Samacheer Kalvi Class 10 Maths Solutions Chapter 7 Mensuration Exercise 7.3

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.3

Question 1. A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.
Answer: First, let's find the radius. The diameter of the vessel is 14 cm, so the radius \( r \) is half of that, which is 7 cm. This radius is for both the hemisphere and the cylinder. The total height of the vessel is 13 cm. Since the hemisphere has a height equal to its radius (7 cm), the height of the cylindrical part \( h \) is the total height minus the hemisphere's height, so \( 13 - 7 = 6 \) cm. The capacity of the vessel is the sum of the volume of the cylinder and the volume of the hemisphere. To find the capacity, we calculate the volume of the cylinder and the volume of the hemisphere, then add them together.
Volume of the cylinder \( = \pi r^2 h \)
Volume of the hemisphere \( = \frac{2}{3} \pi r^3 \)
Total Capacity \( = \pi r^2 h + \frac{2}{3} \pi r^3 \)
\( = \pi r^2 (h + \frac{2}{3} r) \)
\( = \frac{22}{7} \times 7^2 (6 + \frac{2}{3} \times 7) \)
\( = 22 \times 7 (6 + \frac{14}{3}) \)
\( = 22 \times 7 (\frac{18}{3} + \frac{14}{3}) \)
\( = 22 \times 7 (\frac{18+14}{3}) \)
\( = 22 \times 7 \times \frac{32}{3} \)
\( = \frac{154 \times 32}{3} \)
\( = \frac{4928}{3} \)
\( = 1642.67 \text{ cm}^3 \)
So, the capacity of the vessel is approximately \( 1642.67 \text{ cm}^3 \).
In simple words: The vessel holds water like a cylinder with a bowl on top. We first find the radius and the cylinder's height. Then, we add the volumes of the cylinder and the hemisphere to get the total capacity.

🎯 Exam Tip: Remember that when a hemisphere is mounted on a cylinder, the radius of both shapes is usually the same, and the hemisphere's height contributes to the total height of the object.

Question 2. Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of the model that Nathan made.
Answer: The model is made up of a cylinder with two cones attached at its ends. The diameter of the model is 3 cm, so the radius \( r \) is \( \frac{3}{2} \) cm or 1.5 cm. The total length (height) of the model is 12 cm. Each cone has a height \( H \) of 2 cm. Since there are two cones, one at each end, their combined height is \( 2 + 2 = 4 \) cm. So, the height of the cylindrical part \( h \) is the total length minus the heights of the two cones, which is \( 12 - 4 = 8 \) cm. To find the total volume, we add the volume of the cylinder and the volumes of the two cones.
Volume of the cylinder \( = \pi r^2 h \)
Volume of two cones \( = 2 \times \frac{1}{3} \pi r^2 H \)
Total Volume \( = \pi r^2 h + 2 \times \frac{1}{3} \pi r^2 H \)
\( = \pi r^2 (h + \frac{2}{3} H) \)
\( = \frac{22}{7} \times (1.5)^2 (8 + \frac{2}{3} \times 2) \)
\( = \frac{22}{7} \times \frac{9}{4} (8 + \frac{4}{3}) \)
\( = \frac{22}{7} \times \frac{9}{4} (\frac{24}{3} + \frac{4}{3}) \)
\( = \frac{22}{7} \times \frac{9}{4} \times \frac{28}{3} \)
\( = 22 \times \frac{9}{4} \times \frac{4}{3} \) (because \( \frac{28}{7} = 4 \))
\( = 22 \times 3 \times 2 \) (because \( \frac{9}{3} = 3 \) and \( \frac{4}{4} = 1 \))
\( = 66 \text{ cm}^3 \)
So, the volume of the model Nathan made is \( 66 \text{ cm}^3 \).
In simple words: The model looks like a pill, with a cylinder in the middle and two cones on the ends. We find the cylinder's height by subtracting the cone heights from the total. Then, we add the volume of the cylinder and both cones to get the total volume.

🎯 Exam Tip: Always draw a simple diagram for combined shapes to visualize the dimensions and ensure you correctly identify the height of each component part.

