SEBA Class 10 Mathematics Chapter 13 Surface Area Volumes

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SEBA Class 10 Mathematics Chapter 13 Surface Area Volumes

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 13 Surface Area Volumes Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 13 Surface Area Volumes Solutions for All Subject, You can practice these here.

Surface Area Volumes

Chapter – 13

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area

of the resulting cuboids. 

Ans: 

Let side of cube = x cm 

Volume of cube = 64cm³

{Volume of cube = (side) } 

X = 4 cm 

∴ side of cube = 4cm 

When cube are joined end to end cuboid is formed whose 

Length = 2 cm = (4) = 8 m 

With X = cm = 4 cm height = x cm = 4cm 

∴  surface area of cuboids = 2(lb + bh lh)

= 2 ( 4 x 4 x 4 x 8 x 4 x 8)

= (16+ 32+32)

2(16+64)

= 2x 80 = 160 cm²

2. 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. [Use 𝜋 = 22/7 ]

Ans: 

It can be observed that radius (r) of the cylindrical part and the

the hemispherical part is the same (i.e., 7 cm).

Height of hemispherical part = Radius = 7 cm

Height of cylindrical part (h) = 13 −7 = 6 cm

Inner surface area of the vessel = CSA of cylindrical part + CSA 

of hemispherical part 

= 2 𝜋rh + 2𝜋r²

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.[Use 𝜋 = 22 7 ]

Ans: 

It can be observed that the radius of the conical part and the hemispherical

part is same (i.e., 3.5 cm).

Height of hemispherical part = Radius (r) = 3.5 = 7/2 cm

Height of conical part (h) = 15.5 −3.5 = 12 cm 

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. [Use 𝜋 = 22 7 ]

Ans: From the figure, it can be observed that the greatest diameter possible for such a hemisphere is equal to the cube’s edge, i.e., 7cm.

Radius (r) of hemispherical part = 7/2 = 3.5cm

Total surface area of solid = Surface area of cubical part + CSA of

hemispherical part − Area of base of hemispherical part

= 6 (edge)² +  2πr² – πr² = (edge)² + πr²

Total surface area of solid = 6(7)2 + 22/7 x 7/2 x 7/2

= 294 + 38.5 = 332.5 cm²

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. 

Ans: Diameter of hemisphere = Edge of cube = l

Radius of hemisphere = l/2

Total surface area of solid = Surface area of cubical part + CSA of

hemispherical part − Area of base of hemispherical part

= 6 (edge)² +  2πr² – πr² = (edge)² + πr²

6. A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. [Use 𝜋 = 22/ 7 ]. 

Ans: 

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part 

= Diameter of the capsule /2 = 5/2

Length of cylindrical part (h) = length of the entire capsule – 2 x r = 14 – 5 = 9 cm 

Surface area of capsule = 2×CSA of hemispherical part + CSA of cylindrical part 

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2 . (Note that the base of the tent will not be covered with canvas.)[Use 𝜋 = 22 7 ]

Ans:  

Given that,

Height (h) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (l) of conical part = 2.8 m

Area of canvas used = CSA of conical part + CSA of cylindrical part 

= πrl + 2πrh

= π x 2 x 2.8 + 2π x 2 x 2.1 

= 44 m²

Cost of 1 m²

canvas = Rs 500

Cost of 44 m²

canvas = 44 × 500 = 22000

Therefore, it will cost Rs 22000 for making such a tent. 

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical

a cavity of the same height and same diameter is hollowed out. Find the total

surface area of the remaining solid to the nearest cm2 . [Use 𝜋 =22/7]

.

Ans: 

Given that,

Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Therefore, radius (r) of the cylindrical part = 0.7 cm 

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base 

The total surface area of the remaining solid to the nearest cm2 is 18 cm²

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in given 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 the article. [Use 𝜋 = 22 7 ]

Ans: 

Given that,

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm

Height of cylindrical part (h) = 10 cm

Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part 

2πrh + 2 +2πr²

= 2π x 3.5 x 10 + 2 x 2π x 3.5 x 3.5 

= 70π + 49π

= 119π

= 17 x 22 = 374cm²

1. A solid is in the shape of a cone standing on a hemisphere with both their

radii being equal to 1 cm and the height of the cone is equal to its radius.

Find the volume of the solid in terms of π. 

