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13. Surface Areas and Volumes

Exercise 13.1

Chapter 13. Surface Areas and Volumes Exercise 13.1 solved solution

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

Solution: 

The volume of cube = 64 cm³

Joining two surfaces;

l = 4 + 4 = 8 cm 

b = 4 cm 

h = 4 cm 

The Surface Area of such formed cuboid = 2(lb + bh + lh)

= 2(8×4 + 4×4 + 8×4)

​= 2(32 + 16 + 32)

= 2×80

= 160 cm²

Therefore, The area obtained from this cuboid is 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.

Solution: 

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.

Solution:

The radius of hemisphererr = 3.5 cm

The radius of conical partr = 3.5 cm

The height of conical partrh = 15.5 – 3.5 = 12 cm

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.

Solution:

The edge of cubical block = 7 cm

The maximum diameter of the blockd = 7 cm

The total surface area of block = The Area of cuboidal block + The Area of hemisphere – Area of a circle covered with hemisphere  

= 294 + 38.5 = 332.5 cm²

Therefore, total surface area of block = 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.

Solution:

Let the diameter of hemisphered = l unit

And the edge of cubea = l unit

(Hence, diameter of hemisphere and the edge of cube are same)

Total surface Area of remaining solid = Area of cuboidal block + Area of hemisphere – Area of a circle covered with hemisphere  

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

The height of cylinder = length of capsule – 2(2.5 mm)

= 14 – 5 [Hence diameter = 2.5 mm]

= 9 mm

Total Surface area of capsule = 2(curve surface area of hemisphere) + curve surface area of cylinder

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. (Note that the base of the tent will not be covered with canvas.)

Solution:

 

The diameter of the cylindrical part of tent = 4 m

So, radius = 2 m

The height of cylindrical part = 2.1 m

The slant height of cone = 2.8 m

And radius = 2 m

The area of required canvas = the curved surface area of cylindrical part + curved surface area of conical part

= 2πrh + πrl

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.

Solution:

The height of cylinder h = 2.4 cmcm

The diameter of cylinder = 1.4 cm

So, radius of cylinder = 0.7 cm

The height of hollowed out = 2.4 cm

Radius = 0.7 cm  

Total surface area of remaining solid = Curved surface area of cylinder + curved surface area of cone + Area of the bottom of solid

 = 2πrh + πrl + πr²

 = πr(2h + l + r)

The total surface area of the remaining solid to the nearest cm² = 18 cm² Answer

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. 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: 

The height of cylinder = 10 cm

The radius (r) of base = 3.5 cm

The radius (r) of hemisphere = 3.5 cm

Total surface area = Curved surface area of cylinder + Curved surface area of top of hemisphere + Curved surface area of bottom of hemisphere

= 2πrh + 2πr² + 2πr²

= 2πr(h h + rr + rr )

= 2πr(hh + 2rr)

Total surface area is 374 cm²

13. Surface Areas and Volumes

Exercise 13.2

Exercise 13.2

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Q1. 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:

The radius of cone r = 1 cm

The height of cone = 1 cm

The radius of hemisphere r = 1 cm

The volume of solid is 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 by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution: 

The height of the cone h = 2 cm

The radius of cylinder r = 1.5 cm

The height of cylinder H = 12 – 2 – 2 = 8 cm

Volume of air contained in the model = 2(volume of cone) + volume of cylinder

Thus, the volume of air contained in the model is 66 cm³

Q3. A 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 (see Fig. 13.15).

Solution: 

Diameter of hemispherical end = 2.8 cm

Then the radius of the hemispherical end r = 1.4 cm

Length of whole Gulab Jamun l = 5 cm

So the length of the cylindrical part h = 5 – (1.4 + 1.4)

                            = 5 – 2.8 cm

                            = 2.2 cm

Volume of all 45 gulab jamuns = 45(volume of hemisphere + volume of cylinder + volume of hemisphere)

Sugar syrup = 30% of 1127.280 cm3
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= 338.1840 cm3
Hence, the quantity of syrup in 45 Gulab Jamuns is 338 cm3.

Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 13.16).

Solution: 

The length of the arc l = 15 cm

The breadth of cuboid b = 10 cm

The height of the cuboid h = 3.5 cm

The radius of conical part (r) = 0.5 cm

Height (h) = 1.4 cm

Volume of whole wood of Pen holder = Volume of cuboid – 4 (Volume of conical pit)

= 525 - 1.47 cm³

= 523.53 cm³

The volume of the whole lotus wood is 523.53 cm3.

Q5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, 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 into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution: 

Height of cone (h) = 8 cm

Radius of cone (R) = 5 cm

Radius of led balls (r) = 0.5 cm

Let the number of balls put in the vessel = n

Hence, the number of led balls is 100.

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)

Solution: 

Height (H) of the obtuse cylinder = 220 cm

Diameter (d) = 24 cm

Thus, radius (R) = 12 cm

Height of slender cylinder (h) = 60 cm

radius (r) = 8 cm

Now the volume of the iron pillar = pR²H + pr²h

                       = p(R²H + r²h)

                       = 3.14 (12×12×220 + 8×8×60)

                       = 3.14 (31680 + 3840)

                       = 3.14 (35520)

                       = 111532.8 cm³

The mass of Iron = 111532.8 cm³ × 8

               = 892262.4 g

Hence, the mass of the iron pillar is 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.

Solution: 

Height of cone of solid (h) = 120 cm
Radius of cone of solid (r) = 60 cm
Radius of hemisphere of solid (r) = 60 cm
Height of large cylinder (H) = 180 cm
Radius of the large cylinder (r) = 60 cm
Volume of remaining water = volume of large cylinder – volume of solid

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.

