Question

A metal pipe is 77 cm long. The inner diameter of a cross
section is 4 cm. the outer diameter being 4.4 cm
(see Fig. 13.11). Find its
(i) inner curved surface area,
(1) outer curved surface area.
(iii) total surface area,​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Height of metal pipe, (h) = 77 cm
• Inner diameter of a cross section = 4 cm
• Outer diameter of a cross section = 4.4 cm

$\underline{\bf{Explanation\::}}$

As we know that formula of the curved surface area of cylinder;

$\boxed{\bf{C.S.A = 2 \pi rh\:\:\:(sq.unit)}}$

• (i) Inner curved surface area :

A/q

→ Inner radius of a cross section (r) = Diameter/2

→ Inner radius of a cross section (r) = 4/2

→ Inner radius of a cross section (r) = 2 cm

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = 2\pi rh}$

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = 2\times \dfrac{22}{7} \times 2 \times 77}$

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = \dfrac{44}{\cancel{7}} \times 2 \times \cancel{77}}$

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = 44 \times 2 \times 11}$

$\mapsto\bf{C.S.A \:_{(metal\:pipe)} = 968\:cm^{2}}$

• (ii) Outer curved surface area :

→ Outer radius of a cross section (R) = Diameter/2

→ Outer radius of a cross section (R) = 4.4/2

→ Outer radius of a cross section (R) = 2.2 cm

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = 2\pi Rh}$

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = 2\times \dfrac{22}{7} \times 2.2 \times 77}$

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = \dfrac{44}{\cancel{7}} \times 2.2 \times \cancel{77}}$

$\mapsto\tt{C.S.A \:_{(metal\:pipe)} = 44 \times 2.2 \times 11}$

$\mapsto\bf{C.S.A \:_{(metal\:pipe)} = 1064.8\:cm^{2}}$

• (iii) Total surface area :

As we know that formula of the T.S.A of cylinder;

$\boxed{\bf{T.S.A= C.S.A_{(inner)} + C.S.A_{(outer)} + 2 \times Area\:of\:base}}$

So,

⇒ Area of base = Area of circle with greater radius - Area of circle with smaller radius

⇒ Area of base = πR² - πr²

⇒ Area of base = π(2.2)² - π(2)²

⇒ Area of base = π(2.2 × 2.2 - 2 × 2)

⇒ Area of base = 22/7 (4.84 - 4)

⇒ Area of base = 22/7 × 0.84

⇒ Area of base = 22 × 0.12

⇒ Area of base = 2.64 cm

Now,

→ Total surface area of cylinder = 968 cm² + 1064.8 cm² + 2 × 2.64 cm

→ Total surface area of cylinder = 2032.8 cm² + 5.28 cm

→ Total surface area of cylinder = 2038.08 cm² .

Answer 2

Answer:

Inner Surface area of the Pipe = 968 Square cm.

Outer Surface area of the pipe = 1064.8 Square cm.

Total Surface Area = = 2038.08 Square cm.

Step-by-step explanation:

A pipe will have two layers as shown in picture.  

There is a inner Cylinder and Outer Cylinder.

Inner Radius = Inner Diameter / 2 = 4 / 2 = 2 cm = r

Outer Radius = Outer Diameter / 2 = 4.4/2 = 2.2 cm = R

Height of the pipe = 77 cm = h

Inner Surface area of the Pipe = 2∏rh = 2*22*2*77/7 = 968 Square cm.

Outer Surface area of the pipe = 2∏Rh = 2*22*2.2*77/7 = 1064.8 Square cm.

Total Surface area = Inner Surface area + Outer Surface area + 2 * Cross section surface area.  

The cross section suface area is present at two ends of the pipe.  

This is the area covered between 2 circles.  

Total Surface area = Inner Surface area + Outer Surface area + 2 * Cross section surface area.  

= 968 + 1064.8 + 2*∏(R^2 – r^2)

= 2032.8 + 2*22*(2.2*2.2 – 2*2)/7

= 2032.8 + 5.28

= 2038.08 Square cm.