Ans: 

Given that, Height (h) of conical part = Radius(r) of conical part = 1 cm Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm Volume of solid = Volume of conical part + Volume of hemispherical part 

= ⅓ πr²h + ⅔ πr³

= ⅓ π. 12.1 + ⅔ 𝜋. 1³ = 𝜋 𝑐𝑚³

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. 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 air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) [Use 𝜋 = 22 7 ] 

Ans: Radius of  cone = Radius of cylinder (R) = 3/2  cm

(R) = 3/2  cm

Height of each cone(h) = 2cm ….Height of cylinder = 12 – 2 – 2 = 8c

Volume of air in cylinder = volume of cylinder + 2(volume of cone)

3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find

approximately how much syrup would be found in 45 gulab jamuns, each

shaped like a cylinder with two hemispherical ends with length 5 cm and

diameter 2.8 cm (see the given figure). [Use 𝜋 = 22/7] 

Ans:  

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm

Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part

= 5 − 2 × 1.4 = 2.2 cm

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of

hemispherical part 

Volume of 45 gulab jamuns = 45×25.05 = 1,127.25 cm³

Volume of sugar syrup = 30% of volume.

30/100 x 1,127.25

= 338.17cm³

= 338 cm³

4. A pen stand made of wood is in the shape of a cuboid with four conical

depressions to hold pens. The dimensions of the cuboids are 15 cm by 10

cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth

is 1.4 cm. Find the volume of wood in the entire stand (see the following figure). [Use 𝜋 = 22/7]

Ans: Length of cuboid (L) = 15 cm 

Width of cuboid (b) = 10cm

Height of cuboid (H) = 3.5 cm 

Height of conical cavity (r) = 0.5 cm 

Radius of conical cavity (h) = 1.4 cm

Hence Volume of wood in pen stands = volume cuboid – 4 (volume of cone)

Hence , volume of wood in pen stand = 523.53

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. 

Ans:  Radius of cone (R ) = 5cm

Height of cone (H) = 8cm

Radius  of each  spherical leads shot (r) = 0.5 cm 

According to question 

Hence , the number of lead shorts dropped in the vessel is 100

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter

24 cm, which is surmounted by another cylinder of height 60 cm and radius

8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately

8 g mass. [Use π = 3.14] 

Ans: 

From the figure, it can be observed that

Height (h¹) of larger cylinder = 220 cm

Radius (r¹) of larger cylinder = 24/2 = 12 cm

Height (h²) of smaller cylinder = 60 cm

Radius (r²) of smaller cylinder = 8 cm 

7. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter o the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3 . Check whether she is correct, taking the above as the inside measurements, and π = 3.14. 

Ans: 

Radius of cone = radius of hemisphere 

=  radius cylinder 

R = 60cm 

Height of cylinder vessel = πR²H 

= 22/7 x 60  x 60  x 180 = 2036571.4 cm 

Volume of solid inserted in cylinder = volume of hemisphere 

= ⅔  πR³  + ⅓   π R² h 

Volume of water flows out = 9055142.86 cm³ 

∴ Volume of water left in cylinder = volume of cylinder – volume of solid inserted in the vessel 

= (203657.14 – 905142.86) cm²

= 1131428.5 cm² = 1131428.5/ 100 x 100 x 100 m³ = 1.131 m³

Hence , volume of water left in cylinder = 1.131m³

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3 . Check whether she is correct, taking the above as the inside measurements, and π = 3.14. 

Ans: 

Height (h) of cylindrical part = 8 cm

Radius (r²) of cylindrical part = 1 cm

Radius (r¹) spherical part = 4.25 cm

Volume of vessel = Volume of sphere + Volume of cylinder

Hence, she is wrong.  

1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a

cylinder of radius 6 cm. Find the height of the cylinder. 

Ans: Radius (r¹) of hemisphere = 4.2 cm

Radius (r²) of cylinder = 6 cm

Let the height of the cylinder be h.

The object formed by recasting the hemisphere will be the same in Volume.

Volume of sphere = Volume of cylinder 

Hence, the height of the cylinder so formed will be 2.74 cm. 

2. Metallic spheres of radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.  

Ans: Radius (r¹) of 1st sphere = 6 cm

Radius (r²) of 2nd sphere = 8 cm

Radius (r³) of 3rd sphere = 10 cm

Let the radius of the resulting sphere be r.

