# Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

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

The diagram is given as:

Given,

The Volume (V) of each cube is = 64 cm3

This implies that a3 = 64 cm3

∴ a = 4 cm

Now, the side of the cube = a = 4 cm

Also, the length and breadth of the resulting cuboid will be 4 cm each. While its height will be 8 cm. So, the surface area of the cuboid = 2(lb+bh+lh)

= 2(8×4+4×4+4×8) cm2

= 2(32+16+32) cm2

= (2×80) cm2 = 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.

The diagram is as follows:

Now, the given parameters are:

The diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

Also, the height of the cylinder = h = (13-7) = 6 cm

And, the radius of the hollow hemisphere = 7 cm

Now, the inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part

(2πrh+2πr2) cm2  =2πr(h+r)cm2

2×(22/7)×7(6+7) cm2 = 572 cm2

## 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.

The diagram is as follows:

Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7/2 cm

The total height of the toy is given as 15.5 cm. So, the height of the cone (h) = 15.5-3.5 = 12 cm

∴ Thecurvedsurfaceareaofcone=πrl

(22/7)×(7/2)×(25/2) = 275/2 cm2

Also,thecurvedsurfaceareaofthehemisphere=2πr2

2×(22/7)×(7/2)2

= 77 cm2

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Now, the Total surface area of the toy = CSA of cone + CSA of hemisphere

= (275/2)+77 cm2

= (275+154)/2 cm2

= 429/2 cm2  = 214.5cm2

So, the total surface area (TSA) of the toy is 214.5cm2

## 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.

It is given that each side of cube is 7 cm. So, the radius will be 7/2 cm.

We know,

The total surface area of solid (TSA) = surface area of cubical block + CSA of hemisphere – Area of base of hemisphere

∴ TSA of solid = 6×(side)2+2πr2-πr2

= 6×(side)2+πr2

= 6×(7)2+(22/7)×(7/2)×(7/2)

= (6×49)+(77/2)

= 294+38.5 = 332.5 cm2

So, the surface area of the solid is 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.

The diagram is as follows:

Now, the diameter of hemisphere = Edge of the cube = l

So, the radius of hemisphere = l/2

∴ The total surface area of solid = surface area of cube + CSA of hemisphere – Area of base of hemisphere

TSA of remaining solid = 6 (edge)2+2πr2-πr2

= 6l2 πr2

= 6l2+π(l/2)2

= 6l2+πl2/4

= l2/4(24+π) sq. units

## 6. 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.

Two hemisphere and one cylinder are shown in the figure given below.

Here, the diameter of the capsule = 5 mm

∴ Radius = 5/2 = 2.5 mm

Now, the length of the capsule = 14 mm

So, the length of the cylinder = 14-(2.5+2.5) = 9 mm

∴ The surface area of a hemisphere = 2πr2 = 2×(22/7)×2.5×2.5

= 275/7 mm2

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Now, the surface area of the cylinder = 2πrh

= 2×(22/7)×2.5×9

(22/7)×45 = 990/7 mm2

Thus, the required surface area of medicine capsule will be

= 2×surface area of hemisphere + surface area of the cylinder

= (2×275/7) × 990/7

(550/7) + (990/7) = 1540/7 = 220 mm2

## 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.)

It is known that a tent is a combination of cylinder and a cone.

From the question we know that

Diameter = 4 m

Slant height of the cone (l) = 2.8 m

Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m

Height of the cylinder (h) = 2.1 m

So, the required surface area of tent = surface area of cone + surface area of cylinder

=πrl+2πrh

=πr(l+2h)

= (22/7)×2(2.8+2×2.1)

= (44/7)(2.8+4.2)

= (44/7)×7 = 44 m2

∴ Thecostofthecanvasofthetentattherateof₹500perm2 will be

= Surface area × cost per m2

44×500=₹22000

So, Rs. 22000 will be the total cost of the canvas.

## 8. 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.

The diagram for the question is as follows:

From the question we know the following:

The diameter of the cylinder = diameter of conical cavity = 1.4 cm

So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

Also, the height of the cylinder = height of the conical cavity = 2.4 cm

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Now, the TSA of remaining solid = surface area of conical cavity + TSA of the cylinder

=πrl+(2πrh+πr2)

=πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm2

So, the total surface area of the remaining solid is 17.6 cm2

## Solution:

Here r = 1 cm and h = 1 cm. The diagram is as follows.

