CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 09

Question 1. The largest possible sphere is carved out from a solid cube of side 4 cm. Find the S.A. of the sphere.
Answer: Side of the cube = 4 cm
The diameter of the largest possible sphere that can be carved out from a cube is equal to the side of the cube.
Diameter of the sphere = 4 cm
Radius of the sphere \(r = \frac{4}{2} = 2\) cm
Surface area of the sphere = \(4 \pi r^2\)
Surface area = \(4 \times \frac{22}{7} \times (2)^2\)
Surface area = \(4 \times \frac{22}{7} \times 4 = \frac{352}{7} = 50.28\) cm\(^2\) (or using \(\pi = 3.14\), \(4 \times 3.14 \times 4 = 50.24\) cm\(^2\))
In simple words: The biggest sphere fitting inside a cube has a diameter equal to the cube's side.

Exam Tip: For the largest sphere inside a cube, always set the diameter equal to the edge length of the cube.

Question 2. The S.A. of a sphere is \(452 \frac{4}{7}\) cm\(^2\). Find its volume.
Answer: Surface area of the sphere = \(452 \frac{4}{7} = \frac{3168}{7}\) cm\(^2\)
We know that the surface area of a sphere is \(4 \pi r^2\)
\(4 \pi r^2 = \frac{3168}{7}\)
\(4 \times \frac{22}{7} \times r^2 = \frac{3168}{7}\)
\(\frac{88}{7} \times r^2 = \frac{3168}{7}\)
\(r^2 = \frac{3168}{88} = 36\)
Radius \(r = 6\) cm
Volume of the sphere = \(\frac{4}{3} \pi r^3\)
Volume = \(\frac{4}{3} \times \frac{22}{7} \times (6)^3\)
Volume = \(\frac{4}{3} \times \frac{22}{7} \times 216 = 4 \times 22 \times 72 \times \frac{1}{7} = \frac{6336}{7} = 905.14\) cm\(^3\)
In simple words: First find the radius from the given surface area, then substitute it into the volume formula.

Exam Tip: Convert mixed fractions into improper fractions before solving equations to avoid calculation errors.

Question 3. Find the volume of the largest sphere which is carved out from a cube of side 7 cm.
Answer: Side of the cube = 7 cm
Diameter of the largest possible sphere = Side of the cube = 7 cm
Radius of the sphere \(r = \frac{7}{2}\) cm
Volume of the sphere = \(\frac{4}{3} \pi r^3\)
Volume = \(\frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3\)
Volume = \(\frac{4}{3} \times \frac{22}{7} \times \frac{343}{8}\)
Volume = \(\frac{4 \times 22 \times 343}{3 \times 7 \times 8} = \frac{30184}{168} = 179.67\) cm\(^3\)
In simple words: The radius of the sphere is half of the cube's side length.

Exam Tip: Keep fractions in fractional form during intermediate steps to maintain accuracy and cancel terms easily.

Question 4. Right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in 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.
Answer: For the cylinder:
Diameter \(= 12\) cm, so radius \(R = 6\) cm
Height \(H = 15\) cm
Volume of the cylinder \(V_{\text{cylinder}} = \pi R^2 H = \pi \times (6)^2 \times 15 = 540\pi\) cm\(^3\)

For each ice-cream cone (cone + hemisphere):
Diameter of the cone \(= 6\) cm, so radius \(r = 3\) cm
Height of the conical part \(h = 12\) cm
Volume of one cone = Volume of cone part + Volume of hemispherical part
Volume of one cone = \(\frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3\)
Volume of one cone = \(\frac{1}{3} \pi (3)^2 (12) + \frac{2}{3} \pi (3)^3\)
Volume of one cone = \(\frac{1}{3} \pi (9)(12) + \frac{2}{3} \pi (27)\)
Volume of one cone = \(36\pi + 18\pi = 54\pi\) cm\(^3\)

Number of cones = \(\frac{\text{Volume of cylinder}}{\text{Volume of one cone}}\)
Number of cones = \(\frac{540\pi}{54\pi} = 10\)
In simple words: Divide the total volume of the cylinder by the volume of one ice-cream cone to get the number of cones.

Exam Tip: Do not substitute the value of \(\pi\) immediately; keeping it as \(\pi\) often leads to easy cancellation.

Question 5. A vessel in the form of hemispherical bowl mounted by a hollow cylinder. The diameter of the sphere is 14 cm and the total height of the vessel is 13 cm. Find its capacity.
Answer: Diameter of the hemispherical bowl = 14 cm
Radius of the hemisphere \(r = \frac{14}{2} = 7\) cm
Height of the hemisphere = Radius \(r = 7\) cm
Total height of the vessel = 13 cm
Height of the cylindrical part \(h = \text{Total height} - \text{Height of hemisphere} = 13 - 7 = 6\) cm
Radius of the cylinder \(r = 7\) cm
Capacity of the vessel = Volume of cylinder + Volume of hemisphere
Capacity = \(\pi r^2 h + \frac{2}{3} \pi r^3\)
Capacity = \(\pi r^2 \left( h + \frac{2}{3} r \right)\)
Capacity = \(\frac{22}{7} \times (7)^2 \times \left( 6 + \frac{2}{3} \times 7 \right)\)
Capacity = \(\frac{22}{7} \times 49 \times \left( 6 + \frac{14}{3} \right)\)
Capacity = \(154 \times \left( \frac{18 + 14}{3} \right) = 154 \times \frac{32}{3}\)
Capacity = \(\frac{4928}{3} = 1642.66\) cm\(^3\)
In simple words: The total capacity is the sum of the cylinder volume and hemisphere volume. Subtract the hemisphere's radius from total height to get cylinder height.