Question 3. From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm³.
Answer: We start with a solid cylinder. Its height \( h \) is 2.4 cm and its diameter is 1.4 cm. This means its radius \( r \) is half of the diameter, so \( \frac{1.4}{2} = 0.7 \) cm. A cone with the exact same height and diameter is carved out from this cylinder. This means the cone also has a height of 2.4 cm and a radius of 0.7 cm. To find the volume of the remaining solid, we subtract the volume of the carved-out cone from the volume of the original cylinder. Carving out a shape means removing its volume.
Volume of the cylinder \( = \pi r^2 h \)
Volume of the cone \( = \frac{1}{3} \pi r^2 h \)
Volume of the remaining solid \( = \text{Volume of cylinder} - \text{Volume of cone} \)
\( = \pi r^2 h - \frac{1}{3} \pi r^2 h \)
\( = \pi r^2 h (1 - \frac{1}{3}) \)
\( = \frac{2}{3} \pi r^2 h \)
Now, we plug in the values:
\( = \frac{2}{3} \times \frac{22}{7} \times (0.7)^2 \times 2.4 \)
\( = \frac{2}{3} \times \frac{22}{7} \times 0.49 \times 2.4 \)
\( = \frac{2}{3} \times \frac{22}{7} \times \frac{49}{100} \times \frac{24}{10} \)
\( = \frac{2 \times 22 \times 7 \times 24}{3 \times 7 \times 1000} \)
\( = \frac{2 \times 22 \times 7 \times 8}{1000} \) (after cancelling 3 and 24)
\( = \frac{2464}{1000} \)
\( = 2.464 \text{ cm}^3 \)
Rounding to the nearest cm³, the volume of the remaining solid is \( 2.46 \text{ cm}^3 \).
In simple words: We have a cylinder, and we scoop out a cone from it. Both shapes have the same size base and height. To find what's left, we take the cylinder's volume and subtract the cone's volume.

🎯 Exam Tip: When a shape is 'carved out' or 'removed', it means you should subtract its volume from the original solid. Look for common factors in the volume formulas to simplify calculations.

Question 4. A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.
Answer: When an object is placed in water, the volume of water displaced is equal to the volume of the object submerged in the water. In this case, the entire solid (cone on a hemisphere) is placed in a cylinder full of water, so the displaced water volume equals the total volume of the solid. The solid is made of a cone and a hemisphere. The cone has a height \( h \) of 12 cm and a radius \( r \) of 6 cm. The hemisphere also has a radius \( r \) of 6 cm. The total height of the combined solid is the height of the cone plus the radius of the hemisphere, which is \( 12 + 6 = 18 \) cm. This matches the height of the cylinder, meaning the solid fits perfectly inside. The radius of the cylinder is also 6 cm, matching the solid. Therefore, the volume of water displaced is the volume of the solid.
Volume of the cone \( = \frac{1}{3} \pi r^2 h \)
Volume of the hemisphere \( = \frac{2}{3} \pi r^3 \)
Volume of the solid \( = \text{Volume of cone} + \text{Volume of hemisphere} \)
\( = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 \)
\( = \frac{1}{3} \pi r^2 (h + 2r) \)
Now, substitute the given values:
\( = \frac{1}{3} \times \frac{22}{7} \times 6^2 (12 + 2 \times 6) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 36 (12 + 12) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 36 \times 24 \)
\( = \frac{22}{7} \times 12 \times 24 \) (after cancelling 3 with 36)
\( = \frac{22 \times 288}{7} \)
\( = \frac{6336}{7} \)
\( \approx 905.14 \text{ cm}^3 \)
The volume of the water displaced is approximately \( 905.14 \text{ cm}^3 \).
In simple words: When we put a shape into water, the water that spills out is exactly the same amount as the space the shape takes up. Here, the shape is like an ice cream cone on a half-ball. We find the total volume of this shape by adding the cone's volume and the half-ball's volume. This total volume is how much water is displaced.

🎯 Exam Tip: Remember Archimedes' principle: the volume of water displaced is always equal to the volume of the submerged object. Check if the solid's dimensions allow it to be fully submerged in the cylinder.

