Q. 193.7( 12 Votes )

# An open metal buc

Answer : Given: radius of upper circular end of frustum = R = 45 cm

Radius of lower circular end of frustum = radius of cylindrical base = r = 25 cm

Height of bucket = hb = 40 cm

Height of cylindrical base = hc = 6 cm

From the figure height of frustum = hf = hb - hc

= 40 - 6

hf = 34 cm Volume of cylinder = πr2hc

Curved surface area of cylinder = 2πrhc

curved surface area of frustum = π(R + r)l cm2

Where l = slant height  = √(400 + 1156)

= 39.44 cm

Area of metallic sheet used = curved surface area of frustum + curved surface area of base

cylinder + area of base circle of cylinder

now, curved surface area of frustum = π × (R + r) × l cm2

= (22/7) × (45 + 25) × 39.44 cm2

= 22 × 10 × 39.44 cm2

= 8676.8 cm2

Curved surface area of base cylinder = 2πrhc

= 2 × (22/7) × 25 × 6

= 942.85 cm2

Surface area of base circle of cylinder = πr2

= (22/7) × 252

= 1964.28 cm2

Area of metallic sheet used = 8676.8 + 942.85 + 1964.28

= 11583.93 cm2

Therefore, area of metallic sheet used to make the bucket is 11583.93 cm2 i.e. 1.158393 m2

Volume of water bucket can hold = volume of bas cylinder + volume of frustum

Volume of base cylinder = πr2hc

= (22/7) × 252 × 6

= 11785.71 cm3  = 35.62 × 3775

= 134465.5 cm3

volume of water bucket can hold = 11785.71 + 134465.5

= 146251.21 cm3

Now 1 litre is 1000 cm3

146251.21 cm3 = 146251.21/1000 = 146.25121 litres

Volume of water bucket can hold = 146.25121 litres

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