Question

Easy question this time :)

An open container made up of a metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper ends as 4 cm and 10 cm respectively. Find the cost of oil which can completely fill the container at the rate of rs. 50 per litre. Also, find the cost of metal used, if it costs rs. 5 per 100 cm². ​

Answer 1

Answer:

Rs. 24.49

Hope it helps you

Answer 2

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

An open container made up of a metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper ends as 4 cm and 10 cm respectively.

It means,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{h \: = \: 8 \: cm} \\ \\ &\sf{r \: = \: 4 \: cm} \\ \\ &\sf{R \: = \: 10 \: cm} \end{cases}\end{gathered}\end{gathered}$

We know,

Volume of frustum of height h and radius of ends are r andR respectively is given by

$\\ \red{\boxed{\tt{ \: Volume_{(frustum)} \: = \: \frac{\pi \: h}{3}[ {R}^{2} + Rr + {r}^{2}] \: \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:Volume_{(frustum)} = \dfrac{22 \times 8}{7 \times 3}[ {10}^{2} + 10 \times 4 + {4}^{2}]$

$\rm :\longmapsto\:Volume_{(frustum)} = \dfrac{22 \times 8}{7 \times 3}[100 + 40 + 16]$

$\rm :\longmapsto\:Volume_{(frustum)} = \dfrac{22 \times 8}{7 \times 3} \times 156$

$\bf\implies \:Volume_{(frustum)} = 1307.43 \: {cm}^{3}$

$\bf\implies \:Volume_{(frustum)} = 1.30743 \: litres$

So, Cost of oil at the rate of Rs 50 per litre is

$\rm \:  =  \: 1.30743 \times 50$

$\rm \:  =  \: Rs \: 65.3715$

$\rm \:  =  \: Rs \: 65.37 \: approx.$

Now, To find the cost of metal sheet, we have to first the area of frustum.

So, Amount of sheet required is equal to Curved Surface Area of frustum + Area of base of radius 4 cm

We know,

Slant height (l) of frustum of height h and radius r and R respectively is given by

$\purple{\rm :\longmapsto\:l = \sqrt{ {h}^{2} + {(R - r)}^{2} }}$

$\purple{\rm :\longmapsto\:l = \sqrt{ {8}^{2} + {(10 - 4)}^{2} }}$

$\purple{\rm :\longmapsto\:l = \sqrt{ {8}^{2} + {6}^{2} }}$

$\purple{\rm :\longmapsto\:l = \sqrt{64 + 36}}$

$\purple{\rm :\longmapsto\:l = \sqrt{100}}$

$\purple{\rm :\longmapsto\:l = 10 \: cm}$

Now, Surface Area of Frustum

$\rm :\longmapsto\:Surface \: Area_{(frustum)} = \pi(R + r)l + \pi {r}^{2}$

$\rm \:  =  \: \pi\bigg[(R + r)l + {r}^{2}\bigg]$

$\rm \:  =  \: \dfrac{22}{7} \bigg[(10+ 4)10 + {4}^{2}\bigg]$

$\rm \:  =  \: \dfrac{22}{7} \bigg[140 + 16\bigg]$

$\rm \:  =  \: \dfrac{22}{7} \bigg[156\bigg]$

$\rm \:  =  \: 490.29 \: {cm}^{2}$

Cost of metal used, if it costs Rs. 5 per 100 cm² is

$\rm \:  =  \: 490.29 \: \times 0.05$

$\rm \:  =  \: Rs \: 24.51 \: approx$