MP Board Class 10 Maths Chapter 12 Surface Areas and Volumes Solutions
Chapter 12: Surface Areas and Volumes - Step-by-Step Solutions
Welcome, MP Board Class 10 students! Chapter 12 deals with finding the surface areas and volumes of combinations of fundamental 3D solids such as cubes, cuboids, cylinders, cones, spheres, and hemispheres. Master these concepts with complete NCERT/MP Board step-by-step solutions below.
Important Formulas
- Cuboid: TSA = 2(lb + bh + hl), Volume = l × b × h
- Cube: TSA = 6a2, Volume = a3
- Cylinder: CSA = 2πrh, TSA = 2πr(r + h), Volume = πr2h
- Cone: Slant Height (l) = √(r2 + h2), CSA = πrl, TSA = πr(r + l), Volume = (1/3)πr2h
- Sphere: Surface Area = 4πr2, Volume = (4/3)πr3
- Hemisphere: CSA = 2πr2, TSA = 3πr2, Volume = (2/3)πr3
Exercise 12.1 Solutions
Question 1
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = a3 = 64 cm3 ⇒ Edge of cube (a) = 4 cm.
When two cubes are joined end to end, dimensions of the resulting cuboid are:
Length (l) = 4 + 4 = 8 cm, Breadth (b) = 4 cm, Height (h) = 4 cm.
Surface Area of Cuboid = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 2(80) = 160 cm2.
Question 2
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.
Solution:
Radius of hemisphere (r) = 14 / 2 = 7 cm.
Height of hemisphere = radius = 7 cm.
Height of cylinder (h) = Total height - Radius of hemisphere = 13 - 7 = 6 cm.
Inner surface area of vessel = CSA of cylinder + CSA of hemisphere
= 2πrh + 2πr2 = 2πr(h + r) = 2 × (22/7) × 7 × (6 + 7) = 44 × 13 = 572 cm2.
Question 3
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius (r) = 3.5 cm.
Height of cone (h) = Total height - Radius = 15.5 - 3.5 = 12 cm.
Slant height of cone (l) = √(r2 + h2) = √(3.52 + 122) = √(12.25 + 144) = √(156.25) = 12.5 cm.
Total surface area of toy = CSA of cone + CSA of hemisphere
= πrl + 2πr2 = πr(l + 2r) = (22/7) × 3.5 × (12.5 + 2 × 3.5) = 11 × (12.5 + 7) = 11 × 19.5 = 214.5 cm2.
Question 4
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.
Solution:
Greatest diameter of hemisphere = side of cube = 7 cm.
Radius of hemisphere (r) = 7 / 2 = 3.5 cm.
Surface Area of solid = Surface area of cube + CSA of hemisphere - Base area of hemisphere
= 6a2 + 2πr2 - πr2 = 6a2 + πr2
= 6(7)2 + (22/7) × (3.5)2 = 6(49) + 38.5 = 294 + 38.5 = 332.5 cm2.
Question 5
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.
Solution:
Edge of cube = l, Radius of hemisphere (r) = l/2.
Surface area of remaining solid = TSA of cube - Base area of hemisphere + CSA of hemisphere
= 6l2 - π(l/2)2 + 2π(l/2)2 = 6l2 + π(l/2)2 = 6l2 + πl2/4 = (l2 / 4)(24 + π) sq units.
Question 6
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:
Radius (r) = 5 / 2 = 2.5 mm.
Length of cylinder (h) = Total length - 2(r) = 14 - 5 = 9 mm.
Surface area of capsule = CSA of cylinder + 2 × CSA of hemisphere
= 2πrh + 2 × (2πr2) = 2πr(h + 2r) = 2 × (22/7) × 2.5 × (9 + 5) = (110 / 7) × 14 = 220 mm2.
Question 7
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas at Rs 500 per m2.
Solution:
Cylinder: h = 2.1 m, r = 4 / 2 = 2 m. Cone: l = 2.8 m, r = 2 m.
Area of canvas = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h + l)
= (22/7) × 2 × (2(2.1) + 2.8) = (44/7) × (4.2 + 2.8) = (44/7) × 7 = 44 m2.
Cost of canvas = 44 × 500 = Rs 22,000.
Question 8
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Radius (r) = 1.4 / 2 = 0.7 cm, Height (h) = 2.4 cm.
Slant height of cone (l) = √(r2 + h2) = √(0.72 + 2.42) = √(0.49 + 5.76) = √(6.25) = 2.5 cm.
Total surface area = CSA of cylinder + Base area of cylinder + CSA of cone
= 2πrh + πr2 + πrl = πr(2h + r + l) = (22/7) × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8.0 = 17.6 cm2 ≈ 18 cm2.
Question 9
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
Radius (r) = 3.5 cm, Height (h) = 10 cm.
TSA of article = CSA of cylinder + 2 × CSA of hemisphere
= 2πrh + 2(2πr2) = 2πr(h + 2r) = 2 × (22/7) × 3.5 × (10 + 2 × 3.5) = 22 × 17 = 374 cm2.
