Question

When the temperature of a metal cylinder is raised from 0.0°C to 110°C, its length increases by 0.22%. Find the percent change in density.

Answer 1

Volume of cylinder = pi*r*2*L

As, from the above formula,volume is directly proportional to length, 
So, if we increase in length also increases in volume by 0.22%

we know 

density=mass/volume

As, density is inversely proportional to volume it means increasing in volume decreases the density by 50.22%

Answer 2

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 %).