Extra Practice Questions
Surface Areas and Volumes (R.D. Sharma Level – Class 9)
1. The curved surface area of a right circular cone is 12320 cm². If the radius of its base is 56 cm, find its height. (Use π = 22/7)
Given: CSA = 12320 cm², r = 56 cm.
CSA = πrl => 12320 = (22/7) × 56 × l => 12320 = 22 × 8 × l => l = 12320 / 176 = 70 cm.
h = √(l² - r²) = √(70² - 56²) = √((70-56)(70+56)) = √(14 × 126) = √1764 = 42 cm.
Answer: The height of the cone is 42 cm.
2. A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls each of diameter 0.6 cm. Find the number of small balls thus obtained.
Volume of large sphere = (4/3)πR³ = (4/3)π(3)³ = 36π cm³.
Diameter of small ball = 0.6 cm, so radius r = 0.3 cm.
Volume of one small ball = (4/3)πr³ = (4/3)π(0.3)³ = (4/3)π(0.027) = 0.036π cm³.
Number of balls = (Volume of large sphere) / (Volume of small ball) = (36π) / (0.036π) = 1000.
Answer: 1000 small balls can be obtained.
… (18 more questions follow)
3. The surface areas of two spheres are in the ratio 1:4. Find the ratio of their volumes.
Let radii be r₁ and r₂. Given: (4πr₁²) / (4πr₂²) = 1/4.
(r₁/r₂)² = 1/4 => r₁/r₂ = 1/2.
Ratio of volumes = ((4/3)πr₁³) / ((4/3)πr₂³) = (r₁/r₂)³.
Ratio = (1/2)³ = 1/8.
Answer: The ratio of their volumes is 1:8.
4. How many meters of cloth 5m wide will be required to make a conical tent, the radius of whose base is 7m and height is 24m?
r = 7 m, h = 24 m. First, find slant height l.
l = √(7² + 24²) = √(49+576) = √625 = 25 m.
Area of cloth required = CSA of cone = πrl = (22/7) × 7 × 25 = 550 m².
Length of cloth = Area / Width = 550 / 5 = 110 m.
Answer: 110 m of cloth is required.
5. The volumes of two spheres are in the ratio 64:27. Find the ratio of their surface areas.
Let radii be r₁ and r₂. Given: ((4/3)πr₁³) / ((4/3)πr₂³) = 64/27.
(r₁/r₂)³ = 64/27 => r₁/r₂ = ∛(64/27) = 4/3.
Ratio of surface areas = (4πr₁²) / (4πr₂²) = (r₁/r₂)².
Ratio = (4/3)² = 16/9.
Answer: The ratio of their surface areas is 16:9.
6. A corn cob, shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm² of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob.
r = 2.1 cm, h = 20 cm.
l = √((2.1)² + 20²) = √(4.41 + 400) = √404.41 ≈ 20.11 cm.
CSA = πrl = (22/7) × 2.1 × 20.11 = 6.6 × 20.11 = 132.726 cm².
Number of grains = Area × 4 = 132.726 × 4 ≈ 530.9.
Answer: Approximately 531 grains would be on the cob.
7. A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π.
Let sphere radius be r, cube side be a.
Given: 4πr² = 6a² => a² = (4πr²)/6 = (2πr²)/3 => a = r * √(2π/3).
Volume of sphere V₁ = (4/3)πr³.
Volume of cube V₂ = a³ = [r * √(2π/3)]³ = r³ * (2π/3) * √(2π/3).
Ratio V₁/V₂ = [(4/3)πr³] / [r³ * (2π/3) * √(2π/3)]
= (4π/3) / [(2π/3) * √(2π/3)] = 2 / √(2π/3) = 2 * √(3/(2π)) = √(12/(2π)) = √(6/π).
The ratio is √6 : √π. (Proved)
8. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and the level of water is raised by 6.75 cm. Find the radius of the ball.
The volume of the spherical ball is equal to the volume of the water raised in the cylindrical tub.
Volume of water raised = πR²h, where R=12 cm (radius of tub) and h=6.75 cm (height rise).
Volume = π × 12² × 6.75 = 972π cm³.
Let the radius of the ball be r. Volume of ball = (4/3)πr³.
Equating volumes: (4/3)πr³ = 972π => (4/3)r³ = 972 => r³ = (972*3)/4 = 729.
r = ∛729 = 9 cm.
Answer: The radius of the ball is 9 cm.
9. The radii of the bases of a cylinder and a cone are in the ratio 3:4 and their heights are in the ratio 2:3. Find the ratio of their volumes.
Let radii be 3x and 4x. Let heights be 2y and 3y.
Volume of cylinder (V₁) = πr₁²h₁ = π(3x)²(2y) = 18πx²y.
Volume of cone (V₂) = (1/3)πr₂²h₂ = (1/3)π(4x)²(3y) = 16πx²y.
Ratio V₁/V₂ = (18πx²y) / (16πx²y) = 18/16 = 9/8.
Answer: The ratio of their volumes is 9:8.
10. The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Let original diameter be d, radius be r. Original SA = 4πr².
New diameter = d – 0.25d = 0.75d. New radius = 0.75r.
