Telangana SCERT Solution Class X (10) Maths Chapter 10 Mensuration Exercise 10.4
1) A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6cm. find the height of the cylinder
Ans:
In sphere’s radius r = 4.2cm
Volume V = 4/3 π r3
= 4/3 x π x 4.2 x 4.2 x 4.2 cm3
In Cylinder
Radius r = 6cm
Let height be h
Volume v = π r2 h
= π x 6 x 6 x h cm3
According to the Problem,
Volume of the sphere = Volume of the sphere
4/3 x π x 4.2 x 4.2 x 4.2 = π x 6 x 6 x h
4/3 x 4.2 x 4.2 x 4.2 = 6 x 6 x h
4/3 x 0.7 x 0.7 x 4.2 = h
4 x 0.7 x 0.7 x 1.4 = h
2.744 = h
Therefore, Height of the cylinder = 2.744 cm
2) Three metallic spheres of radii 6 cm, 8cm and 10cm respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.
Ans:
Given,
The radii of the spheres
r1 = 6cm
r2 = 8cm
r3 = 10 cm
Let the radius of the resulting sphere = R
Here, Volume of the three spheres = Volume of the resulting sphere
4/3 π r31 + 4/3 π r32 + 4/3 π r32 = 4/3 π r3
= 4/3 π ( r31 + r32 + r33) = 4/3 π r3
= r31 + r32 + r33 = R3
= 63 + 83 + 103 = R3
1738 = R3
R = ∛1738
R = 12
Therefore, Radius of the resulting sphere = 12
3) A 20cm deep well of diameter 7m is dug and the earth got by digging is every spread out to form a rectangular platform of base 22m x 14m and the height of the Platform
Ans:
In Cylindrical Well’s diameter d = 7cm
radius r = d/2 = 7/2cm
Depth (Height) h = 20m
Volume = π r2 h
= 22/7 x 7/2 x 7/2 x 20
= 22 x 7/2 x 7/2 x 20
= 22 x 5 x 7 cm3
The earth from dragging is evenly spread to form of plat of length 22m and breadth 14m
Let the height of the platform = h
Volume of the cylindrical well = Volume of the platform (Cuboid)
……> 22 x 7 x 5 = l b h
…….> 22 x 7 x 5 = 22 x 14 x h
…..> h = 22 x 7 x 5/ 22 x 14
……> h = 5/2
h = 2.5m
Therefore, Height of the platform is 2.5m
4) A well of the diameter 14m is dug 15m deep. The earth taken out of it has been spread evenly to form circular embankment all around the well of width 7m. Find the height of the embankment.
Ans:
In Cylindrical Well’s
Diameter d = 14m
Radius r = d/2
= 14/2
r = 7m
Depth (height) h = 15m
Volume = π r2 h
= π x 7 x 7 x 15 m3
The earth is dug and form the embankment of width 7m circular ring.
Inner radius of the Circular ring = 7m
Outer radius R = r + w
= 7 + 7
= 14m
Volume V = π (r2 – r2) h
= π ( 142 – 72) h
= π (14 + 7) (14 – 7) h
= π x 21 x 7 x h m3
Volume of the cylinder = volume of the embankment
π x 7 x 7 x 15 = π x 21 x 7 x h
7 x 15 = 21 x h
7 x 15 /21 = h
15 / 3 = h
h = 5
Therefore, Height of the embankment = 5m
5) A Container shaped a right circular cylinder having diameter 12cm and height 15cm is full of ice Cream. The ice cream is to be filled into cones of height 12cm and diameter 6cm making a hemispherical shape on the top. Find the number of the such cone which can be filled with ice cream.
