15 extra multiple-choice questions for Mensuration (10th Standard Mathematics, Samacheer Kalvi), beyond the ones printed in the textbook — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
If the radius of a right circular cylinder is doubled and its height is halved, how does its curved surface area change?
- A. It is halved
- B. It remains unchangedCorrect
- C. It is doubled
- D. It is quadrupled
Explanation. The curved surface area of a cylinder is given by \(2\pi r h\). When the radius \(r\) is doubled (\(2r\)) and the height \(h\) is halved (\(h/2\)), the new curved surface area is \(2\pi (2r)(h/2) = 2\pi r h\), which remains unchanged.
Q2
What is the curved surface area of a solid right circular cone whose base radius is \(7\text{ cm}\) and height is \(24\text{ cm}\)?
- A. \(168\pi\text{ cm}^2\)
- B. \(175\pi\text{ cm}^2\)Correct
- C. \(185\pi\text{ cm}^2\)
- D. \(576\pi\text{ cm}^2\)
Explanation. The slant height of the cone is \(l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{625} = 25\text{ cm}\). The curved surface area is given by \(\pi r l = \pi \times 7 \times 25 = 175\pi\text{ cm}^2\).
Q3
If the numerical value of the volume of a solid hemisphere is equal to its total surface area, then what is its radius?
- A. \(2\text{ units}\)
- B. \(3\text{ units}\)
- C. \(4.5\text{ units}\)Correct
- D. \(9\text{ units}\)
Explanation. For a solid hemisphere, the volume is \(\frac{2}{3}\pi r^3\) and the total surface area is \(3\pi r^2\). Setting them numerically equal gives \(\frac{2}{3}\pi r^3 = 3\pi r^2\), which simplifies to \(\frac{2}{3}r = 3 \implies r = 4.5\text{ units}\).
Q4
If the ratio of the surface areas of two spheres is \(4:9\), then what is the ratio of their volumes?
- A. 2:3
- B. 4:9
- C. 8:27Correct
- D. 16:81
Explanation. The ratio of the surface areas is \(r_1^2 : r_2^2 = 4 : 9 \implies r_1 : r_2 = 2 : 3\). Since the volume of a sphere is proportional to \(r^3\), the ratio of their volumes is \(r_1^3 : r_2^3 = 2^3 : 3^3 = 8 : 27\).
Q5
The volume of a right circular cylinder is \(360\pi\text{ cm}^3\). If its height is \(10\text{ cm}\), what is its curved surface area?
- A. \(60\pi\text{ cm}^2\)
- B. \(72\pi\text{ cm}^2\)
- C. \(120\pi\text{ cm}^2\)Correct
- D. \(240\pi\text{ cm}^2\)
Explanation. From the volume formula, \(360\pi = \pi r^2 \times 10 \implies r^2 = 36 \implies r = 6\text{ cm}\). The curved surface area is \(2\pi r h = 2\pi \times 6 \times 10 = 120\pi\text{ cm}^2\).
Q6
If a right circular cylinder and a cone have the same base radius and the same height, and the volume of the cylinder is \(120\text{ cm}^3\), then what is the volume of the cone?
- A. \(40\text{ cm}^3\)Correct
- B. \(60\text{ cm}^3\)
- C. \(240\text{ cm}^3\)
- D. \(360\text{ cm}^3\)
Explanation. The volume of a cone is exactly one-third of the volume of a cylinder with the same base radius and height. Thus, the cone's volume is \(\frac{1}{3} \times 120\text{ cm}^3 = 40\text{ cm}^3\).
Q7
The outer and inner radii of a right circular hollow cylinder are \(5\text{ cm}\) and \(3\text{ cm}\) respectively, and its height is \(10\text{ cm}\). Find its total surface area.
- A. \(160\pi\text{ cm}^2\)
- B. \(176\pi\text{ cm}^2\)
- C. \(192\pi\text{ cm}^2\)Correct
- D. \(224\pi\text{ cm}^2\)
Explanation. The total surface area of a hollow cylinder is given by \(2\pi (R + r)(h + R - r)\). Substituting \(R=5, r=3, h=10\), we get \(T.S.A. = 2\pi(8)(12) = 192\pi\text{ cm}^2\).
Q8
A frustum of a cone has a height of \(4\text{ cm}\). If the radii of its circular bases are \(6\text{ cm}\) and \(3\text{ cm}\), what is its slant height?
