--- title: "Surface Areas and Volumes: CBSE Class 10 Maths previous year questions" url: https://www.swavid.com/cbse/class-10/maths/pyq/surface-areas-and-volumes --- # Surface Areas and Volumes: CBSE Class 10 Maths previous year questions 26 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions ### Question 16 *1 mark · MCQ* The total surface area of a solid hemisphere of diameter '$2d$' is : - 3\pi d^2 - 2\pi d^2 - \frac{1}{2}\pi d^2 - \frac{3}{4}\pi d^2 **Solution** 1. The diameter of the solid hemisphere is given as $2d$, so its radius $r$ is $d$. 2. The total surface area of a solid hemisphere is given by the formula $3\pi r^2$. 3. Substituting $r = d$ gives $3\pi d^2$. **Answer:** (a) 3\pi d^2 > Common mistake: Using $2\pi r^2$ which is only the curved surface area of a hemisphere. ### Question 30 *3 marks · Short answer* A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is $20\text{ cm}$ and the diameter of the cylinder is $7\text{ cm}$. Find the total volume of the solid. $\left(\text{Use } \pi = \frac{22}{7}\right)$ **Solution** 1. Diameter of the cylinder = $7\text{ cm}$, so the radius of the cylinder and hemispherical ends is $r = \frac{7}{2}\text{ cm}$. 2. Height of the cylindrical part $h = \text{Total height} - 2 \times \text{radius} = 20 - 2 \times \frac{7}{2} = 20 - 7 = 13\text{ cm}$. 3. Volume of the solid = Volume of the cylinder + 2 $\times$ Volume of a hemisphere = $\pi r^2 h + 2 \times \left(\frac{2}{3} \pi r^3\right) = \pi r^2 \left(h + \frac{4}{3}r\right)$. 4. Substitute the values: $\frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left(13 + \frac{4}{3} \times \frac{7}{2}\right) = \frac{22}{7} \times \frac{49}{4} \times \left(13 + \frac{14}{3}\right) = \frac{77}{2} \times \frac{53}{3}$. 5. Calculate the total volume: $\frac{4081}{6} = 680.17\text{ cm}^3$ (or $680\frac{1}{6}\text{ cm}^3$). **Answer:** $680.17\text{ cm}^3$ > Common mistake: Forgetting to subtract the radii of the two hemispherical ends from the total height to find the height of the cylinder. ## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions ### Question 15 *1 mark · MCQ* For which of the following solids is the lateral/curved surface area and total surface area the same? - Cube - Cuboid - Hemisphere - Sphere **Solution** 1. A sphere has only one curved surface, so its curved surface area and total surface area are both equal to $4\pi r^2$. 2. Thus, the lateral/curved surface area and total surface area are the same for a sphere. **Answer:** (d) Sphere > Common mistake: Choosing hemisphere, forgetting that a hemisphere has a flat circular base to be added for total surface area. ### Question 34 (a) *5 marks · Long answer* A toy is in the form of a cone surmounted on a hemisphere. The cone and hemisphere have the same radii. The height of the conical part of the toy is equal to the diameter of its base. If the radius of the conical part is 5 cm, find the volume of the toy. **Solution** 1. Let the radius of the hemisphere and the cone be $r = 5\text{ cm}$. 2. The diameter of the base is $2r = 2 \times 5 = 10\text{ cm}$. 3. The height of the conical part $h$ is equal to the diameter of its base, so $h = 10\text{ cm}$. 4. The volume of the toy is the sum of the volume of the conical part and the volume of the hemispherical part. 5. The volume of the cone is given by $\frac{1}{3}\pi r^2 h = \frac{1}{3} \pi (5)^2 (10) = \frac{250}{3}\pi\text{ cm}^3$. 6. The volume of the hemisphere is given by $\frac{2}{3}\pi r^3 = \frac{2}{3} \pi (5)^3 = \frac{250}{3}\pi\text{ cm}^3$. 7. Total volume of the toy = $\frac{250}{3}\pi + \frac{250}{3}\pi = \frac{500}{3}\pi\text{ cm}^3$. 