Question

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill up the container, at the rate of Rs. 40 per litre. (Use \( \pi = 3.14 \))

Hint:

To calculate the volume of a frustum of a cone, use the formula \( V = \frac{1}{3} \pi h \left( r_1^2 + r_2^2 + r_1 r_2 \right) \), where \( r_1 \) and \( r_2 \) are the radii of the upper and lower ends, and \( h \) is the height.

Answer 1

The volume \( V \) of a frustum of a cone is given by the formula: \[ V = \frac{1}{3} \pi h \left( r_1^2 + r_2^2 + r_1 r_2 \right), \]
where:
- \( h = 24 \, \text{cm} \) is the height, - \( r_1 = 20 \, \text{cm} \) is the radius of the upper end, - \( r_2 = 8 \, \text{cm} \) is the radius of the lower end. Substitute the given values into the formula: \[ V = \frac{1}{3} \times 3.14 \times 24 \left( 20^2 + 8^2 + 20 \times 8 \right). \] Simplify: \[ V = \frac{1}{3} \times 3.14 \times 24 \left( 400 + 64 + 160 \right) = \frac{1}{3} \times 3.14 \times 24 \times 624. \] \[ V = \frac{1}{3} \times 3.14 \times 14976 = 15792.48 \, \text{cm}^3. \] Since 1 litre = 1000 cm³, the volume in litres is: \[ \text{Volume in litres} = \frac{15792.48}{1000} = 15.79248 \, \text{litres}. \] The cost of the milk is: \[ \text{Cost} = 40 \times 15.79248 = 631.6992 \, \text{rupees}. \] Thus, the cost of the milk is approximately \( \text{Rs.} 631.70 \).

Conclusion: The cost of milk to completely fill the container is approximately \( \text{Rs.} 631.70 \).