Telangana SCERT Solution Class X (10) Maths Chapter 10 Mensuration Exercise 10.3
Exercise 10.3
1) An Iron Pillar consists of a cylindrical Portion of 2.8m height and 20cm in diameter and a cone of 42cm height surmounting it. Find the weight of the Pillar if 1cm3 of iron weights 7.5 kg
Ans:
Cone:
Diameter d = 20cm
Radius = r = d/2= 20/2 = 10cm
Height h = 42 cm
Volume V = 1/3 π r2h
= 1/3 x 22/7 x 10 x 10 x 42
= 1/3 x 22 x 10 x 10 x 6
= 22 x 10 x 10 x 2
= 22 x 100 x 2
= 22 x 200
= 4400cm3
Cylinder:
Radius r = 10cm
Height h = 2.8 cm
= 2.8 x 100
h = 280 cm
Volume V = π r2h
= 22/7 x 10 x 10 x 280
= 22 x 10 x 10 x 40
= 22 x 100 x 40
= 22 x 4000
= 88000 cm3
Volume of the Iron Pillar = Volume of Cone + Volume of Cylinder
= 4400cm3 + 88000 cm3
= 92400 cm3
Weight of the iron Pillar
1cm3 …..> 7.5 gr
92400 cm3……..> 7
92400 x 7.5 = 693000 gr
= 693000/1000
= 693 kg
2) A toy is made in the form of hemisphere Surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 3/2 of the hemisphere. Calculate the height of the cone and the surface of the area of the toy correct to 2 places of decimal
Ans:
hemisphere
Radius r = 7cm
Volume = 2/3 π r3
= 2/3 x π x 7 x 7 x 7cm3
Cone:
Radius r = 7cm
Height = h
Volume V = 1/3 π r2h
= 1/3 π x 7 x 7 x h
Accordingly to the Problem
Volume of Cone = 3/2 Volume of Hemisphere
1/3 π x 7 x7 x h = 3/2 x 2/3 x π x 7 x 7 x 7
h = 7 x 3
h = 21 cm
Surface Area of the toy = CSA of Hemisphere + CSA of Cone
= 2 π r2 + π r l
= 2 x 22/7 x 7 x 7 + 22/7 x 7 x √r2 + h2
= 2 x 22/7 x 7 x 7 + 22/7 x 7 x √(7)2 + (21)2
= 2 x 22 x 7 + 22 x √ 49 + 441
= 308 + 22 x √ 490
= 308 + 22 x √ 49 x 10
= 308 + 22 x 7 √10
= 308 + 154 + (3.16)
= 306 + 48664
= 794.64cm2
3) Find the Volume of the largest right circular cone that can be cut out of a cube whose edge is 7cm
Ans:
Diameter d =7cm
Radius r = 7/2 cm
Height h = 7 cm
Volume = 1/3 π r2 h
= 1/3 x 22/7 x 7/2 x 7/2 x 7
= 11 X 49 / 6
= 539 / 6
= 89.83 cm3
Therefore, Volume of the cone is 89.83
4) A Cylinder tub of height is 9.8cm is fall of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the mug. The radius of the hemisphere is 3.5cm and height of Conical part 5cm. Find the Volume of Water left in the Tub
Ans:
Cylindrical Tubs
Radius r = 5cm
height (h) 9.8 cm
Volume v = π r2 h
= 22/7 x 5 x 5 x 9.8
= 22 x 5 x 5 x 98/10
= 22 x 5 x 5 x 14/10
= 22 x 5 x 14/2
= 770 cm3
Volume of Hemisphere + Volume of Cone
= 2/3 π r3 + 1/3 π r2 h
= 1/3 π r2 (2r + h)
= 1/3 x 22/7 x 3.5 x 3.5 x (7+5)
= 1/3 x 22/7 x 3.5 x 3.5 x 12
= 22/7 x 3.5 x 3.5 x 4
= 22 x 0.5 x 3.5 x 4
= 154 cm3
Volume of the Water left in the tub = Volume of the Cylinder – Volume of Solid
= 770 cm3 – 154 cm3
= 616 cm3
5) In the adjacent figure, the height of a solid Cylinder is 10 cm and Diameter is 7cm . Two Equal conical holes of radius 3cm and height 4cm are cut off as shown the figure. Find the volume of the remaining solid
Ans:
Cylindrical Solid’s
diameter d = 7cm
radius r = d/2 = 7/2 cm
height h = 10cm
volume v = π r2 h
= 22/7 x 7/2 x 7/2 x 10
= 11 x 7/2 x 10
= 11 x 7 x 5
= 385cm3
In Cone’s
Radius r = 3cm
Height h = 4cm
Volume V = 1/3 π r2 h
= 1/3 x 22/7 x 3 x 3 x 4
= 22/7 x 3 x 4
= 264/7
= 37.71 cm3
If two equal conical holes are cut off then volume of the remaining solid
= Volume of the Cylinder – 2 x volume of the cable
= 385 – 2(37.71)
= 385 – 75.42
= 309.58 cm3
6) Spherical Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7cm, which contains some water, Find the number of marbles that Should be dropped into the beaker, so that water level rises by 5.6 cm
Ans:
In Spherical marbles,
Diameter D= 1.4 cm
Radius r = d/2
= 1.4/2
= 0.7 cm
Volume V = 4/3 π r3
= 4/3 x π x 0.7 x 0.7 x 0.7
If the marbles have dropped into the beaker, then water level raised = 5.6cm
Number of marbles have dropped = X
Volume of Water level raised = π r2 h
= π x 3.5 x 3.5 x 5.6 cm3
therefore, X volume of marbles = Volume of the water raised
X = Volume of the water / Volume of marbles
= π x 3.5 x 3.5 x 5.6/ 4 /3 x π x 0.7 x 0.7 x 0.7
= π x 5 x 5 x 5.6 / 4/3 x π x 0.7
= 3 x 5 x 5 x 5.6 / 4 x 0.7
= 3 x 5 x 5 x 1.4 / 0.7
= 3 x 5 x 5 x 2
= 150
Therefore, Number of marbles dropped into beaker = 150
7) A pen stand is Stand of wood in the shape of Cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10cm by 10cm by 3.5cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the Volume of wood of wood in the entire stand.
Ans:
In Cuboid
Length = 15 cm
breadth = 10 cm
Height = 3.5 cm
Volume V= lbh
= 15 x 10 x 3.5
= 525 cm3
In Cone’s
Radius r = 0.5 cm
Height h = 1.4 cm
Volume = 1/3 π r2 h
= 1/3 x 22/7 x 0.5 x 0.5 x 1.4
= 1/3 x 22/7 x 5/10 x 5/10 x 1.4/10
= 1/3 x 22 x 5/10 x 5/10 x 2/10
= 1/3 x 22 x ½ x ½ x 1/5
= 1/3 x 11 x ½ x 1/5
= 11/30 cm3
Volume of the Wooden pen Stand = Volume of Cuboid – 3 x Volume of Cone
= 525 – 3 x 11/30
= 525 – 11/10
= 525 – 1.1
= 523.9 cm3
Therefore, The Volume of Wood in the entire Stand = 523.9 cm3
Here is your solution of Telangana SCERT Class 10 Math Chapter 10 Mensuration Exercise 10.3
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