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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.

The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______

(1) π cm^{2}

(2) 2π cm^{2}

(3) 3π cm^{2}

(4) 2 cm^{2}

Answer:

(2) 2π cm^{2}

Hint:

C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm^{2} = 2π cm^{2}

Question 2.

The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.

(1) \(\frac{3}{2} \pi h\)

(2) \(\frac{2}{3} \pi h^{2}\)

(3) \(\frac{3}{2} \pi h^{2}\)

(4) \(\frac{2}{3} \pi h\)

Answer:

(3) \(\frac{3}{2} \pi h^{2}\)

Hint:

T.S.A = 2πr(h + r)

[radius is half of the height]

= \(2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)\)

= \(=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}\) sq. units

Question 3.

Base area of a right circular cylinder is 80 cm^{2}. If its height is 5 cm, then the volume is equal to _______

(1) 400 cm^{3}

(2) 16 cm^{3}

(3) 200 cm^{3}

(4) \(\frac{400}{3}\) cm^{3}

Answer:

(1) 400 cm^{3}

Hint:

Volume of a cylinder = πr^{2}h cu. units

[Base area (πr^{2}) = 80 cm^{2} = 80 × 5 cm^{3} = 400 cm^{3}

Question 4.

If the total surface area of a solid right circular cylinder is 200π cm^{2} and its radius is 5 cm, then the sum of its height and radius is ______

(1) 20 cm

(2) 25 cm

(3) 30 cm

(4) 15 cm

Answer:

(1) 20 cm

Hint:

T.S.A of a cylinder = 200π cm^{2}

2πr (h + r) = 200π

2 × 5 (h + r) = 200

(h + r) = 20 cm

Question 5.

The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______

(1) πa^{2}b sq.cm

(2) 2πab sq.cm

(3) 2π sq.cm

(4) 2 sq.cm

Answer:

(2) 2πab sq.cm .

Hint:

C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.

Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm^{3}, then the volume of the cone is equal to _______

(1) 1200 cm^{3}

(2) 360 cm^{3}

(3) 40 cm^{3}

(4) 90 cm^{3}

Answer:

(3) 40 cm^{3}

Hint:

Volume of the cone = \(\frac{1}{3}\) × volume of the cylinder

= \(\frac{1}{3}\) × 120 cm^{3}

= 40 cm^{3}

Question 7.

If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is

(1) 10 cm

(2) 20 cm

(3) 30 cm

(4) 96 cm

Answer:

(1) 10 cm

Hint:

Slant height of a cone

Question 8.

If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______

(1) 1200π cm^{2}

(2) 600π cm^{2}

(3) 300π cm^{2}

(4) 600 cm^{2}

Answer:

(2) 600π cm^{2}

Hint:

Circumference (2πr) = 120π cm

Slant height (l) = 10 cm;

Curved surface area of a cone = πrl sq. units

= \(\frac{120 \pi}{2}\) × 10 cm^{2} = 600π cm^{2}

Question 9.

If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______

(1) 6 cm

(2) 8 cm

(3) 10 cm

(4) 12 cm

Answer:

(4) 12 cm

Hint:

Volume of a cone = 48π cm^{3}

[Base area (πr^{2}) = 12π]

\(\frac{1}{3}\) πr^{2}h = 48π

\(\frac{1}{3}\) × 12π × h = 48π

[Substitute πr^{2} = 12π]

h = \(\frac{48}{4}\) = 12 cm

Question 10.

If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______

(1) 240 cm^{3}

(2) 120 cm^{3}

(3) 80 cm^{3}

(4) 480 cm^{3}

Answer:

(3) 80 cm^{3}

Hint:

Volume of a cone (V) = \(\frac{1}{3}\) πr^{2}h sq. units

Base area (πr^{2}) = 48 sq. cm

V = \(\frac{1}{3}\) × 48 × 5 = 80 cm^{3}

Question 11.

