Answer: The percent change in density is (-0.21 %).
Explanation:
Density before raising the temperature:
\(\rho_1=\frac{M}{V_1}=\frac{M}{\pi r^2h}\)
Temperature of a metal cylinder is raised from 0.0°C to 110°C.
\(h'=h+0.22\%\times h=h+0.0022h=1.0022 h\)
\(\rho_2=\frac{M}{V_2}=\frac{M}{\pi r^2h'}=\frac{M}{\pi r^2(1.0022)h}\)
\(\text{percentage change}=\frac{rho_2-rho_1}{\rho_1}\times 100\)
\(\text{percentage change in density}=\frac{\frac{M}{\pi r^2(1.0022)h}-\frac{M}{\pi r^2h}}{\frac{M}{\pi r^2h}}\times 100=-0.21\%\)
The negative sign indicates that density had been decreased.
The percent change in density is (-0.21 %).
