Samacheer Kalvi Class 10 Maths Solutions Chapter 7 Mensuration Exercise 7.4

Question 1. An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.
Answer:
Given:
Sphere - Radius \( r_1 = 12 \) cm
Cylinder - Radius \( r_2 = 8 \) cm

When a sphere is melted to form a cylinder, their volumes are equal. This is a key principle in mensuration.
Volume of cylinder = Volume of sphere
\( \pi r_2^2 h_2 = \frac{4}{3} \pi r_1^3 \)
\( \pi \times 8 \times 8 \times h_2 = \frac{4}{3} \pi \times 12 \times 12 \times 12 \)
Divide both sides by \( \pi \):
\( 8 \times 8 \times h_2 = \frac{4}{3} \times 12 \times 12 \times 12 \)
\( 64 \times h_2 = 4 \times 4 \times 12 \times 12 \)
\( 64 \times h_2 = 2304 \)
\( h_2 = \frac{2304}{64} \)
\( h_2 = 36 \) cm
Therefore, the height of the cylinder made is 36 cm.
In simple words: When you melt a sphere and make a cylinder from it, the amount of material stays the same. So, we make the volume of the sphere equal to the volume of the cylinder and then solve to find the height.

🎯 Exam Tip: Remember that when an object is reshaped, its volume remains constant. This is the fundamental concept for solving such problems. Use the correct formulas for each shape.

Question 2. Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
Answer:
For the rectangular tank:
Length (l) = 50 m = \( 50 \times 100 = 5000 \) cm
Width (b) = 44 m = \( 44 \times 100 = 4400 \) cm
Level of water (h) = 21 cm
Volume of water needed in the tank = \( l \times b \times h \)
\( = 5000 \times 4400 \times 21 \) cm\(^3 \)

For the pipe (which acts like a cylinder):
Diameter = 14 cm, so Radius (r) = \( \frac{14}{2} = 7 \) cm
Speed of water = 15 km/hr
To match units, convert speed to cm/hr:
\( = 15 \times 1000 \times 100 \) cm/hr
\( = 1500000 \) cm/hr
This speed represents the length (H) of the water column flowing out of the pipe in one hour.

Volume of water flowing from the pipe in one hour = \( \pi r^2 H \)
\( = \frac{22}{7} \times 7 \times 7 \times 1500000 \) cm\(^3 \)
\( = 22 \times 7 \times 1500000 \) cm\(^3 \)

Time taken to fill the tank = \( \frac{\text{Volume of the tank}}{\text{Volume of water flowing in one hour}} \)
\( = \frac{5000 \times 4400 \times 21}{22 \times 7 \times 1500000} \)
We can simplify the numbers by cancelling common factors.
\( = \frac{5 \times 44 \times 21}{22 \times 7 \times 15} \)
\( = \frac{1 \times 2 \times 3}{1 \times 1 \times 3} \)
\( = 2 \) hours
The time in which the level of water in the tanks will rise by 21 cm is 2 hours.
In simple words: First, we find out how much water the tank needs. Then, we find out how much water the pipe can give in one hour. By dividing the total water needed by the water per hour, we get the time it takes to fill the tank. Remember to keep all units consistent (like using cm for everything).

🎯 Exam Tip: Always convert all measurements to a consistent unit (like cm) at the beginning of the problem to avoid errors. Pay close attention to whether a diameter or radius is given.