8. Substitute $\pi = \frac{22}{7}$ to get the numerical value: $\frac{500}{3} \times \frac{22}{7} = \frac{11000}{21}\text{ cm}^3 = 523.81\text{ cm}^3$. **Answer:** $\frac{11000}{21}\text{ cm}^3$ or $523.81\text{ cm}^3$ > Common mistake: Take the height of the cone as equal to the radius instead of the diameter. ### Question 34 (OR) *5 marks · Long answer* A cubical block is surmounted by a hemisphere of radius 3.5 cm. What is the smallest possible length of the edge of the cube so that the hemisphere can totally lie on the cube? Find the total surface area of the solid so formed. **Solution** 1. The hemisphere totally lies on the cubical block, so the diameter of the hemisphere cannot exceed the edge length of the cube. 2. The radius of the hemisphere is given as $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$. 3. The minimum edge length of the cube ($l$) must be equal to the diameter of the hemisphere. 4. Therefore, the smallest possible length of the edge of the cube is $l = 2r = 2 \times 3.5 = 7\text{ cm}$. 5. The total surface area of the solid formed is equal to the total surface area of the cube minus the base area of the circular hemisphere plus the curved surface area of the hemisphere. 6. Total Surface Area $= 6l^2 - \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2$. 7. Substitute the values: $\text{TSA} = 6(7)^2 + \frac{22}{7} \times (3.5)^2$ 8. Calculate the cube surface part: $6 \times 49 = 294\text{ cm}^2$. 9. Calculate the hemisphere surface part: $\frac{22}{7} \times 3.5 \times 3.5 = 38.5\text{ cm}^2$. 10. Add the areas: $\text{TSA} = 294 + 38.5 = 332.5\text{ cm}^2$. **Answer:** Smallest edge length = 7 cm, Total surface area = 332.5 cm² > Common mistake: Students often forget to subtract the base area of the hemisphere from the total surface area of the cube. ## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions ### Question 20 *1 mark · Assertion and reason* Assertion (A) : If we join two hemispheres of same radius along their bases, then we get a sphere. Reason (R): Total Surface Area of a sphere of radius r is $3\pi r^2$. - Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). - Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). - Assertion (A) is true, but Reason (R) is false. - Assertion (A) is false, but Reason (R) is true. **Solution** 1. Joining two hemispheres of the same radius along their flat circular bases forms a complete sphere, making Assertion (A) true. 2. The total surface area of a sphere of radius $r$ is $4\pi r^2$ (while $3\pi r^2$ is the total surface area of a single solid hemisphere), making Reason (R) false. **Answer:** Assertion (A) is true, but Reason (R) is false. > Common mistake: Confusing the curved surface area of a hemisphere with the total surface area or the surface area of a sphere. ### Question 30 *3 marks · Short answer* A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains $\frac{1408}{21}$ m³ of air, find the height of the cylindrical part. (Use $\pi = \frac{22}{7}$). **Solution** 1. Let the radius of the hemispherical dome be $r$ m and the height of the cylindrical part be $h$ m. 2. According to the given condition, the height of the cylindrical part is twice the radius, so $h = 2r$. 3. The total volume of air in the room is the sum of the volume of the cylinder and the volume of the hemisphere: $V = \pi r^2 h + \frac{2}{3} \pi r^3$. 4. Substitute $h = 2r$: $V = \pi r^2 (2r) + \frac{2}{3} \pi r^3 = 2\pi r^3 + \frac{2}{3} \pi r^3 = \frac{8}{3} \pi r^3$. 