Surface Areas and Volumes — Class 10 Maths Notes

Surface Areas and Volumes · Class 10 Maths · 18 topics.

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Topics covered in Surface Areas and Volumes

  1. 1.Introduction of Surface Areas & Volumes

    Surface areas and volumes are important concepts in mathematics, especially in geometry. They help us understand how much space an object occupies (volume) and how much material is needed to cover it (surface area). Let's explore these concepts with examples and activities.

    1. Surface Area: This is the total area that the surface of an object covers. It's like measuring how much wrapping paper you need to cover a gift box.

      • Example: Consider a cube with each side measuring 2 cm. To find the surface area, you calculate the area of each face (side × side) and then add them up. Since a cube has 6 faces, the surface area is 6 times the area of one face.

    2. Volume: This represents the space an object occupies. It's like figuring out how much water you can pour into a container without spilling.

      • Example: For the same cube with a side of 2 cm, the volume is calculated by multiplying the length, width, and height. So, it's side × side × side.

    Real-Life Application:

    • Surface Area: In real life, you use surface area when you need to paint walls, wrap gifts, or even when baking (like rolling dough to a certain size).
    • Career Connection: Architects, engineers, and designers often work with surface area calculations.

    Volume:

    • Real-Life Application: Cooking often involves measuring volume, like the amount of water in a pot. Even when you fill a fuel tank, you're dealing with volume.
    • Career Connection: Chefs, scientists, and engineers frequently calculate volumes.

    Activity to Try:

    • Take a small box at home. Measure its length, width, and height. Now, try to calculate its surface area and volume. This will help you understand these concepts better.

    Use in Career:

    • If you become an architect, engineer, or designer, you'll often use these concepts to design buildings, products, and even in graphic design.

  2. 2.Surface Area of a Combination of Solids

    Calculating the surface area of a combination of solids involves adding up the surface areas of each individual solid in the combination. Let's break down this concept with an example and a simple activity.

    Example: Imagine you have a solid that is made up of a cylinder and a cone, where the base of the cone is attached to one end of the cylinder.

    1. Calculate the Surface Area of the Cylinder:

      • If the cylinder has a radius r and height ℎh, its surface area is the sum of the area of its two circular bases and the area of its curved surface.
      • The area of one base is 2πr2, and since there are two bases, their combined area is 222πr2.
      • The area of the curved surface is the circumference of the base (which is 22πr) times the height ℎh, so 2ℎ2πrh.
      • Therefore, the total surface area of the cylinder is 22+2ℎ2πr2+2πrh.

    2. Calculate the Surface Area of the Cone:

      • If the cone has the same radius r and a slant height l, its surface area is the sum of the area of its base and its curved surface.
      • The base area is 2πr2, and the curved surface area is πrl.
      • So, the total surface area of the cone is 2+πr2+πrl.

    3. Combine the Surface Areas:

      • Since the base of the cone is attached to the cylinder, that area is not exposed and should not be counted twice.
      • Therefore, the combined surface area is the sum of the cylinder's surface area and the curved surface area of the cone: 22+2ℎ+2πr2+2πrh+πrl.

    Activity to Try:

    • Take a cylindrical container and a cone (like a party hat). Measure their dimensions - radius and height for the cylinder, and radius and slant height for the cone. Now, try calculating the surface area for each and then add them together, remembering not to double-count the base of the cone that's attached to the cylinder.

    Real-Life Application:

    • In real life, this concept is used in designing and manufacturing various objects, such as containers, rockets, and architectural structures.

    Career Connection:

    • This knowledge is particularly useful in fields like mechanical engineering, architecture, and product design.

  3. 3.The decorative block shown (in given Fig. ) is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.

    To find the total surface area of the decorative block that consists of a cube and a hemisphere, we can follow a few simple steps. Let's solve this together: 1. Calculate the Total Surface Area of the Cube:

      • The formula is 6×edge26×edge2.
      • With an edge of 5 cm5 cm, the surface area of the cube is 6×5×5=150 cm26×5×5=150 cm2. 2. Adjust for the Area Covered by the Hemisphere on the Cube:
      • The hemisphere sits on the cube, covering a circular area.
      • This area should be subtracted from the cube's surface area.
      • The area of this circle (base of the hemisphere) is 2πr2, where r is the radius of the hemisphere (4.2 cm4.2 cm diameter, so 2.1 cm2.1 cm radius).