Question 5. A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?
Answer: The capsule is a combination of a cylinder and two hemispheres at its ends. The amount of medicine it can hold is its total volume. The diameter of the capsule is 3 mm, so its radius \( r \) is \( \frac{3}{2} \) mm or 1.5 mm. This radius applies to both the cylinder and the hemispheres. The total length of the capsule is 12 mm. Each hemisphere has a "height" equal to its radius (1.5 mm). So, the height of the cylindrical part \( h \) is the total length minus the radii of the two hemispheres: \( 12 - (1.5 + 1.5) = 12 - 3 = 9 \) mm. Two hemispheres combine to form one full sphere.
Volume of the cylinder \( = \pi r^2 h \)
Volume of two hemispheres (one sphere) \( = \frac{4}{3} \pi r^3 \)
Total Volume of the capsule \( = \pi r^2 h + \frac{4}{3} \pi r^3 \)
\( = \pi r^2 (h + \frac{4}{3} r) \)
Now, substitute the values:
\( = \frac{22}{7} \times (1.5)^2 (9 + \frac{4}{3} \times 1.5) \)
\( = \frac{22}{7} \times 2.25 (9 + 2) \)
\( = \frac{22}{7} \times 2.25 \times 11 \)
\( = \frac{22 \times 2.25 \times 11}{7} \)
\( = \frac{544.5}{7} \)
\( \approx 77.7857 \text{ mm}^3 \)
Rounding to one decimal place, the capsule can hold approximately \( 77.8 \text{ mm}^3 \) of medicine.
In simple words: The capsule looks like a cylinder with a half-ball on each end. To find how much medicine it holds, we add the volume of the cylinder and the volume of both half-balls (which is the same as one whole ball). First, we figure out the cylinder's height by subtracting the half-ball parts from the total length.

🎯 Exam Tip: When calculating the volume of a combined solid, ensure you correctly determine the individual heights or lengths of each component part, especially for the central cylindrical section.

Question 6. As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.
Answer: We have a cubical block with a side length \( a = 7 \) cm. A hemisphere is placed on top of it, "surmounting" it. This means the diameter of the hemisphere is equal to the side of the cube, so its diameter is 7 cm. The radius \( r \) of the hemisphere is therefore \( \frac{7}{2} \) cm or 3.5 cm. To find the total surface area of this new solid, we need to consider the visible parts. It includes the total surface area of the cube, minus the area of the top face where the hemisphere sits, plus the curved surface area of the hemisphere. Another way to think about it is the surface area of 5 faces of the cube, plus the area of the top face that is not covered by the hemisphere, plus the curved surface area of the hemisphere. It's often simpler to use the formula: (Total Surface Area of Cube) + (Curved Surface Area of Hemisphere) - (Area of the base of the hemisphere). This is because the base of the hemisphere covers a part of the cube's top surface, so that area is not exposed.
Total Surface Area of cube \( = 6a^2 \)
Curved Surface Area of hemisphere \( = 2\pi r^2 \)
Area of base of hemisphere \( = \pi r^2 \)
Surface area of the solid \( = 6a^2 + 2\pi r^2 - \pi r^2 \)
\( = 6a^2 + \pi r^2 \)
Substitute the values:
\( = 6 \times (7)^2 + \frac{22}{7} \times (\frac{7}{2})^2 \)
\( = 6 \times 49 + \frac{22}{7} \times \frac{49}{4} \)
\( = 294 + \frac{22 \times 7}{4} \)
\( = 294 + \frac{154}{4} \)
\( = 294 + 38.5 \)
\( = 332.5 \text{ cm}^2 \)
The surface area of the solid is \( 332.5 \text{ cm}^2 \).
In simple words: Imagine a box with a half-ball stuck on top. We want to paint the outside. We paint all sides of the box, but on the top, we only paint the part around the half-ball, not under it. Then we also paint the curved part of the half-ball. We use a formula that adds the area of the box (mostly) to the curved area of the half-ball, subtracting where they join.

🎯 Exam Tip: When combining solids for surface area calculations, remember to subtract any areas that are hidden or covered by the joining of the two shapes.