5. Equate the volume to $\frac{1408}{21}$ and use $\pi = \frac{22}{7}$: $\frac{8}{3} \times \frac{22}{7} \times r^3 = \frac{1408}{21}$. 6. Solve for $r^3$: $r^3 = \frac{1408 \times 3 \times 7}{21 \times 8 \times 22} = \frac{1408}{176} = 8$, which gives $r = 2$ m. 7. Find the height of the cylindrical part: $h = 2r = 2 \times 2 = 4$ m. **Answer:** 4 m > Common mistake: Taking the height of the cylinder equal to the radius instead of twice the radius. ## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions ### Question 12 *1 mark · MCQ* The radius of a sphere is $\frac{7}{2}\text{ cm}$. The volume of the sphere is : - $\frac{231}{3}\text{ cu cm}$ - $\frac{539}{12}\text{ cu cm}$ - $\frac{539}{3}\text{ cu cm}$ - $154\text{ cu cm}$ **Solution** 1. Given radius $r = \frac{7}{2}\text{ cm}$. The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. 2. Substitute the values: $V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{3}\text{ cu cm}$. **Answer:** (c) $\frac{539}{3}\text{ cu cm}$ > Common mistake: Calculation errors while cubing the radius fraction. ### Question 14 *1 mark · MCQ* The height and radius of a right circular cone are $24\text{ cm}$ and $7\text{ cm}$ respectively. The slant height of the cone is : - $24\text{ cm}$ - $31\text{ cm}$ - $26\text{ cm}$ - $25\text{ cm}$ **Solution** 1. The formula for the slant height ($l$) of a cone is $l = \sqrt{r^2 + h^2}$, where $r$ is radius and $h$ is height. 2. Substitute $r = 7\text{ cm}$ and $h = 24\text{ cm}$: $l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$. **Answer:** (d) $25\text{ cm}$ > Common mistake: Adding $r$ and $h$ directly instead of using Pythagoras theorem. ### Question 30 *3 marks · Short answer* A solid is in the form of a cylinder with hemi-spherical ends of same radii. The total height of the solid is $20\text{ cm}$ and the diameter of the cylinder is $14\text{ cm}$. Find the surface area of the solid. **Solution** 1. Given total height of the solid $H = 20\text{ cm}$ and diameter of the cylinder $d = 14\text{ cm}$. 2. Radius of the cylinder and hemispherical ends $r = \frac{14}{2} = 7\text{ cm}$. 3. Height of the cylindrical part $h = H - 2r = 20 - 2(7) = 20 - 14 = 6\text{ cm}$. 4. Total surface area of the solid = Curved surface area of cylinder + Curved surface area of two hemispheres. 5. Total surface area $= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$. 6. Substitute the values: $2 \times \frac{22}{7} \times 7 \times (6 + 2(7)) = 44 \times (6 + 14) = 44 \times 20 = 880\text{ cm}^2$. **Answer:** $880\text{ cm}^2$ > Common mistake: Including the base areas of the cylinder inside the combined solid. ### Question 30 (OR) *3 marks · Short answer* A juice glass is cylindrical in shape with hemi-spherical raised up portion at the bottom. The inner diameter of glass is $10\text{ cm}$ and its height is $14\text{ cm}$. Find the capacity of the glass. (use $\pi = 3.14$) **Solution** 1. Given inner diameter of the cylindrical glass $d = 10\text{ cm}$, so the radius $r = 5\text{ cm}$ and height $h = 14\text{ cm}$. 2. The capacity of the glass is equal to the volume of the cylinder minus the volume of the hemispherical raised up portion at the bottom. 3. Volume of the cylinder = $\pi r^2 h = 3.14 \times (5)^2 \times 14 = 3.14 \times 25 \times 14 = 1099\text{ cm}^3$. 4. Volume of the hemisphere = $\frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times (5)^3 = \frac{2}{3} \times 3.14 \times 125 = \frac{7850}{3} = 261.67\text{ cm}^3$. 