# KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3

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### Karnataka State Syllabus Class 10 Maths SolutionsChapter 15 Surface Areas and Volumes Ex 15.3

(Take π =227 unless stated otherwise)

Question 1.
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:
Let the height of cylinder be ‘h’ cm.
Radius of cylinder, r = 6 cm.
i) Volume of metallic sphere, V = 43πr3

ii) Volume of Cylinder, V = πr2h
227 × 36 × h = 310.46

∴ Height of cylinder is 2.74 cm.

Question 2.
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 the resulting sphere be ‘r’ cm.
(i) Volume of 3 spheres having radii 8 cm and 10 cm.

(i) Volume of the resulting Sphere, V

∴ Volume of the resulting sphere = Volume of three spheres

∴ Radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with a 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:
Radius of well, ‘r’ = 3.5 m.
Height of well, h = 20 m.

(i) Volume of soil dug out of well, V = πr2h (∵ cylinder sphere)
= 227×(3.5)2×20
= =227×3510×3510×20
= 770m3
(ii) Length of platform, l = 22 m.
Height, h =?
Volume of Platform, V= l × b × h (∵ Cuboid)
= 22 × 14 × h
∴ Volume of Platform = Volume of soil dug out of well.

∴ Height of platform is 2.5 m

Question 4.
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:
Let the height of the embankment be ‘h’ m.

Dimeter of well, d = 3 m.
Radius of well, r =1.5 m.
Height of well, h = 14 m.
Volume of an embankment = Volume of soil dug from well.

∴ Height of embankment = 1.125 m.

Question 5.
A container shaped like a right cylinder diameter 12height 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 cones which can be filled with ice cream.

Solution:
Diameter of right circular cylinder is 12 cm ➝ r = 6 cm.
Height: 15 cm. ➝ h = 15 cm.
Diameter of Cone, 6 cm. ➝ r = 3 cm.
Height 12 cm. ➝ h = 12 cm.
(i) Volume of the ice cream which has cone-shaped and hemisphere.

(ii) Volume of cylindrical vessel, V = πr2h

Let number of cones be ‘n’,
n × 54π = π × 36 × 15
54n = 36 × 15
n=36×1554=54054
∴ n = 10
Ice cream can be filled in 10 cones.

Question 6.
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:
Diameter of the silver coins is 1.75 cm.
Thickness = 2 mm.
Measurement of cuboid: 5.5 × 10 × 3.5 cm.
∴ Let the required silver coins be ‘n’,
Volume of ‘n’ coins = Volume of cuboid
πr2h = l × b × h

∴ n = 400
∴ 400 silver coins are required.

Question 7.
A cylindrical bucket, 32 cm high and with a 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:
Height of cylindrical bucket, h = 32 cm.
Height of conical heap of sand = 24 cm.
Slant height, l =?
Volume of the conical heap of sand = Volume of Cylinder.

Radius of Cone, r = 36 cm.
Slant height of cone, l = h2+r2−−−−−−√

∴ Radius of cone, r = 36 cm.
Slant height of cone, h = 1213−−√ cm

Question 8.
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:
Breadth of a canal, b = 6 m.
Height, h = 1.5 m
Speed of water =10 km/hr.
Quantity of water = 8 cm.
Quantity of water flowing in 30 minutes?
Volume of water flowing in 60 minutes,
= (10 × 1000) × 6 × 1.5 cm3.
( ∵ Speed = 10 km/hr.; 10 × 1000 m/hr.)
Volume of water flowing in 30 minutes,
= 10000×9×3060m3
∴ Let the area of land be ‘x’ sq.m.

x= 562500 sq.m.

Question 9.
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:
Internal diameter of a pipe = 20 cm.
Radius of pipe =10 cm. = 110m
Diameter of cylindrical tank = 10 m; r=5 m.
Depth of cylindrical tank = 2 m.
Speed of water is 3 km/hr.
Time required to fill the water tank =?
Speed of water = 3 km/hr.
= 300060 m/minute
= 50 m/min.
Let the time required to fill the tank be n’ minutes.
∴ Flowing water in ‘n’ minutes = Volume of the water tank

∴ n = 100 minutes
∴ Time required to fill the water tank = 100 minutes

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