The cost of polishing a closed container, which is of the shape of the frustum of a cone, is Rs 440 at the rate of Re 1 per 100 cm². The respective radii of the upper and the lower bases of the frustum are 1 m and 40 cm.What is the height of the container?

Answer :

Given : The cost of polishing a closed container, which is of the shape of the frustum of a cone, is Rs 440 at the rate of Re 1 per 100 cm² .

So,
Total surface area of frustum of cone  = 440 $\times$ 100   =  44000 cm²

We know
Total surface area of frustum of cone  = $\pi \left( R^2 + r^{ 2 } +\hspace{0.17em}\left(R + r \right)\sqrt{{\left(R - r \right)}^2 + h^2}\right)$

Here R  =  1 m  =  100 cm
And
r  =  40 cm , SO we get

$$\begin{gathered} \Rightarrow \frac{22}{7}\left( {100}^2 + {40}^{ 2 } +\hspace{0.17em}\left(100 + 40 \right)\sqrt{{\left(100 - 40 \right)}^2 + h^2}\right) = 44000 \\ \Rightarrow \frac{1}{7}\left( {100}^2 + {40}^{ 2 } +\hspace{0.17em}\left(100 + 40 \right)\sqrt{{\left(100 - 40 \right)}^2 + h^2}\right) = 2000 \\ \Rightarrow \left( 10000+ 1600 +\hspace{0.17em}\left(140 \right)\sqrt{{\left(60 \right)}^2 + h^2}\right) = 2000 \times 7 \\ \Rightarrow \left( 11600 +\hspace{0.17em}\left(140 \right)\sqrt{{\left(60 \right)}^2 + h^2}\right) = 14000 \\ \Rightarrow \left(140 \right)\sqrt{3600+ h^2} = 2400 \\ Taking whole square on both hand side ,we get \\ \Rightarrow {\left[ \left(140 \right)\sqrt{3600+ h^2}\right]}^2 = {\left( 2400\right)}^2 \\ \Rightarrow 19600 \left(3600 +\hspace{0.17em}{h}^2\right) = 5760000 \\ \Rightarrow \left(3600 +\hspace{0.17em}{h}^2\right) = \frac{5760000}{19600} \\ \Rightarrow \left(3600 +\hspace{0.17em}{h}^2\right) = \frac{14400}{49} \\ \Rightarrow h^2 = \frac{14400}{49} - 3600 \\ \Rightarrow h^2 = \frac{14400 - 176400}{49} \\ \Rightarrow h^2 =\frac{-162000}{49} \\ SO, There is some information is wrong Please look into it and get back to us , So we can help you precisely. \end{gathered}$$