Question

1. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of 20 per litre. Also find the cost of metal sheet used to make the container, if it costs 8 per 100 cm. (Take pi = 3.14)​

Answer 1

$\huge{\bf{\underline{\underline{\red{Answer :}}}}}$

Given ,

• Height of a cone - 16 cm
• First Radius [$\bold{r_{1}}$] - 8 cm
• Second Radius [ $\bold{r_{2}}$ ] - 20 cm

We know this formula :

$\implies \bold{\blue{\frac{\pi }{3} \ x \ h [ \bold{ (r_1)^{2} \ + \ (r_2)^{2} \ + \ r_{1} \ r_{2} }}}} ]$

$\implies \bold{\frac{3.14}{3} \ x \ 16 [ (8)^{2} \ + \ ( 20)^{2} \ + \ 8 \ x \ 20 ]}$

$\implies \bold{\green{\frac{3.14 \ x \ 16 \ x \ ( 64 \ + \ 400 \ + \ 160) }{3}}}$

$\implies \bold{\frac{3.14 \ x \ 16 \ x \ 624 }{3} \bold{cm^{3} }}$

We know that :

∵

$\implies \bold { 1 \ m^{3} = \bold{ \frac{1}{100} }} \ Litre$

$\implies \bold{\frac{3.14 \ x \ 16 \ x \ 624 }{3 \ x \ 1000} \ Litre }$

Given that : The cost of 1 litre milk = 20 Ruppes

So , the cost of $\bold{\frac{3.14 \ x \ 16 \ x \ 624 }{3 \ x \ 1000} \ Litre }$ milk = $\bold{\frac{3.14 \ x \ 16 \ x \ 20 \ x \ 624 }{3 \ x \ 1000} \ }$ Ruppes

$\implies \bold{\frac{626995.2}{3000}}$

⇒209 ruppes

Now the Curved surface area of frustum :

$\implies \bold{\bf{\boxed{ \pi ( {r_{1} \ + \ r_{2} ) l \ + \ \pi ( r_1)^{2}}}}}$

$\implies \bold{ 3.14 ( 8 \ + \ 20 ) \ x \ 20 \ + \ 3.14 \ x \ 64}$

$\implies \bold{3.14(28 \ x \ 20 \ + \ 64)}$

$\implies \bold{3.14 \ x \ 624 }$

$\implies \bold{1959.36 \ cm^{2} }$

So Given the cost of metal sheet = 8 Rupees \ 100 m²

Cost of metal sheet required = $\bold{\bf{\frac{8 \ x \ 1959.36}{100} }}$

=> 156.75 Rupees approx ......Answer