# Class 10 RD Sharma Solutions – Chapter 16 Surface Areas and Volumes – Exercise 16.2 | Set 1

### Question 1. Consider a tent cylindrical in shape and surmounted by a conical top having height 16 m and radius as common for all the surfaces constituting the whole portion of the tent which is equal to 24 m. Height of the cylindrical portion of the tent is 11 m. Find the area of Canvas required for the tent.

**Solution:**

According to the question

Diameter of the cylinder = 24 m,

So the radius(r) = 24/2 = 12 m,

Height of the cylindrical part (H1) = 11m

and the total height of the shape = 16 m

So, the height of the cone (h)= 16 – 11 = 5m

Now, first we find the slant height of the cone

So, according to the formula of slant height

l = √r

^{2}+ h^{2}l = √12

^{2}+ 5^{2}= 13mNow, we find the curved surface area of the cone

A1 = πrl

= 22/7 × 6 × 13 ……(1)

Now, we find the curved surface area of the cylinder

A2 = 2πrH1

= 2π(12)(11) ……(2)

Now we find the total area of canvas required for tent,

So,

A = A1 + A2

= 22/7 × 12 × 13 + 2 × 22/7 × 12 × 11

= 490 + 829.38 = 1319.8 = 1320 m

^{2}

Hence, the total canvas required for tent is 1320 m^{2}

### Question 2. Consider a Rocket. Suppose the rocket is in the form of a Circular Cylinder Closed at the lower end with a Cone of the same radius attached to its top. The Cylindrical portion of the rocket has radius say, 2.5 m and the height of that cylindrical portion of the rocket is 21m. The Conical portion of the rocket has a slant height of 8m, then calculate the total surface area of the rocket and also find the volume of the rocket.

**Solution: **

According to the question

Radius of the cylindrical part of the rocket(r) = 2.5 m,

Height of the cylindrical part of the rocket (h) = 21 m,

Slant Height of the conical surface of the rocket(l) = 8 m,

Now, we find the curved surface area of the cone

A1 = πrl

= π(2.5)(8)

= π x 20 ……(1)

Now, we find the curved surface area of the cone

A2 = 2πrh + πr

^{2}= (2π × 2.5 × 21) + π (2.5)

^{2}= (π × 10.5) + (π × 6.25) ……(2)

Now we find the total curved surface area

A = A1 + A2

= (π x 20) + (π x 10.5) + (π x 6.25)

= 62.83 + 329.86 + 19.63 = 412.3 m

^{2}Hence, the total curved surface area of the conical surface is 412.3 m

^{2}Let us considered H be the height of the conical portion in the rocket,

So, l

^{2}= r^{2}+ H^{2}H

^{2}= l^{2}– r^{2}h = √8

^{2}– 2.5^{2}= 23.685 mNow we find the volume of the conical surface of the rocket

V1 = 1/3πr

^{2}H= 1/3 × 22/7 × (2.5)

^{2}× 23.685 ……(3)Now we find the volume of the cylindrical part

V2 = πr

^{2}hV2 = 22/7 × 2.5

^{2}× 21Therefore, the total volume of the rocket

V = V1 + V2

V = 461.84 m

^{2}

Hence, the total volume of the Rocket is 461.84 m^{2}

### Question 3. Take a tent structure in vision being cylindrical in shape with height 77 dm and is being surmounted by a cone at the top having height 44 dm. The diameter of the cylinder is 36 m. Find the curved surface area of the tent.

**Solution: **

According to the question

Height of the tent = 77 dm,

Height of a surmounted cone = 44 dm,

Diameter of the cylinder = 36 m,

So, the radius of the cylinder(r) = 36/2 = 18 m

Therefore, the height of the cylindrical Portion(h) = 77 – 44 = 33 dm = 3.3 m

Let us considered l be the slant height of the cone,

So,l

^{2}= r^{2}+ h^{2}l

^{2}= 18^{2}+ 3.3^{2}l

^{2}= 324 + 10.89 = 334.89 = 18.3 mNow we find the curved surface area of the cylinder

A1 = 2πrh

= 2π184.4 m

^{2}……(1)Now we find the curved surface area of the cone

A2 = πrh

= π × 18 × 18.3 ……(2)

Hence, the total curved surface of the tent

A = A1 + A2

Put all the values from eq(1) and (2)

A = (2π x 18 × 4.4) + (π x 18 × 18.3) = 1532. 46 m

^{2}

Hence, the total Curved Surface Area is 1532.46 m^{2}

### Question 4. A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy.

**Solution:**

According to the question

The height of the cone (h) = 4 cm

Diameter of the cone (d) = 6 cm,

So, radius of the cone (r) = 3

Also radius of the cone = radius of the hemisphere.

