A bucket in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm , respectively. Find the capacity of the bucket. Also find the cost of milk which can completely fill the bucket at the rate of ₹ 40 per litre. Use π = 22 7

Given that, a bucket in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm , respectively. Capacity of the bucket can be obtained by finding the volume of the frustum of cone. We know that, volume of the frustum of cone = 1 3 πh r 1 2 + r 2 2 + r 1 r 2 Where, h is the height the frustum of the cone, r 1 , r 2 are the radii of both ends of the frustum. Using the formula mentioned above, Capacity = 1 3 × 22 7 × 30 20 2 + 10 2 + 20 × 10 = 22 7 × 10 400 + 100 + 200 = 22 7 × 10 × 700 = 22 × 10 × 100 = 22000 cm 3 = 22 litre [ ∵ 1 litre = 1000 cm 3 ] Cost of 22 litre milk = Cost of 1 litre milk × 22 = ₹ 40 × 22 = ₹ 880 Hence, the capacity of the bucket is 22 litre and the cost of milk is ₹ 880 .

A cone of base radius 4 cm is divided into two parts by drawing a plane through the mid-point of its height and parallel to its base. Compare the volume of the two parts.

Given, radius of base of cone = 4 cm The cone is divided in two parts through the height of the cone. Let, the height of the cone is h cm . So, the volume of the cone = 1 3 × π × 4 2 × h cm 3 = 16 πh 3 cm 3 ∴ Height of the smaller cone is h 2 cm and radius is 4 2 cm = 2 cm ∴ Volume of smaller cone = 1 3 π × 2 2 × h 2 cm 3 = 2 πh 3 cm 3 ∴ Volume of lower frustum of the cone = Total volume of cone - Volume of shorter cone = 16 πh 3 - 2 πh 3 cm 3 = 16 πh - 2 πh 3 cm 3 = 14 πh 3 cm 3 ∴ Ratio of two parts of the cone = 2 πh 3 : 14 πh 3 = 2 : 14 = 1 : 7 The ratio of volumes of two parts of the cone is 1 : 7 .

The diameters of the lower and upper ends of a bucket in the form of frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm , find (i) the area of the metal sheet used to make the bucket. (ii) Why we should avoid the bucket made by ordinary plastic? [Use π = 3 . 14 ]

(i) Given, the diameter of both ends are d 1 = 10 cm and d 2 = 30 cm Hence, radii are r 1 = d 1 2 = 10 2 = 5 cm and r 2 = d 2 2 = 30 2 = 15 cm Also, given height, h = 24 cm Now, slant height of frustum, l = h 2 + r 2 - r 1 2 = ( 24 ) 2 + ( 15 - 5 ) 2 = 576 + 100 = 676 = 26 cm ∴ Area of the metal sheet required to make the bucket = Curved surface area of the frustum + Base area = π r 1 + r 2 l + π r 1 2 = 3 . 14 ( 5 + 15 ) × 26 + 3 . 14 × 5 2 = 1632 . 8 + 78 . 5 = 1711 . 3 cm 2 And, (ii) We should avoid the bucket made by ordinary plastic, because it has maximum chance for breaking the bucket.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 c m , and its base is of radius 3.5 c m . Find the total surface area of the article.

From the given diagram it can be clearly observed that radius of hemisphere is same as radius of cylinder. Given radius of cylindrical part = radius of hemispherical part = r = 3 . 5 cm . Also, height of the cylinder h = 10 cm . The total surface area of the article = ( C S A of cylinder) + ( 2 × CSA of hemisphere) = ( 2 π r h ) + ( 2 × 2 π r 2 ) = ( 2 π × 3 .5 × 10 ) + ( 2 × 2 π × 3 .5 × 3 .5 ) = 70 π + 49 π = 119 × 22 7 = 17 × 22 = 374 cm 2

A heap of rice in the form of a cone of base diameter 24 m and height 3 . 5 m . Find the volume of rice. How much canvas cloth is required to just cover the heap?

Given that a heap of rice is in the form of a cone. Diameter of the base of the heap = d = 24 m So, radius of the base of the heap = 24 2 = 12 m Height of the heap = h = 3 . 5 m We know that volume of cone V = π r 2 h 3 So, volume of the heap = 22 7 × 12 × 12 × 3 . 5 3 m 3 = 528 m 3 Given that canvas cloth is required to just cover the heap. It covered the only curved surface area of heap. Slant height of cone = l = h 2 + r 2 So, slant height of the heap = 3 . 5 2 + 12 2 = 12 . 5 m Now, curved surface area of cone S = π r l So, curved surface area of the heap = 22 7 × 12 × 12 . 5 m = 471 . 42 m 2 Hence, the volume of the heap is 528 m 3 and curved surface area of the heap is 471 . 42 m 2 .

MEDIUM

10th CBSE

IMPORTANT

Earn 100

The tent is in the shape of a cylinder surrounded by a conical top. If the height and diameter of cylindrical part are $9 m$ and $30 m$ respectively and height of conical part is $8 m$ with same diameter as that of the cylindrical part, then find the cost of the canvas bought for the tent at the rate $₹ 200$ per $s q m$ , if $30 s q m$ canvas was wasted during stitching.

Important Questions on Surface Areas and Volumes

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2022 (Delhi - II)>Q 12

A solid piece of metal in the form of a cuboid of dimensions

11 cm \times 7 cm \times 7 cm

is melted to form

' n'

number of solid spheres of radii

\frac{7}{2} cm

each. Find the value of

n

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Delhi - I)>Q 14

A student was asked to make a model shaped like a cylinder with two cones attached to its end by using a thin aluminium sheet. The diameter of the model is $3 c m$ and its total length is $12 c m$ . If each cone has a height of $2 c m$ , find the volume of air contained in the model.

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Delhi - I)>Q 15

A student was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminium sheet. The diameter of the model is

3 cm

and its total length is

12 cm

. If each cone has a height of

2 cm

, find the volume of air contained in the model.

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Delhi - II)>Q 16

From a solid cylinder of height $20 cm$ and diameter $12 cm$ , a conical cavity of height $8 cm$ and radius $6 cm$ is hallowed out. Find the total surface area of the remaining solid.

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Delhi - III)>Q 17

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is $10 cm$ and its base is of radius $3.5 cm$ , find the total surface area of article.

Question Image

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Outside Delhi)>Q 18

The volume of a right circular cone whose area of the base is

156 {cm}^{2}

and the vertical height is

8 cm

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Outside Delhi)>Q 19

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $7 cm$ and the height of the cone is equal to its diameter. Find the volume of the solid

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 13 - Surface Areas and Volumes>CBSE-2023 (Outside Delhi)>Q 20

A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being $3.5 c m$ and the total height of the solid is $9.5 c m$ . Find the volume of the solid.

Important Questions on Surface Areas and Volumes

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