Assertion (A) : A semi-circular sheet of metal of diameter 14 cm is bent into an open conical cone, then height of the cone is $\dfrac{7}{2}\sqrt{3}$ cm.

Reason (R) : When a sector is bent into an open cone, the radius of the sector becomes the slant height of the cone and the length of the arc becomes the circumference of the circular base of the cone.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

We know that,

When a sector is bent into an open cone, the radius of the sector becomes the slant height of the cone and the length of the arc becomes the circumference of the circular base of the cone.

Given,

A semi-circular sheet of metal of diameter 14 cm is bent into an open conical cone.

Radius of semi-circle (r) = $\dfrac{14}{2}$ = 7 cm.

Circumference of semi-circle = πr = $\dfrac{22}{7} \times 7$ = 22 cm.

Circumference of base of cone = Circumference of semi-circle = 22 cm.

Let radius of cone be R cm.

⇒ 2πR = 22

⇒ R = $\dfrac{22}{2π} = \dfrac{22}{2 \times \dfrac{22}{7}} = \dfrac{22 \times 7}{2 \times 22} = \dfrac{7}{2}$ cm.

Slant height of cone (l) = Radius of semi-circular sheet = 7 cm.

Let height of cone be h cm.

By formula,

⇒ l² = r² + h²

⇒ 7² = $\Big(\dfrac{7}{2}\Big)^2$ + h²

⇒ 49 = $\dfrac{49}{4}$ + h²

⇒ h² = $49 - \dfrac{49}{4}$

⇒ h² = $\dfrac{196 - 49}{4}$

⇒ h² = $\dfrac{147}{4}$

⇒ h = $\sqrt{\dfrac{147}{4}}$

⇒ h = $\dfrac{7\sqrt{3}}{2}$ cm.

Hence, Option 3 is the correct option.