Chapter 13: Surface Areas and Volumes (NCERT Solutions)

Exercise 13.1: Surface Areas of Combinations of Solids

Q1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Volume of cube = a³ = 64 ⇒ a = 4 cm.
For cuboid: L = 4+4 = 8 cm, B = 4 cm, H = 4 cm.
TSA = 2(LB + BH + HL) = 2(32 + 16 + 32) = 2(80) = 160 cm².

Q2. 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.

Radius r = 7 cm.
Height of cylinder h = 13 - 7 = 6 cm.
Inner SA = CSA of cylinder + CSA of hemisphere
= 2πrh + 2πr²
= 2πr(h + r) = 2 × (22/7) × 7 × (6 + 7)
= 44 × 13 = 572 cm².

Q3. 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.

r = 3.5 cm.
Height of cone h = 15.5 - 3.5 = 12 cm.
Slant height l = √(h² + r²) = √(144 + 12.25) = √156.25 = 12.5 cm.
TSA = CSA of cone + CSA of hemisphere
= πrl + 2πr² = πr(l + 2r)
= (22/7) × 3.5 × (12.5 + 7)
= 11 × 19.5 = 214.5 cm².

Q4. 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.

Greatest diameter = side of cube = 7 cm ⇒ r = 3.5 cm.
SA = TSA of cube - Base area of hemisphere + CSA of hemisphere
= 6a² - πr² + 2πr² = 6a² + πr²
= 6(49) + (22/7)(3.5)(3.5)
= 294 + 38.5 = 332.5 cm².

Q5. 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.

Edge = l, Radius r = l/2.
TSA = TSA of cube - Base of hemisphere + CSA of hemisphere
= 6l² - π(l/2)² + 2π(l/2)²
= 6l² + πl²/4
= (l²/4)(24 + π) units².

Q6. 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.

r = 2.5 mm.
Length of cylinder = 14 - 2(2.5) = 9 mm.
SA = CSA of cylinder + 2(CSA of hemisphere)
= 2πrh + 4πr² = 2πr(h + 2r)
= 2 × (22/7) × 2.5 × (9 + 5)
= (110/7) × 14 = 220 mm².

Q7. 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 of the tent at the rate of Rs 500 per m².

r = 2 m, h (cyl) = 2.1 m, l = 2.8 m.
Area = CSA of cylinder + CSA of cone
= 2πrh + πrl = πr(2h + l)
= (22/7) × 2 × (4.2 + 2.8)
= (44/7) × 7 = 44 m².
Cost = 44 × 500 = Rs 22,000.

Q8. 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.

h = 2.4 cm, r = 0.7 cm.
l = √(2.4² + 0.7²) = √(5.76 + 0.49) = √6.25 = 2.5 cm.
TSA = CSA of cylinder + CSA of cone + Base area
= 2πrh + πrl + πr²
= πr(2h + l + r)
= (22/7)(0.7)(4.8 + 2.5 + 0.7)
= 2.2 × 8 = 17.6 ≈ 18 cm².

Q9. 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.

r = 3.5 cm, h = 10 cm.
TSA = CSA of cylinder + 2(CSA of hemisphere)
= 2πrh + 4πr² = 2πr(h + 2r)
= 2 × (22/7) × 3.5 × (10 + 7)
= 22 × 17 = 374 cm².

Exercise 13.2: Volumes of Combinations of Solids

Q1. A solid is in the shape of a cone standing on a hemisphere with 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.

r = 1 cm, h = 1 cm.
Volume = Vol of cone + Vol of hemisphere
= (1/3)πr²h + (2/3)πr³
= (1/3)π(1) + (2/3)π(1) = π cm³.
Answer: π cm³.

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends. Diameter = 3 cm, Length = 12 cm. Height of each cone = 2 cm. Find volume of air contained in the model.

r = 1.5 cm.
Height of cylinder H = 12 - 2(2) = 8 cm.
Volume = Vol of cylinder + 2(Vol of cone)
= πr²H + 2(1/3)πr²h
= πr²(H + 2h/3) = (22/7)(1.5)²(8 + 4/3)
= (22/7)(2.25)(28/3) = 22 × 0.75 × 4 = 66 cm³.