    3. 3. Add the Curved Surface Area of the Hemisphere:

      • The formula for the curved surface area (CSA) of a hemisphere is 222πr2.
      • Add this to the adjusted surface area of the cube.

    4. 4. Calculate the Total Surface Area:

      • Total surface area = Surface area of cube - Area of the hemisphere base + Curved surface area of the hemisphere.
      • Using =227π=722​, the calculation is 150−×(2.1)2+2×(2.1)2150−π×(2.1)2+2π×(2.1)2.
      • This results in 150+13.86=163.86 cm2150+13.86=163.86 cm2.

    So, the total surface area of the block is 163.86 cm2163.86 cm2.

  4. 4.2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

    To solve this, let's go step by step:

    Step 1: Determine the Side Length of Each Cube

    The volume of a cube is given by the formula Volume=side3Volume=side3. Given that each cube has a volume of 64 cm³, we find the side length by taking the cube root of 64.

    Step 2: Calculate the Dimensions of the Resulting Cuboid

    When two cubes are joined end to end, the length of the resulting cuboid becomes the sum of the side lengths of the two cubes, while the width and height remain the same as a single cube.

    Step 3: Find the Surface Area of the Cuboid

    The surface area of a cuboid is calculated using the formula Surface Area=2(+ℎ+ℎ)Surface Area=2(lw+lh+wh), where l, w, and ℎh are the length, width, and height of the cuboid, respectively.

    Let's calculate the side length of each cube and then find the surface area of the resulting cuboid.

    The surface area of the resulting cuboid, formed by joining two cubes (each with a volume of 64 cm³) end to end, is approximately 160 cm2160 cm2.

    Here's the step-by-step breakdown of the calculation:

    1. Determine the Side Length of Each Cube:

      • Given volume =64 cm3V=64 cm3, the side length a of each cube is found using =3V=a3. The cube root of 64 is 4, so =4 cma=4 cm.

    2. Calculate the Dimensions of the Resulting Cuboid:

      • When two cubes are joined end to end, the length of the resulting cuboid is the sum of the side lengths of the two cubes, =2=8 cmL=2a=8 cm. The width W and height H are the same as the side of the cube, 4 cm4 cm.

    3. Find the Surface Area of the Cuboid:

      • The surface area is calculated using Surface Area=2(+ℎ+ℎ)Surface Area=2(lw+lh+wh), which gives 2(8×4+8×4+4×4)=160 cm22(8×4+8×4+4×4)=160 cm2

  5. 5.A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.


    To find the inner surface area of a vessel made up of a hollow hemisphere and a hollow cylinder, we need to calculate the surface areas of both parts and add them together. Let's go through the steps:

    Step 1: Calculate the Inner Surface Area of the Hollow Hemisphere

    • The inner surface area of a hemisphere is given by the formula 222πr2, where r is the radius of the hemisphere.
    • The diameter of the hemisphere is 14 cm, so the radius r is half of that, which is 7 cm (=142r=214​).
    • So, the inner surface area of the hemisphere is 2×722π×72.

    Step 2: Calculate the Inner Surface Area of the Hollow Cylinder

    • The inner surface area of a cylinder includes only the lateral surface area, given by 2ℎ2πrh, where r is the radius and ℎh is the height of the cylinder.
    • The height of the cylinder is the total height of the vessel minus the radius of the hemisphere. The total height of the vessel is 13 cm, and the radius of the hemisphere is 7 cm, so the height of the cylinder is 13−7=613−7=6 cm.
    • Thus, the inner surface area of the cylinder is 2×7×62π×7×6.

    Step 3: Add the Surface Areas of Both Parts

    • The total inner surface area of the vessel is the sum of the inner surface areas of the hemisphere and the cylinder.

