A big iron boiler tank opened on top is cylindrical in shape. It's 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= $\frac{22}{7}$ )

192

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202

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198

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208

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Solution

The correct option is C
198

Surface area of the boiler = Curved Surface area + Bottom area
$= 2 π r h + π r^{2}$
$= 2 \times \frac{22}{7} \times 7 \times 10 + \frac{22}{7} \times 7^{2}$
= 594 sq. meter
Total no. of days $= \frac{T o t a l S u r f a c e A r e a o f t a n k}{A r e a c o r r o d e d p e r d a y} = \frac{594}{3} = 198$

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A big iron boiler tank opened on top is cylindrical in shape. It's 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)

The correct option is C 198 Surface area of the boiler = Curved Surface area + Bottom area =2πrh+πr2 =2×227×7×10+227×72 = 594 sq. meter Total no. of days =Total Surface Area of tankArea corroded per day=5943=198

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