Now, Volume of solid = Volume of conical part + Volume of hemispherical part

We know the volume of cone = ⅓ πr2h

And,

The volume of hemisphere = ⅔πr3

So, volume of solid will be

= π cm3

## Solution:

Given,

Height of cylinder = 12–4 = 8 cm

Height of cone = 2 cm

Now, the total volume of the air contained will be = Volume of cylinder+2×(Volume of cone)

∴ Total volume = πr2h+[2×(⅓ πr2h )]

= 18 π+2(1.5 π)

= 66 cm3.

## Solution:

It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends. So, the total height of a gulab jamun = 5 cm.

Diameter = 2.8 cm

∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm

=2.2 cm

Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres

= πr2h+(4/3)πr3

= 4.312π+(10.976/3) π

= 25.05 cm3

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

We know that the volume of sugar syrup = 30% of total volume

So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm3)

= 45×7.515 = 338.184 cm3

## Solution:

Volume of cuboid = length x width x height

We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm

So, the volume of the cuboid = 15x10x3.5 = 525 cm3

Here, depressions are like cones and we know, Volume of cone = (⅓)πr2h

Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm

∴ Volume of 4 cones = 4x(⅓)πr2h

= 1.46 cm2

Now, volume of wood = Volume of cuboid – 4 x volume of cone

= 525-1.46 = 523.54 cm2

## Solution:

For the cone, Radius = 5 cm, Height = 8 cm Also,

Radius of sphere = 0.5 cm

The diagram will be like

It is known that,

Volume of cone = volume of water in the cone

= ⅓πr2h = (200/3)π cm3

Now,

Total volume of water overflown= (¼)×(200/3) π =(50/3)π

= (4/3)πr3

= (1/6) π

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Now,

The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot

= (50/3)π/(⅙)π

= (50/3)×6 = 100

## Solution:

Given, the height of the big cylinder (H) = 220 cm

Radius of the base (R) = 24/12 = 12 cm

So, the volume of the big cylinder = πR2H

= π(12)2 × 220 cm3

= 99565.8 cm3

Now, the height of smaller cylinder (h) = 60 cm

Radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr2h

= π(8)2×60 cm3

= 12068.5 cm3

∴ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder

= 99565.8 + 12068.5

=111634.5 cm3

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

We know,

Mass = Density x volume

So, mass of the pole = 8×111634.5

= 893 Kg (approx.)

## Solution:

Here, the volume of water left will be = Volume of cylinder – Volume of solid

Given,

Radius of cone = 60 cm, Height of cone = 120 cm Radius of cylinder = 60 cm Height of cylinder = 180 cm Radius of hemisphere = 60 cm

Now,

Total volume of solid = Volume of Cone + Volume of hemisphere

Volume of cone = π×122×103cm3 = 144×103π cm3

So, Total volume of solid = 144×103π cm3 -(⅔)×π×103 cm3

Volume of hemisphere = (⅔)×π×103 cm3

Volume of cylinder = π×602×180 = 648000 = 648×103  π cm3

Now, volume of water left will be = Volume of cylinder – Volume of solid

= (648-288) × 103×π = 1.131 m3

## Solution:

Given,

For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

Now, volume of this vessel = Volume of cylinder + Volume of sphere

= π×(1)2×8+(4/3)π(1)3

= 346.51 cm3

## Solution:

t is given that radius of the sphere (R) = 4.2 cm  Also, Radius of cylinder (r) = 6 cm

Now, let height of cylinder = h

It is given that the sphere is melted into a cylinder. So, Volume of Sphere = Volume of Cylinder

∴ (4/3)×π×R3 = π×r2×h.

⇒ h = 2.74 cm

## Solution:

For Sphere 1:

∴ Volume (V1) = (4/3)×π×r13

For Sphere 2:

∴ Volume (V2) = (4/3)×π×r23

For Sphere 3:

∴ Volume (V3) = (4/3)× π× r33

Also, let the radius of the resulting sphere be “r” Now, Volume of resulting sphere = V1+V2+V3

(4/3)×π×r3 = (4/3)×π×r13+(4/3)×π×r23 +(4/3)×π×r33

r3 = 63+83+103

r3 = 1728

r = 12 cm

## Solution:

It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m

Also, Depth (h) = 20 m

Volume of the earth dug out will be equal to the volume of the cylinder

∴ Volume of Cylinder = π×r2×h

= 22×7×5 m3

Let the height of the platform = H

Volume of soil from well (cylinder) = Volume of soil used to make such platform

π×r2×h = Area of platform × Height of the platform

We know that the dimension of the platform is = 22×14

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

So, Area of platform = 22×14 m2

∴ π×r2×h = 22×14×H

⇒ H = 2.5 m

## Solution:

The shape of the well will be cylindrical as given below.

