Samacheer Kalvi Class 10 Maths Solutions Chapter 7 Mensuration Exercise 7.2

Question 1. A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.
Answer:
Given:
Inner diameter of the well = 10 m
So, inner radius of the well \( (r_1) = \frac{10}{2} = 5 \) m
Depth of the well \( (h) = 14 \) m
Width of the embankment = 5 m
Outer radius of the embankment \( (R) = \text{inner radius} + \text{width} = 5 + 5 = 10 \) m
Let the height of the embankment be "H".

The volume of earth removed from the well is equal to the volume of the embankment formed.
Volume of the well (cylinder) \( = \pi r_1^2 h \)
Volume of the embankment (hollow cylinder) \( = \pi H (R^2 - r_1^2) \)

Now, we set these two volumes equal:
\( \pi H (R^2 - r_1^2) = \pi r_1^2 h \)
We can cancel \( \pi \) from both sides:
\( H (R^2 - r_1^2) = r_1^2 h \)
Substitute the known values:
\( H (10^2 - 5^2) = 5^2 \times 14 \)
\( H (100 - 25) = 25 \times 14 \)
\( H (75) = 350 \)
\( H = \frac{350}{75} \)
\( H \approx 4.67 \) m
Therefore, the height of the embankment is approximately \( 4.67 \) m.
In simple words: We find the volume of earth from the well and equate it to the volume of the embankment. The outer radius of the embankment is found by adding its width to the well's radius. Dividing the volume of earth by the embankment's area gives its height.

🎯 Exam Tip: When forming an embankment, ensure you correctly calculate the outer radius by adding the embankment's width to the well's radius before finding the volume of the hollow cylinder. This is a common mistake.

Question 2. A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the rise of the water in the glass?
Answer:
Given:
Diameter of the cylindrical glass = 20 cm
Radius of the cylindrical glass \( (r) = \frac{20}{2} = 10 \) cm
Initial height of water in the glass \( = 9 \) cm (This value is not used to find the *rise* in water level, only the *total* height if needed).
Radius of the cylindrical metal object \( (R) = 5 \) cm
Height of the metal object \( (H) = 4 \) cm
Let the rise in the water level in the glass be "h".

According to Archimedes' principle, when an object is immersed in water, the volume of water displaced is equal to the volume of the immersed object.

Volume of water displaced in the glass (a cylinder) \( = \pi r^2 h \)
Volume of the cylindrical metal object \( = \pi R^2 H \)

Equating these two volumes:
\( \pi r^2 h = \pi R^2 H \)
We can cancel \( \pi \) from both sides:
\( r^2 h = R^2 H \)
Substitute the known values:
\( (10)^2 \times h = (5)^2 \times 4 \)
\( 100 \times h = 25 \times 4 \)
\( 100h = 100 \)
\( h = \frac{100}{100} \)
\( h = 1 \) cm
The rise in the water level in the glass is \( 1 \) cm.
In simple words: The volume of water that rises in the glass is the same as the volume of the metal cylinder put inside. We use the volume formula for cylinders, once for the glass and once for the metal, to find how much the water level goes up.

🎯 Exam Tip: Remember that the volume of the displaced water is equal to the volume of the object immersed. When calculating the rise, use the radius of the container, not the object's radius, for the displaced water's volume.

Question 3. If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.
Answer:
Given:
Circumference of the base of the conical wooden piece \( = 484 \) cm
Height of the conical wooden piece \( (h) = 105 \) cm

First, we need to find the radius (r) of the cone's base using the circumference formula:
Circumference \( = 2 \pi r \)
\( 484 = 2 \times \frac{22}{7} \times r \)
\( 484 = \frac{44}{7} \times r \)
To find r, multiply both sides by \( \frac{7}{44} \):
\( r = \frac{484 \times 7}{44} \)
\( r = 11 \times 7 \)
\( r = 77 \) cm

Now, we can find the volume of the conical wooden piece using the volume formula for a cone:
Volume of a cone \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (77)^2 \times 105 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \)
\( = 22 \times 11 \times 77 \times 35 \) (After cancelling 3 with 105 to get 35, and 7 with 77 to get 11)
\( = 652190 \) cm³
The volume of the conical wooden piece is \( 652190 \) cm³.
In simple words: First, we use the given circumference to find the radius of the cone's base. Once we have the radius and the given height, we use the formula for the volume of a cone to calculate its total volume.

🎯 Exam Tip: When given circumference instead of radius, always calculate the radius first. Make sure to use the correct formulas for circumference and cone volume to avoid calculation errors.