Question 3. A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius x r units. Find the height of water in the cylindrical flask.
Answer:
For the conical flask:
Radius = r units
Height = h units
Volume of the conical flask = \( \frac{1}{3} \pi r^2 h \) cubic units

For the cylindrical flask:
Radius = \( xr \) units
Let the height of water in the cylindrical flask be \( H \) units.
Volume of the cylindrical flask = \( \pi (\text{radius})^2 H \)
\( = \pi (xr)^2 H \)

Since the water from the conical flask is poured into the cylindrical flask, their volumes are equal:
Volume of cylindrical flask = Volume of conical flask
\( \pi (xr)^2 H = \frac{1}{3} \pi r^2 h \)
\( \pi x^2 r^2 H = \frac{1}{3} \pi r^2 h \)
Divide both sides by \( \pi r^2 \) (assuming \( r \neq 0 \)):
\( x^2 H = \frac{1}{3} h \)
Now, solve for \( H \):
\( H = \frac{h}{3x^2} \) units
The height of the water in the cylindrical flask is \( \frac{h}{3x^2} \) units.
In simple words: We have water in a cone and pour it into a cylinder. The amount of water stays the same, so their volumes are equal. We use the formulas for cone and cylinder volume, set them equal, and then figure out the height of the water in the cylinder.

🎯 Exam Tip: When transferring liquid between containers, the volume of the liquid remains unchanged. This equality of volumes is the basis for solving such problems. Simplify the equations by canceling common terms early.

Question 4. A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
Answer:
For the cone:
Diameter = 14 cm, so Radius (r) = \( \frac{14}{2} = 7 \) cm
Height (h) = 8 cm
Volume of the cone = \( \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \pi \times 7^2 \times 8 \) cm\(^3 \)

For the hollow sphere:
External diameter = 10 cm, so External radius (R) = \( \frac{10}{2} = 5 \) cm
Let the internal radius be \( x \) cm.
Volume of the hollow sphere = \( \frac{4}{3} \pi (R^3 - x^3) \)
\( = \frac{4}{3} \pi (5^3 - x^3) \) cm\(^3 \)

Since the cone is melted to form the hollow sphere, their volumes are equal:
Volume of hollow sphere = Volume of cone
\( \frac{4}{3} \pi (5^3 - x^3) = \frac{1}{3} \pi \times 7^2 \times 8 \)
Multiply both sides by \( \frac{3}{\pi} \):
\( 4 (5^3 - x^3) = 7^2 \times 8 \)
\( 4 (125 - x^3) = 49 \times 8 \)
\( 4 (125 - x^3) = 392 \)
Divide both sides by 4:
\( 125 - x^3 = \frac{392}{4} \)
\( 125 - x^3 = 98 \)
Rearrange to solve for \( x^3 \):
\( x^3 = 125 - 98 \)
\( x^3 = 27 \)
Take the cube root of both sides:
\( x = \sqrt[3]{27} \)
\( x = 3 \) cm

The internal radius of the hollow sphere is 3 cm. Therefore, the internal diameter is \( 2 \times 3 = 6 \) cm.
In simple words: A cone is melted and turned into a hollow ball (sphere). The amount of material stays the same, so the volume of the cone and the volume of the hollow sphere are equal. We use their volume formulas, put in the numbers we know, and then calculate the unknown inner radius of the sphere. Finally, double the radius to find the diameter.

🎯 Exam Tip: Remember that "hollow" means you need to find the difference between the volumes of two spheres (external and internal). Be careful with cube roots and ensure all calculations are accurate.

Question 5. Seenu's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m × 1.5 m × 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Answer:
For the cuboid sump:
Length (l) = 2 m = \( 2 \times 100 = 200 \) cm
Breadth (b) = 1.5 m = \( 1.5 \times 100 = 150 \) cm
Height (h) = 1 m = \( 1 \times 100 = 100 \) cm
Initial volume of water in the cuboid sump = \( l \times b \times h \)
\( = 200 \times 150 \times 100 \) cm\(^3 \)
\( = 30,00,000 \) cm\(^3 \)

For the cylindrical overhead tank:
Radius (r) = 60 cm
Height (h) = 105 cm
Volume of the cylindrical tank = \( \pi r^2 h \)
\( = \frac{22}{7} \times 60 \times 60 \times 105 \) cm\(^3 \)
\( = 22 \times 3600 \times 15 \) cm\(^3 \) (since \( \frac{105}{7} = 15 \))
\( = 22 \times 54000 \) cm\(^3 \)
\( = 11,88,000 \) cm\(^3 \)
This is the volume of water transferred to the overhead tank.

