RD Chapter 16 Surface Areas and Volumes Ex 16.2 Solutions

Question - 1 : - A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.

Answer - 1 : -

Given,

The diameter of thecylinder (also the same for cone) = 24 m.

So, its radius (R)= 24/2 = 12 m

The height of theCylindrical part (H₁) = 11m

So, Height of the conepart (H₂) = 16 – 11 = 5 m

Now,

Vertex of the coneabove the ground = 11 + 5 = 16 m

Curved Surface area ofthe Cone (S₁) = πRL = 22/7 × 12 × L

The slant height (L)is given by,

L = √(R² +H₂²) = √(12² + 5²) = √169

L = 13 m

So,

Curved Surface Area ofCone (S₁) = 22/7 × 12 × 13

And,

Curved Surface Area ofCylinder (S₂) = 2πRH₁

S₂=2π(12)(11) m²

Thus, the area ofCanvas required for tent

S = S₁ +S₂= (22/7 × 12 × 13) + (2 × 22/7 × 12 × 11)

S = 490 + 829.38

S = 1319.8 m²

S = 1320 m²

Therefore, the area ofcanvas required for the tent is 1320 m²

Question - 2 : - A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket.

Answer - 2 : -

Given,

Radius of thecylindrical portion of the rocket (R) = 2.5 m

Height of thecylindrical portion of the rocket (H) = 21 m

Slant Height of theConical surface of the rocket (L) = 8 m

Curved Surface Area ofthe Cone (S₁) = πRL = π(2.5)(8)= 20π

And,

Curved Surface Area ofthe Cone (S₂) = 2πRH + πR²

S₂=(2π × 2.5 × 21) + π (2.5)²

S₂ =(π × 105) + (π × 6.25)

Thus, the total curvedsurface area S is

S = S₁ +S₂

S = (π20) + (π105) +(π6.25)

S = (22/7)(20 + 105 +6.25) = 22/7 x 131.25

S = 412.5 m²

Therefore, the totalSurface Area of the Conical Surface = 412.5 m²

Now, calculating thevolume of the rocket

Volume of the conicalpart of the rocket (V₁) = 1/3 × 22/7 × R² × h

V₁ = 1/3× 22/7 × (2.5)² × h

Let, h be the heightof the conical portion in the rocket.

We know that,

L² = R²+h²

h² = L²–R² = 8² – 2.5²

h = 7.6 m

Using the value of h,we will get

Volume of the conicalpart (V₁) = 1/3 × 22/7 × 2.5² × 7.6 m² =49.67 m²

Next,

Volume of theCylindrical Portion (V₂) = πR²h

V₂ = 22/7× 2.5² × 21 = 412.5 m²

Thus, the total volumeof the rocket = V₁ + V₂

V = 412.5 + 49.67 =462.17 m²

Hence, the totalvolume of the Rocket is 462.17 m²

Question - 3 : - A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m²

Answer - 3 : -

Given,

Height of the tent =77 dm

Height of a surmountedcone = 44 dm

Height of theCylindrical Portion = Height of the tent – Height of the surmounted Cone

= 77 – 44

= 33 dm = 3.3 m

And, given diameter ofthe cylinder (d) = 36 m

So, its radius (r) ofthe cylinder = 36/2 = 18 m

Let’s consider L asthe slant height of the cone.

Then, we know that

L² = r² +h²

L² =18² + 3.3²

L²=324 + 10.89

L² = 334.89

L = 18.3 m

Thus, slant height ofthe cone (L) = 18.3 m

Now, the CurvedSurface area of the Cylinder (S₁) = 2πrh

S₁ =2π (184.4) m²

And, the CurvedSurface area of the cone (S₂) = πrL

S₂= π× 18 × 18.3 m²

So, the total curvedsurface of the tent (S) = S₁+ S₂

S = S₁ +S₂

S = (2π18 × 4.4)+ (π18 × 18.3)

S = 1533.08 m²

Hence, the totalCurved Surface Area (S) = 1533.08 m²

Next,

The cost of 1 m² canvas= Rs 3.50

So, 1533.08 m² ofcanvas will cost = Rs (3.50 x 1533.08)

= Rs 5365.8

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.

Answer - 4 : -

Given that,

The height of the cone(h) = 4 cm

Diameter of the cone(d) = 6 cm

So, its radius (r) = 3

Let, ‘l’ be the slantheight of cone.

Then, we know that

l² = r² +h²

l² = 3² +4² = 9 + 16 = 25

l = 5 cm

Hence, the slantheight of the cone (l) = 5 cm

So, the curved surfacearea of the cone (S₁) = πrl

S₁ = π(3)(5)

S₁ =47.1 cm²

And, the curvedsurface area of the hemisphere (S₂) = 2πr²

S₂ = 2π(3)²

S₂ =56.23 cm²

So, the total surfacearea (S) = S₁+ S₂

S = 47.1 + 56.23

S = 103.62 cm²

Therefore, the curvedsurface area of the toy is 103.62 cm²

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).

