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