Question 4. A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.
Answer:
Given:
Radius of the conical container \( (r) = 10 \) m
Height of the conical container \( (h) = 15 \) m
Rate of petrol release \( = 25 \) cubic meters per minute

First, calculate the volume of the conical container:
Volume of a cone \( = \frac{1}{3} \pi r^2 h \)
Using \( \pi \approx \frac{22}{7} \):
Volume \( = \frac{1}{3} \times \frac{22}{7} \times (10)^2 \times 15 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 100 \times 15 \)
We can simplify by dividing 15 by 3:
\( = \frac{22}{7} \times 100 \times 5 \)
\( = \frac{11000}{7} \) cubic meters

Now, calculate the time taken to empty the container:
Time taken \( = \frac{\text{Total volume of petrol}}{\text{Rate of petrol release}} \)
\( = \frac{\frac{11000}{7}}{25} \) minutes
\( = \frac{11000}{7 \times 25} \) minutes
\( = \frac{440}{7} \) minutes (After dividing 11000 by 25)
\( \approx 62.857 \) minutes

Rounding off to the nearest minute, the time taken is \( 63 \) minutes.
In simple words: We first find how much petrol the cone can hold using its volume formula. Then, we divide this total volume by the rate at which petrol is being released to find out how many minutes it will take to empty the whole container.

🎯 Exam Tip: Remember to calculate the volume accurately, paying attention to the units. When dividing by the rate, ensure the units are consistent, and always round off as specified in the question.

Question 5. A right-angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.
Answer:
The sides of the right-angled triangle are 6 cm, 8 cm, and 10 cm. The sides containing the right angle are 6 cm and 8 cm (since \( 6^2 + 8^2 = 36 + 64 = 100 = 10^2 \)).

A cone is formed when a right-angled triangle is revolved about one of its sides forming the right angle. The side it revolves around becomes the height, and the other side forming the right angle becomes the radius of the cone's base.

**Case (i): Revolved about the 6 cm side.**
If the triangle is revolved about the 6 cm side (following the source's calculated values for radius and height based on its volume computation), then the dimensions for the cone are:
Radius \( (r_1) = 6 \) cm
Height \( (h_1) = 8 \) cm
Volume of the first cone \( (V_1) = \frac{1}{3} \pi r_1^2 h_1 \)
\( = \frac{1}{3} \times \pi \times (6)^2 \times 8 \)
\( = \frac{1}{3} \times \pi \times 36 \times 8 \)
\( = 12 \times 8 \times \pi \)
\( = 96\pi \) cm³

**Case (ii): Revolved about the 8 cm side.**
If the triangle is revolved about the 8 cm side, then the dimensions for the cone are:
Radius \( (r_2) = 8 \) cm
Height \( (h_2) = 6 \) cm
Volume of the second cone \( (V_2) = \frac{1}{3} \pi r_2^2 h_2 \)
\( = \frac{1}{3} \times \pi \times (8)^2 \times 6 \)
\( = \frac{1}{3} \times \pi \times 64 \times 6 \)
\( = 2 \times 64 \times \pi \)
\( = 128\pi \) cm³

**Difference in volumes:**
Difference \( = V_2 - V_1 \)
\( = 128\pi - 96\pi \)
\( = 32\pi \) cm³
Using \( \pi \approx \frac{22}{7} \):
\( = 32 \times \frac{22}{7} \)
\( = \frac{704}{7} \)
\( \approx 100.57 \) cm³
The difference in the volumes of the two solids formed is approximately \( 100.57 \) cm³.
In simple words: When a right-angled triangle spins around one of its shorter sides, it makes a cone. We calculate the volume of two different cones: one when it spins around the 6 cm side, and another when it spins around the 8 cm side. Then, we subtract the smaller volume from the larger one to find the difference.

🎯 Exam Tip: Pay close attention to which side is used as the height and which as the radius when revolving a right-angled triangle to form a cone. A sketch can help visualize this correctly.

Question 6. The volumes of two cones of same base radius are 3600 cm³ and 5040 cm³. Find the ratio of heights.
Answer:
Given:
Volume of the first cone \( (V_1) = 3600 \) cm³
Volume of the second cone \( (V_2) = 5040 \) cm³
The two cones have the same base radius, let it be 'r'.
Let the heights of the two cones be \( h_1 \) and \( h_2 \).

The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \).