    Let's perform the calculations:

    The total inner surface area of the vessel, which is composed of a hollow hemisphere and a hollow cylinder, is approximately 571.77 cm2571.77 cm2. This calculation is the sum of the inner surface areas of both the hemisphere and the cylinder

  6. 6.A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.


    To find the total surface area of the toy, which consists of a cone mounted on a hemisphere both having the same radius, we can follow these steps:

    Step 1: Calculate the Surface Area of the Hemisphere

    • The formula for the surface area of a hemisphere is 222πr2, where r is the radius.
    • Given the radius of the hemisphere is 3.5 cm, the surface area of the hemisphere is 2×3.522π×3.52.

    Step 2: Calculate the Surface Area of the Cone

    • For the cone, we only need to calculate the lateral (or slant) surface area because the base of the cone is not exposed (it's attached to the hemisphere). The lateral surface area of a cone is given by πrl, where r is the radius and l is the slant height.
    • The slant height l can be found using the Pythagorean theorem in the right triangle formed by the radius, height, and slant height of the cone. The total height of the toy is 15.5 cm, which includes the radius of the hemisphere, so the height ℎh of the cone is 15.5−3.5=12 cm15.5−3.5=12 cm. Then =3.52+122l=3.52+122​.
    • The lateral surface area of the cone is ×3.5×π×3.5×l.

    Step 3: Add the Surface Areas of Both Parts

    • The total surface area of the toy = Surface area of the hemisphere + Lateral surface area of the cone.

    Let's calculate these values.

    The total surface area of the toy, which is a combination of a cone mounted on a hemisphere (both having the same radius of 3.5 cm), is approximately 214.41 cm2214.41 cm2.

  7. 7.A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

    To determine the greatest diameter that the hemisphere can have when surmounted on a cubical block with a side of 7 cm, we need to consider the width of the cube. The hemisphere will sit on top of the cube, and its diameter cannot exceed the width of the cube, otherwise, it would extend beyond the cube's sides.

    Greatest Diameter of the Hemisphere

    • Since the cube has a side length of 7 cm, the greatest diameter of the hemisphere is also 7 cm. This is because the diameter of the hemisphere must be equal to or less than the width of the cube to fit properly on it.

    Surface Area of the Solid

    The total surface area of the solid (the cubical block with the hemisphere surmounted on it) is the sum of:

    1. The surface area of the cube, excluding the face on which the hemisphere sits.
    2. The surface area of the hemisphere.

    Surface Area of the Cube

    • The cube has 6 faces, each with an area of 7 cm×7 cm7 cm×7 cm.
    • However, the area of the top face, where the hemisphere sits, should be excluded. So, the surface area of the cube to be considered is the area of 5 faces.
    • Surface area of 5 faces of the cube = 5×7×7 cm25×7×7 cm2.

    Surface Area of the Hemisphere

    • The surface area of a hemisphere is given by 222πr2.
    • The radius of the hemisphere is half its diameter, which is 7227​ cm.
    • So, the surface area of the hemisphere is 2×(72)2 cm22π×(27​)2 cm2.

    Let's calculate the total surface area of the solid.

    The greatest diameter the hemisphere can have when surmounted on a cubical block with a side of 7 cm is 7 cm. The total surface area of the solid (the cube with the hemisphere surmounted on it) is approximately 322.0 cm2322.0 cm2.

  8. 8.A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter L of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

    To determine the surface area of the remaining solid after a hemispherical depression is cut out from one face of a cubical wooden block, we need to follow these steps:

    Step 1: Calculate the Surface Area of the Cube Excluding the Face Where the Hemisphere is Cut Out

    • A cube has six faces, and the surface area of each face is given by side2side2.
    • Since the depression is cut out from one face, we only count the surface area of the remaining five faces.
    • If the edge of the cube is l, then the surface area of the cube excluding one face is 5×25×l2.

    Step 2: Calculate the Curved Surface Area of the Hemisphere

    • The curved surface area of a hemisphere is given by 222πr2, where r is the radius.
    • Since the diameter of the hemisphere is equal to the edge of the cube, =2r=2l​.
    • So, the curved surface area of the hemisphere is 2×(2)22π×(2l​)2.