Given, Depth (h1) of well = 14 m

Diameter of the circular end of the well =3 m

So, Radius (r1) = 3/2 m

Width of the embankment = 4 m

From the figure, it can be said that the embankment will be a cylinder having an outer radius

(r2) as 4+(3/2) = 11/2 m and inner radius (r1) as 3/2m

Now, let the height of embankment be h2

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

π×r12×h = π×(r22-r12) × h2

Solving this, we get,

The height of the embankment (h2) as 1.125 m.

## Solution:

Number of cones will be = Volume of cylinder / Volume of ice cream cone

For the cylinder part, Radius = 12/2 = 6 cm Height = 15 cm

∴ Volume of cylinder = π×r2×h = 540π

For the ice cone part,

Radius of conical part = 6/2 = 3 cm

Height = 12 cm

Radius of hemispherical part = 6/2 = 3 cm

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Now,

Volume of ice cream cone = Volume of conical part + Volume of hemispherical part

= (⅓)×π×r2×h+(⅔)×π×r3

= 36π +18π

= 54π

∴ Number of cones = (540π/54π)

= 10

## Solution:

It is known that the coins are cylindrical in shape. So, height (h1) of the cylinder = 2 mm = 0.2 cm

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

Now, the number of coins to be melted to form the required cuboids be “n”

So, Volume of n coins = Volume of cuboids n × π × r2 × h1  = l × b × h

n×π×(0.875)2×0.2 = 5.5×10×3.5

Or, n = 400

## Solution:

The diagram will be as-

Given,

Height (h1) of cylindrical part of the bucket = 32 cm Radius (r1) of circular end of the bucket = 18 cm Height of the conical heap ((h2) = 24 cm

Now, let “r2” be the radius of the circular end of the conical heap.

We know that volume of the 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

π×r12×h1 = (⅓)×π×r22×h2

π×182×32 = (⅓)×π ×r22×24

Or, r2= 36 cm

And,

Slant height (l) = √(362+242) = 12√13 cm.

## Solution:

It is given that the canal is the shape of a cuboid with dimensions as: Breadth (b) = 6 m and Height (h) = 1.5 m

It is also given that

The speed of canal = 10 km/hr

Length of canal covered in 1 hour = 10 km Length of canal covered in 60 minutes = 10 km Length of canal covered in 1 min = (1/60)x10 km

Length of canal covered in 30 min (l) = (30/60)x10 = 5km = 5000 m

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

We know that the canal is cuboidal in shape. So, Volume of canal = lxbxh

= 5000x6x1.5 m3

= 45000 m3

Now,

Volume of water in canal = Volume of area irrigated

= Area irrigated x Height

So, Area irrigated = 56.25 hectares

∴ Volume of canal = lxbxh

45000 = Area irrigatedx8 cm

45000 = Area irrigated x (8/100)m

Or, Area irrigated = 562500 m2 = 56.25 hectares.

## Solution:

Consider the following diagram-

Volume of water that flows in t minutes from pipe = t×0.5π m3

Volume of water that flows in t minutes from pipe = t×0.5π m3

Radius (r2) of circular end of cylindrical tank =10/2 = 5 m

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

t×0.5π = π×r22×h2

Or, t = 100 minutes

## Solution:

Radius (r1) of the upper base = 4/2 = 2 cm Radius (r2) of lower the base = 2/2 = 1 cm Height = 14 cm

Now, Capacity of glass = Volume of frustum of cone

So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)

= (⅓)×π×(14)(22+12+ (2)(1))

∴ The capacity of the glass = 102×(⅔) cm3

## Solution:

Given,

Slant height (l) = 4 cm

Circumference of upper circular end of the frustum = 18 cm ∴

2πr1 = 18

Or, r1 = 9/π

Similarly, circumference of lower end of the frustum = 6 cm ∴

2πr2 = 6

Or, r2 = 6/π

Now, CSA of frustum = π(r1+r2) × l

= π(9/π+6/π) × 4

= 12×4 = 48 cm2

## Solution:

Given,

For the lower circular end, radius (r1) = 10 cm For the upper circular end, radius (r2) = 4 cm Slant height (l) of frustum = 15 cm

Now,

The area of material to be used for making the fez = CSA of frustum + Area of upper circular end

CSA of frustum = π(r1+r2)×l

= 210π

And, Area of upper circular end = πr22

= 16π

∴ The area of material used = 710 × (2/7) cm2

## Solution:

Given,

r1  = 20 cm,

r2  = 8 cm and h = 16 cm

∴ Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

It is given that the rate of milk = Rs. 20/litre

So, Cost of milk = 20×volume of the frustum

= Rs. 209

Now, slant height will be

So, CSA of the container = π(r1+r2)×l

= 1758.4 cm2

Hence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)

= 1758.4+201 = 1959.4 cm2

∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157

## Solution:

The diagram will be as follows

Consider AEG

Radius (r1) of upper end of frustum = (10√3)/3 cm Radius (r2) of lower end of container = (20√3)/3 cm Height (r3) of container = 10 cm

Now,

Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Solving this we get,

Volume of the frustum = 22000/9 cm3

The radius (r) of wire = (1/16)×(½) = 1/32 cm

Now,

Let the length of wire be “l”.

