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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 the total surface area.
Given :
Thickness of metallic pipe = 0.7 cm
Inner Radius (r) = 3.5 cm
External Radius (R) = $3.5 + 0.7 = 4.2 $ cm
Length of pipe (h) = 5 d m = 50 cm (1 d m = 10 cm)
To do:
We have to find the total surface area.
Solution :
To find the total surface area of pipe, find all the surfaces on pipe and add.
Total surface area of pipe = Inner surface area $+$ External surface area $+$ Surface Areas of top and bottom
Inner Surface area : Curved surface area of inner pipe = 2πrh.
$2\ π\ r\ h\ =\ 2\ \times \ \frac{22}{7} \ \times \ 3.5\ \times \ 50\ $
$=\ 2\ \times \ \frac{22}{2} \ \times \ 50$
$=22\ \times \ 50\ \ =\ 1100$ cm 2
Inner Surface area = 1100 cm 2
External Surface Area : Curved surface area of inner pipe = 2πRh
$2\ π\ R\ h\ =\ 2\ \times \ \frac{22}{7} \times \ 4.2\ \times \ 50$
$2πRh=2\times 22 \times 0.6 \times 50$
External Surface Area = 1320 cm 2
Areas of top and bottom :
Area of top surface = Area of Outer circle $-$ Area of inner Circle
Area of top surface = $π\ R^{2} \ -\ π\ r^{2}$
Take π as common,
Area of top surface = $π\ \left( \ R^{2} \ -\ \ r^{2} \ \right)$
Area of top surface = Area of bottom surface
Surface areas of two circles = Area of top surface $+$ Area of bottom surface
Areas of top and bottom =$2\ π\ \left( \ R^{2} \ -\ \ r^{2} \ \right)$
$$\displaystyle 2\ π\ \left( \ R^{2} \ -\ \ r^{2} \ \right) \ =\ 2\ π\ ( R\ +\ r) \ ( R\ -\ r) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ a^{2} \ -\ b^{2} \ =\ ( a+b)( a-b)\right]$$
$2\ π\ ( R\ +\ r) \ ( R\ -\ r) \ \ \ =\ 2\ \times \ \frac{22}{7}( 4.2\ +\ 3.5) \ ( 4.2\ -\ 3.5) \ $
$\ 2\ π\ ( R\ +\ r) \ ( R\ -\ r) \ \ \ =\ 2\ \times \ \frac{22}{7}( 7.7) \ ( 0.7) \ \ \ \ \ \ \ \ \ $
$2\ π\ ( R\ +\ r) \ ( R\ -\ r) \ \ \ =\ 33.88$ cm 2
Areas of top and bottom = 33.88 cm 2
Total Surface Area of Pipe = $1100\ +\ 1320\ +\ 33.88\ =\ 2453.88$ cm 2
Total Surface Area of Pipe = 2453.88 cm 2