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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 \( \pi=\frac{22}{7} \) ]

Ask by Frazier Nichols. in Gambia
Feb 03,2025

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Tutor-Verified Answer

Answer

The radius of the sphere is approximately 14 cm.

Solution

Given: - Diameter of the cylindrical iron block = 31.5 cm - Height of the cylindrical iron block = 18 cm - \( 82\% \) of the melted metal was used to mould a sphere To find the radius of the sphere, we need to follow these steps: 1. Calculate the volume of the cylindrical iron block. 2. Calculate the volume of the sphere using the volume of the cylindrical iron block. 3. Find the radius of the sphere. Let's start by calculating the volume of the cylindrical iron block: The formula for the volume of a cylinder is: \[ V = \pi r^2 h \] where: - \( V \) is the volume of the cylinder - \( r \) is the radius of the cylinder - \( h \) is the height of the cylinder Given that the diameter of the cylinder is 31.5 cm, the radius is half of the diameter: \[ r = \frac{31.5}{2} = 15.75 \, \text{cm} \] Now, we can calculate the volume of the cylindrical iron block: \[ V = \pi \times (15.75)^2 \times 18 \] Next, we will calculate the volume of the sphere using the volume of the cylindrical iron block: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \] where: - \( V_{\text{sphere}} \) is the volume of the sphere - \( r \) is the radius of the sphere We are given that \( 82\% \) of the melted metal was used to mould the sphere. Therefore: \[ V_{\text{sphere}} = 0.82 \times V \] Now, we can find the radius of the sphere by solving the equation: \[ \frac{4}{3} \pi r^3 = 0.82 \times V \] Let's calculate the volume of the cylindrical iron block and then find the radius of the sphere. Calculate the value by following steps: - step0: Calculate: \(\pi \times 15.75^{2}\times 18\) - step1: Convert the expressions: \(\pi \left(\frac{63}{4}\right)^{2}\times 18\) - step2: Multiply the terms: \(\frac{3969\pi }{16}\times 18\) - step3: Reduce the numbers: \(\frac{3969\pi }{8}\times 9\) - step4: Multiply: \(\frac{3969\pi \times 9}{8}\) - step5: Multiply: \(\frac{35721\pi }{8}\) Calculate or simplify the expression \( 0.82 * 14027.603897 \). Calculate the value by following steps: - step0: Calculate: \(0.82\times 14027.603897\) - step1: Multiply the numbers: \(11502.63519554\) Calculate or simplify the expression \( \\frac{4}{3} * \\pi * r^3 = 11502.63519554 \). Solve the equation \( \frac{4}{3} * \pi * r^3 = 11502.63519554 \). Solve the equation by following steps: - step0: Solve for \(r\): \(\frac{4}{3}\pi r^{3}=11502.63519554\) - step1: Multiply: \(\frac{4\pi }{3}r^{3}=11502.63519554\) - step2: Multiply by the reciprocal: \(\frac{4\pi }{3}r^{3}\times \frac{3}{4\pi }=11502.63519554\times \frac{3}{4\pi }\) - step3: Multiply: \(r^{3}=\frac{1725395279331}{200000000\pi }\) - step4: Simplify the equation: \(\sqrt[3]{r^{3}}=\sqrt[3]{\frac{1725395279331}{200000000\pi }}\) - step5: Calculate: \(r=\sqrt[3]{\frac{1725395279331}{200000000\pi }}\) - step6: Simplify the root: \(r=\frac{\sqrt[3]{8626976396655\pi ^{2}}}{1000\pi }\) The radius of the sphere is \( r = \frac{\sqrt[3]{8626976396655\pi ^{2}}}{1000\pi} \) cm. To find the radius of the sphere to the nearest whole number, we can calculate the value of \( r \) and round it to the nearest whole number. Calculate the value by following steps: - step0: Calculate: \(\frac{\sqrt[3]{8626976396655\pi ^{2}}}{1000\pi }\) The radius of the sphere is approximately 14 cm when rounded to the nearest whole number.

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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: \[ \boxed{14} \text{ cm} \]

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