AP 10th Class Maths 12th Chapter Surface Areas and Volumes Exercise 12.1 Solutions
Surface Areas and Volumes Class 10 Exercise 12.1 Solutions - 10th Class Maths 12.1 Exercise Solutions
Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Given volume of cube = 64 cm3
Let a3 = 64 = 43
So, side of a cube a = 4 cm
Two cubes are joined end to end.
So, it becamecuboid.
length (l) =4 + 4= 8 cm
breadth (b) = 4 cm
height (h) = 4 cm
Surface area of the cuboid = 2(lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2 × 80 = 160 cm2
Therefore, surface area of the cuboid
= 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:
Given diameter of hemisphere (or) cylinder is d = 14 cm
Then radius r = d/2 = 14/2 = 7 cm
Height of the vessel = 13 cm
Then height of cylinder = 13 - 7 = 6 cm
Inner surface area of the vessel = Surface area of cylinder + Surface area of hemisphere
= 2πrh + 2πr2
= 2πr (h + r)
= 2 × 22/7 × 7 × (6+7) = 572 cm
= 2 × 22 × 13 = 572 cm2
Therefore, inner surface area of vessel = 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:
Given radius of the cone (or) hemisphere r = 3.5 cm
height of the toy H = 15.5 cm
height of the cone h = H - r = 15.5 - 3.5
h = 12 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
l2 = r2 + h2
l = √(3.5)2+122
= √12.25+144
= √156.25
So, l = 12.5 cm
Total surface area of toy = πrl + 2πr2
=πr (l + 2r)
= 22/7 × 3.5 (12.5 + 2 × 3.5)
= 22 × 0.5 × 19.5 = 214.5 cm2
Therefore total surface area of toy = 214.5 cm2
Question 4.
A cubical block of side 7 cm is sur-mounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Given side of the cubical block s = 7 cm
radius of the hemisphere r = 7/2 cm
Total surface area of the solid = Surface area of the cube + Curved surface area of hemisphere - Base area of the hemisphere.
= 6.s2 + 2πr2 - πr2
= 6.s2 + πr2
= 6 × 49 + 11 × 3.5
Therefore, total surface area of solid = 332.5 cm2
Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter / of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Given length of cubical wooden block = l cm.
radius of the hemisphere = l/2 cm
Hemisphere is cutout from the block.
So, surface area of remaining block = Total surface area of cube + Curved surface area of hemisphere - Area of the top of the hemispherical part.
= 6 (side)2 + 2πr2 - πr2 = 6l2 + πr2
= 6l2 + π(l/2)2
= (6/1 + π/4)l2
= (24+π/4) l2 sq.units
Therefore, surface area of remaining (24+π/4) l2 sq.units
Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (as shown in figure.) The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:
Given radius of the cylindrical part r = 5 mm
Length of the capsule =14 mm
radius of the hemisphere = 5/2 = 2.5 mm
length of the cylindrical part =14 - 5 = 9 mm
Surface area of capsule = Surface area of cylinder + 2 × Surface area of hemisphere
= 2πrh + 2 × 2πr2 = 2πr (h + 2r)
Therefore, surface area of capsule = 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 of the tent at the rate of ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Solution:
Given radius of the cylinder
radius of the cone = d/2
r = 42/ = 2 m
height of the cylinder h = 2.1 m
slanted height of the cone l = 2.8 m
Area of canvas used to make tent = Surface area of cylinder + surface area of cone
= 2πrh + πrl = nr (2h + l)
= 22/7 × 2(2 × 2.1 × 2.8)
Surface area of tent = 44 m2
Cost of canvas per m2 = ₹ 500
∴ Cost of canvas per 44 m2 = 500 × 44 = ₹ 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:
Given height of the cylinder h = 2.4 cm
diameter of cylinder (d) = 1.4 cm
Radius of the cylinder (or) cone
r = d/2 = 1.4/2 = 0.7 cm
height of the cone h = 2.4 cm
slant height of the cone
l = √r2+h2
= √(0.7)2+(2.4)2
= 2.5 cm
If conical part removed from cylinder, then total surface area of remaining part = Surface area of cylinder + Surface area of cone + Base area of cylinder
= 2πrh + πrl + 2πr
= πr (2h + l + r)
= 22/7 × 0.7 (2 × 2.4 + 2.5 + 0.7)
= 22/7 × 0.7 × 8
= 17.6 cm2
Therefore, total surface area of remaining part = 17.6 cm2
Question 9.
A wooden article was made by scoop-ing out a hemisphere from each end of a solid cylinder, as shown in Figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the to-tal surface area of the article.