    Step 3: Calculate the Total Surface Area of the Remaining Solid

    • The total surface area of the remaining solid is the sum of the surface area of the cube (excluding the face with the hemisphere) and the curved surface area of the hemisphere.

    Calculation

    Let's assume the edge of the cube l is a known value. We will calculate the total surface area using these steps:

    1. Surface Area of the Cube Excluding One Face: 5×25×l2

    2. Curved Surface Area of the Hemisphere: 2×(2)22π×(2l​)2

    3. Total Surface Area of the Remaining Solid: 5×2+2×(2)25×l2+2π×(2l​)2

    Let's perform the calculation using a specific value of l.

    The total surface area of the remaining solid, after a hemispherical depression is cut out from one face of a cubical wooden block with an edge length of 7 cm, is approximately 321.97 cm2321.97 cm2.

    The calculation steps are as follows:

    1. Surface Area of the Cube Excluding One Face: 5×72=245 cm25×72=245 cm2

    2. Curved Surface Area of the Hemisphere: 2×(72)2≈76.97 cm22π×(27​)2≈76.97 cm2

    3. Total Surface Area of the Remaining Solid: 245 cm2+76.97 cm2≈321.97 cm2245 cm2+76.97 cm2≈321.97 cm2

    This approach calculates the surface area of the remaining parts of the cube and the exposed hemispherical surface

  9. 9.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.



    To find the surface area of a capsule, which is a combination of a cylinder and two hemispheres on either end, we can follow these steps:

    Curved Surface Area of the Cylinder (CSA)

    The curved surface area of the cylindrical part of the capsule is given by the formula: CSAcylinder=2ℎCSAcylinder​=2πrh where r is the radius and ℎh is the height of the cylindrical part.

    The radius r of the capsule is half of the diameter, so: =5 mm2=2.5 mmr=25 mm​=2.5 mm

    The height ℎh of the cylindrical part is the total length of the capsule minus the diameter of one of the hemispheres (since there are two hemispheres, one diameter is subtracted, not two): ℎ=14 mm−5 mm=9 mmh=14 mm−5 mm=9 mm

    Now, we calculate the CSA of the cylinder: CSAcylinder=2××2.5 mm×9 mmCSAcylinder​=2×π×2.5 mm×9 mm CSAcylinder=2××22.5 mm2CSAcylinder​=2×π×22.5 mm2 CSAcylinder=45 mm2CSAcylinder​=45π mm2

    Curved Surface Area of One Hemisphere (CSA)

    The curved surface area of one hemisphere is given by: CSAhemisphere=22CSAhemisphere​=2πr2

    Calculating this for one hemisphere: CSAhemisphere=2××(2.5 mm)2CSAhemisphere​=2×π×(2.5 mm)2 CSAhemisphere=2××6.25 mm2CSAhemisphere​=2×π×6.25 mm2 CSAhemisphere=12.5 mm2CSAhemisphere​=12.5π mm2

    Since there are two hemispheres, we multiply this by 2: Total CSAhemispheres=2×12.5 mm2Total CSAhemispheres​=2×12.5π mm2 Total CSAhemispheres=25 mm2Total CSAhemispheres​=25π mm2

    Total Surface Area of the Capsule

    Adding the curved surface area of the cylinder and the total curved surface area of the two hemispheres gives us the total surface area of the capsule: Total Surface Area=CSAcylinder+Total CSAhemispheresTotal Surface Area=CSAcylinder​+Total CSAhemispheres​ Total Surface Area=45 mm2+25 mm2Total Surface Area=45π mm2+25π mm2 Total Surface Area=70 mm2Total Surface Area=70π mm2

    Using the value of ≈3.14159π≈3.14159, we can now find the numerical value: Total Surface Area≈70×3.14159 mm2Total Surface Area≈70×3.14159 mm2 Total Surface Area≈219.91 mm2Total Surface Area≈219.91 mm2

    So, the total surface area of the medicine capsule is approximately 219.91 mm2219.91 mm2.

  10. 10.A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, (as shown in Fig). If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.