The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______

(1) 4 : 1

(2) 1 : 4

(3) 2 : 1

(4) 1 : 2

Answer:

(3) 2 : 1

Hint:

h_{1} : h_{2} = 1 : 2

r_{1} : r_{2} = 2 : 1

Ratio of their volumes

= \(\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}\)

= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.

If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________

(1) 8π cm^{2}

(2) 16 cm^{2}

(3) 12π cm^{2}

(4) 16π cm^{2}

Answer:

(4) 16π cm^{2}

Hint:

C.S.A of a sphere = 4πr^{2} sq. units

[radius = 2 cm]

= 4 × π × 22 cm^{2}

= 16π cm^{2}

Question 13.

The total surface area of a solid hemisphere of diameter 2 cm is equal to _______

(1) 12 cm^{2}

(2) 12π cm^{2}

(3) 4π cm^{2}

(4) 3π cm^{2}

Answer:

(4) 3π cm^{2}

Hint:

Radius of a hemisphere = \(\frac{2}{2}\) = 1 cm

Total surface area of a hemisphere = 3πr^{2} sq. units = 3 × π × 12 cm^{2} = 3π cm^{2}

Question 14.

If the volume of a sphere is \(\frac{9}{16} \pi\) cu.cm, then its radius is ________

(1) \(\frac{4}{3}\) cm

(2) \(\frac{3}{4}\) cm

(3) \(\frac{3}{2}\) cm

(4) \(\frac{2}{3}\) cm

Answer:

(2) \(\frac{3}{4}\) cm

Hint:

Volume of the sphere = \(\frac{9}{16} \pi\)

Question 15.

The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______

(1) 81 : 625

(2) 729 : 15625

(3) 27 : 75

(4) 27 : 125

Answer:

(4) 27 : 125

Hint:

Ratio of their surface area = 9 : 25

Question 16.

The total surface area of a solid hemisphere whose radius is a units, is equal to ________

(1) 2πa^{2} sq. units

(2) 3πa^{2} sq. units

(3) 3πa sq. units

(4) 3a^{2} sq. units

Answer:

(2) 3πa^{2} sq. units

Hint:

T.S.A. of a solid hemisphere = 3πr^{2} sq. units

= 3 × π × a × a sq.units

= 3πa^{2} sq. units

Question 17.

If the surface area of a sphere is 100π cm^{2}, then its radius is equal to ______

(1) 25 cm

(2) 100 cm

(3) 5 cm

(4) 10 cm

Answer:

(3) 5 cm

Hint:

Surface area of a sphere = 100π cm^{2}

4πr^{2} = 100π

r^{2} = 25

r = √25 = 5 cm

Question 18.

If the surface area of a sphere is 36π cm^{2}, then the volume of the sphere is equal to _______

(1) 12π cm^{3}

(2) 36π cm^{3}

(3) 72π cm^{3}

(4) 108π cm^{3}

Answer:

(2) 36π cm^{3}

Hint:

Surface area of a sphere = 36π cm^{2}

4πr^{2} = 36π

r^{2} = 9

r = 3 cm

Volume of a sphere = \(\frac{4}{3} \pi r^{3}\) cu. units

= \(\frac{4}{3} \pi\) × 3 × 3 × 3 cm^{3} = 36π cm^{3}

Question 19.

If the total surface area of a solid hemisphere is 12π cm^{2} then its curved surface area is equal to ______

(1) 6π cm^{2}

(2) 24π cm^{2}

(3) 36π cm^{2}

(4) 8π cm^{2}

Answer:

(4) 8π cm^{2}

Hint:

T.S.A of a hemisphere = 12π cm^{2}

3πr^{2} = 12π

r^{2} = 4

r = 2

Curved surface area of a hemisphere = 2πr^{2} = 2 × π × 4 = 8π cm^{2}

Question 20.

If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____

(1) 1 : 8

(2) 2 : 1

(3) 1 : 2

(4) 8 : 1

Answer:

(1) 1 : 8

Hint:

\(r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2\)

II. Answer the following questions:

Question 1.

Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.

Answer:

Given, Circumference of the base of a cylinder = 110 cm

2πr = 110 ……. (1)

Curved surface area = 4400 cm^{2}

2πrh = 4400 cm^{2} ……. (2)

From (1) & (2), \(\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}\)

Height of the cylinder (h) = 40 cm

From (1), 2πr = 110

2 × \(\frac{22}{7}\) × r = 110

r = \(\frac{35}{2}\)

We know that, diameter (d) = 2 × radius

d = 2 × \(\frac{35}{2}\) = 35 cm

Diameter of the Circular cylinder = 35 cm

Question 2.

A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.

Answer:

Given, Radius of a cylinder (r) = 50 cm = 0.5 m

Height of a cylinder (h) = 3.5 m

Curved surface area of a pillar = 2πrh sq. units

Curved surface area of 12 pillars = 12 × 2πrh

= 12 × 2 × \(\frac{22}{7}\) × 0.5 × 3.5 m^{2}

= 132 sq. m.

Cost for painting the lateral surface of pillars per metre = ₹ 20

Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.

The total surface area of a solid right circular cylinder is 231 cm^{2}. Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.

Answer:

Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm

Curved surface area = \(\frac{2}{3}\) × T.S.A = \(\frac{2}{3}\) × 231 = 154 cm^{2}

2πrh = 154 cm^{2} …… (1)

Total surface area = 231 cm^{2}

2πr (h + r) = 231

2πrh + 2πr^{2}= 231

154 + 2πr^{2} = 231 [from (1)]

2πr^{2} = 231 – 154 = 77

Radius of the cylinder = 3.5 cm

Height of the cylinder = 7 cm

Question 4.

The total surface area of a solid right circular cylinder is 1540 cm^{2}. If the height is four times the radius of the base, then find the height of the cylinder.

Answer:

Given, Let the radius of the cylinder be ‘r’

Height of a cylinder (h) = 4r (by given condition)

Total surface area = 1540 cm^{2}

2πr(h + r) = 1540 cm^{2}

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.

If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.

Answer:

Given, In the figure, OAB is a cone and OC ⊥ AB

∠AOC = \(\frac{60^{\circ}}{2}\) = 30°

In the right ∆OAC, tan 30° = \(\frac{\mathrm{AC}}{\mathrm{OC}}\)

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.

The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.

Answer:

Given, Radius of a sector (r) = 21 cm

The angle of the sector (θ) = 180°

Let “R” be the radius of the cone.

Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.

If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.

Answer:

Given, the Curved surface area of a solid hemisphere = 2772 cm^{2}

2πr^{2} = 2772

Total surface area = 3πr^{2} sq. units

= 3 × \(\frac{22}{7}\) × 21 × 21

= 4158 sq.cm

Aliter:

C.S.A of a hemisphere = 2772 cm^{2}

2πr^{2} = 2772 cm^{2}

πr^{2} = \(\frac{2772}{2}\) = 1386 cm

T.S.A of a hemisphere = 3πr^{2} sq.units = 3 × 1386 cm^{2} = 4158 cm^{2}

Question 8.

An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.

Answer:

Given, Circumference of the dome = 17.6 m

2πr = 17.6

\(r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}\)

Curved surface area of the dome = 2πr^{2} sq. units

= 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m^{2}

= 49.28 m^{2}

Cost of painting for one sq.metre = ₹ 5

Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.

Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.

Answer:

Given, Height of a cylinder (h) = 4.5 cm

Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.

A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.

Answer:

After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm^{3}

Question 11.

The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.

Answer:

Given, Height of the wooden solid cone (h) = 12 m

Circumference of the base = 44 m

2πr = 44

r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m

Volume of the wooden solid = \(\frac{1}{3} \pi r^{2} h\) cu. units

= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}\)

= 88 × 7

= 616 m^{3}

Volume of the solid = 616 m^{3}

Question 12.

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.

Answer:

Given, Edge of the cube = 14 cm

The largest circular cone is cut out from the cube.

