A milk container is made of metal sheet in the shape of frustum

Question:

A milk container is made of metal sheet in the shape of frustum of cone whose volume is $10459 \frac{3}{7} \mathrm{~cm}^{3}$. The radii of its lower and upper circular ends are $8 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. Find the cost of metal sheet used in making the container at the rate of Rs. $1.40$ per $\mathrm{cm}^{2}$.

(Use $\pi=22 / 7$ )

Solution:

Let the depth of the container is cm. The radii of the top and bottom circles of the container are r1 =20cm and r2 =8cm respectively.

The volume/capacity of the container is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(20^{2}+20 \times 8+8^{2}\right) \times h$

$=\frac{1}{3} \times \frac{22}{7} \times 624 \times h$

$=\frac{22}{7} \times 208 \times h \mathrm{~cm}^{3}$

Given that the capacity of the bucket is $10459 \frac{3}{7} \mathrm{~cm}^{3}$. Thus, we have

$\frac{22}{7} \times 208 \times h=10459 \frac{3}{7}$

$\Rightarrow h=\frac{73216}{22 \times 208}$

$\Rightarrow h=16 \mathrm{~cm}$

Hence, the height of the container is 16 cm.

The slant height of the container is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{(20-8)^{2}+16^{2}}$

$=\sqrt{400}$

 

$=20 \mathrm{~cm}$

The surface area of the used metal sheet to make the container is

$S_{1}=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{2}^{2}$

$=\pi \times(20+8) \times 20+\pi \times 8^{2}$

$=\pi \times 28 \times 20+64 \pi$

$=624 \pi \mathrm{cm}^{2}$

The cost to make the container is $=624 \pi \times 1.4=624 \times \frac{22}{7} \times 1.4=R s .2745 .6$

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