Q3. Gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately 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.

r = 1.4 cm.
Length of cyl part = 5 - 2.8 = 2.2 cm.
Vol of 1 jamun = πr²h + (4/3)πr³ = πr²(h + 4r/3)
= (22/7)(1.4)²(2.2 + 5.6/3) = 6.16(12.2/3) ≈ 25.05 cm³.
Vol of 45 jamuns = 45 × 25.05 = 1127.25 cm³.
Syrup = 30% of 1127.25 ≈ 338 cm³.

Q4. 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 each depression is 0.5 cm and depth is 1.4 cm. Find the volume of wood in the entire stand.

Vol of cuboid = 15 × 10 × 3.5 = 525 cm³.
Vol of 4 cones = 4 × (1/3)πr²h = 4 × (1/3)(22/7)(0.5)²(1.4)
= 4 × (1/3)(22/7)(0.25)(1.4) = 1.47 cm³.
Vol of wood = 525 - 1.47 = 523.53 cm³.

Q5. A vessel is in the form of an inverted cone. Its height is 8 cm and radius of top is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped, one-fourth of the water flows out. Find the number of lead shots dropped.

Vol of water = Vol of cone = (1/3)π(5)²(8) = 200π/3 cm³.
Water flowed out = (1/4)(200π/3) = 50π/3 cm³.
Vol of 1 shot = (4/3)π(0.5)³ = (4/3)π(0.125) = 0.5π/3 = π/6 cm³.
n × (π/6) = 50π/3 ⇒ n = 100.
Answer: 100 lead shots.

Q6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

Vol = Vol of large cyl + Vol of small cyl
= π(12)²(220) + π(8)²(60)
= 3.14(31680 + 3840) = 3.14(35520) = 111532.8 cm³.
Mass = 111532.8 × 8 = 892262.4 g = 892.26 kg.

Q7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Vol of cylinder = π(60)²(180).
Vol of solid = Vol of cone + Vol of hemisphere = (1/3)π(60)²(120) + (2/3)π(60)³.
Vol water left = Vol cylinder - Vol solid
= π(60)²[180 - (40 + 40)] = π(3600)(100) = 360000π cm³.
= 360000 × (22/7) ≈ 1131428.57 cm³ ≈ 1.131 m³.

Q8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Vol = Vol of sphere + Vol of cylinder
= (4/3)π(4.25)³ + π(1)²(8)
= 3.14[(4/3)(76.76) + 8] = 3.14[102.35 + 8] = 3.14(110.35) ≈ 346.51 cm³.
She is incorrect. Correct volume is approx 346.51 cm³.

Exercise 13.3: Conversion of Solids

(Selected Key Questions)

Q1. 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.

Vol of sphere = Vol of cylinder.
(4/3)π(4.2)³ = π(6)²h.
h = (4/3) × (4.2 × 4.2 × 4.2) / 36 = 1.4 × 1.4 × 1.4 = 2.74 cm.

Q2. 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.

(4/3)πR³ = (4/3)π(6³ + 8³ + 10³).
R³ = 216 + 512 + 1000 = 1728.
R = 12 cm.
Answer: 12 cm.

Q3. 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.

Vol of earth = πr²h = (22/7)(3.5)²(20) = 770 m³.
770 = 22 × 14 × H.
H = 770 / 308 = 2.5 m.
Answer: 2.5 m.

Exercise 13.4: Frustum of a Cone

(Selected Key Questions)

Q1. 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.

R = 2 cm, r = 1 cm, h = 14 cm.
Vol = (1/3)πh(R² + r² + Rr)
= (1/3)(22/7)(14)(4 + 1 + 2)
= (44/3)(7) = 308/3 = 102.67 cm³.

Surface Areas and Volumes - RD Sharma Important Questions

Q1. A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into 12 toys in the shape of a right circular cone mounted on a hemisphere of same radius. Find the radius of the hemisphere and the total height of the toy if the height of the cone is 3 times the radius.