Radius of the cone (r) = \(\frac{14}{2}\) = 7 cm

Height of the cone (h) = 14 cm

Volume of a cone

Volume of a cone = 718.67 cm^{3}

Question 13.

The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = \(\frac{22}{7}\))

Answer:

Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm

R = r + w = 5 + 0.25 = 5.25 cm

Now, outer surface area of the bowl = 2πR^{2}

= 2 × \(\frac{22}{7}\) × 5.25 × 5.25

= 173.25 sq. cm

Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.

Volume of a hollow sphere is \(\frac{11352}{7}\) cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = \(\frac{22}{7}\))

Answer:

Let R and r be the outer and inner radii of the hollow sphere respectively.

Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.

How many litres of water will a hemispherical tank whose diameter is 4.2 m?

Answer:

Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m

Volume of the hemisphere

= \(\frac{2}{3} \pi r^{3}\) cu.units

= \(\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}\)

= 19.404 m^{3}

= 19.404 x 1000 lit

= 19,404 litres

III. Answer the following questions.

Question 1.

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer:

For cylindrical part:

Radius (r) = 7 cm

Height (h) = 6 cm

Curved surface area = 2πrh = 2 × \(\frac{22}{7}\) × 7 × 6 cm^{2} = 264 cm^{2}

For hemispherical part:

Radius (r) = 7 cm

Surface area (h) = 2πr^{2}

= 2 × \(\frac{22}{7}\) × 7 × 7 cm^{2}

= 308 cm^{2}

Total surface area = (264 + 308) = 572 cm^{2}

Question 2.

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Answer:

Question 3.

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Answer:

For cylinderical part:

Height (h) = 2.4 cm

Diameter (d) = 1.4 cm

Radius (r) = 0.7 cm

Total surface area of the cylindrical part

For conical part:

Base area (r) = 0.7 cm

Height (h) = 2.4 cm

Question 4.

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Answer:

Diameter of the cylindrical well = 7 m

Radius of the cylinder (r) = \(\frac{7}{2}\) m

Depth of the well (h) = 20 m

Volume = πr^{2}h

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}\)

= 22 × 7 × 5 m^{3}

Volume of the earth taken out = 22 × 7 × 5 m^{3}

Now this earth is spread out to form a cuboidal platform having

Length (l) = 22 m

Breadth (b) = 14 m

Let ‘h’ be the height of the platform.

Volume of the platform = 22 × 14 × h m^{3}

Volume of the platform = Volume of the earth taken out

22 × 14 × h = 22 × 7 × 5

\(h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}\)

Thus, the required height of the platform is 2.5 m.

Question 5.

The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]

Answer:

Let the radii of circular ends are R and r [R > r]

Perimeter of circular ends are 207.24 cm and 169.56 cm

2πR = 207.24 cm

The whole surface area of the frustum = π [(R^{2} + r^{2}) + (R + r) l]

Required whole surface area of the frustum

= 3.14 [33^{2} + 27^{2} + (33 + 27) × 10] cm^{2}

= 3.14 [1089 + 729 + 600] cm^{2}

= 3.14 [2418] cm^{2}

= 7592.52 cm^{2}

Question 6.

A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = \(\frac {22}{7}\))

Answer:

Let h_{1} be the length of the pipe

Let R and r be the outer and inner radii of the pipe respectively.

Iron slab:

Volume = lbh = 55 × 40 × 15 cm^{3}

Iron pipe:

Outer diameter, 2R = 8 cm

Outer radius, R = 4 cm

Thickness, w = 1 cm

Inner radius, r = R – w = 4 – 1 = 3 cm

Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.

The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.

Answer:

Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.

A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.

Answer:

Given, Total height of solid = 13.5 cm

Diameter of the cylinder (d) = 28 m

Height of a cylinder (h) = 3 m

Height of a conical portion = 13.5 – 3 = 10.5 m

From the diagram, Radius of a cone = Radius of a cylinder

Radius (r) = 14 m