Exercise 12.2 Solutions
Question 1
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.
Solution:
Radius (r) = 1 cm, Height of cone (h) = 1 cm.
Volume of solid = Volume of cone + Volume of hemisphere
= (1/3)πr2h + (2/3)πr3 = (1/3)π(1)2(1) + (2/3)π(1)3 = (1/3)π + (2/3)π = π cm3.
Question 2
Rachel, an engineering student, made a model shaped like a cylinder with two cones attached at its ends. Diameter is 3 cm and length is 12 cm. Height of each cone is 2 cm. Find the volume of air contained in the model.
Solution:
Radius (r) = 3/2 = 1.5 cm. Height of cone (h1) = 2 cm.
Height of cylinder (h2) = 12 - (2 + 2) = 8 cm.
Volume = Volume of cylinder + 2 × Volume of cone
= πr2h2 + 2 × (1/3)πr2h1 = πr2[h2 + (2/3)h1]
= (22/7) × (1.5)2 × [8 + (4/3)] = (22/7) × 2.25 × (28/3) = 22 × 0.75 × 4 = 66 cm3.
Question 3
A gulab jamun contains sugar syrup up to about 30% of its volume. Find how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Solution:
Radius (r) = 1.4 cm. Height of cylinder (h) = 5 - (1.4 + 1.4) = 2.2 cm.
Volume of 1 gulab jamun = Volume of cylinder + 2 × Volume of hemisphere
= πr2h + (4/3)πr3 = πr2[h + (4/3)r] = (22/7) × (1.4)2 × [2.2 + (4/3)(1.4)]
= 6.16 × [2.2 + 1.867] = 6.16 × 4.067 ≈ 25.05 cm3.
Volume of 45 gulab jamuns = 45 × 25.05 = 1127.25 cm3.
Syrup volume = 30% of 1127.25 = 0.30 × 1127.25 ≈ 338 cm3.
Question 4
A pen stand made of wood is in the shape of a cuboid with four conical depressions. Dimensions of cuboid are 15 cm by 10 cm by 3.5 cm. Radius of depression is 0.5 cm and depth is 1.4 cm. Find volume of wood.
Solution:
Volume of cuboid = 15 × 10 × 3.5 = 525 cm3.
Volume of 4 conical depressions = 4 × (1/3)πr2h = 4 × (1/3) × (22/7) × (0.5)2 × 1.4 = 1.47 cm3.
Volume of wood in stand = 525 - 1.47 = 523.53 cm3.
Question 5
An inverted cone of height 8 cm and radius 5 cm is filled with water. Lead shots of radius 0.5 cm are dropped into it, causing one-fourth of water to flow out. Find number of lead shots dropped.
Solution:
Volume of water in cone = (1/3)πR2H = (1/3)π(5)2(8) = (200/3)π cm3.
Volume of water displaced = (1/4) × (200/3)π = (50/3)π cm3.
Volume of 1 spherical lead shot = (4/3)πr3 = (4/3)π(0.5)3 = (0.5/3)π cm3.
Number of lead shots = Displaced volume / Volume of 1 shot = [(50/3)π] / [(0.5/3)π] = 50 / 0.5 = 100.
Question 6
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by another cylinder of height 60 cm and radius 8 cm. Find mass of pole, given 1 cm3 of iron has 8g mass (Use π = 3.14).
Solution:
Large cylinder: R = 12 cm, H = 220 cm. Volume = πR2H = 3.14 × 144 × 220 = 99475.2 cm3.
Small cylinder: r = 8 cm, h = 60 cm. Volume = πr2h = 3.14 × 64 × 60 = 12057.6 cm3.
Total Volume = 99475.2 + 12057.6 = 111532.8 cm3.
Mass = 111532.8 × 8 g = 892262.4 g = 892.26 kg.
Question 7
A solid consisting of a cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a cylinder full of water touching the bottom. Find volume of water left in cylinder if radius is 60 cm and height is 180 cm.
Solution:
Volume of outer cylinder = πR2H = π × (60)2 × 180 = 648000π cm3.
Volume of solid = Vol of cone + Vol of hemisphere = (1/3)πr2h + (2/3)πr3
= (1/3)π(60)2(120) + (2/3)π(60)3 = 144000π + 144000π = 288000π cm3.
Volume of water left = 648000π - 288000π = 360000π cm3 = 360000 × 3.14159 / 1000000 ≈ 1.131 m3.
Question 8
A spherical vessel has a cylindrical neck 8 cm long, 2 cm in diameter; diameter of spherical part is 8.5 cm. A child finds its volume to be 345 cm3. Check whether she is correct (π = 3.14).
Solution:
Cylindrical neck: r = 1 cm, h = 8 cm. Volume = πr2h = 3.14 × 12 × 8 = 25.12 cm3.
Spherical part: R = 4.25 cm. Volume = (4/3)πR3 = (4/3) × 3.14 × (4.25)3 ≈ 321.39 cm3.
Total Volume = 25.12 + 321.39 = 346.51 cm3.
Hence, she is incorrect. The correct volume is 346.51 cm3.
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