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Question

216 identical solid spheres of radius 7cm are melted to form a big sphere, find the diameter of the big sphere.

The correct answer is
84 cm

Calculating Big Sphere Diameter from Melting Small Spheres

This problem involves understanding the principle of volume conservation. When identical solid spheres are melted and reformed into a larger sphere, the total volume of the material remains the same. We need to find the diameter of the resulting larger sphere.

Understanding Volume Conservation Principle

The core idea is that the total volume of the 216 small spheres is equal to the volume of the single large sphere formed after melting.

Formulas Needed

The volume ($V$) of a sphere with radius ($r$) is given by the formula: $ V = \frac{4}{3}\pi r^3 $

Step-by-Step Calculation

  1. Identify Given Information:
    • Number of small spheres, $N = 216$.
    • Radius of each small sphere, $r = 7$ cm.
  2. Calculate the Volume of One Small Sphere: Using the formula $V_{small} = \frac{4}{3}\pi r^3$: $ V_{small} = \frac{4}{3}\pi (7 \text{ cm})^3 $ $ V_{small} = \frac{4}{3}\pi (343) \text{ cm}^3 $
  3. Calculate the Total Volume of Small Spheres: Multiply the volume of one small sphere by the total number of spheres: $ V_{total} = N \times V_{small} $ $ V_{total} = 216 \times \left( \frac{4}{3}\pi (7)^3 \right) \text{ cm}^3 $ $ V_{total} = 216 \times \frac{4}{3}\pi \times 343 \text{ cm}^3 $
  4. Relate Total Volume to the Big Sphere's Volume: Let the radius of the big sphere be $R$. Its volume is $V_{big} = \frac{4}{3}\pi R^3$. Since volume is conserved: $ V_{big} = V_{total} $ $ \frac{4}{3}\pi R^3 = 216 \times \left( \frac{4}{3}\pi (7)^3 \right) $
  5. Solve for the Radius (R) of the Big Sphere: Cancel out the common term $\frac{4}{3}\pi$ from both sides: $ R^3 = 216 \times (7)^3 $ To find $R$, take the cube root of both sides: $ R = \sqrt[3]{216 \times 7^3} $ We know that $\sqrt[3]{216} = 6$ (since $6^3 = 216$) and $\sqrt[3]{7^3} = 7$. $ R = \sqrt[3]{6^3} \times \sqrt[3]{7^3} $ $ R = 6 \times 7 $ $ R = 42 \text{ cm} $
  6. Calculate the Diameter of the Big Sphere: The diameter ($D$) is twice the radius ($D = 2R$): $ D = 2 \times R $ $ D = 2 \times 42 \text{ cm} $ $ D = 84 \text{ cm} $

Conclusion

By applying the principle of volume conservation and the formula for the volume of a sphere, we determined that the diameter of the large sphere formed is 84 cm.

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Important Questions from Mensuration 3D (Notes)

  1. The height of a cylinder is 14cm and its curved surface area is 264cm². The volume of the cyclinder (in cm³) is:
    ($\pi=\frac{22}{7}$)
  2. A cylindrical rod has an outer curved surface area of \(7500 \text{ cm}^2\). If the length of the rod is 92 cm, then the outer radius (in cm) of the rod, rounded off to two places of decimal, is:
    \(\left(\text{Take } \pi = \frac{22}{7}\right)\)
  3. A number of 512 identical small spheres are cast from a sphere of radius 40 cm, with the total volume of the small spheres being equal to the volume of the larger sphere. The diameter (in cm) of each of the small spheres is:
  4. There is a wooden block in the form of a cube whose each side is 8 meters long. 

    The maximum possible number of cylinders with a diameter of 1 meter and a height of 4 meters were cut from this block. The cylinders are to be painted at the rate of ₹14 per square meter.
     

    What is the total amount (in ₹) needed to paint all the cylinders if we paint the entire surface of each cylinder? (Take $\pi = \frac{22}{7}$)

  5. If the lateral surface area of a cylinder is $140.1 \text{ cm}^2$ and its height is $3 \text{ cm}$, then find its volume. (Use $\pi = 3.14$ and round off to two decimal places.)
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