Question

A petrol tank is in the form of a frustum of a cone height \[20m\] with diameter of its lower and upper ends as \[20m\] and \[50m\] . Find the cost of petrol which can fill the tank completely at the rate of rupees seventy per liter. Also find the surface area of the tank.

Answer 1

Hint: Here the given question needs to find the volume occupied by petrol in the tank and then multiply it by cost of per liter petrol in order to gain total cost of fuel, and in the second part of the question it needs to find the surface area of the fuel tank, and surface area means the area accomplished by the surfaces of the shape.

Complete step by step solution:
Here the given question is divided into two part, in the first part we need to calculate the volume of the frustum tank in order to calculate fuel cost, and in the next part we need to find the surface area of the tank, on solving we get:

Given the
Height of the frustum tank \[h = 20m\]
Diameter of the lower part of frustum tank \[r = 20m\]
Diameter of the upper part of the frustum tank \[R = 50m\]
Volume of frustum is given by the formula:
 \[ \Rightarrow V = \dfrac{\pi }{3}h\left( {{R^2} + {r^2} + R \times r} \right)\]
Now putting the values in the above formula we get:
 \[ \Rightarrow V = \dfrac{\pi }{3}20\left( {{{50}^2} + {{20}^2} + 50 \times 20} \right)\]
Now further solving the obtained equation we get
 \[
   \Rightarrow V = \dfrac{\pi }{3}20\left( {{{50}^2} + {{20}^2} + 50 \times 20} \right) \\
   \Rightarrow V = \dfrac{\pi }{3}20\left( {2500 + 400 + 1000} \right) \\
   \Rightarrow V = \dfrac{\pi }{3}20\left( {3900} \right) \\
   \Rightarrow V = \dfrac{\pi }{3} \times 20 \times 3900 \\
   \Rightarrow V = 3.141 \times 20 \times 1300 \\
   \Rightarrow V = 81666{m^3} \\
 \]
Hence we get the volume of the frustum \[V = 81666{m^3}\]
Now we will convert the volume of the frustum in to liter \[1000{m^3} = 1litre\] , we get
 \[
   \Rightarrow 81666{m^3} = \dfrac{{81666}}{{1000}}litre\,\,\,\,\,\,\left( {\because 1000{m^3} = 1litre} \right) \\
 = 81.66 litre \;
 \]
Hence cost of fuel:
 \[ \Rightarrow 81.66litre \times 70\dfrac{{Rs}}{{litre}} = Rs 5716.2\]
Now the second part of question is to find the surface area of the frustum tank:
 \[
   \Rightarrow SA = \pi (r + R)\sqrt {{{(R - r)}^2} + {h^2}} \\
   \Rightarrow SA = \pi (20 + 50)\sqrt {{{(50 - 20)}^2} + {{20}^2}} \\
   \Rightarrow SA = \pi (20 + 50)\sqrt {900 + 400} \\
   \Rightarrow SA\, = \pi (70)\sqrt {1300} \\
   \Rightarrow SA = 3.141 \times 70 \times 36.05 \\
   \Rightarrow SA = 7962\;{m^2} \;
 \]
Hence this is our required surface area.

Note: Here the given is of geometry and need to be solved by the formulas of the geometric figure, here for frustum if the formulas are not remembered then we can also find it by using the complete cone from which the upper section is cut and we obtain the frustum shape.