# NCERT Exemplar Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 12.4 in Chapter 12 - Surface Areas and Volumes

Question 1 Surface Areas and Volumes - Exercise 12.4

A milk container of height 16 cm is made of metal sheet in the form of a frustrum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ₹22 per L, which the container can hold.

Here, r1 = 8 cm

r2 = 20 cm

h = 16 cm

∴ Volume of milk = Volume of frustrum as it is filled completely

=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right) \\ =\frac{1}{3} \times \frac{22}{7} \times 16\left[8^{2}+20^{2}+8 \times 20\right] \\ =\frac{22 \times 16}{21}[64+400+160] \\ =\frac{352}{21} \times 624=\frac{352 \times 208}{7}=\frac{73216}{7}

= 10459.428 cm3 = 10.459 litre

Volume of milk = 10.459 litre

∴ Cost of milk = ₹22 × 10.459 = ₹230.098

Hence, the cost of milk = ₹230.098

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