Volume of wire = Area of cross-section x Length

= (πr2)xl

= π(1/32)2x l

Now, Volume of frustum = Volume of wire

22000/9 = (22/7)x(1/32)2x l

Solving this we get, l = 7964.44 m

## Solution:

Given that,

Diameter of cylinder = 10 cm

So, radius of the cylinder (r) = 10/2 cm = 5 cm

∴ Length of wire in completely one round = 2πr = 3.14×5 cm = 31.4 cm

It is given that diameter of wire = 3 mm = 3/10 cm

∴ The thickness of cylinder covered in one round = 3/10 m

Hence, the number of turns (rounds) of the wire to cover 12 cm will be-

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds

40 x 31.4 cm = 1256 cm

Radius of the wire = 0.3/2 = 0.15 cm

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

= π(0.15)2×1257.14

= 88.898 cm3

We know,

Mass = Volume × Density

= 88.898×8.88

= 789.41 gm

## Solution:

Draw the diagram as follows:

Let us consider the  ABA Here,

AS = 3 cm, AC = 4 cm

So, Hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’ Now, AD/CA = AB/CB

By putting the value of CA, AB and CB we get, AD = 2/5 cm

We also know, DB/AB = AB/CB So, DB = 9/5 cm As, CD = BC-DB, CD = 16/5 cm

Now, volume of double cone will be

Solving this we get, V = 30.14 cm3

The surface area of the double cone will be

= 52.75 cm2

## Solution:

Given that the dimension of the cistern = 150 × 120 × 110

So, volume = 1980000 cm3

Volume to be filled in cistern = 1980000 – 129600

= 1850400 cm3

Now, let the number of bricks placed be “n”

So, volume of n bricks will be = n×22.5×7.5×6.5

Now as each brick absorbs one-seventeenth of its volume, the volume will be

= n/(17)×(22.5×7.5×6.5)

For the condition given in the question,

The volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern

Or, n×22.5×7.5×6.5 = 1850400+n/(17)×(22.5×7.5×6.5) Solving this we get,

n = 1792.41

## Solution:

From the question, it is clear that

Total volume of 3 rivers = 3×[(Surface area of a river)×Depth] Given,

Surface area of a river = [1072×(75/1000)] km

And,

Depth = (3/1000) km

Now, volume of 3 rivers = 3×[1072×(75/1000)]×(3/1000)

= 0.72 km3

Now, volume of rainfall = total surface area × total height of rain

= 9.7 km3

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

For the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.

But, 9.7 km3  ≠ 0.72 km3

So, the question statement is false.

## Solution:

Given,

Diameter of upper circular end of frustum part = 18 cm

So, radius (r1) = 9 cm

Now, the radius of the lower circular end of frustum (r2) will be equal to the radius of the circular end of the cylinder

So, r2 = 8/2 = 4 cm

Now, height (h1) of the frustum section = 22 – 10 = 12 cm

And,

Height (h2) of cylindrical section = 10 cm (given)

Now, the slant height will be-

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

= π(r1+r2)l+2πr2h2

Solving this we get,

Area of tin sheet required = 782×(4/7) cm2

## Solution:

Consider the diagram

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1  and r2 are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the ΔABG and ΔADF,

Here, DF||BG

Now, by rearranging we get,

### YOU ARE READING: Surface Areas and Volumes Chapter 13 NCERT Solutions For Class 10 CBSE Mathematics

The total surface area of frustum will be equal to the total CSA of frustum + the area of upper circular end + area of the lower circular end

= π(r1+r2)l+πr22+πr12

∴ Surface area of frustum = π[r1+r2)l+r12+r22]

## Solution:

Consider the same diagram as the previous question.

Now, approach the question in the same way as the previous one and prove that

Again,

Now, rearrange them in terms of h and h1

The total volume of frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

= (⅓)πr12h1 -(⅓)πr22(h1 – h)

= (π/3)[r12h1-r22(h1 – h)]

Now, solving this we get,

∴ Volume of frustum of the cone = (⅓)πh(r12+r22+r1r2)