RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2: This exercise focuses on determining the surface areas and volumes of solid combinations. Students can use the RD Sharma Class 10 Solutions to clear up any doubts they have about this chapter. Students can also download the RD Sharma Solutions for Class 10 Maths Chapter 16 Surface Areas And Volumes Exercise 16.2 PDF.
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RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2
Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.2- Important Question with Answers
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.
Solution:
Given,
The diameter of the cylinder (also the same for cone) = 24 m.
So, its radius (R) = 24/2 = 12 m
The height of the Cylindrical part (H1) = 11m
So, Height of the cone part (H2) = 16 – 11 = 5 m
Now,
Vertex of the cone above the ground = 11 + 5 = 16 m
Curved Surface area of the 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 of Cone (S1) = 22/7 × 12 × 13
And,
Curved Surface Area of Cylinder (S2) = 2πRH1
S2 = 2π(12)(11) m2
Thus, the area of Canvas 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 of canvas required for the tent is 1320 m2
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.
Solution:
Given,
Radius of the cylindrical portion of the rocket (R) = 2.5 m
Height of the cylindrical portion of the rocket (H) = 21 m
Slant Height of the Conical surface of the rocket (L) = 8 m
Curved Surface Area of the Cone (S1) = πRL = π(2.5)(8)= 20π
And,
Curved Surface Area of the Cone (S2) = 2πRH + πR2
S2 = (2π × 2.5 × 21) + π (2.5)2
S2 = (π × 105) + (π × 6.25)
Thus, the total curved surface 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 total Surface Area of the Conical Surface = 412.5 m2
Now, calculating the volume of the rocket
Volume of the conical part of the rocket (V1) = 1/3 × 22/7 × R2 × h
V1 = 1/3 × 22/7 × (2.5)2 × h
Let, h be the height of 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 conical part (V1) = 1/3 × 22/7 × 2.52 × 7.6 m2 = 49.67 m2
Next,
Volume of the Cylindrical Portion (V2) = πR2h
V2 = 22/7 × 2.52 × 21 = 412.5 m2
Thus, the total volume of the rocket = V1 + V2
V = 412.5 + 49.67 = 462.17 m2
Hence, the total volume of the Rocket is 462.17 m2
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
Solution:
Given,
Height of the tent = 77 dm
Height of a surmounted cone = 44 dm
Height of the Cylindrical Portion = Height of the tent – Height of the surmounted Cone
= 77 – 44
= 33 dm = 3.3 m
And, given diameter of the cylinder (d) = 36 m
So, its radius (r) of the cylinder = 36/2 = 18 m
Let’s consider L as the 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 of the cone (L) = 18.3 m
Now, the Curved Surface area of the Cylinder (S1) = 2πrh
S1 = 2π (184.4) m2
And, the Curved Surface area of the cone (S2) = πrL
S2 = π × 18 × 18.3 m2
So, the total curved surface of the tent (S) = S1 + S2
S = S1 + S2
S = (2π18 × 4.4) + (π18 × 18.3)
S = 1533.08 m2
Hence, the total Curved Surface Area (S) = 1533.08 m2
Next,
The cost of 1 m2 canvas = Rs 3.50
So, 1533.08 m2 of canvas will cost = Rs (3.50 x 1533.08)
= Rs 5365.8
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:
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 slant height of cone.
Then, we know that
l2 = r2 + h2
l2 = 32 + 42 = 9 + 16 = 25
l = 5 cm
Hence, the slant height of the cone (l) = 5 cm
So, the curved surface area of the cone (S1) = πrl
S1 = π(3)(5)
S1 = 47.1 cm2
And, the curved surface area of the hemisphere (S2) = 2πr2
S2 = 2π(3)2
S2 = 56.23 cm2
So, the total surface area (S) = S1 + S2
S = 47.1 + 56.23
S = 103.62 cm2
Therefore, the curved surface area of the toy is 103.62 cm2
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