Percentage 81914

The carpenter worked on a rotary cylinder with a base radius of 2.5 dm and a height of 2 dm. He reduced the radius by 1 cm by uniform grinding, and the height of the cylinder was preserved. Calculate the percentage by which the volume of the cylinder has shrunk.

Final Answer:

p =  6.1333 %

Step-by-step explanation:

$$\begin{gathered} r_1=2.5\text{dm}\to \text{cm}=2.5\cdot 10\text{cm}=25\text{cm} \\ h=2\text{dm}\to \text{cm}=2\cdot 10\text{cm}=20\text{cm} \\ {r}_2=r_1-1=25-1=24\text{cm} \\ {S}_1=2\pi \cdot r_1^2+2\pi \cdot r_1\cdot h=2\cdot 3.1416\cdot 25^2+2\cdot 3.1416\cdot 25\cdot 20\doteq 7068.5835{\text{cm}}^2 \\ {S}_2=2\pi \cdot r_2^2+2\pi \cdot r_2\cdot h=2\cdot 3.1416\cdot 24^2+2\cdot 3.1416\cdot 24\cdot 20\doteq 6635.0437{\text{cm}}^2 \\ p=100\cdot \frac{S_1-S_2}{S_1}=100\cdot \frac{7068.5835-6635.0437}{7068.5835}=6.1333\% \end{gathered}$$

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