Question

A container opened from the top in the form of a frustum of a cone of height 16cm with the radii of its lower and upper ends are 8 cm and 20cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs.20 per litre use (pie= 3.14)​

Answer 1

Answer:

209 apx

Step-by-step explanation:

h = height = 16 cm

r₁ = radius of upper end = 20 cm

r₂ radius of lower end = 8 cm

Volume of container == πh(r₁² + r₂² + r₁ r²)

=> 3.14 × 16(20)² + (8)² + 20 × 8)

= 16.74 (624)

= 10445.76 cm³

10445.76 /1000 litre

= 10.44576 litre

Cost of 1 litre milk = Rs 20

Cost of 10.44576 litre milk = Rs 20 x 10.44576

Rs 208.91

Answer 2

$\\ \underline{\purple{ \sf \large{Question :- }}}$

A container opened from the top in the form of a frustum of a cone of height 16cm with the radii of its lower and upper ends are 8 cm and 20cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs.20 per litre use (pie= 3.14)

$\\ \underline{\purple{ \sf \large{Solution :- }}}$

In this, question

• R = 20
• r = 8
• h = 16

Now, We have to find the capacity of container

therefore,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \underline{ \boxed{\sf \: Volume \: of \: frustum = \green{ \dfrac{1}{3} \pi h(R^{2} + r ^{2} + Rr)} }}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \dfrac{1}{3} \times 3.14 \times 16 \: ( {20}^{2} + {8}^{2} + 20 \times 8)$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \dfrac{3.14}{3} \times 16 \: ( {20}^{2} + {8}^{2} + 20 \times 8)$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \dfrac{50.24}{3} \times \: ( 400 + 64 + 160)$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \dfrac{50.24}{3} \times 624$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: {50.24}\times 208$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \orange{\frak{\: \: \dfrac{ 10449.92}{1000} \: \: liters} }}$

Now, Cost of milk at Rs 20 per litre

$\\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: 20 \times \dfrac{10449.92}{1000}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: 208.99 \approx \red{ \bold 209 }$

To find a slant height

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {l}^{} = \sqrt{ {16}^{2} + {12}^{2} }$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ 256 + 144 }$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ 400}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\boxed{ \blue{ \frak{{l}^{} = 20}}} }$

Now, we have to find

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \underline{ \boxed{\sf Curved \: surface \: area \: of \: container := \green{ π(R+r)l} }}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \frac{22}{7} \times (20 + 8) \times 20$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \frac{22}{7} \times 2 8\times 20$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \frac{22}{7} \times 560$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \orange{\frak{\: \: { }{1758.4} \: {cm}^{2} } }}$

Now, let's find

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \underline{ \boxed{\sf Area \: of \: bottom\: of\: container= \green{ πr^{2} } }}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \frac{22}{7} \times ( 8) ^{2}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \frac{22}{7} \times 64$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \orange{\frak{\: \: { }{200.96} \: {cm}^{2} } }}$

Now,

Total area of metal required

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies 1758.4 + 200.96$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \underline{ \boxed{ \color{blue}{\frak{\: \: { }{1959.36} \: {cm}^{2} } }}}$

Now, at last

Cost of metal sheet used to manufacture the container at Rs 20 per 100 cm²

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \frac{20}{100} \times 1959.36$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \underline{ \boxed{ \pink{\frak{\: \: { }{ \:Rs \: 391.88} \: } }}}$