Volume of water left in the sump = Initial volume of sump - Volume of overhead tank
\( = 30,00,000 - 11,88,000 \) cm\(^3 \)
\( = 18,12,000 \) cm\(^3 \)
The volume of water left in the sump is 18,12,000 cm\(^3 \).
In simple words: We first find out how much water the underground tank can hold. Then, we find out how much water the overhead tank needs. We subtract the water needed for the overhead tank from the total water in the underground tank to see how much water is left. Always use the same units for all measurements.

🎯 Exam Tip: Ensure all dimensions are in the same unit (e.g., cm) before performing calculations. Remember that the volume of water transferred is equal to the volume of the overhead tank when it is filled.

Question 6. The internal and external diameter of a hollow hemispherical shell is 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder.
Answer:
For the hollow hemispherical shell:
Internal diameter = 6 cm, so Internal radius (r) = \( \frac{6}{2} = 3 \) cm
External diameter = 10 cm, so External radius (R) = \( \frac{10}{2} = 5 \) cm
Volume of a hollow hemisphere = \( \frac{2}{3} \pi (R^3 - r^3) \)
\( = \frac{2}{3} \pi (5^3 - 3^3) \)
\( = \frac{2}{3} \pi (125 - 27) \)
\( = \frac{2}{3} \pi (98) \) cm\(^3 \)

For the solid cylinder:
Diameter = 14 cm, so Radius (r) = \( \frac{14}{2} = 7 \) cm
Let the height of the cylinder be \( h \) cm.
Volume of the cylinder = \( \pi r^2 h \)
\( = \pi \times 7^2 \times h \) cm\(^3 \)

When the shell is melted and recast into a cylinder, their volumes are equal:
Volume of cylinder = Volume of hollow hemispherical shell
\( \pi \times 7^2 \times h = \frac{2}{3} \pi (98) \)
Divide both sides by \( \pi \):
\( 49 h = \frac{2}{3} \times 98 \)
\( h = \frac{2 \times 98}{3 \times 49} \)
Since \( 98 = 2 \times 49 \):
\( h = \frac{2 \times (2 \times 49)}{3 \times 49} \)
\( h = \frac{4}{3} \)
\( h \approx 1.33 \) cm
The height of the cylinder is approximately 1.33 cm.
In simple words: A hollow half-ball is melted and made into a solid cylinder. The amount of material is the same, so we set the volume of the hollow half-ball equal to the volume of the cylinder. Then, we use the given numbers and solve for the cylinder's height.

🎯 Exam Tip: Differentiate between hollow and solid shapes when choosing volume formulas. A hollow hemisphere's volume is the difference between the external and internal hemisphere volumes.

Question 7. 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, then find the thickness of the cylinder.
Answer:
For the solid sphere:
Radius (r) = 6 cm
Volume of the sphere = \( \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \pi \times 6^3 \) cm\(^3 \)

For the hollow cylinder:
External radius (R) = 5 cm
Height (h) = 32 cm
Let the internal radius be \( x \) cm.
The thickness of the cylinder is \( R - x \). This is what we need to find.
Volume of the hollow cylinder = \( \pi h (R^2 - x^2) \)
\( = \pi \times 32 (5^2 - x^2) \) cm\(^3 \)

Since the sphere is melted into the hollow cylinder, their volumes are equal:
Volume of hollow cylinder = Volume of solid sphere
\( \pi \times 32 (5^2 - x^2) = \frac{4}{3} \pi \times 6^3 \)
Divide both sides by \( \pi \):
\( 32 (25 - x^2) = \frac{4}{3} \times 216 \)
\( 32 (25 - x^2) = 4 \times 72 \)
\( 32 (25 - x^2) = 288 \)
Divide both sides by 32:
\( 25 - x^2 = \frac{288}{32} \)
\( 25 - x^2 = 9 \)
Rearrange to solve for \( x^2 \):
\( x^2 = 25 - 9 \)
\( x^2 = 16 \)
Take the square root of both sides:
\( x = \sqrt{16} \)
\( x = 4 \) cm