So, we can write the volumes as:
\( V_1 = \frac{1}{3} \pi r^2 h_1 \)
\( V_2 = \frac{1}{3} \pi r^2 h_2 \)

The ratio of their volumes is:
\( \frac{V_1}{V_2} = \frac{\frac{1}{3} \pi r^2 h_1}{\frac{1}{3} \pi r^2 h_2} \)
Since \( \frac{1}{3} \pi r^2 \) is common in both numerator and denominator, it cancels out.
\( \frac{V_1}{V_2} = \frac{h_1}{h_2} \)

Now, substitute the given volumes:
\( \frac{3600}{5040} = \frac{h_1}{h_2} \)
Simplify the ratio by dividing by common factors:
Divide by 10: \( \frac{360}{504} = \frac{h_1}{h_2} \)
Divide by 4: \( \frac{90}{126} = \frac{h_1}{h_2} \)
Divide by 3: \( \frac{30}{42} = \frac{h_1}{h_2} \)
Divide by 2: \( \frac{15}{21} = \frac{h_1}{h_2} \)
Divide by 3: \( \frac{5}{7} = \frac{h_1}{h_2} \)

So, the ratio of their heights is \( h_1 : h_2 = 5 : 7 \).
In simple words: When two cones have the same base size, their volumes are directly related to their heights. By comparing their given volumes, we can find the ratio of their heights. We simplify the ratio of volumes to its simplest form.

🎯 Exam Tip: For objects with the same base (like cones or cylinders), the ratio of their volumes is directly proportional to the ratio of their heights. Remember to simplify the ratio to its lowest terms.

Question 7. If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.
Answer:
Given:
The ratio of the radii of two spheres \( r_1 : r_2 = 4 : 7 \).

The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \).

Let \( V_1 \) and \( V_2 \) be the volumes of the two spheres.
So, \( V_1 = \frac{4}{3} \pi r_1^3 \)
And \( V_2 = \frac{4}{3} \pi r_2^3 \)

The ratio of their volumes is:
\( \frac{V_1}{V_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} \)
Since \( \frac{4}{3} \pi \) is common in both numerator and denominator, it cancels out.
\( \frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} \)
\( = \left(\frac{r_1}{r_2}\right)^3 \)

Substitute the given ratio of radii:
\( \frac{V_1}{V_2} = \left(\frac{4}{7}\right)^3 \)
\( = \frac{4^3}{7^3} \)
\( = \frac{4 \times 4 \times 4}{7 \times 7 \times 7} \)
\( = \frac{64}{343} \)

So, the ratio of their volumes is \( V_1 : V_2 = 64 : 343 \).
In simple words: For spheres, if you know the ratio of their radii, you can find the ratio of their volumes by cubing the radius ratio. This is because volume depends on the radius multiplied by itself three times.

🎯 Exam Tip: Remember that for any two similar three-dimensional shapes, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions (like radii, heights, or side lengths).

Question 8. A solid sphere and a solid hemisphere have an equal total surface area. Prove that the ratio of their volume is \( 3\sqrt{3}:4 \).
Answer:
Let \( r_1 \) be the radius of the solid sphere.
Let \( r_2 \) be the radius of the solid hemisphere.

The total surface area of a solid sphere \( = 4 \pi r_1^2 \).
The total surface area of a solid hemisphere \( = 2 \pi r_2^2 + \pi r_2^2 = 3 \pi r_2^2 \) (curved surface area + area of the flat base).

Given that their total surface areas are equal:
\( 4 \pi r_1^2 = 3 \pi r_2^2 \)
Cancel \( \pi \) from both sides:
\( 4 r_1^2 = 3 r_2^2 \)
Rearrange to find the ratio of their radii:
\( \frac{r_1^2}{r_2^2} = \frac{3}{4} \)
Take the square root of both sides:
\( \frac{r_1}{r_2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \)
So, \( r_1 : r_2 = \sqrt{3} : 2 \).

Now, let's find the ratio of their volumes:
Volume of the solid sphere \( (V_1) = \frac{4}{3} \pi r_1^3 \).
Volume of the solid hemisphere \( (V_2) = \frac{2}{3} \pi r_2^3 \).

The ratio of their volumes is:
\( \frac{V_1}{V_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{2}{3} \pi r_2^3} \)
Cancel \( \frac{1}{3} \pi \) from both sides:
\( = \frac{4 r_1^3}{2 r_2^3} \)
\( = 2 \frac{r_1^3}{r_2^3} \)
\( = 2 \left(\frac{r_1}{r_2}\right)^3 \)

Substitute the ratio \( \frac{r_1}{r_2} = \frac{\sqrt{3}}{2} \):
\( = 2 \left(\frac{\sqrt{3}}{2}\right)^3 \)
\( = 2 \times \frac{(\sqrt{3})^3}{2^3} \)
\( = 2 \times \frac{3\sqrt{3}}{8} \)
\( = \frac{6\sqrt{3}}{8} \)
Simplify the ratio by dividing by 2:
\( = \frac{3\sqrt{3}}{4} \)

Thus, the ratio of their volumes \( V_1 : V_2 = 3\sqrt{3} : 4 \).
Hence, it is proved.
In simple words: We first use the fact that their total surface areas are equal to find the relationship between their radii. Then, we use this radius relationship in their volume formulas to calculate the ratio of their volumes, simplifying it to the required form. The flat base of the hemisphere adds to its total surface area, making it different from its curved surface area.

