right) \\ & r_{90}=0.02822  m \end{aligned}

Now, area at 90°C,          A =\pi r_{90}^2=2.5015 \times 10^{-3}  m ^2

Height of water column at 90 °C

\begin{aligned} h_2 & =\frac{V_2}{A_2} \\ & =\frac{5.4348 \times 10^4}{2.5015 \times 10^3}=0.2173  m \\ & =21.73  cm \end{aligned}