(a) 36 m<br />(b) 42 m<br />(c) 45 m<br />(d) 39 m<br />18. The advantages of cone bottom tanks are found in nearly every industry,<br />especially where getting every last drop from the tank is important. This type of<br />tank has excellent geometry for draining, especially with high solids content<br />slurries as these cone tanks provide a better full-drain solution. The conical tank<br />eliminates many of the problems that flat base tanks have as the base of the tank is<br />sloped towards the centre giving the greatest possible full-drain system in vertical<br />tank design. Rajesh has been given the task of designing a conical bottom tank for<br />his client. Height of conical part is equal to its radius. Length of cylindrical part is<br />the 3 times of its radius. Tank is closed from top. The cross section of conical tank<br />is given below.<br />읽<br />A<br />Barch<br />351<br />I ATX 3x<br />27T<br />D<br />Y<br />s<br />3<br />Tirch<br />22 x 3xx *3<br />(i) If radius of cylindrical part is taken as 3 meter, what is the volume of<br />above conical tank?<br />(a) 1207 m<br />(67907 m<br />(C) 60 m<br />(d) 300 m<br />(ii) What is the area of metal sheet used to make this conical tank?<br />Assume that tank is covered from top.<br />(a) 21 7+2) i m<br />(b) 9( 7+V2) πm<br />(c) 27( 5+12) a m?<br />(d) 9( 5+ 2) m<br />(iii) What is the ratio of volume of cylindrical part to the volume of conical<br />part?<br />(a) 6 (b) 9<br />(c) 1/6<br />(d) 1/9<br />(iv) The cost of metal sheet is Rs 2000 per square meter and fabrication<br />cost is 1000 per square meter . What is the total cost of tank?
Answers
Step-by-step explanation:
Rajesh has been given the task of designing a conical bottom tank for his client.
Height of the conical part is equal to its radius.
Length of cylindrical part is the 3 times its radius.
The tank is closed from top.
To find:
If the radius of the cylindrical part is taken as 3 meters, what is the volume of above conical tank?
What is the area of metal sheet used to make this conical tank?
What is the ratio of the volume of the cylindrical part to the volume of the conical part?
The cost of the metal sheet is Rs 2000 per square meter and the fabrication cost is 1000 per square meter. What is the total cost of the tank?
Oil is to be filled in the tank. The density of oil is 1050 kg per cubic meter. What is the weight of oil filled in the tank?
Solution:
Finding the volume of the above conical tank:
Radius of the cylindrical part = Radius of the conical part = r = 3 m
Height of the cylindrical part = 3r = 3 × 3 = 9 m
Height of the conical part = r = 3 m
∴ The volume of the conical bottomed tank,
= Volume of cylinder + Volume of cone
= \pi r^2h + \frac{1}{3} \pi r^2 hπr2h+31πr2h
= [\pi \times 3^2\times 9] + [\frac{1}{3} \times \pi \times 3^2 \times 3][π×32×9]+[31×π×32×3]
= \boxed{\bold{90\pi \:m^3}}90πm3
Finding the area of metal sheet used to make this conical tank:
Radius, r = 3 m
∴ Slant height, l = \sqrt{h^2 + r^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \:mh2+r2=32+32=18=32m
∴ The area of the metal sheet used to make this conical tank is,
= [CSA of the cylindrical part] + [Area of the top of the cylindrical part] + [CSA of the conical part]
= 2\pi rh + \pi r^2 + \pi rl2πrh+πr2+πrl
= [2\pi\times 3\times 9] + [\pi \times 3^2] + [\pi\times 3 \times 3\sqrt{2}][2π×3×9]+[π×32]+[π×3×32]
= 54\pi + 9 \pi + 9\sqrt{2} \pi54π+9π+92π
= 63\pi + 9\sqrt{2} \pi63π+92π
= \boxed{\bold{9\pi [7 + \sqrt{2} ]\:m^2}}9π[7+2]m2
Finding the ratio of the volume of the cylindrical part to the volume of the conical part:
∴ The ratio of the volume of the cylindrical part to the volume of the conical part,
= \frac{Volume \:of\: cylinder}{Volume \:of\: cone}VolumeofconeVolumeofcylinder
= \frac{\pi r^2h}{ \frac{1}{3} \pi r^2 h}31πr2hπr2h
= \frac{\pi \times 3^2\times 9}{\frac{1}{3} \times \pi \times 3^2 \times 3}31×π×32×3π×32×9
= \boxed{\bold{9:1}}9:1
Finding the total cost of the tank:
The cost of the metal sheet is = Rs 2000 per square meter
The fabrication cost is = Rs. 1000 per square meter
∴ The total cost of the tank per meter square is = Rs. 2000 + Rs. 1000 = Rs. 3000
∴ The total cost of the conical bottom tank is,
= 3000 \times 9\pi [7 + \sqrt{2}]3000×9π[7+2]
= \boxed{\bold{27000\pi [7 + \sqrt{2}]}}27000π[7+2]
Finding the weight of oil filled in the tank:
The density of the oil = 1050 kg per cubic meter
∴ The weight of the oil to be filled in the conical tank is,
= 90 \pi \times 105090π×1050
= 90 \times \frac{22}{7} \times 105090×722×1050
= 297000 \:kg297000kg
we know 1 kg = 0.001 tonne
= \frac{297000}{1000}1000297000
= \boxed{\bold{297\:tonne}}297tonne
Answer:
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