Vol of cylinder = π(6)²(15) = 540π.
Vol of 1 toy = (1/3)πr²(3r) + (2/3)πr³ = πr³ + (2/3)πr³ = (5/3)πr³.
12 × (5/3)πr³ = 540π.
20r³ = 540 ⇒ r³ = 27 ⇒ r = 3 cm.
Height of toy = 3r + r = 4r = 12 cm.
Radius = 3 cm, Height = 12 cm.

Q2. Water flows at the rate of 10 m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

Radius of pipe = 0.25 cm. Length in x min = 1000x cm.
Vol of water = π(0.25)²(1000x) = 62.5πx.
Vol of cone = (1/3)π(20)²(24) = 3200π.
62.5πx = 3200π.
x = 3200 / 62.5 = 51.2 minutes.
Time = 51 minutes 12 seconds.

Q3. A hemispherical tank full of water is emptied by a pipe at the rate of 25/7 litres per second. How much time will it take to empty half the tank, if it is 3 m in diameter? (Use π = 22/7)

R = 1.5 m.
Vol = (2/3)π(1.5)³ = (2/3)(22/7)(3.375) = 99/14 m³ = 99000/14 litres.
Half Vol to empty = 99000/28 litres.
Rate = 25/7 litres/sec.
Time = (99000/28) / (25/7) = 99000/100 = 990 seconds = 16.5 minutes.

Q4. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre?

Vol of barrel = π(0.25)²(7) = (22/7)(0.0625)(7) = 1.375 cm³.
1 bottle = 1/5 litre = 200 cm³.
Number of barrels = 200 / 1.375 ≈ 145.45.
Words = 145.45 × 3300 = 480,000 words (approx).

Q5. A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of Rs 25 per litre.

Vol = (1/3)πh(R² + r² + Rr)
= (1/3)(3.14)(30)(400 + 100 + 200)
= 31.4 × 700 = 21980 cm³ = 21.98 litres.
Cost = 21.98 × 25 = Rs 549.50.

Q6-Q15. [Additional problems on melting and recasting, flow of water, and frustum surface area/volume]

Key concepts:
• Volume remains constant during recasting.
• Rate of flow problems involve volume per unit time.
• Frustum formulas are critical for bucket/glass problems.

Surface Areas and Volumes - Formulas & PYQs

Key Formulas

1. Cuboid and Cube

Cuboid TSA = 2(lb + bh + hl)
Cuboid Volume = lbh
Cube TSA = 6a²
Cube Volume = a³

2. Cylinder

CSA = 2πrh
TSA = 2πr(h + r)
Volume = πr²h

3. Cone

CSA = πrl (where l = √(h² + r²))
TSA = πr(l + r)
Volume = (1/3)πr²h

4. Sphere and Hemisphere

Sphere SA = 4πr²
Sphere Volume = (4/3)πr³
Hemisphere CSA = 2πr²
Hemisphere TSA = 3πr²
Hemisphere Volume = (2/3)πr³

5. Frustum of a Cone

Volume = (1/3)πh(R² + r² + Rr)
CSA = πl(R + r) where l = √(h² + (R-r)²)
TSA = πl(R + r) + πR² + πr²

Previous Year Questions (CBSE/JKBOSE)

Q1. A solid is in the shape of a cone surmounted on a hemisphere. The radius of each of them is being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid. (CBSE 2020)

Vol = Vol cone + Vol hemisphere.
h_cone = 6 cm. r = 3.5 cm.
= (1/3)π(3.5)²(6) + (2/3)π(3.5)³ = π(3.5)²(2 + 2.33) = 12.25π(4.33) ≈ 166.83 cm³.

Q2. Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is 12√3 cm. Find the edges of the three cubes. (CBSE 2018)

Diagonal = a√3 = 12√3 ⇒ Side a = 12.
Vol = 12³ = 1728.
(3x)³ + (4x)³ + (5x)³ = 1728.
27x³ + 64x³ + 125x³ = 1728 ⇒ 216x³ = 1728 ⇒ x³ = 8 ⇒ x = 2.
Edges: 6 cm, 8 cm, 10 cm.

📱 Practice MCQs for this topic inside our App