NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.4
Question 1

Question. 1

A solid metallic hemisphere of radius 8 cm is melted and recast into a right circular cone of base radius 6 cm. Determine the height of the cone.

Answer:

\(\displaystyle h=\dfrac{256}{9}\,\text{cm}\approx 28.44\,\text{cm}\)

Detailed Answer with Explanation:

Step 1: Understand the problem.

A metallic hemisphere is melted and reshaped into a cone. Since the material is the same, the volume of hemisphere = volume of cone.

Step 2: Write the formula for the volume of a hemisphere.

The volume of a sphere is \(V = \dfrac{4}{3}\pi r^3\).

A hemisphere is half a sphere, so:

\(V_{hemisphere} = \dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3\).

Step 3: Substitute the radius of the hemisphere.

Here, \(r = 8\,\text{cm}\).

So, \(V_{hemisphere} = \dfrac{2}{3}\pi (8)^3 = \dfrac{2}{3}\pi (512) = \dfrac{1024}{3}\pi\,\text{cm}^3\).

Step 4: Write the formula for the volume of a cone.

The volume of a cone is \(V_{cone} = \dfrac{1}{3}\pi R^2 h\),

where \(R\) is the base radius, \(h\) is the height.

Step 5: Substitute the base radius of the cone.

Here, \(R = 6\,\text{cm}\).

So, \(V_{cone} = \dfrac{1}{3}\pi (6)^2 h = \dfrac{1}{3}\pi (36) h = 12\pi h\,\text{cm}^3\).

Step 6: Equate the volumes.

\(V_{hemisphere} = V_{cone}\)

\(\dfrac{1024}{3}\pi = 12\pi h\)

Step 7: Simplify the equation.

Cancel \(\pi\) on both sides:

\(\dfrac{1024}{3} = 12h\)

Multiply both sides by 3:

\(1024 = 36h\)

Divide both sides by 36:

\(h = \dfrac{1024}{36} = \dfrac{256}{9}\,\text{cm}\).

Step 8: Write the final answer in decimal form.

\(h \approx 28.44\,\text{cm}\).

Therefore, the height of the cone is \(28.44\,\text{cm}\).

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