Answer - 5 : -

Given,

Radius of the commonbase (r) = 3.5 cm

Height of thecylindrical part (h) = 10 cm

Height of the conicalpart (H) = 6 cm

Let, ‘l’ be the slantheight of the cone

Then, we know that

l² = r² +H²

l² =3.5² + 6² = 12.25 + 36 = 48.25

l = 6.95 cm

So, the curved surfacearea of the cone (S₁) = πrl

S₁ = π(3.5)(6.95)

S₁ =76.38 cm²

And, the curvedsurface area of the hemisphere (S₂) = 2πr²

S₂ = 2π(3.5)²

S₂ =77 cm²

Next, the curvedsurface area of the cylinder (S₃) = 2πrh

S₂ = 2π(3.5)(10)

S₂ =220 cm²

Thus, the totalsurface area (S) = S₁+ S₂ + S₃

S = 76. 38 + 77 + 220= 373.38 cm²

Therefore, the totalsurface area of the solid is 373.38 cm²

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.

Answer - 6 : -

Given,

Height of theCylindrical portion (H) = 13 cm

Radius of theCylindrical portion (r) = 5 cm

Height of the wholesolid = 30 cm

Then,

The curved surfacearea of the cylinder (S₁) = 2πrh

S₁=2π(5)(13)

S₁=408.2 cm²

Let, ‘L’ be the slantheight of the cone

And, the curvedsurface area of the cone (S₂) = πrL

S₂ =π(6)L

For conical part, wehave

h = 30 – 13 – 5 = 12cm

Then, we know that

L² = r² +h²

L² = 5² +12²

L²=25 + 144

L² =169

L = 13 m

So,

S₂ =π(5)(13) cm²

S₂ =204.28 cm²

Now, the curvedsurface area of the hemisphere (S₃) = 2πr²

S₃ =2π(5)²

S₃ =157.14 cm²

Thus, the total curvedsurface area (S) = S₁+ S₂+ S₃

S = (408.2 + 204.28 +157.14)

S = 769.62 cm²

Therefore, the surfacearea of the toy is 770 cm²

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.

Answer - 7 : -

Given,

The radius of theCylindrical tub (r) = 5 cm

Height of theCylindrical tub (H) = 9.8 cm

Height of the coneoutside the hemisphere (h) = 5 cm

Radius of the hemisphere= 3.5 cm

Now, we know that

The volume of theCylindrical tub (V₁) = πr²H

V₁ =π(5)²9.8

V₁ =770 cm³

And, the volume of theHemisphere (V₂) = 2/3 × π × r³

V₂= 2/3× 22/7 × 3.5³

V₂ =89.79 cm³

And, the volume of theHemisphere (V₃) = 23 × π × r × 2h

V₃= 2/3× 22/7 × 3.5² × 5

V₃=64.14 cm³

Thus, total volume (V)= Volume of the cone + Volume of the hemisphere

= V₂+V₃

V = 89.79 + 64.14 cm³

= 154 cm³

So, the total volumeof the solid = 154 cm³

In order to find thevolume of the water left in the tube, we have to subtract the volume of thehemisphere and the cone from the volume of the cylinder.

Hence, the volume ofwater left in the tube = V₁– V₂

= 770 – 154

= 616 cm³

Therefore, the volumeof water left in the tube is 616 cm³.

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

Answer - 8 : -

Given,

Radius of thecylindrical portion (R) = 20 m

Height of thecylindrical portion (h₁) = 4.2 m

Height of the conicalportion (h₂) = 2.1 m

Now, we know that

Volume of theCylindrical portion (V₁) = πr²h₁

V₁ =π(20)²4.2

V₁=5280 m³

And, the volume of theconical part (V₂) = 1/3 × 22/7 × r² × h²

V₂= 13× 22/7 × 20² × 2.1

V₂=880 m³

Thus, the total volumeof the tent (V) = volume of the conical portion + volume of the Cylindricalportion

V = V₁+V₂

V = 6160 m³

Therefore, volume ofthe tent is 6160 m³

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 length 9 cm. Determine the capacity of the tank.

Answer - 9 : -

Given,

Base diameter of thecylindrical base of the petrol tank = 21 cm

So, its radius (r)= diameter/2 = 21/2 = 10.5 cm

Height of theCylindrical portion of the tank (h₁) = 18 cm

Height of the Conicalportion of the tank (h₂) = 9 cm

Now, we know that

The volume of theCylindrical portion (V₁) = πr²h₁

V₁ =π(10.5)²18

V₁ =6237 cm³

The volume of theConical portion (V₂) = 1/3 × 22/7 × r² × h₂

V₂= 1/3× 22/7 × 10.5² × 9

V₂=1039.5 cm³

Therefore, the totalvolume of the tank (V) = 2 x volume of a conical portion + volume of theCylindrical portion

V = V₁+V₂ = 2 x 1039.5 + 6237

V = 8316 cm³

So, the capacity ofthe tank = V = 8316 cm³

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.

Answer - 10 : -

Given,

Height of the circularCylinder (h₁) = 12 cm

Base radius of thecircular Cylinder (r) = 5 cm

Height of the conicalhole = Height of the circular cylinder, i.e., h₁= h₂=12 cm

And, Base radius ofthe conical hole = Base radius of the circular Cylinder = 5 cm

Let’s consider, L asthe slant height of the conical hole.

Then, we know that

Now,

The total surface areaof the remaining portion in the circular cylinder (V₁) = πr²+2πrh + πrl

V₁=π(5)² + 2π(5)(12) + π(5)(13)

V₁=210 π cm²

And, the volume of theremaining portion of the circular cylinder = Volume of the cylinder – Volume ofthe conical hole

V = πr²h– 1/3 × 22/7 × r² × h

V = π(5)²(12)– 1/3 × 22/7 × 5² × 12

V = 200 π cm²

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RD Chapter 16 Surface Areas and Volumes Ex 16.2 Solutions

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