A solid iron cuboid of dimensions 49 cm \(\times\) 33 cm \(\times\) 24 cm is moulded into a sphere. The radius of the sphere is
21 cm
23 cm
25 cm
19 cm
Step 1: Write the formula for the volume of a cuboid.
\(V_{\text{cuboid}} = \text{length} \times \text{breadth} \times \text{height}\)
Step 2: Put the given values (in cm).
\(V_{\text{cuboid}} = 49 \times 33 \times 24 = 38808 \, \text{cm}^3\)
Step 3: When the cuboid is melted and moulded into a sphere, the volume remains the same.
So, \(V_{\text{cuboid}} = V_{\text{sphere}}\)
Step 4: Formula for the volume of a sphere:
\(V_{\text{sphere}} = \dfrac{4}{3} \pi r^3\)
Step 5: Equating the volumes:
\(38808 = \dfrac{4}{3} \pi r^3\)
Step 6: Simplify.
Multiply both sides by 3:
\(3 \times 38808 = 4 \pi r^3\)
\(116424 = 4 \pi r^3\)
Step 7: Divide both sides by \(4\pi\).
\(r^3 = \dfrac{116424}{4\pi}\)
Taking \(\pi = \dfrac{22}{7}\):
\(r^3 = \dfrac{116424 \times 7}{4 \times 22} = \dfrac{815,000}{88} = 9261\)
Step 8: Find cube root.
\(r^3 = 9261 \implies r = \sqrt[3]{9261} = 21\)
Final Answer: The radius of the sphere is 21 cm.