Question

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A metallic pipe is 0.7 cm thick. Inner radius of the pipe is 3.5 cm and length is 5 dm . Find its TSA (Total Surface Area).

[HINT: Total Surface area= Inner Surface Area + Outer Surface Area + Area of two rims]

Answer 1

Given:

Thickness of the pipe is 0.7 cm

Inner radius (r) is 3.5 cm

length (height) is 5dm = 50 cm

[ as 10 cm = 1 dm ]

Outer radius (R) = (3.5 + 0.7) = 4.2 cm

$\mathbb{TO\: \:BE\: \: FOUND,}$TOBEFOUND, $\bf{The \:total \:surface\: area \:of \:the \:pipe.}$Thetotalsurfaceareaofthepipe.

$\bf{Total \:surface\: area }$Totalsurfacearea

$\bf{= [ inner surface\: area + outer \:surface \:area + area \:of \:two \:rims ] }$=[innersurfacearea+outersurfacearea+areaoftworims]

Total surface area = 2π(r + R) (h + R - r) unit square

Total surface area = 2 × 22/7(3.5+4.2) (50+4.2-3.5) cm²

Total surface area = 44/7 × 7.7 × 50.7 cm²

Total surface area = 2453.88 cm²

Answer 2

✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔ that would be the area of the inner cylinder, plus the area of the outer cylinder, plus the area of the two rings at the end. So, if
R = outer radius
r = inner radius
h = length of pipe

the total area is

2πrh + 2πRh + π(R^2-r^2)
= 2πh(r+R) + π(R+r)(R-r)
= π(R+r)(2h+R-r)
Total surface area = inner curved surface area + outer curved surface area + area of 2 rings 

We will consider , 
R=outer curved surface area 
r=inner curved surface area 
h=given lenght 

=2πRh + 2πrh + πR^2 + πr^2
=π(r +R )+(2h + R-r )
After solving with values we will get the total surface area as 2453.88 cm ^2
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