Let us considered l be the slant height of the cone,

l = √r

^{2}+ h^{2}l = √3

^{2}+ 4^{2}= 5 cmNow we find the curved surface area of the cone

A1 = πrl

= π(3)(5) = 47.1 cm

^{2}Now we find the curved surface area of the hemisphere

A1 = 2πr

^{2}= 2π(3)

^{2}= 56.23 cm^{2}Hence, the total surface area of toy

A = A1 + A2

= 47.1 + 56.23 = 103.62 cm

^{2}

Hence, the curved surface area of the toy is 103.62 cm^{2}

### Question 5. A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7).

**Solution:**

According to the question

The radius of the common base (r) = 3.5 cm,

Height of the cylindrical part (h) = 10 cm,

Height of the conical part (h1) = 6 cm

Let us considered l be the slant height of the cone,

So,

l = √r

^{2}+ h^{2}= √3.5^{2}+ 6^{2}= 48.25 cmNow we find the curved surface area of the cone

A1 = πrl

= π(3.5)(48.25) = 76.408 cm

^{2}Now we find the curved surface area of the hemisphere

A2 = 2πrh

= 2π(3.5) (10) = 220 cm

^{2}Therefore, the total surface area of the solid

A = A1 + A2

= 76.408 + 220 = 373.408 cm

^{2}

Hence, total surface area of the solid is 373.408 cm^{2}.

### Question 6. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical parts are 5cm and 13 cm, respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm.

**Solution: **

According to the question

The height of the cylindrical part(h1) = 13 cm,

Radius of the cylindrical part(r) = 5 cm,

Height of the whole solid(H) = 30 cm,

The height of the conical part,

h2 = 30 – 13 – 5 = 12 cm

Now we find the slant height of the cone

l = √r

^{2}+ h2^{2}= √5^{2}+ 12^{2}= 13 cmNow we find the curved surface area of the cylinder

A1 = 2πrh1

= 2π(5)(13) = 408.2 cm

^{2}Now we find the curved surface area of the cone

A2 = πrl

= π(5)(13) cm

^{2}= 204.1 cm^{2}Now we find the curved surface area of the hemisphere

A3 = 2πr

^{2}= 2π(5)

^{2}= 157 cm^{2}Hence, the total curved surface area of the toy

A = A1 + A2 + A3

= (408.2 + 204.1 + 157) = 769.3 cm

^{2}

Hence, the surface area of the toy is 769.3 cm^{2}

### Question 7. Consider a cylindrical tub having radius as 5 cm and its length 9.8 cm. It is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in tub. If the radius of the hemisphere is 3.5 cm and the height of the cone outside the hemisphere is 5 cm. Find the volume of water left in the tub.

**Solution: **

According to the question

Radius of the Cylindrical tub(r) = 5 cm,

Height of the Cylindrical tub(h1) = 9.8 cm,

Height of the cone outside the hemisphere (h2) = 5 cm,

Radius of the hemisphere = 5 cm.

Now we find the volume of the cylindrical tub

V1 = πr

^{2}h1= π(5)

^{2}9.8 = 770 cm^{3}Now we find the volume of the Hemisphere

V2 = 2/3 × π × r

^{3}= 2/3 × 22/7 × 3.5

^{3}= 89.79 cm^{3}Now we find the volume of the cone

V3= 1/3 × π × r

^{2}× h2= 1/3 × 22/7 × 3.52 × 5 = 64.14 cm

^{3}Hence, The total volume

V = V2 + V3

V = 89.79 + 64.14 = 154 cm

^{3}Hence, the total volume of the solid = 154 cm

^{3}Hence, the volume of water left in the tube V = V1 – V2

V = 770 – 154 = 616 cm

^{3}

Hence, the volume of water left in the tube is 616 cm^{3}.

### Question 8. A circus tent has a cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 cm. The height of the cylindrical and conical portions is 4.2 cm and 2.1 cm. Find the volume of that circus tent.

**Solution: **

According to the question

Radius of the cylindrical part(R) = 20 m,

Height of the cylindrical part(h1) = 4.2 m,

Height of the conical part(h2) = 2.1 m,

Now we find the volume of the cylindrical part

V1 = πr

^{2}h1= π(20)

^{2}4.2 = 5280 m^{3}Now we find the volume of the conical part

V2 = 1/3 × π × r

^{2}× h2= 1/3 × 22/7 × r

^{2}× h2= 13 × 22/7 × 20

^{2}× 2.1 = 880 m^{3}Hence, the total volume of the circus tent

V = V1 + V2 = 6160 m

^{3}

Hence, the volume of the circus tent is 6160 m^{3}

### Question 9. A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with the conical ends, each of axis 9 cm. Determine the capacity of the tank.