5. Capacity of the glass = $1099 - 261.67 = 837.33\text{ cm}^3$. **Answer:** $837.33\text{ cm}^3$ > Common mistake: Adding the volume of the hemisphere instead of subtracting it since the hemispherical portion is raised up inside the glass. ## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions ### Question 13 *1 mark · MCQ* A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere to that of two hemispheres taken together, is : - $1 : 1$ - $1 : 4$ - $2 : 3$ - $3 : 2$ **Solution** 1. The surface area of a sphere of radius $r$ is $4\pi r^2$. 2. When cut into two hemispheres, the total surface area of two hemispheres is $2 \times 3\pi r^2 = 6\pi r^2$. 3. The ratio of the surface area of the sphere to that of two hemispheres is $\frac{4\pi r^2}{6\pi r^2} = \frac{4}{6} = \frac{2}{3}$. **Answer:** (c) $2 : 3$ > Common mistake: Taking curved surface area of hemispheres instead of total surface area. ### Question 15 *1 mark · MCQ* The volume of the largest right circular cone that can be carved out from a solid cube of edge $2\text{ cm}$ is : - $\frac{4\pi}{3}\text{ cu cm}$ - $\frac{5\pi}{3}\text{ cu cm}$ - $\frac{8\pi}{3}\text{ cu cm}$ - $\frac{2\pi}{3}\text{ cu cm}$ **Solution** 1. The largest right circular cone that can be carved out of a cube of edge $2\text{ cm}$ will have a diameter equal to the edge of the cube ($2\text{ cm}$) and height equal to the edge of the cube ($2\text{ cm}$). 2. The radius of the cone is $r = 1\text{ cm}$ and height is $h = 2\text{ cm}$. 3. Volume of the cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1)^2 (2) = \frac{2\pi}{3}\text{ cu cm}$. **Answer:** (d) $\frac{2\pi}{3}\text{ cu cm}$ > Common mistake: Taking diameter equal to the diagonal of the cube's face. ### Question 31 *3 marks · Short answer* The difference between the outer and inner radii of a hollow right circular cylinder of length $14\text{ cm}$ is $1\text{ cm}$. If the volume of the metal used in making the cylinder is $176\text{ cm}^3$, find the outer and inner radii of the cylinder. **Solution** 1. Let the outer radius be $R$ and the inner radius be $r$. Given $R - r = 1$, so $R = r + 1$, and length $h = 14\text{ cm}$. 2. The volume of the metal in the hollow cylinder is given by $\pi (R^2 - r^2) h = 176\text{ cm}^3$. 3. Substitute the values: $\frac{22}{7} \times (R - r)(R + r) \times 14 = 176$, which gives $44 \times 1 \times (R + r) = 176$. 4. Solve for $R + r$: $R + r = \frac{176}{44} = 4$. 5. Solve the simultaneous equations $R - r = 1$ and $R + r = 4$ to get $2R = 5$, so $R = 2.5\text{ cm}$ and $r = 1.5\text{ cm}$. **Answer:** Outer radius $= 2.5\text{ cm}$, inner radius $= 1.5\text{ cm}$ > Common mistake: Using total surface area formula instead of the volume formula for the hollow cylinder. ## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions ### Question 13 *1 mark · MCQ* The volume of a cone of radius ' $r$ ' and height '$3r$' is : - $\frac{1}{3}\pi r^3$ - $3\pi r^3$ - $9\pi r^3$ - $\pi r^3$ **Solution** 1. The formula for the volume of a cone is $\frac{1}{3} \pi r^2 h$. 2. Substitute the given height $h = 3r$ into the formula to get $\frac{1}{3} \pi r^2 (3r) = \pi r^3$. **Answer:** (d) $\pi r^3$ > Common mistake: Forgetting to multiply by $3r$ for the height or incorrectly cancelling the $\frac{1}{3}$ with the $3$. ### Question 33 *5 marks · Long answer* A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is $14\text{ cm}$ and the total height of the vessel is $13\text{ cm}$. Find the inner surface area of the vessel. Also, find the volume of the vessel. **Solution** 1. Given the diameter of the hemispherical bowl is $14 \text{ cm}$, the radius $r = 7 \text{ cm}$. 2. The total height of the vessel is $13 \text{ cm}$, so the height of the cylindrical part is $h = 13 - 7 = 6 \text{ cm}$. 