🎯 Exam Tip: Be very careful with the formulas for total surface area and volume of both spheres and hemispheres. A common mistake is to forget the base area for a solid hemisphere's total surface area.

Question 9. The outer and the inner surface areas of a spherical copper shell are \( 576\pi \) cm² and \( 324\pi \) cm² respectively. Find the volume of the material required to make the shell.
Answer:
Given:
Outer surface area of the spherical shell \( = 576\pi \) cm²
Inner surface area of the spherical shell \( = 324\pi \) cm²

Let the outer radius be R and the inner radius be r.

For the outer surface area:
\( 4 \pi R^2 = 576\pi \)
Divide both sides by \( 4\pi \):
\( R^2 = \frac{576}{4} \)
\( R^2 = 144 \)
\( R = \sqrt{144} \)
\( R = 12 \) cm

For the inner surface area:
\( 4 \pi r^2 = 324\pi \)
Divide both sides by \( 4\pi \):
\( r^2 = \frac{324}{4} \)
\( r^2 = 81 \)
\( r = \sqrt{81} \)
\( r = 9 \) cm

The volume of the material required to make the shell is the difference between the volume of the outer sphere and the volume of the inner sphere.
Volume of a spherical shell \( = \frac{4}{3} \pi (R^3 - r^3) \)
\( = \frac{4}{3} \times \frac{22}{7} (12^3 - 9^3) \)
\( = \frac{88}{21} (1728 - 729) \)
\( = \frac{88}{21} \times 999 \)
\( = \frac{88 \times 333}{7} \) (Since \( 999/21 = 333/7 \))
\( = \frac{29304}{7} \)
\( \approx 4186.2857 \) cm³

Rounding to two decimal places, the volume of the material required is \( 4186.29 \) cm³.
In simple words: First, we find the outer radius and inner radius of the shell using their given surface areas. Then, we use these radii to calculate the volume of the whole outer sphere and the empty inner space. The amount of material is the difference between these two volumes.

🎯 Exam Tip: Remember that a spherical shell means a hollow sphere. The volume of material is found by subtracting the inner volume from the outer volume, and not by calculating the volume of a hemisphere.

Question 10. A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of Rs. 40 per litre.
Answer:
Given:
Height of the frustum \( (h) = 16 \) cm
Radius of the upper end \( (R) = 20 \) cm
Radius of the lower end \( (r) = 8 \) cm
Cost of milk \( = \) Rs. 40 per litre

First, calculate the volume of the frustum using its formula:
Volume of a frustum \( = \frac{1}{3} \pi h (R^2 + Rr + r^2) \)
Using \( \pi \approx \frac{22}{7} \):
\( = \frac{1}{3} \times \frac{22}{7} \times 16 ((20)^2 + (20 \times 8) + (8)^2) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 16 (400 + 160 + 64) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 16 (624) \)
\( = \frac{22 \times 16 \times 624}{21} \)
\( = \frac{22 \times 16 \times 208}{7} \) (After dividing 624 by 3)
\( = \frac{73216}{7} \) cm³
\( \approx 10459.428 \) cm³

Next, convert the volume from cubic centimeters to litres:
We know that \( 1000 \) cm³ \( = 1 \) litre.
Volume in litres \( = \frac{10459.428}{1000} \)
\( \approx 10.459 \) litres

Finally, calculate the cost of the milk:
Cost \( = \) Volume in litres \( \times \) Cost per litre
Cost \( = 10.459 \times 40 \)
Cost \( = 418.376 \)

Rounding to two decimal places, the cost of the milk is Rs. 418.38.
In simple words: We use a special formula to find the volume of the frustum container, which is like a cone with the top cut off. Then, we change this volume from cubic centimeters to liters and multiply it by the cost of milk per liter to get the total cost. Frustums are often found in real-world objects like buckets, lamp shades, or drinking glasses.

🎯 Exam Tip: Double-check the frustum volume formula, especially the \( (R^2 + Rr + r^2) \) part. Also, remember the correct conversion factor between cubic centimeters and litres for accurate final cost calculation.