Ans:
In Cylindrical container
Diameter d = 12cm
Radius r = 12/2 = 6cm
height h = 15cm
Volume v = π r2 h
= π x 6 x 6 x 15
= 540 π cm3
In Con’s
Diameter d = 6cm
Radius r = 6/2 = 3 cm
Height h = 12 cm
Volume V = 1/3 π r2 h
= 1/3 x π x 3 x 3 x 12
= 36 π
In hemisphere
Radius r = 3cm
Volume V = 2/3 x π x 3 x3 x 3
= 16 π cm3
Volume of Ice Cream = Volume of Hemisphere + Volume of the cone
= 36 π + 18 π
= 54 π
X = Volume of cylindrical Container Volume of Ice Cream
= 540 π / 54 π
X = 10
Therefore, Number of Ice Cream Cones = 10
6) How many silver Coins, 1.75cm in Diameter and thickness 2mm, need to be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm
Ans:
In Silver Coins, (cylinder)
Diameter d = 1.75cm
Radius r = d/2 = 1.75/2
= 0.875 cm
Thickness (Height) h = 2mm
= 2/10
= 0.2 cm
Volume = π r2 h
= 22 / 7 x 0.875 x 0.875 x 0.2
In Cuboid dimension
5.5 cm, 10cm and 3.5cm
Volume = 5.5 x 10 x 3.5 cm3
= 55/10 x 10 x 35/10
Number of Silver coins be X
X= Volume of silver coin = Volume of cuboid
X = Volume of Cuboid/ Volume of Silver coin
= 55 x 35/100
22/7 x 875/1000 x 875/1000 x 2/10
= 55 x 35/10 x 7/22 x 1000/875 x 1000 x 875 x 10/2
= 55 x 35 x 7/22 x 1000/875 x 1000/875 x ½
= 5 x 35 x 7/2 x 1000/875 x 1000/875 x ½
= 5 x 35 x 1/2 x 1000/125 x 1000/875 x ½
= 5 x 35 x ½ x 8/1 x 1000/875 x ½
= 5 x 35 x 4 x 1000/875 x ½
= 35 x 1000/175
= 1000/5 x 2
= 200 x 2
= 400
7) A vessel is in the form of an inverse cone. Its height is 8cm and the radius of its top is 5cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into vessel, ¼ of the water flows out. Find the number of lead shorts dropped into the vessel.
Ans:
In cone’s
Radius r = 5cm
height h = 8cm
Volume V = 1/3 π r2 h
= 1/3 x π x 5 x 5 x 8 cm3
In Spherical Lead shots
Radius r = 0.5 cm
= ½ cm
Volume V = 4/3 π r3
= 4/3 x π x ½ x ½ x ½
= π/6 cm3
Let the number of lead shots dropped into the Vessel be X
X x Volume of lead shots = ¼ Volume of the cone
2 x π/6 = ¼ x 1/3 x π x 5 x 5 x 8
x = 5 x 5 x2 x2
x = 100
Therefore, Number of lead shot’s dropped = 100
8) A Solid metallic sphere of diameter 28cm is melted and recast into a number of smaller cones, each of diameter 42/3 cm and height 3 cm. Find the Number of cones so formed
Ans:
In Metallic sphere’s
Diameter d = 28 cm
Radius r = d/2 = 28/2 = 14 cm
Volume V = 4/3 π r3
= 4/3 x π x 14 x 14 x 14 cm3
In cone’s
Diameter d = 4 – 2/3cm
= 14/3 cm
Radius r = d/2 = 14/ 3/2
= 14/3 x 2
= 14/6
= 7/3
Height v = 1/3 π r2 h
= 1/3 x π x 7/3 x 7/3 x 3
= π x 7/3 x 7/3 cm3
Let, the number of cone’s formed be X
X x Volume of the cone = Volume of the sphere
..> X x π x 7/3 x 7/3 = 4/3 x π x 14 x 14 x 14
..> X x 7/3 x 7/3 = 4/3 x 14 x 14 x 14
..> X = 4/3 x 14/7 x 14/7 x 14 x 3 x 3
= 4 x 2 x 2 x 14 x 3
X = 672
Therefore, Number of Cones has formed = 672
Here is your solution of Telangana SCERT Class 10 Math Chapter 10 Mensuration Exercise 10.4
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