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Question: The radii of the ends of a bucket \[30cm\] high is \[21cm\] and \[7cm\]. Find its capacity in litres...

The radii of the ends of a bucket 30cm30cm high is 21cm21cm and 7cm7cm. Find its capacity in litres and the amount of sheet required to make this bucket.
A. 20.02 liters ; 3067cm220.02{\text{ liters ; 3067}}c{m^2}
B. 20.02 liters ; 3867cm220.02{\text{ liters ; 3867}}c{m^2}
C. 25.02 liters ; 3067cm225.02{\text{ liters ; 3067}}c{m^2}
D. 10.02 liters ; 3067cm210.02{\text{ liters ; 3067}}c{m^2}

Explanation

Solution

Hint: Here the volume of the bucket is equal to the capacity of the bucket in litres. We can find the amount of sheet required to make this bucket by calculating the total surface area of the bucket.

Complete step-by-step answer:

Given,
Height of the bucket h=30cmh = 30cm
Radius of upper end of the bucket R=21cmR = 21cm
Radius of lower end of the bucket r=7cmr = 7cm
We know that volume of the bucket = 13πh(R2+r2+Rr)cm3\dfrac{1}{3}\pi h\left( {{R^2} + {r^2} + Rr} \right)c{m^3}

=13π(30)((21)2+(7)2+(21)(7)) =13π(30)(441+49+147) =13π(30)(637) =13×227(19110) =42042021=20020cm3  = \dfrac{1}{3}\pi (30)\left( {{{\left( {21} \right)}^2} + {{\left( 7 \right)}^2} + \left( {21} \right)\left( 7 \right)} \right) \\\ = \dfrac{1}{3}\pi \left( {30} \right)\left( {441 + 49 + 147} \right) \\\ = \dfrac{1}{3}\pi \left( {30} \right)\left( {637} \right) \\\ = \dfrac{1}{3} \times \dfrac{{22}}{7}\left( {19110} \right) \\\ = \dfrac{{420420}}{{21}} = 20020c{m^3} \\\

Since, 1cm3=11000 litres1c{m^3} = \dfrac{1}{{1000}}{\text{ litres}}
The capacity of the bucket is 200201000 litres = 20.02 litres\dfrac{{20020}}{{1000}}{\text{ litres }} = {\text{ 20}}{\text{.02 litres}}
We know that slant height of the bucket l=h2+(Rr)2l = \sqrt {{h^2} + {{\left( {R - r} \right)}^2}}

l=302+(217)2 l=900+(14)2 l=900+196 l=1096 l=33.10cm  l = \sqrt {{{30}^2} + {{\left( {21 - 7} \right)}^2}} \\\ l = \sqrt {900 + {{\left( {14} \right)}^2}} \\\ l = \sqrt {900 + 196} \\\ l = \sqrt {1096} \\\ \therefore l = 33.10cm \\\

Total area of the metal sheet required to make the bucket = πl(R+r)+πr2\pi l\left( {R + r} \right) + \pi {r^2}

=π×33.10(21+7)+π×72 =π×33.10(28)+π×49 =π×926.8+π×49 =π(926.8+49) =227×975.8 =21467.67 =3066.83067cm2  = \pi \times 33.10\left( {21 + 7} \right) + \pi \times {7^2} \\\ = \pi \times 33.10\left( {28} \right) + \pi \times 49 \\\ = \pi \times 926.8 + \pi \times 49 \\\ = \pi \left( {926.8 + 49} \right) \\\ = \dfrac{{22}}{7} \times 975.8 \\\ = \dfrac{{21467.6}}{7} \\\ = 3066.8 \approx 3067c{m^2} \\\

Therefore, the amount of sheet required to make the bucket is 3067cm2{\text{3067}}c{m^2}.
Thus, the capacity of the bucket is 20.02 litres20.02{\text{ litres}} and the amount of sheet required to make the bucket is 3067cm2{\text{3067}}c{m^2}.
So, the correct option is A. 20.02 liters ; 3067cm220.02{\text{ liters ; 3067}}c{m^2}.

Note: The height given in the problem is the altitude of the bucket. To find the surface area we have to consider the slant height which is given by l=h2+(Rr)2l = \sqrt {{h^2} + {{\left( {R - r} \right)}^2}} . We have converted the volume of the bucket in to capacity of the bucket in litres by using the conversion 1cm3=11000 litres1c{m^3} = \dfrac{1}{{1000}}{\text{ litres}}.