    To find the total surface area of the wooden article, which is a cylinder with hemispherical depressions scooped out from both ends, we need to calculate the curved surface area of the cylinder and add it to the curved surface area of the hemispheres. Since the hemispheres are scooped out, the base area of the hemispheres will not be included in the total surface area. Here's how we can calculate it:

    Step 1: Curved Surface Area of the Cylinder (CSA of Cylinder)

    The curved surface area of the cylinder is calculated using the formula: CSA of Cylinder=2ℎCSA of Cylinder=2πrh where r is the radius and ℎh is the height of the cylinder. For this cylinder: =3.5 cmr=3.5 cm ℎ=10 cmh=10 cm So, the CSA of the cylinder is: CSA of Cylinder=2×3.5×10CSA of Cylinder=2π×3.5×10

    Step 2: Curved Surface Area of the Hemispheres (CSA of Hemispheres)

    The curved surface area of a hemisphere is given by the formula: CSA of Hemisphere=22CSA of Hemisphere=2πr2 However, we have two hemispheres, and we will calculate the total curved surface area for both: Total CSA of Hemispheres=2×(CSA of Hemisphere)Total CSA of Hemispheres=2×(CSA of Hemisphere) Total CSA of Hemispheres=2×(22)Total CSA of Hemispheres=2×(2πr2)

    Step 3: Total Surface Area of the Article

    The total surface area of the article is the sum of the CSA of the cylinder and the total CSA of the hemispheres. The bases of the hemispheres are part of the cylinder's surface area and thus not included in the final calculation. Therefore: Total Surface Area=CSA of Cylinder+Total CSA of HemispheresTotal Surface Area=CSA of Cylinder+Total CSA of Hemispheres

    Now let's calculate these values.

    The total surface area of the wooden article, which is a cylinder with hemispherical depressions scooped out from both ends, is approximately 373.85 cm2373.85 cm2.

    Here is the detailed calculation:

    1. Curved Surface Area of the Cylinder: CSA of Cylinder=2×3.5×10CSA of Cylinder=2π×3.5×10 CSA of Cylinder=70 cm2CSA of Cylinder=70π cm2

    2. Curved Surface Area of One Hemisphere: CSA of Hemisphere=2×(3.5)2CSA of Hemisphere=2π×(3.5)2 CSA of Hemisphere=2×12.25CSA of Hemisphere=2π×12.25 CSA of Hemisphere=24.5 cm2CSA of Hemisphere=24.5π cm2

    3. Total Curved Surface Area of Both Hemispheres: Total CSA of Hemispheres=2×24.5Total CSA of Hemispheres=2×24.5π Total CSA of Hemispheres=49 cm2Total CSA of Hemispheres=49π cm2

    4. Total Surface Area of the Article: Total Surface Area=70+49Total Surface Area=70π+49π Total Surface Area=119 cm2Total Surface Area=119π cm2 Total Surface Area≈373.85 cm2Total Surface Area≈373.85 cm2

    This calculation includes the curved surface area of the cylinder and the outer curved surface area of the two hemispheres. The flat circular bases of the hemispheres are not included in the total surface area since they are part of the surface area of the cylinder and are removed when the hemispheres are scooped out.

  11. 11.Volume of a Combination of Solids

    Let's talk about the volume of a combination of solids in a way that's easy to understand with some examples and how it can be useful in real life.

    Understanding Volume of Combined Solids

    First, what are solids? Solids are three-dimensional objects like cubes, cylinders, spheres, etc. Each of these has a volume, which is the amount of space it occupies.

    Now, imagine you have a toy that's made by combining different shapes, like a cylinder and a cone. The volume of this combined solid is simply the sum of the volumes of the individual shapes.

    How to Calculate

    1. Find the volume of each individual shape: Use the formulas for each shape. For example, the volume of a cylinder is 2ℎπr2h (where r is the radius and ℎh is the height) and the volume of a cone is 132ℎ31​πr2h.

