(1) Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm3. (Mass = Volume x Density).
(2) The outer and inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.
(3) The volume of a solid hemisphere is 1152 Π cu.cm. Find its curved surface area.
(4) Find the volume of the largest right circular cone that can be cut of a cube whose edge is 14 cm.
(5) The radius of a spherical balloon increase from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.
(6) A ball of gold has a radius of 9cm. The density of gold is 19.3g/cm³. Work out the mass of the ball.
(7) A hemispherical bowl has a radius of 10 cm.
(a) Calculate the volume of the bowl.
(b) A cylinder of radius 7 cm and height h cm has the same volume as the bowl. Calculate the value of h .
(8) The total surface area of a hemisphere is given as 618 cm².
(a) Find the radius of the hemisphere.
(b) Find the volume of the hemisphere.
(c) Find the external surface area of the hemisphere if it is hollow. Give your answers to 3 significant figures. Take π = 3.142
(9) Given that the volume of a sphere is 5276 cm³, find its radius and surface area. Give your answers to 3 significant figures. Take π = 3.142
(10) A hemisphere has diameter 22.4 cm. Find the volume and closed surface area. Give your answers to 3 significant figures. Take π = 3.142
(1) Solution :
radius of spherical ball = 0.7 cm
Volume of one spherical ball = (4/3) Π r3
= (4/3) (22/7) ⋅ 0.7 ⋅ 0.7 ⋅ 0.7
= 4.312/3
= 1.437
Volume of 200 steel spherical ball = 200 ⋅ 1.437
= 287.46 cm3
1 cm³ = 7.95 g
Therefore mass of 200 spherical ball bearings
= 287.46 (7.95)
= 2285.307 gram
1000 gram = 1 kg
= 2285.307/1000
= 2.29 kg
Volume of 200 spherical balls = 2.29 kg.
(2) Solution :
From this information we have to find the volume
Outer radius (R) = 12 cm
Inner radius (r) = 10 cm
Volume of hollow sphere = (4/3) Π (R3-r3)
= (4/3)(22/7) (123-103)
= (88/21) (1728-1000)
= (88/21) (728)
= (64064/21)
= 3050.67 cm3
Volume of hollow sphere is 3050.67 cm3.
(3) Solution :
Volume of hollow sphere = 1152 Π
(2/3) Π r3 = 1152 Π
r3 = 1152 Π (3/2Π)
r3 = (576 x 3)/2
r3 = 1728
r = ∛1728
r = 12 cm
Curved surface area = 2Πr2
= 2Π(12)2
= 2Π(144)
= 288Π cm3
Curved surface area = 288Π cm3
(4) Solution :
Since it is cube length of all sides will be equal that is 14 cm. Diameter and height of cone are 14 cm.
r = 14/2 ==> 7
h = 14 cm
Volume of cone = (1/3) Π r2 h
= (1/3) ⋅ (22/7) ⋅ 72 ⋅ 14
= (1/3) ⋅ 22 ⋅ 49 ⋅ 2
= (49 ⋅ 44)/3
= 2156/3
= 718.67 cm3
Volume of cone is 718.67 cm3.
(5) Solution :
Let r₁ and r₂ are the radii of two spherical balloon
r1 : r2 = 7 : 14
Volume of one spherical balloon = (4/3) Π r3
(4/3) Π (7)3 : (4/3) Π 143
73 : 143
7 ⋅ 7 ⋅ 7 : 14 ⋅ 14 ⋅ 14
1 : 8
So, the required ratio is 1 : 8.
(6) Solution :
radius = 9 cm
Density = 19.3 g/cm³
Mass = volume x density
volume of hemisphere = (2/3) Π r3
= (2/3) x 3.14 x 93
= 1526.04 cm³
Mass = 1526.04 x 19.3
= 29452.57 g
1000 gram = 1 kg
= 29452.57/1000
= 29.45 kg
(7) Solution :
a)
r = 10 cm
Volume of bowl = (2/3) Π r3
= (2/3) Π (10)3
= (2/3) x 3.14 x 1000
= 2093.3 cm3
(b) Radius = 7 cm
height = h cm
Volume of cylinder = Π r2 h
= 3.14(7)2 h
= 153.86 h cm3
(8) Solution :
The total surface area of a hemisphere = 618 cm².
(a) 3 Π r2 = 618 cm²
Applying the value of π, we get
3 x 3.142 x r2 = 618
r2 = 618 / (3 x 3.142)
r2 = 65.56
r = √65.56
r = 8.09 cm
(b) volume of the hemisphere = (2/3) Π r3
= (2/3) x 3.14 x 8.093
= 1108.36 cm3
Volume of hemisphere is 1108.36 cm3
(c) External surface area = 2 Π r2
= 2 x 3.14 x 8.092
= 411.01 cm2
So, the external surface area is 411.01 cm2
(9) Solution :
Volume of a sphere = 5276 cm³
(2/3) Π r3 = 5276
(2/3) x 3.14 x r3 = 5276
r3 = (5276/3.14) x (3/2)
r3 = 2520.38
r = 13.60 cm
So, the radius of the hemisphere is 13.60 cm
(10) Solution :
diameter of hemisphere = 22.4 cm
radius (r) = 22.4/2
= 11.2 cm
Volume of hemisphere = (2/3) Π r3
= (2/3) x 3.14 x 11.23
= 2940.98
Approximately the volume is 2941 cm3
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 01, 24 12:04 PM
Oct 01, 24 11:49 AM
Sep 30, 24 11:38 AM