Question

The conical tank eliminates many of the problems that flat base tanks have as the base of the tank is sloped towards the centre giving the greatest possible full-drain system in vertical tank design.
Rajesh has been given the task of designing a conical bottom tank for his client. Height of conical part is equal to its radius. Length of cylindrical part is the 3 times of its radius. Tank is closed from top.
The cross section of conical tank is given below. On the basis of the above information, answer any four of the following questions:
i. If radius of cylindrical part is taken as 3 meter, what is the volume of above conical tank ? (a) 120π m3 (b) 90π m3 (c) 60π m3 (d) 30π m3
ii. What is the area of metal sheet used to make this conical tank ? Assume that tank is covered from top. When radius is 3 meter.
(a)27(7 + √2) (b) 9(7 + √2) (c) 27(5 + √2) (d) 9(5 + √2)
iii. What is the ratio of volume of cylindrical part to the volume of conical part?
(a) 6 (b) 9 (c) 1 / 6 (d) 1 / 9
iv. The cost of metal sheet is Rs 2000 per square meter and fabrication cost is 1000 per square meter. What is the total cost of tank ?
(a)27000(7 + √2) (b) 54000(7 + √2) (c) 27000(5 + √2) (d) 54000(5 + √2)
v. A 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 tank ?
(a) 195 Tonne (b) 200 Tonne (c) 297 Tonne (d) 174 Tonne

Answer 1

Given:

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

= $\pi r^2h + \frac{1}{3} \pi r^2 h$

= $[\pi \times 3^2\times 9] + [\frac{1}{3} \times \pi \times 3^2 \times 3]$

= $\boxed{\bold{90\pi \:m^3}}$

Finding the area of metal sheet used to make this conical tank:

Radius, r = 3 m

∴ Slant height, l = $\sqrt{h^2 + r^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \:m$

∴ 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]

= $2\pi rh + \pi r^2 + \pi rl$

= $[2\pi\times 3\times 9] + [\pi \times 3^2] + [\pi\times 3 \times 3\sqrt{2}]$

= $54\pi + 9 \pi + 9\sqrt{2} \pi$

= $63\pi + 9\sqrt{2} \pi$

= $\boxed{\bold{9\pi [7 + \sqrt{2} ]\:m^2}}$

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,

=  $\frac{Volume \:of\: cylinder}{Volume \:of\: cone}$

= $\frac{\pi r^2h}{ \frac{1}{3} \pi r^2 h}$

= $\frac{\pi \times 3^2\times 9}{\frac{1}{3} \times \pi \times 3^2 \times 3}$

= $\boxed{\bold{9:1}}$

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,

= $3000 \times 9\pi [7 + \sqrt{2}]$

= $\boxed{\bold{27000\pi [7 + \sqrt{2}]}}$

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,

= $90 \pi \times 1050$

= $90 \times \frac{22}{7} \times 1050$

= $297000 \:kg$

we know 1 kg = 0.001 tonne

= $\frac{297000}{1000}$

= $\boxed{\bold{297\:tonne}}$

-------------------------------------------------------------------------------

Also View:

A design for a traffic cone is shown below. In order for it to have a solid foundation, the designers want to make the base and cone out of solid plastic. If the plastic is $0.75 per cubic cm, how much will it cost them to make 5 cones?

brainly.in/question/27019371  

Rachel an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. the diameter of the model is 3cm and its length is 12cm. If each come has a height of 2cm. find the volume of air contained in the model that Rachel made.

brainly.in/question/14315681

Mathematics teacher of a school took her 10th standard students to show Tajmahal.  

The teacher said in this monument one can find a combination of solid figures.  

(i)Write the formula to find the volume of a hemispherical dome.  

(ii)Find the lateral surface area of 4 pillars if the height of each pillar is 20m and radius of the base is 1.4m.  

(iii)How much is the volume of 2 small hemispheres each of radius 2.1m?  

(iv)How much cloth material will be required to cover 1 big dome of radius 4.2m?  

(v)What is the ratio of the sum of volumes of four hemispheres of radius 1m each to the volume of a sphere of radius 2m?

brainly.in/question/30261412