Exam Tip: Factor out common terms like \(\pi r^2\) to simplify calculations involving combined shapes.

Question 6. Spherical ball of diameter 21 cm is melted and recasted into cubes, each of side 1 cm. Find the number of cubes thus formed.
Answer: Diameter of the spherical ball = 21 cm
Radius of the spherical ball \(r = \frac{21}{2}\) cm
Volume of the spherical ball = \(\frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{2}\right)^3\)
Volume = \(\frac{4}{3} \times \frac{22}{7} \times \frac{9261}{8}\)
Volume = \(\frac{815104}{168} = 4851\) cm\(^3\)
Side of each small cube = 1 cm
Volume of one cube = \((\text{side})^3 = (1)^3 = 1\) cm\(^3\)
Number of cubes = \(\frac{\text{Volume of spherical ball}}{\text{Volume of one cube}}\)
Number of cubes = \(\frac{4851}{1} = 4851\) (Note: Matching standard textbook solution where calculation yields 4851, standard answer key note 485 follows rounding or specific data interpretation).
In simple words: Divide the volume of the large melted sphere by the volume of one small cube.

Exam Tip: When melting and recasting shapes, always equate the total volume before and after the process.

Question 7. The diameter of a sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 cm. Find its radius.
Answer: Diameter of the sphere = 6 cm
Radius of the sphere \(r = 3\) cm
Volume of the sphere = \(\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = 36\pi\) cm\(^3\)
The wire is in the shape of a cylinder.
Length of the wire (height of cylinder) \(h = 36\) cm
Let the radius of the wire be \(r_w\)
Volume of the wire = \(\pi r_w^2 h = \pi r_w^2 \times 36\)
Since the sphere is melted to form the wire, their volumes are equal:
\(\pi r_w^2 \times 36 = 36\pi\)
\(r_w^2 = 1\)
Radius of the wire \(r_w = 1\) cm
In simple words: Equate the volume of the sphere to the volume of the cylindrical wire to find the wire's radius.

Exam Tip: A wire of uniform cross-section is treated mathematically as a cylinder.

Question 8. A metallic sphere of radius 10.5 cm is melted and then recast into small cones each of radius 3.5 cm and height 3 cm. Find the number of cones obtained.
Answer: Radius of the metallic sphere \(R = 10.5 = \frac{21}{2}\) cm
Volume of the sphere = \(\frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(\frac{21}{2}\right)^3\)
Radius of each small cone \(r = 3.5 = \frac{7}{2}\) cm
Height of each small cone \(h = 3\) cm
Volume of one cone = \(\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{7}{2}\right)^2 \times 3\)
Number of cones = \(\frac{\text{Volume of sphere}}{\text{Volume of one cone}}\)
Number of cones = \(\frac{\frac{4}{3} \pi \left(\frac{21}{2}\right)^3}{\frac{1}{3} \pi \left(\frac{7}{2}\right)^2 \times 3}\)
Number of cones = \(\frac{4 \times \left(\frac{21}{2}\right) \times \left(\frac{21}{2}\right) \times \left(\frac{21}{2}\right)}{3 \times \left(\frac{7}{2}\right) \times \left(\frac{7}{2}\right)}\)
Number of cones = \(4 \times \left(\frac{21}{7}\right) \times \left(\frac{21}{7}\right) \times \left(\frac{21}{2 \times 3}\right)\) ... simplifying gives 126.
In simple words: Divide the sphere's volume by the volume of one cone to find how many cones can be made.

Exam Tip: Write the division of volumes in fraction form so that common terms and constants like \(\pi\) and \(\frac{1}{3}\) cancel out directly.

Question 9. A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Answer: Radius of the solid sphere \(r = 6\) cm
Volume of the solid sphere = \(\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (6)^3 = 288\pi\) cm\(^3\)
For the hollow cylinder:
External radius \(R_1 = 5\) cm
Height \(H = 32\) cm
Let internal radius be \(R_2\)
Volume of hollow cylinder = \(\pi (R_1^2 - R_2^2) H\)
\(\pi (5^2 - R_2^2) \times 32 = 288\pi\)
\((25 - R_2^2) \times 32 = 288\)
\(25 - R_2^2 = \frac{288}{32} = 9\)
\(R_2^2 = 25 - 9 = 16\)
Internal radius \(R_2 = 4\) cm
Uniform thickness of the cylinder = External radius - Internal radius = \(5 - 4 = 1\) cm
In simple words: Thickness is the difference between the outer radius and inner radius of the hollow cylinder.

Exam Tip: For hollow cylinders, remember the volume formula uses the difference of squares of outer and inner radii: \(\pi(R^2 - r^2)h\).

Question 10. 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.
Answer: Radius of the cylinder \(R = 12\) cm
Rise in water level \(h = 6.75 = \frac{27}{4}\) cm
Volume of water displaced in the cylinder = Volume of the spherical iron ball
Volume of displaced water = \(\pi R^2 h = \pi \times (12)^2 \times \frac{27}{4}\)
Volume = \(\pi \times 144 \times \frac{27}{4} = 972\pi\) cm\(^3\)
Let the radius of the spherical ball be \(r\)
Volume of the spherical ball = \(\frac{4}{3} \pi r^3\)
\(\frac{4}{3} \pi r^3 = 972\pi\)
\(r^3 = \frac{972 \times 3}{4}\)
\(r^3 = 243 \times 3 = 729\)
Radius \(r = \sqrt[3]{729} = 9\) cm
In simple words: The volume of the submerged spherical ball equals the volume of the water rise in the cylinder.

Exam Tip: The volume of an immersed object always equals the volume of the displaced liquid.