Question 7. A right circular cylinder just encloses a sphere of radius r units. Calculate
(i) the surface area of the sphere
(ii) the curved surface area of the cylinder
(iii) the ratio of the areas obtained in (i) and (ii).
Answer: A right circular cylinder "just encloses" a sphere. This means the sphere touches the top, bottom, and sides of the cylinder. Therefore, the radius of the cylinder is the same as the radius of the sphere, \( r \). Also, the height of the cylinder \( h \) must be equal to the diameter of the sphere, which is \( 2r \).
(i) The surface area of a sphere with radius \( r \) is a standard formula.
Surface area of sphere \( = 4\pi r^2 \) sq. units
(ii) The curved surface area (CSA) of a cylinder is given by \( 2\pi r h \). Since the cylinder just encloses the sphere, the height \( h \) of the cylinder is equal to the diameter of the sphere, which is \( 2r \).
Curved Surface Area of cylinder \( = 2\pi r h \)
\( = 2\pi r (2r) \)
\( = 4\pi r^2 \) sq. units
(iii) To find the ratio, we compare the surface area of the sphere to the curved surface area of the cylinder.
Ratio \( = \frac{\text{Surface area of sphere}}{\text{Curved surface area of cylinder}} \)
\( = \frac{4\pi r^2}{4\pi r^2} \)
\( = 1:1 \)
The ratio of the surface area of the sphere to the curved surface area of the cylinder is 1:1. This shows a special relationship between these two shapes when one perfectly encloses the other.
In simple words: If a sphere fits perfectly inside a cylinder, touching all sides, top, and bottom, then the cylinder's radius is the same as the sphere's, and the cylinder's height is twice the sphere's radius. We find the area of the sphere and the curved area of the cylinder. When we compare them, they are exactly the same, so their ratio is 1 to 1.

🎯 Exam Tip: When a shape 'just encloses' another, it implies specific relationships between their dimensions (e.g., radius and height). Make sure to use these relationships correctly in your formulas.

Question 8. A shuttlecock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the shuttlecock is 7 cm. Find its external surface area.
Answer: The shuttlecock is made of two parts: a frustum of a cone and a hemisphere. The hemisphere is mounted at the bottom of the frustum. The smaller diameter of the frustum's base is 2 cm, which means the radius of the hemisphere \( r_H \) is \( \frac{2}{2} = 1 \) cm. The height of this hemisphere is also its radius, 1 cm. The larger diameter of the frustum is 5 cm, so its larger radius \( R \) is \( \frac{5}{2} = 2.5 \) cm. The smaller radius of the frustum \( r_F \) is 1 cm. The total height of the shuttlecock is 7 cm. So, the height of the frustum \( h_F \) is the total height minus the hemisphere's height: \( 7 - 1 = 6 \) cm. To find the external surface area, we need to add the curved surface area of the frustum and the curved surface area of the hemisphere. We do not include the base of the frustum or the top of the hemisphere because they are joined together inside the shuttlecock.
First, calculate the slant height \( l \) of the frustum:
\( l = \sqrt{(R - r_F)^2 + h_F^2} \)
\( = \sqrt{(2.5 - 1)^2 + 6^2} \)
\( = \sqrt{(1.5)^2 + 36} \)
\( = \sqrt{2.25 + 36} \)
\( = \sqrt{38.25} \)
\( \approx 6.185 \text{ cm} \)
Now, calculate the Curved Surface Area (CSA) of the frustum:
CSA of frustum \( = \pi (R + r_F) l \)
\( = \frac{22}{7} (2.5 + 1) \times 6.185 \)
\( = \frac{22}{7} \times 3.5 \times 6.185 \)
\( = 22 \times 0.5 \times 6.185 \)
\( = 11 \times 6.185 \)
\( \approx 67.098 \text{ cm}^2 \) (Rounded to 67.1 cm² as in source)
Next, calculate the Curved Surface Area (CSA) of the hemisphere:
CSA of hemisphere \( = 2\pi r_H^2 \)
\( = 2 \times \frac{22}{7} \times (1)^2 \)
\( = \frac{44}{7} \)
\( \approx 6.285 \text{ cm}^2 \) (Rounded to 6.28 cm² as in source)
Finally, calculate the Total External Surface Area:
Total External Surface Area \( = \text{CSA of frustum} + \text{CSA of hemisphere} \)
\( = 67.1 + 6.28 \)
\( = 73.38 \text{ cm}^2 \)
Rounding to two decimal places, the external surface area of the shuttlecock is approximately \( 73.39 \text{ cm}^2 \).
In simple words: The shuttlecock is shaped like a cone with its top cut off (a frustum) sitting on a half-ball. To find its outer painted area, we add the curved surface area of the cone part and the curved surface area of the half-ball part. We first need to find the slant height of the frustum using its heights and radii.

🎯 Exam Tip: For external surface area, remember to only include the parts that are exposed to the outside. Internal joining surfaces are not counted. Pay close attention to distinguishing between diameters and radii.