- A. \(3\text{ cm}\)
- B. \(4\text{ cm}\)
- C. \(5\text{ cm}\)Correct
- D. \(7\text{ cm}\)
Explanation. The slant height of a frustum is given by \(l = \sqrt{h^2 + (R - r)^2}\). Substituting \(h=4, R=6, r=3\) yields \(l = \sqrt{4^2 + (6 - 3)^2} = \sqrt{16 + 9} = 5\text{ cm}\).
Q9
A solid metallic sphere of radius \(6\text{ cm}\) is melted and recast into a solid cylinder of radius \(8\text{ cm}\). What is the height of the cylinder?
- A. \(3.5\text{ cm}\)
- B. \(4.5\text{ cm}\)Correct
- C. \(6\text{ cm}\)
- D. \(9\text{ cm}\)
Explanation. The volume of the sphere is \(\frac{4}{3}\pi(6)^3 = 288\pi\text{ cm}^3\). Since volume remains unchanged, the cylinder's volume is \(\pi (8)^2 h = 64\pi h\). Solving \(64\pi h = 288\pi\) gives \(h = 4.5\text{ cm}\).
Q10
If the radius of a sphere is doubled, then its surface area increases by what percentage?
- A. 100\%
- B. 200\%
- C. 300\%Correct
- D. 400\%
Explanation. Surface area is proportional to \(r^2\). Doubling the radius increases the area to \(2^2 = 4\) times the original, which represents an increase of \(300\%\).
Q11
A solid cone has a base radius of \(5\text{ cm}\) and a slant height of \(13\text{ cm}\). What is its total surface area?
- A. \(65\pi\text{ cm}^2\)
- B. \(90\pi\text{ cm}^2\)Correct
- C. \(120\pi\text{ cm}^2\)
- D. \(180\pi\text{ cm}^2\)
Explanation. The total surface area of a solid cone is given by \(T.S.A. = \pi r(l + r)\). Substituting \(r=5\text{ cm}\) and \(l=13\text{ cm}\) gives \(T.S.A. = \pi \times 5 \times (13 + 5) = 90\pi\text{ cm}^2\).
Q12
What is the ratio of the curved surface area to the total surface area of a solid hemisphere?
- A. 1:2
- B. 2:3Correct
- C. 3:4
- D. 2:1
Explanation. For a solid hemisphere of radius \(r\), the curved surface area is \(2\pi r^2\) and the total surface area is \(3\pi r^2\). The ratio is \(2\pi r^2 : 3\pi r^2 = 2 : 3\).
Q13
A sphere of radius \(r\) is inscribed inside a right circular cylinder of radius \(r\) and height \(2r\). What is the ratio of the volume of the cylinder to the volume of the sphere?
- A. 1:1
- B. 3:2Correct
- C. 4:3
- D. 2:3
Explanation. The volume of the cylinder is \(\pi r^2(2r) = 2\pi r^3\). The volume of the inscribed sphere is \(\frac{4}{3}\pi r^3\). The ratio is \(2\pi r^3 : \frac{4}{3}\pi r^3 = 3:2\).
Q14
If the volume of a sphere is \(\frac{32}{3}\pi\text{ cm}^3\), then what is its surface area?
- A. \(8\pi\text{ cm}^2\)
- B. \(12\pi\text{ cm}^2\)
- C. \(16\pi\text{ cm}^2\)Correct
- D. \(32\pi\text{ cm}^2\)
Explanation. Using the volume formula \(V = \frac{4}{3}\pi r^3 = \frac{32}{3}\pi\), we get \(r^3 = 8 \implies r = 2\text{ cm}\). The surface area is \(4\pi r^2 = 4\pi(2)^2 = 16\pi\text{ cm}^2\).
Q15
A toy is formed by surmounting a cone of height \(4\text{ cm}\) on a solid hemisphere of radius \(3\text{ cm}\). Find the total volume of the toy.
- A. \(18\pi\text{ cm}^3\)
- B. \(24\pi\text{ cm}^3\)
- C. \(30\pi\text{ cm}^3\)Correct
- D. \(36\pi\text{ cm}^3\)
Explanation. The total volume is the sum of the volumes: \(V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi(27) + \frac{1}{3}\pi(9)(4) = 18\pi + 12\pi = 30\pi\text{ cm}^3\).
