3. A cylindrical iron block of diameter 31.5 cm and height 18 cm was completely melted. \( 82 \% \) of the melted metal was used to mould a sphere. Calculate, correct to the nearest whole number, the radius of the sphere. [Take \( \pi=\frac{22}{7} \) ]
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To solve the problem, we first need to find the volume of the cylindrical iron block. The formula for the volume \( V \) of a cylinder is given by: \[ V = \pi r^2 h \] Where \( r \) is the radius and \( h \) is the height. The diameter of the cylinder is 31.5 cm, so the radius \( r \) is: \[ r = \frac{31.5}{2} = 15.75 \text{ cm} \] The height \( h \) is 18 cm. Now we can calculate the volume: \[ V = \pi (15.75)^2 (18) \] \[ V = \frac{22}{7} (15.75)^2 (18) = \frac{22}{7} \times 248.0625 \times 18 \] \[ V \approx \frac{22}{7} \times 4465.125 \approx 14243.15 \text{ cm}^3 \] Next, we find the volume of the melted iron that was used to mould the sphere. Since \( 82\% \) of the volume is used, we calculate: \[ \text{Volume for sphere} = 0.82 \times 14243.15 \approx 11667.78 \text{ cm}^3 \] The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] Setting this equal to the volume used for the sphere: \[ \frac{4}{3} \pi r^3 = 11667.78 \] Substituting \( \pi \): \[ \frac{4}{3} \times \frac{22}{7} r^3 = 11667.78 \] \[ \frac{88}{21} r^3 = 11667.78 \] \[ r^3 = 11667.78 \times \frac{21}{88} \approx 2768.88 \] Now, we take the cube root to find the radius \( r \): \[ r \approx \sqrt[3]{2768.88} \approx 14.05 \text{ cm} \] Rounding to the nearest whole number, the radius of the sphere is approximately: \[ \boxed{14} \text{ cm} \]