Solution: 

Diameter of circular part = 8.5 cm

Radius of circular part (R) = 8.5/2 cm

Height of cylindrical neck (h) = 8 cm

Diameter of neck (d) = 2 cm

Therefore, radius (r) = 1 cm

Volume of water that can be filled = Volume of sphere + Volume of cylinder

Hence the measurement taken by the child is not correct.

13. Surface Areas and Volumes

Exercise 13.3

Exercise 13.1 Chapter 13. Surface Areas and Volumes

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NCERT Solution for class 10 maths in English Medium:

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.

Solution:

Radius of sphere (r) = 4.2 cm 

Radius of cylinder (R) = 6 cm

Let the height of the cylinder = h cm

Since the sphere is cast into a cylinder, therefore

Volume of cylinder = volume of sphere 

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.

Solution:

Let the radius of large solid sphere = R cm

GIven : r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm

The radius of new sphere is 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.

Solution:

Diameter of well = 7 m

Thus, radius of the well (r) = 3.5 cm

Depth of well (h) = 20 m

Length (l) = 22 m and breadth (b) = 14 m of the platform.

Let the height of platform = h m

Volume of platform = Volume of earth taken out from the well

l × b × h =  πr²h

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The height of platform = 2.5 m

Q4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:

​

Diameter of the well = 3 m

Radius of well (r) = 3/2 m = 1.5 m

Depth of well (H) = 14 m

Width of circular ring around the well = 4 m

Thus, outer radius of the ring (R) = 4 m + 1.5 = 5.5 m

Let the height of the circular embankment = h m

Volume of circular platform = Volume of earth taken out from the well

⇒ πR²h - πr²h = πr²H

⇒ πh(R² - r²) = πr²H

⇒ h (R² - r²) = r²H

⇒ h [(5.5)² - (1.5)²] = 1.5 × 1.5 × 14

⇒ h (5.5 +1.5) (5.5 - 1.5) = 1.5 × 1.5 × 14  [ a² - b² = (a + b) (a - b) ]

⇒ h = 1.125 m

The height of ring embankment = 1.125 m

Q5. 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 which can be filled with ice cream.

Solution:

The diameter of cylindrical vessel = 12 cm

Radius of vessel R = 6 cm

Height of vessel H = 15 cm 

The number of such cones which can be filled with ice-cream is 10.
Q6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution:

Hence, Number of coins is 400

Q7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution: 

Radius of cylindrical bucket R = 18 cm

and height H = 32 cm

Height of conical pile = 24 cm

Volume of cylindrical bucket = πR²H

Q8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution:

Length of canal in 1 hour l = 10km = 10000 m

The breadth of canal, b = 6 m

Depth of canal, h = 1.5 m

The volume of water in canal in 1 hour =  l × b × h

       = 10000 × 6 × 1.5 m³

       = 90000 m³

   Area = 562500 m²

Therefore, 562500 m² areas are required for irrigation.

Q9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution:

Diameter of tank = 10 m

Radius of tank = 5 m

depth of tank h = 2 m

Diameter of pipe = 20 cm

Radius of pipe = 10 cm = 0.1 m

Length of pipe in 1 hour = 3 km = 3000 m

Now volume of water in pipe in 1 hour = πr²h

                = π × 0.1 × 0.1 × 3000

                = π × 30 m³

Volume of water that can be filled in the tank = πr²h

               = π × 5 × 5 × 2

Taken time to fill the tank 

Hence 100 minutes to take to fill the tank.

13. Surface Areas and Volumes

Exercise 13.4

13. Surface Areas and Volumes Exercise 13.4

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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.

Solution: 

Diameter of upper end D = 4 cm

Radius of upper end R = 2 cm

Diameter of lower end d = 2 cm

Radius of lower end r = 1 cm

Q2. 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.

Solution: 

Slant height (l ) of the frustum of the cone = 4 cm

Perimeter of upper end = 18 cm

Thus, the curved surface area of the frustum is 48 cm².              

Q3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). 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 for making it.

Solution: 

Slant height (l) of the cap = 15 cm

Radius of open end (R) = 10 cm

Radius of upper end (r) = 4 cm

Area of material used for making it = CSA of Frustum + Area of upper part

Q4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π = 3.14)

Solution: 

Height of the container (h) = 16 cm

Radius (R) of the top of the container = 20 cm

Radius of the lower end of the container (r) = 8 cm

Capacity in litre= 10.45 litre (Approx)

The cost of milk = 20 × 10.45 = ₹209.00

Area of used sheet = πl (R­ + r) + πr²

= 3.14 × 20 (20­ + 8) + 3.14 × 8 × 8

= 3.14 × 20 (28) + 3.14 × 64

= 3.14 (560 + 64)

= 3.14 (624)

= 3.14 (624)

= 1959.36 cm² 

The cost of sheet at the rate of ₹ 8 per 100 cm²

Hence, the cost of the metal sheet is ₹156.75.

Q5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height 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.

Solution:   

AD = 20 cm

Then, AG = 10 cm (cut into two parts at the middle)

∠BAC = 60^o 

AD bisects ∠BAC

Thus, ∠CAD = 30^o 

In right angled triangle ΔAGF

Similarly, in right angled triangle DADC,

Lets the length of wire be H  

Volume of the wire = Volume of the frustum obtained

H = 7964.44 m

Hence, the length of the wire is 7964.44 m.

13. Surface Areas and Volumes

Exercise 13.5