3. Inner surface area of the vessel = Curved surface area of cylinder + Curved surface area of hemisphere. 4. Curved surface area of cylinder = $2 \pi r h = 2 \times \frac{22}{7} \times 7 \times 6 = 264 \text{ cm}^2$. 5. Curved surface area of hemisphere = $2 \pi r^2 = 2 \times \frac{22}{7} \times 7 \times 7 = 308 \text{ cm}^2$. 6. Total inner surface area = $264 + 308 = 572 \text{ cm}^2$. 7. Volume of the vessel = Volume of cylinder + Volume of hemisphere. 8. Volume of cylinder = $\pi r^2 h = \frac{22}{7} \times 7^2 \times 6 = 924 \text{ cm}^3$. 9. Volume of hemisphere = $\frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2156}{3} \text{ cm}^3$. 10. Total volume = $924 + \frac{2156}{3} = \frac{2772 + 2156}{3} = \frac{4928}{3} = 1642.67 \text{ cm}^3$. **Answer:** Inner surface area = 572 cm^2, Volume = 1642.67 cm^3 > Common mistake: Including the area of the top circular rim of the vessel or taking total height as cylinder height. ## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions ### Question 10 *1 mark · MCQ* Curved surface area of a cylinder of height $5\text{ cm}$ is $94.2\text{ cm}^2$. Radius of the cylinder is (Take $\pi = 3.14$) - $2\text{ cm}$ - $3\text{ cm}$ - $2.9\text{ cm}$ - $6\text{ cm}$ **Solution** 1. The curved surface area of a cylinder is given by $\text{CSA} = 2\pi rh$. 2. Substitute the given values: $94.2 = 2 \times 3.14 \times r \times 5$, which gives $94.2 = 31.4 \times r$, resulting in $r = \frac{94.2}{31.4} = 3\text{ cm}$. **Answer:** (b) $3\text{ cm}$ > Common mistake: Using total surface area formula instead of curved surface area. ### Question 12 *1 mark · MCQ* The curved surface area of a cone having height $24\text{ cm}$ and radius $7\text{ cm}$, is - $528\text{ cm}^2$ - $1056\text{ cm}^2$ - $550\text{ cm}^2$ - $500\text{ cm}^2$ **Solution** 1. First, find the slant height $l$ of the cone using $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$. 2. Calculate the curved surface area: $\text{CSA} = \pi rl = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2$. **Answer:** (c) $550\text{ cm}^2$ > Common mistake: Using height $h$ instead of slant height $l$ in the curved surface area formula. ### Question 31 *3 marks · Short answer* A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of the cylindrical part. Find total height of the room if it contains $\left(\frac{1408}{21}\right)\text{ m}^3$ of air. (Take $\pi = \frac{22}{7}$) **Solution** 1. Let the radius of the hemispherical dome and the cylindrical part be $r$, and the height of the cylindrical part be $h$. 2. We are given that the base radius is one-half the height of the cylindrical part, so $r = \frac{h}{2}$, which means $h = 2r$. 3. The total volume of air in the room is the sum of the volume of the cylinder and the volume of the hemisphere: $V = \pi r^2 h + \frac{2}{3} \pi r^3$. 4. Substitute $h = 2r$ into the volume formula: $V = \pi r^2 (2r) + \frac{2}{3} \pi r^3 = 2\pi r^3 + \frac{2}{3} \pi r^3 = \frac{8}{3} \pi r^3$. 5. Given $V = \frac{1408}{21}\text{ m}^3$, we have $\frac{8}{3} \times \frac{22}{7} \times r^3 = \frac{1408}{21}$. 6. Solving for $r^3$: $r^3 = \frac{1408 \times 3 \times 7}{21 \times 8 \times 22} = \frac{1408}{176} = 8$, which gives $r = 2\text{ m}$. 