Class 9 RD Sharma Solutions Chapter 21 Surface Area and Volume of a Sphere Exercise 21.2 | Set 1 (original) (raw)
Last Updated : 24 Sep, 2024
Chapter 21 of RD Sharma's Class 9 mathematics textbook explores the surface area and volume of a sphere, a fundamental concept in geometry. Exercise 21.2 Set 1 focuses on applying formulas and problem-solving techniques to calculate these measurements for spheres in various contexts, helping students develop a strong understanding of three-dimensional geometry and its real-world applications.
Surface Area and Volume of a Sphere
The surface area of a sphere is the total area of its outer surface, calculated using the formula A = 4πr², where r is the sphere's radius. The volume of a sphere is the amount of space it occupies, given by the formula V = (4/3)πr³. These formulas allow us to quantify the size and capacity of spherical objects.
**Question 1. Find the volume of a sphere whose radius is:
****(i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.**
**Solution:
As we know that,
Volume of a sphere = 4/3πr3 Cubic Units Where r is radius of a sphere
****(i)** Given that, Radius = 2 cm
put in formula and we get,
Volume = 4/3 × 22/7 × (2)3 = 33.52
**Volume = 33.52 cm 3
****(ii)** Given that, Radius = 3.5cm
putting value in formula and we get,
Volume = 4/3×22/7×(3.5)3 = 179.666
**Volume = 179.666 cm 3
****(iii)** Given that, Radius = 10.5 cm
putting this value in formula and we get,
Volume = 4/3×22/7×(10.5)3 = 4851
**Volume = 4851 cm 3
**Question 2. Find the volume of a sphere whose diameter is :
****(i) 14 cm (ii) 3.5 dm (iii) 2.1 m**
**Solution:
As we know that,
Volume of a sphere = 4/3πr3 Cubic Units Where r is radius of a sphere
****(i)** Given that, diameter = 14 cm
So, radius = diameter / 2 = 14/2 = 7cm
putting these value in formula and we get,
Volume = 4/3×22/7×(7)3 = 1437.33
**Volume = 1437.33 cm 3
****(ii)** Given that,
Diameter = 3.5 dm
So, radius = diameter/2 = 3.5/2 = 1.75 dm
putting these value in formula and we get,
Volume = 4/3×22/7×(1.75)3 = 22.46
**Volume = 22.46 dm 3
****(iii)** Given that,
Diameter = 2.1 m
So, radius = diameter/2 = 2.1/2 = 1.05 m
putting these value in formula and we get,
Volume = 4/3×22/7×(1.05)3 = 4.851
**Volume = 4.851 m 3
**Question 3. A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.
**Solution:
Given that,
Radius of hemispherical tank is 2.8 m
Capacity of hemispherical tank is 2/3 πr3 = 2/3×22/7×(2.8)3 m3 = 45.997 m3
[As we know that 1m3 = 1000 liters]
**Therefore, capacity in liters = 45997 liters
**Question 4. A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.
**Solution:
Given that,
Inner radius of a hemispherical bowl is 5 cm
Outer radius of a hemispherical bowl is 5 cm + 0.25 cm = 5.25 cm
As we know that,
Volume of steel used = Outer volume – Inner volume
= 2/3×π×((5.25)3−(5)3) = 2/3×22/7×((5.25)3−(5)3) = **41.282
**Hence Volume of steel used is 41.282 cm 3
**Question 5. How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
**Solution:
Given that,
Edge of a cube = 22 cm,
Diameter of bullet = 2 cm,
So, radius of bullet(r) = 1 cm,
Volume of the cube = (side)3 = (22)3 cm3 = 10648 cm3 and,
Volume of each bullet which will be in spherical in shape = 4/3πr3
= 4/3 × 22/7 × (1)3 = 4/3 × 22/7
= 88/21 cm3
As we know that,
Number of Bullets = (Volume of Cube) / (Volume of Bullet)
= 10648 / (88/21) = **2541
**Hence, 2541 bullets can be made.
**Question 6. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?
**Solution:
Given that,
Volume of laddoo having radius 5 cm (V1) = 4/3×22/7×(5)3 (Using Volume of Sphere formula)