The internal radius of the cylinder is 4 cm.
Thickness of the cylinder = External radius - Internal radius
Thickness = \( R - x = 5 - 4 = 1 \) cm.
The thickness of the cylinder is 1 cm.
In simple words: A solid ball is melted and made into a hollow pipe (cylinder). The amount of material is the same, so their volumes are equal. We know the outer size of the pipe and its height. We use the volume formulas to find the inner size of the pipe, and then we subtract the inner radius from the outer radius to get the thickness.

🎯 Exam Tip: For hollow shapes, remember that volume depends on the difference between the outer and inner volumes. Pay attention to whether radius or diameter is given, and be careful with squares and cubes in the calculations.

Question 8. A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Answer:
Let the height of the cylindrical vessel be \( h \).
The radius of the cylindrical vessel (r) is 50% more than its height:
\( r = h + 50\% \text{ of } h \)
\( r = h + \frac{50}{100} h \)
\( r = h + \frac{1}{2} h \)
\( r = \frac{3}{2} h \)

The diameter is the same for both the hemispherical bowl and the cylindrical vessel. This means their radii are also the same.
Radius of hemispherical bowl = Radius of cylindrical vessel \( = \frac{3}{2} h \)

Volume of the cylindrical vessel (V\(_{cyl} \)) = \( \pi r^2 h \)
Substitute \( r = \frac{3}{2} h \):
\( V_{cyl} = \pi \left(\frac{3}{2} h\right)^2 h \)
\( V_{cyl} = \pi \left(\frac{9}{4} h^2\right) h \)
\( V_{cyl} = \frac{9 \pi h^3}{4} \) cubic units (Equation 1)

Volume of the hemispherical bowl (V\(_{hemi} \)) = \( \frac{2}{3} \pi r^3 \)
Substitute \( r = \frac{3}{2} h \):
\( V_{hemi} = \frac{2}{3} \pi \left(\frac{3}{2} h\right)^3 \)
\( V_{hemi} = \frac{2}{3} \pi \left(\frac{27}{8} h^3\right) \)
\( V_{hemi} = \pi \frac{2 \times 27}{3 \times 8} h^3 \)
\( V_{hemi} = \pi \frac{54}{24} h^3 \)
\( V_{hemi} = \frac{9 \pi h^3}{4} \) cubic units (Equation 2)

From Equation 1 and Equation 2, we see that the volume of the cylindrical vessel is equal to the volume of the hemispherical bowl.
\( V_{cyl} = V_{hemi} \)
Since the volumes are equal, all the juice from the hemispherical bowl can be transferred into the cylindrical vessel.
Percentage of juice transferred = \( \frac{\text{Volume of cylinder}}{\text{Volume of hemisphere}} \times 100\% \)
\( = \frac{\frac{9 \pi h^3}{4}}{\frac{9 \pi h^3}{4}} \times 100\% \)
\( = 1 \times 100\% \)
\( = 100\% \)
Therefore, 100% of the juice can be transferred.
In simple words: We are given a half-ball full of juice and a cylinder. The cylinder's radius is bigger than its height in a special way, and both have the same diameter. We calculate the volume of both the half-ball and the cylinder. Since their volumes turn out to be the same, all the juice from the half-ball can fit perfectly into the cylinder, meaning 100% of it can be transferred.

🎯 Exam Tip: Pay close attention to how the radius and height are related in the cylinder and how the diameter is shared between the shapes. A common mistake is to miss that equal diameter means equal radius. Simplify expressions carefully to compare volumes.