A big iron boiler tank opened on top is cylindrical in shape.It has height is 10 m and radius is 7m. It gets corroded at a rate of 3 sq. meter/day. The number of days in which it totally gets corroded is (Take pi= $\frac{22}{7}$ )

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Solution

Surface area of the boiler = Curved Surface area + Bottom area

= 2πrh + π $r^{2}$ = 594 sq. meter

Total no. of days = $\frac{594}{3}$ = 198

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A big iron boiler tank opened on top is cylindrical in shape.It has height is 10 m and radius is 7m. It gets corroded at a rate of 3 sq. meter/day. The number of days in which it totally gets corroded is: (Take pi= $\frac{22}{7}$ )

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A big iron boiler tank opened on top is cylindrical in shape.It has height is 10 m and radius is 7m. It gets corroded at a rate of 3 sq. meter/day. The number of days in which it totally gets corroded is (Take pi=227)

Surface area of the boiler = Curved Surface area + Bottom area = 2πrh + πr2 = 594 sq. meter Total no. of days = 5943 = 198

A big iron boiler tank opened on top is cylindrical in shape.It has height is 10 m and radius is 7m. It gets corroded at a rate of 3 sq. meter/day. The number of days in which it totally gets corroded is (Take pi= $\frac{22}{7}$ )

Surface area of the boiler = Curved Surface area + Bottom area

= 2πrh + π $r^{2}$ = 594 sq. meter

Total no. of days = $\frac{594}{3}$ = 198