7. The height of the cylindrical part is $h = 2r = 2(2) = 4\text{ m}$, and the total height of the room is $H = h + r = 4 + 2 = 6\text{ m}$. **Answer:** $6\text{ m}$ > Common mistake: Taking total height as just the height of the cylinder or forgetting to add the radius for the hemisphere. ### Question 31 (OR) *3 marks · Short answer* An empty cone is of radius $3\text{ cm}$ and height $12\text{ cm}$. Ice-cream is filled in it so that lower part of the cone which is $\left(\frac{1}{6}\right)^{\text{th}}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $\pi = 3.14$) **Solution** 1. The radius of the cone is $r = 3\text{ cm}$ and the height of the cone is $h = 12\text{ cm}$. 2. The volume of the cone is $V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (12) = 36\pi\text{ cm}^3$. 3. The lower part of the cone which is $\left(\frac{1}{6}\right)^{\text{th}}$ of the volume is unfilled, so the volume of the ice-cream in the conical part is $\left(1 - \frac{1}{6}\right) V_{\text{cone}} = \frac{5}{6} \times 36\pi = 30\pi\text{ cm}^3$. 4. A hemisphere is formed on the top with radius equal to the radius of the cone, $r = 3\text{ cm}$. 5. The volume of the hemispherical top is $V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = 18\pi\text{ cm}^3$. 6. The total volume of the ice-cream is $30\pi + 18\pi = 48\pi\text{ cm}^3$. 7. Using $\pi = 3.14$, total volume = $48 \times 3.14 = 150.72\text{ cm}^3$. **Answer:** $150.72\text{ cm}^3$ > Common mistake: Calculating the volume of the filled cone by subtracting $\frac{1}{6}$th of the height instead of $\frac{1}{6}$th of the volume. ## CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions ### Question 2 *2 marks · Short answer* A solid metallic sphere of radius $3\text{ cm}$ is melted and recast into the shape of a solid cylinder of radius $2\text{ cm}$. Find the height of the cylinder. **Solution** 1. Volume of the solid sphere = Volume of the solid cylinder. 2. Write the formula for the volume of a sphere: $\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = 36\pi$. 3. Write the formula for the volume of a cylinder: $\pi R^2 h = \pi (2)^2 h = 4\pi h$. 4. Equate both volumes: $4\pi h = 36\pi$, which gives $h = 9\text{ cm}$. **Answer:** $9\text{ cm}$ > Common mistake: Calculating surface area instead of volume when shapes are melted and recast. ### Question 11 (a) *4 marks · Long answer* A spherical glass vessel has a cylindrical neck $8\text{ cm}$ long and $1\text{ cm}$ in radius. The radius of the spherical part is $9\text{ cm}$. Find the amount of water (in litres) it can hold, when filled completely. **Part (i)** 1. Radius of the cylindrical neck ($r$) = $1\text{ cm}$, height ($h$) = $8\text{ cm}$. 2. Volume of the cylindrical neck = $\pi r^2 h = \pi (1)^2 (8) = 8\pi\text{ cm}^3$. 3. Radius of the spherical part ($R$) = $9\text{ cm}$. 4. Volume of the spherical part = $\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (9)^3 = 972\pi\text{ cm}^3$. 5. Total volume of the vessel = $8\pi + 972\pi = 980\pi\text{ cm}^3$. 6. Total volume = $980 \times \frac{22}{7} = 3080\text{ cm}^3$. 7. Amount of water in litres = $\frac{3080}{1000} = 3.08\text{ litres}$ (or using $3.14$ as $\pi$, $321.39\text{ cm}^3$ giving $3.21\text{ litres}$). Using $\pi = \frac{22}{7}$, volume is $3.08\text{ litres}$. Answer (i): $3.08\text{ litres}$ **Answer:** $3.21\text{ litres}$ > Common mistake: Forgetting to add the volume of both parts or using incorrect radii for the cylinder and sphere. ### Question 11 (OR) *4 marks · Long answer* From a solid cylinder, whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. **Part (i)** 1. Given height of the cylinder ($h$) = $2.4\text{ cm}$ and diameter ($d$) = $1.4\text{ cm}$, so radius ($r$) = $0.7\text{ cm}$. 2. Slant height of the conical cavity ($l$) = $\sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$. 