# 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 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm^{2}. (Take π = 3.14)

**Solution:**

A figure is drawn below to visualize the container.

As given that the container is in the shape of a frustum of a cone, so volume of milk which can completely fill the container will be equal to volume of frustum of cone.

Therefore, volume of milk = volume of frustum of cone

Let us find the volume of the frustum by using formula;

Volume of frustum of a cone = 1/3 πh (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}), where r_{1 }and r_{2} and h are the radii and height of the frustum of the cone respectively.

As we know the volume of milk, we need to convert it into litre as the cost of milk is in terms of ₹ per litre.

As we know that the container is open from the top the metal sheet will only contain CSA of frustum of the cone and area of the lower circular end.

Therefore,

Area of metal sheet required to make the container = CSA of frustum of the cone + Area of the lower circular end

Let us find the CSA of the frustum by using formulae;

CSA of frustum of a cone = π (r_{1} + r_{2}) l

where r_{1}, r_{2}, h and l are the radii height and slant height of the frustum of the cone respectively.

Area of the lower circular end = πr_{2}^{2}, where r_{2} is the radius of the lower circular end.

Height of the frustum of cone, h = 16 cm

Radius of the upper end, r_{1} = 20 cm

Radius of the lower end, r_{2} = 8 cm

Slant height of the frustum, l = √[(r_{1} - r_{2})^{2} + h^{2}]

l = √[(20 cm - 8 cm)^{2} + (16 cm)^{2}]

= √[144 cm^{2} + 256 cm^{2}]

= √[400 cm^{2}]

= 20 cm

Volume of milk which can completely fill the container = Volume of the frustum of Cone

= 1/3 πh (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

= 1/3 × 3.14 × 16 cm × ((20 cm)^{2} + (8 cm)^{2} + 20 cm × 8 cm

= 1/3 × 3.14 × 16 cm × (400 cm)^{2} + 64 cm² + 160 cm^{2})

= 1/3 × 3.14 × 16 cm × 624 cm^{2}

= 10449.92 cm^{3}

= 10449.92/1000 litre [1000cm^{3} = 1 litre]

= 10.44992 litre

= 10.45 litre

Cost of 1 litre of milk = ₹ 20

Therefore, cost of 10.45 litre of milk = ₹ 20 × 10.45 = ₹ 209

Area of metal sheet required to make the container = CSA of frustum of the cone + Area of lower circular end

= π (r_{1} + r_{2}) l + πr_{2}^{2}

= π [ (r_{1} + r_{2}) l + r_{2}^{2}]

= 3.14 × [ (20 cm + 8 cm) × 20 cm + (8 cm)^{2}]

= 3.14 × [560 cm^{2} + 64 cm^{2}]

= 3.14 × 624 cm^{2}

= 1959.36 cm^{2}

Cost of 100 cm^{2} of metal sheet = ₹ 8

Therefore, cost of 1959.36 cm^{2} of metal sheet = ₹ (8/100) × 1959.36

= ₹ 156.7488

= ₹ 156.75

Thus, the cost of milk which can completely fill the container is ₹ 209 and the cost of the metal sheet required to make the container is ₹ 156.75.

**Video Solution:**

## 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 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm^{2}

### NCERT Solutions for Class 10 Maths - Chapter 13 Exercise 13.4 Question 4 :

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 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm^{2}

The cost of the milk used to completely fill the frustum of a cone shaped container and the cost of metal sheet to make it are ₹ 209 and ₹ 156.75 respectively.