CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 13

Practice MCQs for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

Access targeted multiple-choice questions for Chapter 12 Surface Areas and Volumes designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question: The volume of the largest right circular cone that can be carved out from a solid cube of edge \(2\) cm is
(a) \(\frac{4\pi}{3}\) cu cm
(b) \(\frac{5\pi}{3}\) cu cm
(c) \(\frac{8\pi}{3}\) cu cm
(d) \(\frac{2\pi}{3}\) cu cm
Show Answer & Explanation

Answer: (d) \(\frac{2\pi}{3}\) cu cm

Explanation:
1. The cone has base diameter \(2\) cm (radius \(1\) cm) and height \(2\) cm.
2. \(V = \frac{1}{3}\pi(1)^2(2) = \frac{2\pi}{3}\) cu cm.

Question: A solid sphere is cut into two hemispheres. The ratio of the surface area of the sphere to that of the two hemispheres taken together is
(a) 1 : 1
(b) 1 : 4
(c) 2 : 3
(d) 3 : 2
Show Answer & Explanation

Answer: (c) 2 : 3

Explanation:
1. Sphere: \(S_1 = 4\pi r^2\). Two hemispheres together (including the flat base): \(S_2 = 2 \times 3\pi r^2 = 6\pi r^2\).
2. \(S_1 : S_2 = 4 : 6 = 2 : 3\).

Question: The volume of a right circular cone whose base area is \(156\) cm\(^2\) and vertical height is \(8\) cm is
(a) \(2496\) cm\(^3\)
(b) \(1248\) cm\(^3\)
(c) \(1664\) cm\(^3\)
(d) \(416\) cm\(^3\)
Show Answer & Explanation

Answer: (d) \(416\) cm\(^3\)

Explanation:
1. \(V = \frac{1}{3} \times \text{base area} \times h = \frac{1}{3} \times 156 \times 8 = 416\) cm\(^3\).

Question: If the radius of a sphere is increased by \(100\%\), the volume of the sphere is increased by
(a) \(200\%\)
(b) \(500\%\)
(c) \(700\%\)
(d) \(800\%\)
Show Answer & Explanation

Answer: (c) \(700\%\)

Explanation:
1. New radius \(= 2r\). New volume \(= \frac{4}{3}\pi(2r)^3 = 8 \times \frac{4}{3}\pi r^3\).
2. The volume becomes 8 times, an increase of \(700\%\).

Question: From a solid circular cylinder of height \(10\) cm and base radius \(6\) cm, a right circular cone of the same height and same base is removed. The volume of the remaining solid is
(a) \(280\pi\) cm\(^3\)
(b) \(330\pi\) cm\(^3\)
(c) \(240\pi\) cm\(^3\)
(d) \(440\pi\) cm\(^3\)
Show Answer & Explanation

Answer: (c) \(240\pi\) cm\(^3\)

Explanation:
1. Cylinder \(= \pi(6)^2(10) = 360\pi\). Cone \(= \frac{1}{3}\pi(6)^2(10) = 120\pi\).
2. Remaining \(= 360\pi - 120\pi = 240\pi\) cm\(^3\).

Question: If two solid hemispheres of the same base radius \(r\) are joined together along their bases, the curved surface area of the new solid is
(a) \(4\pi r^2\)
(b) \(6\pi r^2\)
(c) \(3\pi r^2\)
(d) \(8\pi r^2\)
Show Answer & Explanation

Answer: (a) \(4\pi r^2\)

Explanation:
1. Each hemisphere has curved surface area \(2\pi r^2\).
2. Together: \(2\pi r^2 + 2\pi r^2 = 4\pi r^2\).

Question: A right circular cylinder of radius \(r\) and height \(h\) (where \(h > 2r\)) just encloses a sphere of diameter
(a) \(r\)
(b) \(2r\)
(c) \(h\)
(d) \(2h\)
Show Answer & Explanation

Answer: (b) \(2r\)

Explanation:
1. The sphere fits the cylinder's width, so its diameter equals the cylinder's diameter \(2r\).

Question: During conversion of a solid from one shape to another, the volume of the new shape will
(a) increase
(b) decrease
(c) remain unaltered
(d) be doubled
Show Answer & Explanation

Answer: (c) remain unaltered

Explanation:
1. Melting and recasting keep the metal's volume the same.

Question: A solid piece of iron in the form of a cuboid of dimensions \(49\) cm \(\times\) \(33\) cm \(\times\) \(24\) cm is moulded into a solid sphere. The radius of the sphere is
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Show Answer & Explanation

Answer: (a) 21 cm

Explanation:
1. Cuboid volume \(= 49 \times 33 \times 24 = 38808\) cm\(^3\).
2. \(\frac{4}{3} \times \frac{22}{7} r^3 = 38808\), so \(r^3 = 9261 = 21^3\).
3. \(r = 21\) cm.

Question: Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter \(2\) cm and height \(16\) cm. The diameter of each sphere is
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 6 cm
Show Answer & Explanation

Answer: (c) 2 cm

Explanation:
1. Cylinder radius \(1\) cm, height \(16\) cm: volume \(= 16\pi\).
2. \(12 \times \frac{4}{3}\pi r^3 = 16\pi\), so \(r^3 = 1\) and \(r = 1\) cm.
3. Diameter \(= 2\) cm.

Question: In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) circle
(b) frustum of a cone
(c) sphere
(d) hemisphere
Show Answer & Explanation

Answer: (a) circle

Explanation:
1. A cut parallel to the base gives a circle.

Question: The volume of a right circular cylinder of base radius \(7\) cm and height \(10\) cm (take \(\pi = \frac{22}{7}\)) is
(a) \(1540\) cm\(^3\)
(b) \(440\) cm\(^3\)
(c) \(770\) cm\(^3\)
(d) \(220\) cm\(^3\)
Show Answer & Explanation

Answer: (a) \(1540\) cm\(^3\)

Explanation:
1. \(V = \pi r^2 h = \frac{22}{7} \times 49 \times 10 = 1540\) cm\(^3\).

Question: If the radius of the base of a right circular cylinder is halved, keeping the height the same, the ratio of the volume of the reduced cylinder to that of the original cylinder is
(a) 1 : 2
(b) 1 : 4
(c) 2 : 1
(d) 4 : 1
Show Answer & Explanation

Answer: (b) 1 : 4

Explanation:
1. Reduced volume \(= \pi \left(\frac{r}{2}\right)^2 h = \frac{1}{4}\pi r^2 h\).
2. Ratio \(= 1 : 4\).

Question: The volumes of two spheres are in the ratio \(64 : 27\). The ratio of their surface areas is
(a) 4 : 3
(b) 16 : 9
(c) 9 : 16
(d) 3 : 4
Show Answer & Explanation

Answer: (b) 16 : 9

Explanation:
1. \(\frac{r_1^3}{r_2^3} = \frac{64}{27}\), so \(\frac{r_1}{r_2} = \frac{4}{3}\).
2. Surface area ratio \(= \left(\frac{4}{3}\right)^2 = \frac{16}{9}\).

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Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

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FAQs

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