Solution:
Given radius of the cylinder = radius of hemisphere r = 3.5 cm
height of cylinder h = 10 cm
If scooping out two hemispheres from each end of cylinder,
then the total surface area of the article = Surface area of cylinder + Surface area of two hemispheres
= 2πrh + 2?2πr2
= 2πr (h + 2r)
= 2 × 22/7 × 3.5 (10 + 2 × 3.5)
= 2 × 22 × 0.5 × 17
= 374 cm2
Therefore total surface area of the article = 374 cm2
Surface Areas and Volumes Class 10 Exercise 12.2 Solutions - 10th Class Maths 12.2 Exercise 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:
Given radius of hemisphere (or) cone r = 1 cm
height of the cone h = r = 1 cm
Volume of the solid = Volume of hemisphere + Volume of cone
= 2/3 πr3 + 1/3 πr2h
= 1/3 πr2(2r+h)
= 1/3 πr2(2×1 + 1)
= 1/3 π(3)
= π cubic cm
Therefore, volume of the solid = π cm3.
Question 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.)
Solution:
Given radius of cylinder (or) cone
= d/2 = 3/2 = 1.5 cm
height of cone h1 = 2 cm
height of model = 12 cm
height of cylinder h2 = 12 - 2 × 2
= 12 - 4 = 8 cm
Volume of model = Volume of cylinder + 2 x Volume of cone
Volume of the model = 66 cm
Question 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 hemi-spherical ends with length 5 cm and diameter 2.8 cm (as shown in figure).
Solution:
Given length of the Gulab Jamun = 5 cm
radius of cylinder (or) hemisphere (r) d 2.8
= d/2 = 2.8/2 = 1.4 cm
height of cylinder part = 5 - 2 × r
= 5 - 2 × 1.4 h
= 5 - 2.8 = 2.2 cm
Volume of each gulab jamun = Volume of cylinder + Volume of 2 hemispheres
Volume of each gulab jamun = 75.152/3 cm³
Volume of 45 gulab jamun = 45 × 75.152/3
= 1127.28 cm³
Syrup in the container
= 30% of the volume
= 30% of 1127.28 30
= 30/100 × 1127.28
Therefore, volume of sugar syrup = 338.184 cm³
Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimen-sions 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 (as shown in figure.)
Solution:
Given dimensions of cuboid are
15 × 10 × 3.5 cm
Volume of cuboid = 15 × 10 × 3.5 = 525 cm³.
radius of the cone r = 0.5 cm
height of the cone h = 1.4 cm
Volume of each depression (or) cone
= 1/3 πr2h
= 1/3 × 22/7 × 0.5 × 0.5 × 1.4
Volume of 4 depressions
∴ Volume of wood in the entire stand = Volume of cuboid - Volume of 4 depressions
= 525 - 1.47 = 523.53 cm³.
Question 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.
Solution:
Given radius of the cone r = 5 cm
height of the cone h = 8 cm
radius of the sphere r = 0.5 cm
Volume of sphere = 4/3 πr3
= 4/3 π(0.5)3 cm3
Volume of cone = 1/3πr2h
= 1/3π (5)2 × 8
= 1/3π × 25 × 8
= 200/3π cm³
14 of water flows out of the cone
So, volume of water flows out of the cone = 1/4 × 200/3π = 50/3πcm³
Let number of shots = n
Volume of x shots = n × 4/3π (0.5)3
Volume of x shots = Volume of water flows out
n = 100
Therefore, number of shots dropped in the vessel = 100.
Question 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 8g mass. (Use n = 3.14)
Solution:
Given radius of the large cylinder
R = d/2 = 24/2 = 12 cm
height of large cylinder H = 220 cm
radius of small cylinder r = 8 cm
height of small cylinder h = 60 cm
Volume of large cylinder V1 = πR2H
V1 = π × 122 × 220
Volume of small cylinder V2 = πR2H
V2 = π × 82 × 60
Volume of iron pole = V1 + V2 = π × 122 × 220 + π × 82 × 60 = π (144 × 220 + 64 × 60)
= π (31680 + 3840)
= 3.14 × 35520
Volume of iron pole = 111532.8 cm3
Mass of 1 cm3 iron = 8g
Mass of 111532.8 cm3 iron = 111532.8 × 8 = 892262.4 grams
∴ Total mass of iron pole = 892.262 kg.
Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of ra-dius 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:
Given radius of the cylinder R = 60 cm
Height of cylinder H = 180 cm
Radius of the cone r1 = 60 cm
Height of the cone h1 = 120 cm
Radius of hemisphere r2 = 60 cm
Volume of cylinder V = πR²H
= π ? 60² × 180
Volume of cone V1 = 1/3 πr12h1
= 1/3 π × (60)2 × 120
Volume of hemisphere V2 = 2/3 πr23
= 2/3 π × (60)3
Volume of water left out = Volume of cylinder - Volume of cone - Volume of hemisphere
Volume of water left = 1.1314 m
Question 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 n = 3.14.
Solution:
Given radius of cylinder r1 = d/2 = 2/2 =1 cm
height of cylinder h1 = 8 cm
radius of the sphere r2 = d/2 = 8.5/2 cm
Volume of cylindrical part
= πr12h1 = π × 1² × 8 = 8π
Volume of spherical part
Total volume of water in the vessel = Volume of cylinder + Volume of sphere
Total volume of water in the vessel = = 346.5 cm³
But, child found that as 345 cm³ So, child is not correct.
Hence, the volume found by child is not correct.