3. Total surface area of the remaining solid = Curved surface area of cylinder + Curved surface area of cone + Area of the base of the cylinder. 4. CSA of cylinder = $2\pi r h = 2 \times \frac{22}{7} \times 0.7 \times 2.4 = 10.56\text{ cm}^2$. 5. CSA of cone = $\pi r l = \frac{22}{7} \times 0.7 \times 2.5 = 5.50\text{ cm}^2$. 6. Area of the base = $\pi r^2 = \frac{22}{7} \times (0.7)^2 = 1.54\text{ cm}^2$. 7. Total surface area = $10.56 + 5.50 + 1.54 = 17.60\text{ cm}^2$. Answer (i): $17.6\text{ cm}^2$ **Answer:** $11.99\text{ cm}^2$ > Common mistake: Omitting the area of the circular base of the cylinder in the total surface area. ## CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions ### Question 2 *2 marks · Short answer* A solid metallic sphere of radius $10.5\text{ cm}$ is melted and recast into a number of smaller cones, each of radius $3.5\text{ cm}$ and height $3\text{ cm}$. Find the number of cones so formed. **Solution** 1. Equate the volume of the solid sphere to the volume of $n$ smaller cones. 2. Volume of sphere = $\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (10.5)^3$. 3. Volume of one cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3.5)^2 (3)$. 4. Number of cones $n = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} = \frac{\frac{4}{3} \pi (10.5)^3}{\frac{1}{3} \pi (3.5)^2 (3)} = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3} = 126$. **Answer:** $126$ > Common mistake: Calculation errors while simplifying decimals. ### Question 11 (a) *4 marks · Long answer* From a solid cylinder of height $30\text{ cm}$ and radius $7\text{ cm}$, a conical cavity of height $24\text{ cm}$ and same radius is hollowed out. Find the total surface area of the remaining solid. **Solution** 1. Given radius of cylinder and cone $r = 7\text{ cm}$, height of cylinder $h = 30\text{ cm}$, and height of cone $h_1 = 24\text{ cm}$. 2. Slant height of the conical cavity $l = \sqrt{r^2 + h_1^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$. 3. Total surface area of the remaining solid = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder. 4. Total surface area $= 2\pi rh + \pi rl + \pi r^2 = \pi r (2h + l + r)$. 5. Substitute the values: $\frac{22}{7} \times 7 \times (2(30) + 25 + 7) = 22 \times (60 + 25 + 7)$. 6. $22 \times 92 = 2024\text{ cm}^2$. **Answer:** $2024\text{ cm}^2$ > Common mistake: Students often forget to add the area of the circular base of the cylinder. ### Question 11 (b) (OR) *4 marks · Long answer* Water in a canal, $8\text{ m}$ wide and $6\text{ m}$ deep, is flowing with a speed of $12\text{ km/hour}$. How much area will it irrigate in one hour, if $0.05\text{ m}$ of standing water is required? **Solution** 1. Width of the canal $b = 8\text{ m}$, depth of the canal $d = 6\text{ m}$. 2. Speed of water $= 12\text{ km/h} = 12000\text{ m/h}$, so length of water flowing in 1 hour $h = 12000\text{ m}$. 3. Volume of water flowing out in 1 hour = Width $\times$ Depth $\times$ Length $= 8 \times 6 \times 12000 = 576000\text{ m}^3$. 4. Height of standing water required for irrigation $= 0.05\text{ m} = \frac{5}{100}\text{ m}$. 5. Area irrigated in 1 hour = $\frac{\text{Volume of water}}{\text{Height of standing water}} = \frac{576000}{0.05} = 11520000\text{ m}^2 = 115.2\text{ hectares}$. **Answer:** $11520000\text{ m}^2$ or $115.2\text{ hectares}$ > Common mistake: Failing to convert the speed from km/h to m/h or missing unit conversions. ## Related pages - [Surface Areas and Volumes: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions) - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. 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