**Solution: **

According to the question

Base diameter of the cylinder = 21 cm,

Radius (r) = diameter/2 = 25/2 = 11.5 cm,

Height of the cylindrical part of the tank (h1) = 18 cm,

Height of the conical part of the tank (h2) = 9 cm.

Now we find the volume of the cylindrical portion

V1 = πr

^{2}h1= π(11.5)

^{2}18 = 7474.77 cm^{3}Now we find the volume of the conical portion

V2 = 1/3 × 22/7 × r

^{2}× h2= 1/3 × 22/7 × 11.5

^{2}× 9 = 1245.795 cm^{3}Hence, the total the capacity of the tank

V = V1 + V2

= 7474.77 + 1245.795

= 8316 cm

^{3}

Hence, the capacity of the tank is 8316 cm^{3}

### Question 10. A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and base radius of the cone are also the same. Find the whole surface and volume of the remaining Cylinder.

**Solution:**

According to the question

Height of the cone = Height of the cylinder = h = 12 cm,

Radius of the cone = Radius of the cylinder = r = 5 cm.

Let us considered l be the slant height of the cone

So,

l = √r

^{2}+ h^{2}= √5^{2}+ 12^{2}= 13 cmNow we find the total surface area of the remaining part in the circular cylinder

A= πr

^{2}+ 2πrh + πrl= π(5)

^{2}+ 2π(5)(12) + π(5)(13) = 210 π cm^{2}Now we find the volume of the remaining part of the circular cylinder

V = πr

^{2}h – 1/3 πr^{2}h= πr

^{2}h – 1/3 × 22/7 × r^{2}× h= π(5)

^{2}(12) – 1/3 × 22/7 × 5^{2}× 12 = 200 π cm^{2}

Hence, the area of the remaining part is 210 π cm^{2 }and the volume is 200 π cm^{2}

### Question 11. A tent is in the form of a cylinder of diameter 20 m and height 2.5 m surmounted by a cone of equal base and height 7.5 m. Find the capacity of tent and the cost of canvas as well at a price of Rs.100 per square meter.

**Solution: **

According to the question

Diameter of the cylinder = 20 m,

So, the radius of the cylinder = 10 m,

Height of the cylinder (h1) = 2.5 m,

Radius of the cone = Radius of the cylinder = r = 15 m,

Height of the Cone (h2) = 7.5 m.

Let us considered l be the slant height of the cone

So,

l = √r

^{2}+ h2^{2}= √15^{2}+ 7.5^{2}= 12.5 mNow we find the volume of the cylinder

V1 = πr

^{2}h1= π(10)

^{2}2.5 = 250π m^{3}Now we find the volume of the Cone

V2 = 1/3πr

^{2}h2= 1/3 × 22/7 × 10

^{2}× 7.5 = 250π m^{3}Hence, the total capacity of the tent

V = V1 + V2

= 250 π + 250 π = 500π m

^{3}Hence, the total capacity of the tent = V = 4478.5714 m

^{3}Now we find the total area of the canvas required for the tent is

S = 2πrh1 + πrl

= 2(π)(10)(2.5) + π(10)(12.5) = 550 m

^{2}

Hence, the total cost of the canvas is (100 x 550) = Rs. 55000

### Question 12. Consider a boiler which is in the form of a cylinder having length 2 m and there’s a hemispherical ends each of having a diameter of 2 m. Find the volume of the boiler.

**Solution: **

According to the question

Diameter of the hemisphere = 2 m,

So, the radius of the hemisphere (r) = 1 m,

Height of the cylinder (h) = 2 m,

Now we find the volume of the cylinder

V1 = πr

^{2}h= π(1)

^{2 }x 2 = 22/7 × 2 = 44/7m^{3}Since, at each of the ends of the cylinder, two hemispheres are attached.

so, the volume of two hemispheres

V2 = 2 × 2/3πr

^{3}= 2 × 2/3 × 22/7 × 13 = 22/7 × 4/3 = 88/21 m

^{3}Hence, the volume of the boiler is

V = V1 + V2

V = 44/7 + 88/21 = 220/21 m

^{3}

Hence, the volume of the boiler is 220/21 m^{3}

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