    2. Add them up: Simply add the volumes of each shape to get the total volume of the combined solid.

    Real-Life Example

    Imagine you're baking a cake that's shaped like a cylinder with a cone on top (like a hat). To know how much cake mix you need, you calculate the volume of the cylinder and the cone separately and then add them together.

    Application in Real Life and Careers

    This concept is used in many fields:

    • Architecture and Engineering: Designing buildings and structures.
    • Manufacturing: Creating products with different parts.
    • Cooking and Baking: Determining quantities for recipes.

  12. 12.A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π .

    To find the volume of a solid that is a combination of a cone and a hemisphere, we need to calculate the volume of each part separately and then add them together.

    Given:

    1. Radius of both the cone and the hemisphere (r) = 1 cm.
    2. Height of the cone (h) = Radius of the cone = 1 cm.
    3. π is taken as 227722​.

    Volume of the Hemisphere:

    The formula for the volume of a hemisphere is 23332​πr3. Substituting =1r=1 cm and =227π=722​, Volume of Hemisphere=23×227×13=23×227×1Volume of Hemisphere=32​×722​×13=32​×722​×1

    Volume of the Cone:

    The formula for the volume of a cone is 132ℎ31​πr2h. Substituting =1r=1 cm, ℎ=1h=1 cm, and =227π=722​, Volume of Cone=13×227×12×1=13×227×1Volume of Cone=31​×722​×12×1=31​×722​×1

    Total Volume:

    To find the total volume of the solid, we add the volumes of the hemisphere and the cone. Total Volume=Volume of Hemisphere+Volume of ConeTotal Volume=Volume of Hemisphere+Volume of Cone

    Let's calculate these values.

    The volumes of the individual parts and the total volume of the solid are:

    1. Volume of the Hemisphere: 2.0952.095 cubic cm
    2. Volume of the Cone: 1.0481.048 cubic cm
    3. Total Volume of the Solid: 3.1433.143 cubic cm (approximately)

    So, the volume of the solid which is a combination of a cone standing on a hemisphere, with both having radii of 1 cm and the cone's height also being 1 cm, is approximately 3.1433.143 cubic cm

  13. 13.Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find th

    To find the volume of air contained in the model that Rachel made, we need to calculate the volume of the entire model (including the cylinder and the two cones) and then subtract the volume of the aluminium sheet used to make the model. Since the aluminium sheet is thin, we can assume the outer and inner dimensions of the model are nearly the same.

    Given:

    1. Diameter of the model = 3 cm, so the radius (r) = Diameter / 2 = 1.5 cm.
    2. Total length of the model (including the cylinder and the two cones) = 12 cm.
    3. Height of each cone (h) = 2 cm.

    Calculating the Volume:

    1. Volume of the Cylinder:

      • The length of the cylinder is the total length of the model minus the heights of the two cones.
      • Length of the cylinder = Total length - 2 × Height of one cone = 12 cm - 2 × 2 cm = 8 cm.
      • Volume of the cylinder = πr² × Length = π × (1.5 cm)² × 8 cm.

    2. Volume of the Cones:

      • Each cone has a height of 2 cm, and there are two cones.
      • Volume of one cone = 1331​πr²h.
      • Total volume of the two cones = 2 × Volume of one cone = 2 × 1331​π × (1.5 cm)² × 2 cm.

    3. Total Volume of the Model:

      • Total Volume = Volume of the Cylinder + Volume of the Two Cones.

    Let's calculate the volume of the cylinder, the cones, and the total volume of the model.

    The volumes of the different parts of Rachel's model and the total volume of the model are calculated as follows:

    1. Volume of the Cylinder: 56.5756.57 cubic cm (approximately)
    2. Volume of the Two Cones: 9.439.43 cubic cm (approximately)
    3. Total Volume of the Model: 66.0066.00 cubic cm (approximately)

    Therefore, the volume of air contained in the model that Rachel made is approximately 66.0066.00 cubic cm. This volume represents the space inside the model, assuming the aluminium sheet used to construct it is thin enough to not significantly affect the internal dimensions.

  14. 14.A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.


    To find out how much syrup is in 45 gulab jamuns, we first need to calculate the volume of one gulab jamun. Each gulab jamun is shaped like a cylinder with two hemispherical ends.

