10th Maths 13.3
NCERT Class 10th solution of Exercise 13.1
NCERT Class 10th solution of Exercise 13.2
Exercise 13.3
Take π=227 unless stated otherwise.
Q1. 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.
Sol. :
Given :
Radius of Sphere R=4.2 cm
Radius of Cylinder r=6 cm
To find :
Height of the Cylinder
Solve :
According to question
Volume of the Cylinder = Volume of the Sphere
πr2h=43πR3
h=4πR33πr2
h=4R33r2
h=4×(4.2)33(6)2
h=4×4.2×4.2×4.23×6×6
h=4×4.2×0.7×0.73
h=8.2323
h=2.744 cm
Answer :
The height of the cylinder 2.744 cm.
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Q2. 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.
Sol. :
Given :
Radius of metal sphere r1=6,r2=8,r3=10cm
To find :
The radius of the resulting sphere
Solve :
Let Radius of resulting sphere = R
According to question
Vol. of result Sphere = Sum of Vol. of 3 Spheres
43πR3=43πr31+43πr22+43πr33
43πR3=43π(r31+r32+r33)
R3=(r31+r32+r33)
R3=((6)3+(8)3+(10)3)
R3=(216+512+1000)
R3=1728
R3=(12)3
R=12 cm
Answer :
The radius of resulting sphere 12 cm.
Q3. A 20 m deep well with 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.
Sol. :
Given :
Depth (height) of the well (cylinder) h=20cm
Diameter of the well (cylinder) d=7 cm
Radius of the well (cylinder) r=72 cm
To find :
The height of the platform
Solve :
According to question
Vol. of platform(cuboid) = Vol. of well(cylinder)
L×B×H=πr2h
22×14×h=227×(72)2×20
h=22×49×207×4×22×14
h=52
h=2.5 m
Answer :
The height of the platform is 2.5 m.
Q4. 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.
Sol. :
Given :
Diameter of well (cylinder) d=3 m
Radius of well (cylinder) r1=32=1.5 m
Depth (height) of well (cylinder) h=14 m
Width of embankment=4m
Radius of big circle r2=4+1.5=5.5 m
To find :
The height of the embankment.
Solve :
height of the embankment=x
According to question
Volume of embankment = Volume of well
π(r22-r21)x=πr21h
[(5.5)2-(1.5)2]x=(1.5)2×14
[(5.5+1.5)(5.5-1.5)]x=2.25×14
[7×4]x=2.25×14
x=2.25×147×4
x=1.125 m
Answer :
The height of the embankment 1.125 m.
Q5. A container shaped like a right circular cylinder having diameter 12 cm and height 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 which can be filled with ice cream.
Sol. :
Given :
Diameter of Cylinder d=12 cm
Radius of Cylinder r=6 cm
Height of Cylinder h=15 cm
Hight of Cone h′=12 cm
Diameter of Cone d′=6 cm
Radius of Cone & hemisphere r′=3 cm
To find :
The of ice-cream cones.
Solve :
According to question
Volume of ice-cream Cylinder=πr2h
V1=π(6)215
V1=540π
Vol. of ice-cream cone = Vol. of cone
+ Vol. of hemisphere
V2=13π(r′)2h′+23π(r′)3
=π(r′)23(h′+2r′)
=π(3)23(12+2×3)
=3×18π
=54π
Number of ice-cream cones =V1V2
Number of ice-cream cones =540π54π
= 10
Answer :
The number of ice-cream cones 10.
Q6. 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?
Sol. :
Given :
Dimensions of cuboid are 5.5×10×3.5 cm
Thickness (height) of coin h=2 mm=0.2 cm
Diameter of coin d=1.75 cm
Radius of coin r=1.752 cm
To find :
Number of coins
Solve :
Vol. of Cuboid =L×B×H
V1=5.5×10×3.5
=192.5cm3
Vol. of Coin (cylinder) =πr2h
V2=227×(1.752)2×0.2
= 13.47528
Number of coins =V1V2
=192.5×2813.475
=539013.475
=539000013475
=400
Answer :
The number of coins 400.
Q7. A cylindrical bucket, 32 cm height and with 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.
Sol. :
Given :
Height of cylindrical bucket h=32 cm
Radius of cylindrical bucket r=18cm
Height of conical heaph1=24 cm
To find :
The radius & slant height of heap.
Solve :
According to question
Vol. of conical heap = Vol. of Cylindrical bucket
13π(r1)2h1=πr2h
13(r1)2×24=(18)2×32
(r1)2=3×(18)2×3224
(r1)2=(18)2×4
r1=√(18)2×4
r1=18×2
r1=36 cm
slant height l=√(h1)2+(r1)2
l=√(24)2+(36)2
l=√(12)2(22+32)
l=12√4+9
l=12√13
Answer :
The radius of conical heap 36 cm and slant height 12√13 cm.
Q8. 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?
Sol. :
Given :
Cuboid shap canal width B=6 m
Cuboid shap canal depth H=1.5 m
Cuboid shap canal length L = 10km/h×3060
= 5 km = 5000 m
To find :
Area irrigate in 30 minutes.
Solve :
According to question
Vol. of water in field = Vol. of water in canal
l×b×h=L×B×H
Area ×h=5000×6×1.5
Area ×0.08=45000
Area =450000.08
=45000×1008
=22500×25 m2
=562500 m2
=56.25 hectare
Answer :
The area irrigate in 30 minutes 56.25 hectare.
Q9. 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 times will the tank be filled?
Sol. :
Given :
Depth of Cylindrical tank h=2 m
Diameter of Cylinder tank d=10 m
Radius of Cylindrical tank r=5 m
Internal diameter of pipe d1=20cm =0.2 m
Internal radius of pipe r1=0.1 m
The length of water in the pipe in 1 hour
h=3km = 3000 m
To find :
The time fill the tank.
Solve :
Let time to fill is n minutes
According to question
Vol. of tank = n(Vol. of water in pipe in 1 hour)
πr2h=n(πr21h1)
r2h=n(r21h1)
(5)2×2=n((0.1)2×3000)
50=n(30)
n=5030hours
n=53×60 minutes
n=100 minutes
Answer :
The time fill the tank in 100 minutes
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