NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.4
Question 11

Question. 11

Sixteen glass spheres, each of radius 2 cm, are packed into a cuboidal box of internal dimensions \(16\,\text{cm}\times8\,\text{cm}\times8\,\text{cm}\) and the box is then filled with water. Find the volume of water filled.

Answer:

\(\displaystyle 1024-\dfrac{512}{3}\pi\;\text{cm}^3\;\approx 488\,\text{cm}^3\)

Detailed Answer with Explanation:

Step 1: Volume of the cuboidal box

A cuboid's volume is found by multiplying its length, breadth, and height:

\(V_{\text{box}} = 16 \times 8 \times 8 = 1024\,\text{cm}^3\).

Step 2: Volume of one glass sphere

The formula for the volume of a sphere is:

\(V_{\text{sphere}} = \dfrac{4}{3}\pi r^3\).

Here radius \(r = 2\,\text{cm}\).

So, \(V_{\text{sphere}} = \dfrac{4}{3}\pi (2)^3 = \dfrac{4}{3}\pi (8) = \dfrac{32}{3}\pi\,\text{cm}^3\).

Step 3: Volume of 16 spheres

Total volume occupied by 16 spheres:

\(V_{\text{16 spheres}} = 16 \times \dfrac{32}{3}\pi = \dfrac{512}{3}\pi\,\text{cm}^3\).

Step 4: Space left for water

The remaining volume inside the box (after placing the spheres) will be filled with water:

\(V_{\text{water}} = V_{\text{box}} - V_{\text{16 spheres}}\).

\(V_{\text{water}} = 1024 - \dfrac{512}{3}\pi\,\text{cm}^3\).

Step 5: Approximate value

Using \(\pi \approx 3.14\):

\(V_{\text{water}} \approx 1024 - \dfrac{512}{3} (3.14)\).

\(\dfrac{512}{3} \times 3.14 \approx 536\).

\(V_{\text{water}} \approx 1024 - 536 = 488\,\text{cm}^3\).

Final Answer: The box can be filled with about \(488\,\text{cm}^3\) of water.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.4 | Detailed Answers