= 11000/21 cm3
Also, Volume of laddoo having radius 2.5 cm (V2) = 4/3πr3
= 4/3×22/7×(2.5)3 = 1375/21 cm3
**Hence, Number of laddoos of radius 2.5 cm that can be made are = V1/V2 = 11000/1375 = 8
**Question 7. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2cm and 2 cm, find the diameter of the third ball.
**Solution:
Given that,
Volume of lead ball with radius 3/2 cm = 4/3πr3 = 4/3×π×(3/2)3
Lets,
Diameter of first ball (d1) = 3/2cm,
Radius of first ball (r1) = 3/4 cm,
Diameter of second ball (d2) = 2 cm,
Radius of second ball (r2) = 2/2 cm = 1 cm,
Diameter of third ball (d3) = d,
Radius of third ball (r3) = d/2 cm,
As we know that,
Volume of lead ball = 4/3πr13 + 4/3πr23 + 4/3πr33
Volume of lead ball = 4/3π(3/4)3 + 4/3π(2/2)3 + 4/3π(d/2)3
4/3π(3/2)3 = 4/3π[(3/4)3 + (2/2)3 + (d/2)3]
27/8 = 27/64 + 1 + d3/8
d3 = (125 x 8) / 64
d = 10 / 4
**Hence, d = 2.5 cm
**Question 8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder.
**Solution:
Given that,
Radius of sphere = 5cm,
Height of water raised = 5/3cm,
Let us assume that radius of Cylinder is r cm,
As we know that Volume of Sphere = 4/3πr3
= 4/3 × π × (5)3
As we know that, Volume of water raised in cylinder = πr2h
Therefore,
Volume of water rises in cylinder = Volume of sphere
πr2h = 4/3πr3
r2 × 5/3 = 4/3 × π × (5)^3
r2 × 5/3 = 4/3 × 22/7 × 125
r2 = 20 × 5
r = √100
r = 10 cm
**Hence the radius of cylinder is 10 cm.
**Question 9. If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?
**Solution:
Let us assume that v1 and v2 be the volumes of the first and second sphere respectively,
Radius of the first sphere = r,
Radius of the second sphere = 2r
therefore (Volume of first sphere) / (Volume of second sphere)
= 4/3πr3 / 4/3π(2r)3 = 1 / 8
**Hence the ratio is 1 : 8
**Question 10. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
**Solution:
Given that,
Volume of Cone = Volume of Hemisphere
1/3πr2h = 2/3πr3
r2h = 2r3
h = 2r
h/r = 1/1 × 2 = 2
**Hence, Ratio of their heights is 2 : 1
**Question 11. A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.
**Solution:
Given that,
Volume of water in the hemispherical bowl = Volume of water in the cylinder
Let h be the height to which water rises in the cylinder
Inner radii of the bowl = r1 = 3.5 cm
Inner radii of the bowl = r2 = 7 cm
2/3π(r13 )= π(r22)h
h = 2r13 / 3r22
h = 2(3.5)3 / 3(72)
h = 7 / 12 cm
**Hence the height to which the water will rise in the cylinder is 7/12 cm.
**Question 12. A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
**Solution:
Given that,
Height of the cylinder = 2/3 diameter
We know that
Diameter = 2(radius)
h = 2/3 × 2r = 4/3r
Volume of Cylinder = Volume of Sphere
πr2h = 4/3πr3
π × r2 × (4/3r) = 4/3π(4)3
(r)3 = (4)3
r = 4 cm
**Hence, the radius of the base of the Cylinder is 4 cm.
**Question 13. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
**Solution:
Given that,
Volume of water in Hemispherical bowl = Volume of Cylinder
2/3πr13 = πr22h
h = 2x(6)3 / 3x(4)2
h = 9 cm
**Hence the height of water in the cylinder is 9 cm.
**Question 14. A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?
**Solution:
Given that,
Radius of the cylinder = 16 cm,
Let's r be the radius of the iron ball
Then,
Volume of iron ball = Volume of water raised in the hub
4/3 x π x r3 = π x (r)2 x h
4/3 x r3 = (16)2 x 9
r^3 = 1728 = (12)^3
**Hence radius of ball is 12 cm.
**Question 15. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use = 227).
**Solution:
Given that,
Radius of the cylinder = r1 = 12cm,
Raised in raised = r2 = 6.75 cm,
Volume of water raised = Volume of the sphere
π x (r1)2 x h = 4/3 x π x (r2)3
12 x 12 x 6.75 = 4/3 x (r2)3
= (r2)3 = (12 x 12 x 6.75 x 3) / 4
= r2 = 9 cm
**Hence radius of Sphere is 9 cm.
**Question 16. The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.
**Solution:
Given that,
Diameter of a copper sphere = 18 cm,
Radius of the sphere = 9 cm,
Length of the wire = 108 m = 10800 cm,
Volume of cylinder = Volume of sphere
π x (r1)^2 x h = 4/3 x π x (r2)^3
= (r1)^2 x 10800 = 4/3 x 9 x 9 x 9
= (r1)^2 = 0.009
= r1 = 0.3 cm
**Hence Diameter is 0.6 cm.
Summary
Exercise 21.2 Set 1 in Chapter 21 of RD Sharma's Class 9 textbook provides students with a diverse set of problems to practice calculating the surface area and volume of spheres. These exercises help students apply formulas, convert between different measurements, and analyze relationships between radius, surface area, and volume. By working through these problems, students develop critical thinking skills and gain a deeper understanding of spherical geometry.