NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 13.1 — Combination of Solids
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer
Side of cube = ∛64 = 4 cm. Cuboid dimensions: 8 × 4 × 4 cm.
Surface area = 2(8×4 + 4×4 + 8×4) = 2(32 + 16 + 32) = 2 × 80 = 160 cm²
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.
Answer
r = 7 cm. Height of cylinder = 13 – 7 = 6 cm.
Inner SA = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr² = 2π(7)(6) + 2π(49) = 84π + 98π = 182π = 572 cm²
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. Total height = 15.5 cm. Find total surface area of the toy.
Answer
r = 3.5 cm. Height of cone = 15.5 – 3.5 = 12 cm. Slant height l = √(12² + 3.5²) = √(144 + 12.25) = √156.25 = 12.5 cm.
TSA = CSA of cone + CSA of hemisphere = πrl + 2πr² = π(3.5)(12.5) + 2π(3.5)² = 43.75π + 24.5π = 68.25π = 214.5 cm²
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.
Answer
Greatest diameter = side of cube = 7 cm
⇒ r = 3.5 cm.
SA = TSA of cube – base area of hemisphere + CSA of hemisphere
= 6(7²) – πr² + 2πr² = 294 – π(3.5²) + 2π(3.5²) = 294 + π(12.25) = 294 + 38.5 = 332.5 cm²
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.
Answer
Edge = l, r = l/2.
SA = TSA of cube – circular area + CSA of hemisphere = 6l² – π(l/2)² + 2π(l/2)² = 6l² + πl²/4 = l²(24 + π)/4
Exercise 13.2 — Volumes of Combinations
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Answer
V = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π(1)(1) + (2/3)π(1) = π/3 + 2π/3 = π cm³
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter is 3 cm, length of each cone is 2 cm, and total length is 12 cm. Find the volume.
Answer
r = 1.5 cm. Cylinder height = 12 – 2 – 2 = 8 cm. Cone height = 2 cm.
V = πr²h + 2 × (1/3)πr²h_cone = π(2.25)(8) + (2/3)π(2.25)(2) = 18π + 3π = 21π = 66 cm³
A vessel is in the form of an inverted cone. Its height is 8 cm and radius of its top is 5 cm. It is filled with water up to the brim. When lead shots, each a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped.
Answer
Volume of cone = (1/3)π(25)(8) = 200π/3 cm³.
Water overflow = 1/4 of total = 50π/3 cm³.
Volume of each shot = (4/3)π(0.5)³ = (4/3)π(0.125) = π/6 cm³.
Number = (50π/3) / (π/6) = 50 × 6/3 = 100 shots
Exercise 13.3 — Conversion of Solids
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer
Volume of sphere = Volume of cylinder.
(4/3)π(4.2)³ = π(6)²h
⇒ h = 4(4.2)³/(3 × 36) = 4 × 74.088/108 = 2.744 cm
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.
Answer
(4/3)πR³ = (4/3)π(216 + 512 + 1000)
⇒ R³ = 1728
⇒ R = 12 cm
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer
Volume of earth = π(3.5)²(20) = π × 12.25 × 20 = 245π = 770 cm³.
Volume = 22 × 14 × h
⇒ 770 = 308h
⇒ h = 2.5 m
A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly all around the well to a width of 4 m to form an embankment. Find the height of the embankment.
Answer
Volume of well = π(1.5)²(14) = 31.5π m³.
Embankment: outer radius = 1.5 + 4 = 5.5 m. Area = π(5.5² – 1.5²) = π(30.25 – 2.25) = 28π m².
h = 31.5π/28π = 1.125 m
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones.
Answer
Volume of cylinder = π(6)²(15) = 540π cm³.
Volume of each cone + hemisphere = (1/3)π(3)²(12) + (2/3)π(3)³ = 36π + 18π = 54π cm³.
Number = 540π/54π = 10 cones
Exercise 13.4 — Frustum of a Cone
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Answer
R = 2 cm, r = 1 cm, h = 14 cm.
V = (πh/3)(R² + r² + Rr) = (π×14/3)(4 + 1 + 2) = (14π/3)(7) = 98π/3 = 102.67 cm³
A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used.
Answer
Lateral surface + top circle = π(R+r)l + πr² = π(10+4)(15) + π(16) = 210π + 16π = 226π = 710 cm²
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its axis by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Answer
Full cone: h = 20, half-angle = 30°. R = 20 tan 30° = 20/√3 cm.
At h = 10: r = 10/√3 cm.
Volume of frustum = (π×10/3)(R² + r² + Rr) = (10π/3)(400/3 + 100/3 + 200/3) = (10π/3)(700/3) = 7000π/9 cm³.
Wire: radius = 1/32 cm. πr²l = 7000π/9
⇒ l = 7000/(9 × 1/1024) = 7000 × 1024/9 = 796444.4 cm ≈ 7964.4 m
Important Extra Questions
Write all formulas for surface areas and volumes.
Answer
Cube: TSA = 6a², V = a³
Cuboid: TSA = 2(lb+bh+hl), V = lbh
Cylinder: CSA = 2πrh, TSA = 2πr(r+h), V = πr²h
Cone: CSA = πrl, TSA = πr(r+l), V = (1/3)πr²h, l = √(r²+h²)
Sphere: SA = 4πr², V = (4/3)πr³
Hemisphere: CSA = 2πr², TSA = 3πr², V = (2/3)πr³
Frustum: CSA = π(R+r)l, TSA = π[Rl + rl + R² + r²], V = (πh/3)(R² + r² + Rr), l = √[h²+(R–r)²]