    Given:

    1. Length of the cylindrical part of each gulab jamun = 5 cm.
    2. Diameter of each gulab jamun = 2.8 cm, so the radius (r) = Diameter / 2 = 1.4 cm.
    3. Each gulab jamun contains syrup up to about 30% of its volume.

    Calculating the Volume of One Gulab Jamun:

    1. Volume of the Cylindrical Part:

      • Volume of the cylinder = πr² × Length = π × (1.4 cm)² × 5 cm.

    2. Volume of the Hemispherical Ends:

      • There are two hemispherical ends, each with a radius of 1.4 cm.
      • Volume of one hemisphere = 2332​πr³.
      • Total volume of the two hemispheres = 2 × Volume of one hemisphere = 2 × 2332​π × (1.4 cm)³.

    3. Total Volume of One Gulab Jamun:

      • Total Volume = Volume of the Cylindrical Part + Volume of the Hemispherical Ends.

    4. Volume of Syrup in One Gulab Jamun:

      • 30% of the Total Volume of One Gulab Jamun.

    5. Total Volume of Syrup in 45 Gulab Jamuns:

      • 45 × Volume of Syrup in One Gulab Jamun.

    Let's calculate these values.

    The volumes for one gulab jamun and the total syrup in 45 gulab jamuns are as follows:

    1. Volume of the Cylindrical Part of One Gulab Jamun: 13.5513.55 cubic cm (approximately)
    2. Volume of the Hemispherical Ends of One Gulab Jamun: 11.5011.50 cubic cm (approximately)
    3. Total Volume of One Gulab Jamun: 25.0525.05 cubic cm (approximately)
    4. Volume of Syrup in One Gulab Jamun (30% of Total Volume): 7.527.52 cubic cm (approximately)
    5. Total Volume of Syrup in 45 Gulab Jamuns: 338.18338.18 cubic cm (approximately)

    Therefore, approximately 338.18338.18 cubic cm of syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length of 5 cm and a diameter of 2.8 cm.

  15. 15.A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,............

    Que - A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. To find the number of lead shots dropped in the vessel, we need to determine the volume of water that flowed out and then calculate how many lead shots would displace that volume of water.

    Given:

    1. The vessel is an inverted cone with a height (h) of 8 cm and a radius (r) of 5 cm.
    2. The radius of each lead shot, which is a sphere, is 0.5 cm.
    3. One-fourth of the water in the vessel flows out when the lead shots are dropped in.

    Calculating the Volume of Water that Flowed Out:

    1. Volume of the Vessel (Inverted Cone):

      • The formula for the volume of a cone is 132ℎ31​πr2h.
      • Volume of the vessel = 13×52×831​π×52×8 cm³.

    2. Volume of Water that Flowed Out:

      • One-fourth of the vessel's volume = 14×41​× Volume of the vessel.

    Calculating the Volume Displaced by One Lead Shot:

    1. Volume of a Sphere:
      • The formula for the volume of a sphere is 43334​πr3.
      • Volume of one lead shot = 43×0.5334​π×0.53 cm³.

    Finding the Number of Lead Shots:

    1. The total volume of water that flowed out must be equal to the total volume displaced by the lead shots.
    2. Therefore, Number of Lead Shots = Volume of Water that Flowed Out / Volume of One Lead Shot.

    Let's calculate these values.

    The calculations for the vessel and the lead shots are as follows:

    1. Volume of the Vessel (Inverted Cone): 209.52209.52 cubic cm (approximately)
    2. Volume of Water that Flowed Out (One-Fourth of the Vessel's Volume): 52.3852.38 cubic cm (approximately)
    3. Volume of One Lead Shot (Sphere): 0.520.52 cubic cm (approximately)
    4. Number of Lead Shots Dropped in the Vessel: 100100

    Therefore, 100 lead shots, each being a sphere of radius 0.5 cm, were dropped into the vessel to cause one-fourth of the water to flow out. ​

  16. 16.A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass.

    To find the mass of the solid iron pole, we first calculate the volume of each part of the pole and then determine the total mass based on the density of iron.

