1. **Problem Statement:** Three solid spheres with diameters $\frac{3}{2}$ m, 2 m, and $\frac{5}{2}$ m are melted and combined to form a new solid sphere. We need to find the diameter of the new sphere. 2. **Formula Used:** The volume of a sphere is given by: $$V = \frac{4}{3} \pi r^3$$ where $r$ is the radius of the sphere. 3. **Step 1: Calculate the volume of each original sphere.** - Sphere 1 diameter = $\frac{3}{2}$ m, so radius $r_1 = \frac{3}{4}$ m. - Sphere 2 diameter = 2 m, so radius $r_2 = 1$ m. - Sphere 3 diameter = $\frac{5}{2}$ m, so radius $r_3 = \frac{5}{4}$ m. Calculate volumes: $$V_1 = \frac{4}{3} \pi \left(\frac{3}{4}\right)^3 = \frac{4}{3} \pi \frac{27}{64} = \frac{4}{3} \pi \times \frac{27}{64}$$ $$V_2 = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi$$ $$V_3 = \frac{4}{3} \pi \left(\frac{5}{4}\right)^3 = \frac{4}{3} \pi \frac{125}{64} = \frac{4}{3} \pi \times \frac{125}{64}$$ 4. **Step 2: Sum the volumes to get the total volume of the new sphere.** $$V_{total} = V_1 + V_2 + V_3 = \frac{4}{3} \pi \left(\frac{27}{64} + 1 + \frac{125}{64}\right)$$ Combine fractions: $$\frac{27}{64} + \frac{125}{64} = \frac{152}{64} = \frac{19}{8}$$ So: $$V_{total} = \frac{4}{3} \pi \left(\frac{19}{8} + 1\right) = \frac{4}{3} \pi \left(\frac{19}{8} + \frac{8}{8}\right) = \frac{4}{3} \pi \times \frac{27}{8}$$ 5. **Step 3: Let the radius of the new sphere be $R$. Then:** $$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi \times \frac{27}{8}$$ Cancel $\frac{4}{3} \pi$ on both sides: $$\cancel{\frac{4}{3} \pi} R^3 = \cancel{\frac{4}{3} \pi} \times \frac{27}{8}$$ So: $$R^3 = \frac{27}{8}$$ 6. **Step 4: Solve for $R$.** $$R = \sqrt[3]{\frac{27}{8}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2}$$ 7. **Step 5: Find the diameter of the new sphere.** Diameter $D = 2R = 2 \times \frac{3}{2} = 3$ m. **Final Answer:** The diameter of the new sphere is **3 meters**.