Q. 363.8( 4 Votes )

# If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm, find its capacity and total surface area.

Answer :

__Explanation: Here the bucket is in the shape of a frustum. So capacity of bucket will be equal to the volume of the frustum and in order to calculate the total surface area of the bucket we will subtract the top end circular area from the total surface area of the frustum as the bucket is open on top.__

Upper end radius of frustum/bucket = R = 28 cm

Lower end radius of frustum/bucket =r = 7 cm

Height of the frustum/bucket = 28 cm

And we know,

The capacity of bucket = Volume of the bucket

And, Volume of bucket = Volume of Frustum

∴ Capacity of bucket = Volume of frustum

⇒Volume of frustum

Where R = Radius of larger or upper end and r = Radius of smaller or lower end and h = height of frustum π = 22/7

⇒Volume of frustum

= 22 × 4 × 343

∴ Volume of frustum = 30184 cm^{3}

∴ Capacity of bucket = 30184 cm^{3}

T.S.A of bucket = T.S.A of frustum – Area of upper circle →eqn1

Let the slant height of the frustum be ‘ℓ’ cm

So, ℓ^{2} = h^{2} + (R – r)^{2}

⇒ ℓ^{2} = 28^{2} + (28 – 7)^{2}

⇒ ℓ^{2} = 784 + (21)^{2}

⇒ ℓ^{2} = 784 + 441

⇒ ℓ^{2} = 1225

⇒ ℓ = √(1225)

∴ ℓ = 35 cm

⇒ T.S.A of frustum = π(R + r)ℓ + πR^{2} + πr^{2}

= π(28 + 7) × 35 + π(28)^{2} + π(7)^{2}

= 35 × 35π + 784π + 49π

= 1225π + 784π + 49π →eqn2

Area of upper circle = πR^{2}

= π(28)^{2}

= 784π →eqn3

T.S.A of bucket = 1225π + 784π + 49π – 784π (from eqn2 and 3)

⇒ T.S.A of bucket = 1274π

⇒T.S.A of bucket=

⇒ T.S.A of bucket = 182 × 22

∴ T.S.A of bucket = 4004 cm^{2}

__The capacity and total surface area of the bucket is 30184 cm ^{3} and 4004 cm^{2}.__

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