    Given:

    1. First cylinder (lower part of the pole):
      • Height (h1) = 220 cm
      • Base diameter = 24 cm, so the radius (r1) = 12 cm (since radius = diameter / 2)
    2. Second cylinder (upper part of the pole):
      • Height (h2) = 60 cm
      • Radius (r2) = 8 cm
    3. Density of iron ≈ 8 g/cm³

    Calculating the Volume of Each Cylinder:

    1. Volume of the First Cylinder:

      • Volume = πr1²h1
      • Volume = π × 12² × 220 cm³

    2. Volume of the Second Cylinder:

      • Volume = πr2²h2
      • Volume = π × 8² × 60 cm³

    Total Volume of the Pole:

    • Total Volume = Volume of the First Cylinder + Volume of the Second Cylinder

    Calculating the Mass:

    • Mass = Total Volume × Density of Iron
    • Mass = Total Volume × 8 g/cm³

    Let's calculate these values.

    The volumes of each part of the pole and the total mass of the pole are as follows:

    1. Volume of the First Cylinder: 99,565.7199,565.71 cubic cm (approximately)
    2. Volume of the Second Cylinder: 12,068.5712,068.57 cubic cm (approximately)
    3. Total Volume of the Pole: 111,634.29111,634.29 cubic cm (approximately)
    4. Mass of the Pole: 893,074.29893,074.29 grams (approximately)

    Therefore, the mass of the solid iron pole, which consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by another cylinder of height 60 cm and radius 8 cm, is approximately 893,074.29893,074.29 grams. ​

  17. 17.A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³ . Check whether she is correct, taking........

    Que-A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14. To check whether the child's measurement of the volume of the spherical glass vessel is correct, we need to calculate the volume of both the spherical part and the cylindrical neck and then add them together.

    Given:

    1. Cylindrical neck:
      • Height (h) = 8 cm
      • Diameter = 2 cm, so the radius (r₁) = 1 cm (since radius = diameter / 2)
    2. Spherical part:
      • Diameter = 8.5 cm, so the radius (r₂) = 4.25 cm
    3. The child's measured volume = 345 cm³
    4. π = 3.14

    Calculating the Volume:

    1. Volume of the Cylindrical Neck:

      • Volume = 12ℎπr12​h
      • Volume = 3.14 × 1² × 8 cm³

    2. Volume of the Spherical Part:

      • The formula for the volume of a sphere is 43234​πr2
      • Volume = 4334​ × 3.14 × 4.25³ cm³

    Total Volume of the Vessel:

    • Total Volume = Volume of the Cylindrical Neck + Volume of the Spherical Part

    Let's calculate these values.

    The calculated volumes for the parts of the spherical glass vessel are as follows:

    1. Volume of the Cylindrical Neck: 25.1225.12 cubic cm
    2. Volume of the Spherical Part: 321.39321.39 cm³ (approximately)
    3. Total Volume of the Vessel: 346.51346.51 cm³ (approximately)

    Comparing this with the child's measurement of 345345 cm³, we find that the child's measurement is quite close and can be considered correct within a reasonable margin of error. The slight difference could be due to minor variations in measurement or rounding

  18. 18.Quick Revision

    Introduction to Surface Areas & Volumes

    • Surface Area: It's like measuring how much wrapping paper you'd need to cover an object completely. For complex shapes, we just find the surface area of each part and add them up.
    • Volume: This is like figuring out how much water can fit inside a shape. For combined shapes, you calculate the volume of each individual shape and then add or subtract them based on how they are combined.

    Important Formulas:

    1. Surface Area of a Cube: 6×side26×side2
    2. Surface Area of a Cylinder: 2(+ℎ)2πr(r+h) where r is radius and ℎh is height.
    3. Volume of a Cube: side3side3
    4. Volume of a Cylinder: 2ℎπr2h

    Implementing Formulas:

    • To use these formulas, first identify the shape of the object.
    • Measure the necessary dimensions (like radius, height, or side length).
    • Plug these measurements into the formula.
    • Do the math to find the surface area or volume.

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