Samacheer Kalvi 10th Maths Book Solutions Chapter 7 – Mensuration is available here. The Samacheer Kalvi 10th Maths book answers of Chapter 7, available at BYJU’S, contain step by step explanations designed by our Mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 10th Maths solutions. Students can refer to these Samacheer Kalvi Class 10 Maths Textbook Chapter 7 Solutions after solving to the questions to evaluate their performance and thus plan their further studies, accordingly.

Chapter 7 of the Samacheer Kalvi 10th Maths guide will help the students to solve problems related to the surface area and volume of solid shapes, volume and surface area of combined solids, conversion of solids from one shape to another with no change in volume.

### Samacheer Kalvi 10th Maths Chapter 7: Mensuration Book Exercise 7.1 Questions and Solutions

**Question 1: The radius and height of a cylinder are in the ratio 5:7 and its curved surface area is 5500 sq.cm. Find its radius and height.**

**Solution:**

Radius of cylinder (r) = 5x

Height of cylinder (h) = 7x

Curved surface area of cylinder = 5500 sq.cm

2Πrh = 5500

2⋅ (22 / 7) ⋅ 5x ⋅ 7x = 5500

x^{2} = 5500 (7 / 22) ⋅ (1 / 2) ⋅ (1 / 35)

x^{2} = 25

x = 5

Radius = 5(5) = 25 cm

Height = 7(5) = 35 cm

**Question 2: A solid iron cylinder has a total surface area of 1848 sq.m. Its curved surface area is five – sixth of its total surface area. Find the radius and height of the iron cylinder.**

**Solution:**

Total surface area of cylinder = 1848 sq.cm

Curved surface area = (5 / 6) of 1848

= (5 / 6) 1848

2Πrh = 1540—-(1)

2Πr (h + r) = 1848

2Πrh + 2Πr^{2} = 1848

1540 + 2Πr^{2} = 1848

2Πr^{2} = 1848 – 1540

2(22 / 7)r^{2} = 308

r^{2} = 308 (7 / 22) (1 / 2)

r^{2} = 49

r = 7 m

By applying the value of r in (1),

2 ⋅ (22 / 7) ⋅ 7h = 1540

h = 1540 / 44

h = 35 m

**Question 3: The external radius and the length of a hollow wooden log are 16 cm and 13 cm, respectively. If its thickness is 4 cm then find its total surface area.**

**Solution:**

External radius (R) = 16 cm

height of log = 13 cm

Thickness = 4 cm

Thickness = R – r => 4 = 16 – r

Internal radius (r) = 16 – 4 = 12 cm

Total surface area = 2Π (R + r) (R – r + h)

= 2 ⋅ (22 / 7) (16 + 12) (16 – 12 + 13)

= 2 ⋅ (22 / 7) (28) (17)

= 2992 cm^{2}

**Question 4: 4 persons live in a conical tent whose slant height is 19 cm. If each person requires 22 cm ^{2} of the floor area, then find the height of the tent.**

**Solution:**

The slant height of conical tent = 19 cm

The required area of the floor for 1 person = 22 cm^{2}

Area for 4 persons = 22(4) = 88

Πr^{2} = 88

(22 / 7) ⋅ r^{2} = 88

r^{2} = 88 ⋅ (7 / 22)

r = 2√7

height = √(l^{2} – r^{2})

= √(19^{2} – (2√7)^{2}

= √[361 – 4(7)]

h = √333

h = 18.25 cm

### Samacheer Kalvi 10th Maths Chapter 7: Mensuration Book Exercise 7.2 Questions and Solutions

**Question 1: A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.**

**Solution:**

Given the radius of the well = 5 m

Height of the well = 14 m

Width of the embankment = 5 m

Radius of the embankment = 5 + 5 = 10 m

Let h be the height of the embankment.

Hence, the volume of the embankment = volume of the well

That is, π(R^{2} – r^{2})h = πr^{2}h

(10^{2} – 5^{2})h = 5^{2}(14)

(100 – 25)h = 25(14)

h = 25(14) / 75

h = 14 / 3

h = 4.67 m

Hence, the height of the embankment is 4.67 m.

**Question 2: A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed in it completely. Calculate the rise of the water in the glass?**

**Solution:**

Measurement of glass:

Radius of glass = 10 cm

Height of glass = h

Measurement of small metal :

Radius of metal = 5 cm

Height of glass = 4 cm

The quantity of water raised in the cylindrical glass = volume of the small cylindrical metal

πr^{2}h = πr^{2}h

10^{2} h = 5^{2} (4)

h = 100 / 100

h = 1 cm

**Question 3: If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.**

**Solution:**

Circumference of piece = 484 cm

2 π r = 484

2 (22 / 7) r = 484

r = 484 (7 / 22) (1 / 2)

r = 77

Volume of cone = (1 / 3) πr^{2}h

= (1 / 3)⋅ (22 / 7) ⋅ 77^{2} ⋅ (105)

= 652190 cm^{3}

**Question 4: A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cubic meters per minute, in how many minutes will the container be emptied? Round off your answer to the nearest minute.**