The object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres.

Volume of 3 spheres = Volume of resulting sphere 

Therefore, the radius of the sphere so formed will be 12 cm. 

3. 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. [Use 𝜋 = 22/7]

The shape of the well will be

cylindrical. Depth (h) of well = 20 m

Radius (r) of circular end of well = 7/2 cm

Area of platform = Length × Breadth = 22 × 14 m²

Let height of the platform = H

Volume of soil dug from the well will be equal to the volume of soil scattered on the platform. 

Volume of soil from well = Volume of soil used to make such platform 

Therefore, the height of such platform will be 2.5 m

4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been

spread evenly all around it in the shape of a circular ring of width 4 m to

form an embankment. Find the height of the embankment. 

Ans: 

The shape of the well will be cylindrical.

Depth (h¹) of well = 14 m

Radius (r¹) of the circular end of well = 3/2 m

Width of embankment = 4 m

From the figure, it can be observed that our embankment will be in a cylindrical shape having outer radius (r2) as 4+ (3/2) = 11/2 m and inner radius (r¹) as 3/2 m.

Let the height of the embankment be h².

Volume of soil dug from well = Volume of earth used to form embankment 

Therefore, the height of the embankment will be 1.125 m. 

5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.  

Ans: Height (h¹) of cylindrical container = 15 cm

Radius (r¹) of circular end of container = 6 cm

Radius (r²) of circular end of ice-cream cone = 3 cm

Height (h²) of conical part of ice-cream cone = 12 cm

Let n ice-cream cones be filled with ice-cream of the container.

Volume of ice-cream in cylinder = n × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top) 

Therefore, 10 ice-cream cones can be filled with the ice-cream in the container. 

6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm ×10 cm × 3.5 cm? [Use 𝜋 =22/7]

Ans:   

Coins are cylindrical in shape.

Height (h¹) of cylindrical coins = 2 mm = 0.2 cm

Radius (r) of circular end of coins = 1.75/2 = 0.875 cm

Let n coins be melted to form the required cuboids.

Volume of n coins = Volume of cuboids 

Therefore , the number of coins melted to from such a cuboid id 400

7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm. Find the radius and slant height of the heap. 

Ans: 

Height (h¹) of cylindrical bucket = 32 cm

Radius (r¹) of circular end of bucket = 18 cm

Height (h²) of conical heap = 24 cm

Let the radius of the circular end of the conical heap be r².

The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.

Volume of sand in the cylindrical bucket = Volume of sand in conical heap 

8. Water in canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. how much area will it irrigate in 30 minutes, if 8 cm of standing water is it needed? 

Ans: width of canal = 6m

Velocity of water in canal 1.5m 

Volume of water discharge in one hour 

= (Area of cross section) velocity 

= (6 x 1.5 m2) x 10km

= 6 x 1.5 x 10 x 10 x 1000m³

∴ volume of water discharge in ½ hour 4500000m³

Let us suppose area to be irrigate =  (x)m²

∴ volume of water discharge in ½ hour 

9. A farmer connects a pipe of internal diameter 20 cm form a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? 

Ans: 

Consider an area of cross-section of pipe as shown in the figure.

Radius (r¹) of circular end of pipe = 20/200 = 0.1 m

Area of cross-section = 𝜋𝑟¹

2 = 𝜋(0.1)

2 = 0.01𝜋 𝑚³

Speed of water = 3 km/h = 3000/60 = 50 m/min

Volume of water that flows in 1 minute from pipe = 50 × 0.01𝜋 = 0.5π m3

Volume of water that flows in t minutes from pipe = t × 0.5π m

Radius (r²) of circular end of cylindrical tank = 5m

Depth (h²) of cylindrical tank = 2 m

Let the tank be filled completely in t minutes.

Volume of water filled in tank in t minutes is equal to the volume of water

flowed in t minutes from the pipe.

Volume of water that flows in t minutes from pipe = Volume of water in tank Therefore, the cylindrical tank will be filled in 100 minutes. 

1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass. [Use 𝜋 =22/7]

Ans: 

Radius (r¹) of upper base of glass = 4/2 = 2 cm

Radius (r²) of lower base of glass = 2/2 = 1 cm

Capacity of glass = Volume of frustum of cone 

Therefore, the capacity of the glass is 102 ⅔ 𝑐𝑚³

2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. find the surface area of the frustum. 