New SA = 4π(0.75r)² = 4π(0.5625r²) = 0.5625(4πr²).
Decrease in area = Original SA – New SA = (1 – 0.5625) × Original SA = 0.4375 × Original SA.
Percentage decrease = 0.4375 × 100 = 43.75%.
Answer: The curved surface area decreases by 43.75%.
11. A toy is in the form of a cone mounted on a hemisphere of same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy.
Diameter=6cm, so r=3cm. Cone height h=4cm.
Slant height of cone l = √(3²+4²) = √25 = 5 cm.
Surface area of toy = CSA of cone + CSA of hemisphere
= πrl + 2πr² = πr(l+2r)
= π × 3 × (5 + 2*3) = 3π(11) = 33π cm².
Answer: The surface area is 33π cm².
12. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid.
Radius r=7cm. Height of cone h = diameter = 14cm.
Volume = Vol(cone) + Vol(hemisphere) = (1/3)πr²h + (2/3)πr³
= (1/3)π(7²)(14) + (2/3)π(7)³
= (1/3)π(49)(14) + (2/3)π(343) = (686π + 686π)/3 = 1372π/3 cm³.
Answer: The volume is 1372π/3 cm³.
13. The circumference of the base of a 9m high wooden solid cone is 44m. Find the volume.
h=9m. Circumference = 2πr = 44 => 2 × (22/7) × r = 44 => r = 7m.
Volume = (1/3)πr²h = (1/3) × (22/7) × 7² × 9 = 462 m³.
Answer: The volume is 462 m³.
14. If the total surface area of a solid hemisphere is 462 cm², find its volume.
TSA of hemisphere = 3πr² = 462.
3 × (22/7) × r² = 462 => r² = (462 × 7) / 66 = 49 => r = 7 cm.
Volume = (2/3)πr³ = (2/3) × (22/7) × 7³ = 2156/3 ≈ 718.67 cm³.
Answer: The volume is approximately 718.67 cm³.
15. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Equal bases means they have equal radii, r. Let cone height be h. Height of hemisphere is its radius r.
Given: (1/3)πr²h = (2/3)πr³.
Divide by (1/3)πr²: h = 2r.
Ratio of heights = (height of cone) / (height of hemisphere) = h/r = 2r/r = 2/1.
Answer: The ratio of their heights is 2:1.
16. The rainwater from a roof 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the vessel is just full, find the rainfall in cm.
Volume of water from roof = Volume of cylindrical vessel.
Let rainfall height be h. Volume from roof = 22 × 20 × h = 440h m³.
Volume of vessel = πR²H = (22/7) × 1² × 3.5 = 11 m³.
440h = 11 => h = 11/440 = 1/40 m.
Rainfall in cm = (1/40) × 100 = 2.5 cm.
Answer: The rainfall is 2.5 cm.
17. A hemispherical bowl is filled to the brim with a liquid. The liquid is to be filled into cylindrical bottles of diameter 5 cm and height 6 cm. If the diameter of the bowl is 30 cm, find how many bottles are required.
Volume of bowl = (2/3)πR³ = (2/3)π(15)³ = 2250π cm³.
Volume of one bottle = πr²h = π(2.5)²(6) = 37.5π cm³.
Number of bottles = Vol(bowl) / Vol(bottle) = (2250π) / (37.5π) = 60.
Answer: 60 bottles are required.
18. A cone of maximum volume is carved out of a solid hemisphere of radius r. Find the ratio of the volume of the cone to the volume of the original hemisphere.
For maximum volume, the cone’s base will be the same as the hemisphere’s base.
Radius of cone = r. Height of cone = r.
Volume of cone = (1/3)πr²h = (1/3)πr²(r) = (1/3)πr³.
Volume of hemisphere = (2/3)πr³.
Ratio = [(1/3)πr³] / [(2/3)πr³] = 1/2.
Answer: The ratio is 1:2.
19. A right-angled triangle, whose sides are 3 cm and 4 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume of the double cone so formed.
Sides are 3, 4, so hypotenuse is 5. Let the double cone have a common radius r and heights h₁ and h₂. The height of the triangle from the hypotenuse is the radius r.
Area = (1/2)*3*4 = 6. Also, Area = (1/2)*5*r => r = 12/5 = 2.4 cm.
Volume = Vol(cone 1) + Vol(cone 2) = (1/3)πr²h₁ + (1/3)πr²h₂ = (1/3)πr²(h₁+h₂).
Here, h₁+h₂ is the hypotenuse = 5 cm.
Volume = (1/3) × π × (2.4)² × 5 = (1/3) × π × 5.76 × 5 = 9.6π cm³.
Answer: The volume is 9.6π cm³.
20. 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 curved surface area of the frustum.
Let radii be R and r. Slant height l=4cm.
2πR = 18 => R = 9/π cm.2πr = 6 => r = 3/π cm.
CSA of frustum = πl(R+r) = π × 4 × (9/π + 3/π)
= 4π × (12/π) = 48 cm².
Answer: The curved surface area is 48 cm².
