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NCERT Class 10 Mathematics Chapter 12 Surface Area Volumes
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Surface Area Volumes
Chapter – 12
| Exercise 12.1 |
1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area
of the resulting cuboids.
Ans:
Let side of cube = x cm
Volume of cube = 64cm3
{Volume of cube = (side)}
X = 4 cm
∴ side of cube = 4cm
When cube are joined end to end cuboid is formed whose
Length = 2 cm = (4) = 8 m
With X = cm = 4 cm height = x cm = 4cm
∴ surface area of cuboids = 2(lb + bh lh)
= 2 (4x4x4x8x4x8x)
= (16+ 32+32)
2(16+64)
= 2x 80 = 160 cm2
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. [Use 𝜋 = 22/7]
Ans:
It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm).
Height of hemispherical part = Radius = 7 cm
Height of cylindrical part (h) = 13 −7 = 6 cm
Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part = 2 Пrh + 2Пr2
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. [Use 𝜋 = 22 7]
Ans:
It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5 cm).
Height of hemispherical part = Radius (r) = 3.5 = 7/2 cm
Height of conical part (h) = 15.5 −3.5 = 12 cm
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. [Use 𝜋 = 22 7]
Ans: From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm.
Radius (r) of hemispherical part = 7/2 = 3.5cm
Total surface area of solid = Surface area of cubical part + CSA of
hemispherical part − Area of base of hemispherical part
= 6 (edge)2 + 2πr2 – πr2 = (edge)2 + πr2
Total surface area of solid = 6(7)2 + 22/7 x 7/2 x 7/2
= 294 + 38.5 = 332.5 cm2
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.
Ans: Diameter of hemisphere = Edge of cube = l
Radius of hemisphere = l/2
Total surface area of solid = Surface area of cubical part + CSA of
hemispherical part − Area of base of hemispherical part
= 6 (edge)2 + 2πr2 – πr2 = (edge)2 + πr2
6. A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. [Use 𝜋 = 22/ 7]
Ans:
It can be observed that
Radius (r) of cylindrical part = Radius (r) of hemispherical part
= Diameter of the capsule /2 = 5/2
Length of cylindrical part (h) = length of the entire capsule – 2 x r = 14 – 5 = 9 cm
Surface area of capsule = 2 × CSA of hemispherical part + CSA of cylindrical part
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 of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)[Use 𝜋 = 22 7]
Ans:
Given that,
Height (h) of the cylindrical part = 2.1 m
Diameter of the cylindrical part = 4 m
Radius of the cylindrical part = 2 m
Slant height (l) of conical part = 2.8 m
Area of canvas used = CSA of conical part + CSA of cylindrical part
= πrl + 2πrh
= π x 2 x 2.8 + 2π x 2 x 2.1
= 44 m2
Cost of 1 m2
canvas = Rs 500
Cost of 44 m2
canvas = 44 × 500 = 22000
Therefore, it will cost Rs 22000 for making such a tent.
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. [Use 𝜋 =22/7]
Ans:
Given that,
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm
Diameter of the cylindrical part = 1.4 cm
Therefore, radius (r) of the cylindrical part = 0.7 cm
Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base
The total surface area of the remaining solid to the nearest cm2 is 18 cm2
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in given figure. 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. [Use 𝜋 = 22 7]
Ans:
Given that,
Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm
Height of cylindrical part (h) = 10 cm
Surface area of article = CSA of cylindrical part + 2 × CSA of
hemispherical part
2πrh + 2 +2πr2
= 2π x 3.5 x 10 + 2 x 2π x 3.5 x 3.5
= 70π + 49π
= 119π
= 17 x 22 = 374cm2
| Exercise 12.2 |
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.
Ans:
Given that, Height (h) of conical part = Radius(r) of conical part = 1 cm Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm Volume of solid = Volume of conical part + Volume of hemispherical part
= ⅓ πr2h + ⅔ πr3
= ⅓ π. 12.1 + ⅔ 𝜋. 13 = 𝜋 𝑐𝑚3
2. 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.) [Use 𝜋 = 22 7]
Ans: Radius of cone = Radius of cylinder (R) = 3/2 cm
(R) = 3/2 cm
Height of each cone(h) = 2cm ….Height of cylinder = 12 – 2 – 2 = 8c
Volume of air in cylinder = volume of cylinder + 2(volume of cone)
3. 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 the given figure). [Use 𝜋 = 22/7]
Ans: It can be observed that
Radius (r) of cylindrical part = Radius (r) of hemispherical part =
It can be observed that
Radius (r) of cylindrical part = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm
Length of each hemispherical part = Radius of hemispherical part = 1.4 cm
Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part
= 5 − 2 × 1.4 = 2.2 cm
Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part
Volume of 45 gulab jamuns = 45 × 25.05 = 1,127.25 cm3
Volume of sugar syrup = 30% of volume.
30/100 x 1,127.25
= 338.17cm3
= 338 cm3
4. A pen stand made of wood is in the shape of a cuboid with four conical
depressions to hold pens. The dimensions of the cuboids 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 the following
figure). [Use 𝜋 = 22/7]
Ans: Length of cuboid (L) = 15 cm
Width of cuboid (b) = 10cm
Height of cuboid (H) = 3.5 cm
Height of conical cavity (r) = 0.5 cm
Radius of conical cavity (h) = 1.4 cm
Hence Volume of wood in pen stands = volume cuboid – 4 (volume of cone)
Hence, volume of wood in pen stand = 523.53
5. 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.
Ans: Radius of cone (R) = 5cm
Height of cone (H) = 8cm
Radius of each spherical leads shot (r) = 0.5 cm
According to question
Hence, the number of lead shorts dropped in the vessel is 100
6. 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
8 g mass. [Use π = 3.14]
Ans:
From the figure, it can be observed that
Height (h1) of larger cylinder = 220 cm
Radius (r1) of larger cylinder = 24/2 = 12 cm
Height (h2) of smaller cylinder = 60 cm
Radius (r2) of smaller cylinder = 8 cm
7. 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.
Ans:
Radius of cone = radius of hemisphere
= radius cylinder
R = 60cm
Height of cylinder vessel = πR2H
= 22/7 x 60 x 60 x 180 = 2036571.4 cm
Volume of solid inserted in cylinder = volume of hemisphere
= ⅔ πR3 + ⅓ π R2 h
Volume of water flows out = 9055142.86 cm3
∴ Volume of water left in cylinder = volume of cylinder – volume of solid inserted in the vessel = (203657.14 – 905142.86) cm2
= 1131428.5 cm2 = 1131428.5/ 100 x 100 x 100 m3 = 1.131 m3
Hence, volume of water left in cylinder = 1.131m3
8. 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.
Ans:
Height (h) of cylindrical part = 8 cm
Radius (r2) of cylindrical part = 1 cm
Radius (r1) spherical part = 4.25 cm
Volume of vessel = Volume of sphere + Volume of cylinder
Hence, she is wrong.