Ans: 

Perimeter of upper circular end of frustum = 18

2πr¹ =18 

R¹ = 9/π

Perimeter of lower end of frustum = 6 cm

2πr² = 6 

R² = 3/π

Slant height (l) of frustum = 4

CSA of frustum = π (r¹ + r²) ^l

= 48cm²

Therefore, the curved surface area of the frustum is 48 cm².

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure given below). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material use for making it. [Use 𝜋 = 22 7 ].

Ans: 

Radius (r²) at upper circular end = 4 cm

Radius (r¹) at lower circular end = 10 cm

Slant height (l) of frustum = 15 cm

Area of material used for making the fez = CSA of frustum + Area of upper circular end 

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 Rs.20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs.8 per 100 cm2 . [Take π = 3.14].

Ans: 

Radius (r¹) of upper end of container = 20 cm

Radius (r²) of lower end of container = 8 cm

Height (h) of container = 16 cm

Slant height of the frustum =

Capacity of container = Volume of frustum 

Cost of 1 litre milk = Rs 20

Cost of 10.45 litre milk = 10.45 × 20 = Rs 209

Area of metal sheet used to make the container

Therefore, the cost of the milk which can completely fill the container is Rs 209 and the cost of the metal sheet used to make the container is Rs 156.75. 

5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1/16 cm, find the length of the wire. [Use 𝜋 = 22/7]

Ans: 

1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³

Ans: 

It can be observed that 1 round of wire will cover 3 mm height of cylinder. 

Number of round = height of cylinder/ Diameter = 12/0.3

Length of wire required 1 round = circumference of base of cylinder 

= 2πr = 2π x 5 = 10π

Length of wire in 40 rounds = 40 × 10π 

= 400 x 22/7 = 8800/7

= 1257.14 cm = 12.57 m

Radius of wire = 0.15 cm

Volume of wire = Area of cross-section of wire × Length of wire

= π(0.15)2 × 1257.14

= 88.898 cm³

Mass = Volume × Density

= 88.898 × 8.88

= 789.41 gm 

2. A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse 𝐴𝐶 = √3² + 4² = √25 = 5 cm

Area of ∆ABC 

= ½ x AB X AC

= ½ AC X OB = ½ X 4 X 3 

½ X 5 X OB = 6

OB = 12/5 2.4 cm

Volume of double cone = volume 1 + volume of cone 2 

3. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Ans: Volume of cistern = 150 × 120 × 110 = 1980000 cm³

Volume to be filled in cistern = 1980000 − 129600 = 1850400 cm³

Let n numbers of porous bricks were placed in the cistern.

Volume of n bricks = n × 22.5 × 7.5 × 6.5 = 1096.875n

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks

Therefore, 1792 bricks were placed in the cistern. 

4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2 , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Ans: Rainfall in the river valley = 10 cm = 0.1 m

Area of the river valley = 97280 km2 = 97280000000 m²

Volume of the water = 97280000000×0.1 = 9728000000 m³

Now volume of one river water = 1072000×75×3 = 241200000 m³

Now volume of 3 rivers water = 3×241200000 = 843600000 m

So, 9728000000 m3 ≈ 843600000 m

5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion

attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given figure).

Ans: 

Radius (r¹) of upper circular end of frustum part 9 cm.

Radius (r²) of lower circular end of frustum part = Radius of circular end of cylindrical part = 4 cm

Height (h¹) of frustum part = 22 − 10 = 12 cm

Height (h²) of cylindrical part = 10 cm

Slant height (l) of frustum part

6. Derive the formula for the curved surface area and total surface area of the frustum of the cone. 

Ans:   

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.

Let r¹ and r² be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In ∆ABG and ∆ADF, DF||BG

∴ ∆ABG ∼ ∆ADF 

CSA of frustum DECB =  π( r ¹ – r ²)l

Total  surface area frustum = CSA OF Frustum + Area of upper 

• Area of two circle end 

7. Derive the formula for the volume of the frustum of a cone.

Ans: 

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.

Let r¹ and r² be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In ∆ABG and ∆ADF, DF||BG

∴ ∆ABG ∼ ∆ADF

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