Directions: In the following question, two statements are numbered as Quantity I and Quantity II. On solving these statements, we get quantities I and II respectively. Solve both quantities and choose the correct option.

**Quantity I:** A solid sphere of radius 7 cm is melted and converted into a right circular cone of vertical angle 30°. Find the height (in cm) of the cone. (Take ∛12 = 2.29)

**Quantity II: **Moment of inertia of a square sheet is 72 cm^{4}. Calculate its side (in cm) if the radius of gyration of the square sheet is 9 cm.

(**Use****:** Moment of inertia = Area × (radius of gyration)^{2})

Option 1 : Quantity > Quantity II

**Solving for Quantity I:**

**Given:**

Radius of solid sphere = 7 cm

The vertical angle of cone = 30°

**Formula used:**

The volume of Sphere = 4/3 × π × R^{3}

The volume of Cone = 1/3 × π × r^{2} × h

**Calculations:**

By using trigonometry,

tan 30° = r/h

⇒ 1/√3 = r/h

⇒ r = h/√3

Volume of melted sphere = Volume of cone

⇒ 4/3 × π × R^{3} = 1/3 × π × r^{2} × h

⇒ 4 × (7)^{3} = (h/√3)^{2} × h

⇒ 4 × 343 = h^{3}/3

⇒ h^{3} = 12 × 343

⇒ h = ∛(12 × 343)

⇒ h = 7∛12

⇒ h = 7 × 2.29 cm

Height of the cone is 16.03 cm.

**Solving for Quantity II:**

**Given:**

Moment of inertia of a square sheet is 72 cm^{4}.

The radius of gyration of the rectangular sheet is 3 cm.

Moment of inertia = Area × (radius of gyration)^{2}

**Calculations:**

Moment of inertia = Area × (radius of gyration)^{2}

⇒ 256 = Area × 2^{2}

⇒ Area = 256/4

⇒ Area = 64 cm^{2}

⇒ Area = (Side)^{2}

⇒ Side = 8 cm

**∴**** Quantity I > Quantity II**

** **

Vertical angle is the angle made between height and slant height, not between slant height and base.