**Solution:**

The volume of petrol in the container = (1 / 3) πr^{2}h

Radius = 10 m, height = 15 m

Releasing rate = 25 cubic meter

Number of minutes = (1 / 3) π 10^{2 }(15) / 25

= (1 / 3) (22 / 7) 10^{2 }(15) / 25

= 62.85

= 63 minutes

### Samacheer Kalvi 10th Maths Chapter 7: Mensuration Book Exercise 7.3 Questions and Solutions

**Question 1: A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.**

**Solution:**

Height of vessel = 13 cm

Radius of hemisphere + height of cylinder = 13

7 + h = 13

h = 13 – 7 = 6 cm

Volume of vessel

= Volume of hemisphere + volume of the cylinder

= (2 / 3)πr^{3} + πr^{2}h

= πr^{2 }[(2 / 3) r + h]

= (22 / 7) 7^{2 }[(2 / 3) 7 + 6]

= (22 / 7) 49 (32 / 3)

= 1642.67 cm^{3}

**Question 2: Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of the model that Nathan made.**

**Solution:**

Volume of model = 2 volume of cones + volume of cylinder

= 2 (1 / 3) πr^{2}h + πr^{2}h

Height of the model = 12

2(height of cone) + height of cylinder = 12

2(2) + h = 12

height of cylinder = 8 cm

= πr^{2} [(2 / 3) 2 + 8]

= (22 / 7) (3 / 2)^{2} [(4 / 3) + 8]

= (22 / 7) (9 / 4) (28 / 3)

Volume of model = 66 cm^{3}

**Question 3: From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm ^{3}.**

**Solution:**

The volume of remaining solid = Volume of the cylinder – Volume of the cone

= πr^{2}h – (1 / 3) πr^{2}h

= πr^{2}h[1 – (1 / 3)]

= (22 / 7) (0.7)^{2} (2.4) (2 / 3)

= (22 / 7) (0.7)^{2} (2.4) (2 / 3)

= 2.46 cm^{3}

**Question 4: A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.**

**Solution:**

The volume of water displaced = Volume of water in the cylinder – (Volume of cone + Volume hemisphere)

= πr^{2}h – [(1 / 3) πr^{2}h + (2 / 3)πr^{3}]

= πr^{2}h – (1 / 3) πr^{2}h – (2 / 3)πr^{3}

= πr^{2} (h – (1 / 3)h – (2 / 3)r)

= (22 / 7) 6^{2} (18 – (1 / 3)(12) – (2 / 3)(6))

= (22 / 7) 36 (18 – 4 – 4)

= (22 / 7) 36 (10)

= 1131.42 cm^{3}

### Samacheer Kalvi 10th Maths Chapter 7: Mensuration Book Exercise 7.4 Questions and Solutions

**Question 1: An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.**

**Solution:**

The volume of sphere = Volume of cylinder

(4 / 3) πr^{3} = πr^{2}h

Radius of sphere = 12 cm

Radius of cylinder = 8 cm

Height of cylinder = h

(4 / 3) 12^{3} = 8^{2}h

h = (4 / 3) (12⋅12⋅12) / (8⋅8)

h = 36 cm

**Question 2: Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.**

**Solution:**

Quantity of water flowing out = quantity of water in a rectangular tank

Speed of water = 15 km per hour = 15000 m

Diameter of pipe = 14 cm = (14 / 100) m

The radius of pipe = 7 / 100

Length of rectangular tank = 50 m

Width of tank = 44 m

Height of tank = 21 cm = (21 / 100) m

Area of cross-section ⋅ Time ⋅ Speed = l ⋅ w ⋅ h

πr^{2} ⋅ Time ⋅ 15000 = (50) ⋅ (44) ⋅ (21/100)

(22 / 7) ⋅ (7 / 100)^{2} ⋅ Time ⋅ 15000 = (50) ⋅ (44) ⋅ 21

Time = (50⋅44⋅21⋅100⋅100⋅7) / (15000⋅100⋅7⋅ 7⋅22)

Time = 2 hours

**Question 3: A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius xr units. Find the height of water in the cylindrical flask.**

**Solution:**

The volume of water poured from the conical tank = Volume of water in the cylindrical flask

(1 / 3)πr^{2}h = πr^{2}h

Radius of cylinder = xr

(1 / 3)πr^{2}h = π(xr)^{2}h

(r^{2}h / 3) = (x^{2}r^{2})h

h = r^{2}h / 3x^{2}r^{2}

Height of cylinder = h / 3x^{2}

**Question 4: A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal radius.**

**Solution:**

Radius of cone = 7 cm

Height of cone = 8 cm

Radius of sphere = 5 cm

The volume of cone = Volume of a sphere

(1 / 3)πr^{2}h = (4 / 3)π(R^{3} – r^{3})

7^{2}(8) = 4 (5^{3} – r^{3})

(125 – r^{3}) = 49(2)

125 – r^{3} = 98

r^{3} = 125 – 98

r^{3} = 27

r = 3

Hence, the required internal radius is 3 cm.