A metallic spherical shell with internal and external diameters 4 cm and 8 cm is recast into a cone of base diameter 8 cm. The height of the cone is
12 cm
14 cm
15 cm
18 cm
Step 1: Write the given data.
Step 2: Find the volume of the spherical shell.
Volume of shell = Volume of outer sphere − Volume of inner sphere.
Outer radius = 4 cm, Inner radius = 2 cm.
So,
\[ V_{shell} = \dfrac{4}{3}\pi \big( R^3 - r^3 \big) = \dfrac{4}{3}\pi (4^3 - 2^3). \]
\(4^3 = 64, \; 2^3 = 8 \).
Therefore, \( V_{shell} = \dfrac{4}{3}\pi (64 - 8) = \dfrac{4}{3}\pi (56) = \dfrac{224}{3}\pi \; \text{cm}^3. \)
Step 3: Volume of the cone formed.
Formula: \( V_{cone} = \dfrac{1}{3} \pi r^2 h \)
Here, \(r = 4\,\text{cm}, h = ?\)
So, \( V_{cone} = \dfrac{1}{3} \pi (4^2) h = \dfrac{1}{3} \pi (16) h = \dfrac{16}{3}\pi h \).
Step 4: Equating the volumes.
Since the shell is melted to form the cone:
\( V_{shell} = V_{cone} \)
\( \dfrac{224}{3}\pi = \dfrac{16}{3}\pi h \)
Step 5: Simplify and solve for \(h\).
Cancel \( \dfrac{\pi}{3} \) on both sides:
\(224 = 16h \)
\( h = \dfrac{224}{16} = 14 \)
Final Answer: Height of the cone = 14 cm.