Samacheer Kalvi Class 10 Maths Solutions Chapter 7 Mensuration Exercise 7.5

Multiple Choice Questions

Question 1. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is
(a) \( 60\pi \text{ cm}^2 \)
(b) \( 136\pi \text{ cm}^2 \)
(c) \( 120\pi \text{ cm}^2 \)
(d) \( 136\pi \text{ cm}^2 \)
Answer: (d) \( 136\pi \text{ cm}^2 \)
In simple words: To find the curved surface area of a cone, we use the formula \( \pi rl \). First, we find the slant height (l) using the height and radius with the Pythagorean theorem. Once we have the slant height, we can easily calculate the curved surface area.

🎯 Exam Tip: Remember that the diameter is twice the radius, so always halve the given diameter to find the radius before calculations. Also, be careful with units; here, it's \( \text{cm}^2 \).

Question 2. If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is
(a) \( 4\pi r^2 \text{ sq. units} \)
(b) \( 6\pi r^2 \text{ sq. units} \)
(c) \( 3\pi r^2 \text{ sq. units} \)
(d) \( 8\pi r^2 \text{ sq. units} \)
Answer: (a) \( 4\pi r^2 \text{ sq. units} \)
In simple words: When two halves of a sphere (hemispheres) are put together by their flat bases, they form a complete, solid sphere. The curved surface area of this new solid is simply the curved surface area of a full sphere. A sphere is a perfectly round 3D shape, like a ball.

🎯 Exam Tip: Visualizing the shapes helps a lot! Think about what happens when you combine two objects and what surfaces remain exposed.

Question 3. The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be
(a) 12 cm
(b) 10 cm
(c) 13 cm
(d) 5 cm
Answer: (a) 12 cm
In simple words: For a right circular cone, the radius, height, and slant height form a right-angled triangle. We can find any of these three measurements if we know the other two by using the Pythagoras theorem.

🎯 Exam Tip: Recognize the relationship \( l^2 = h^2 + r^2 \) for a right cone. Often, common Pythagorean triplets like (5, 12, 13) appear in these problems.

Question 4. If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is
(a) 1:2
(b) 1:4
(c) 1:6
(d) 1:8
Answer: (b) 1:4
In simple words: The volume of a cylinder depends on the square of its radius. So, if you make the radius half, the new volume becomes one-fourth of the old volume. The height remaining the same does not change this squared effect of the radius.

🎯 Exam Tip: Remember the volume formula \( V = \pi r^2 h \). If a quantity is squared in the formula, changing it by a factor will change the volume by that factor squared.

Question 5. The total surface area of a cylinder whose radius is \( \frac{1}{3} \) of its height is
(a) \( \frac{9 \pi h^{2}}{8} \text{ sq. units} \)
(b) \( 24\pi h^{2} \text{ sq.units} \)
(c) \( \frac{8 \pi h^{2}}{8} \text{ sq.units} \)
(d) \( \frac{56 \pi h^{2}}{8} \text{ sq.units} \)
Answer: (c) \( \frac{8 \pi h^{2}}{8} \text{ sq.units} \)
In simple words: We find the total surface area by adding the areas of the two circular bases and the curved side. First, we use the given relationship between the radius and height to express everything in terms of height. Then we substitute these into the total surface area formula for a cylinder.

🎯 Exam Tip: Always write down the given relationship (like \( r = \frac{1}{3} h \)) and the formula for Total Surface Area (TSA = \( 2\pi r(h+r) \)) clearly before substituting values. This helps avoid errors.

Question 6. In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is
(a) \( 5600\pi \text{ cm}^3 \)
(b) \( 11200\pi \text{ cm}^3 \)
(c) \( 56\pi \text{ cm}^3 \)
(d) \( 3600\pi \text{ cm}^3 \)
Answer: (a) \( 5600\pi \text{ cm}^3 \)
In simple words: To find the volume of material in a hollow cylinder, we subtract the volume of the inner empty space from the volume of the outer cylinder. This is done using the height and the difference of the squares of the external and internal radii. A useful trick is that the difference of squares can be factored into \( (R+r)(R-r) \), which simplifies calculations.

🎯 Exam Tip: For hollow shapes, the volume of the material is always the volume of the outer shape minus the volume of the inner empty space. Don't forget to use \( R^2 - r^2 = (R+r)(R-r) \) for easier calculation.

Question 7. If the radius of the base of a cone is tripled and the height is doubled then the volume is
(a) made 6 times
(b) made 18 times
(c) made 12 times
(d) unchanged
Answer: (b) made 18 times
In simple words: The volume of a cone is found using the formula \( \frac{1}{3}\pi r^2 h \). If the radius (r) is tripled, its squared value becomes 9 times larger. If the height (h) is doubled, it becomes 2 times larger. So, the total volume changes by \( 9 \times 2 \), which means it increases 18 times.

🎯 Exam Tip: Pay close attention to powers in formulas. If a variable is squared (like \( r^2 \)), doubling it means the squared term increases by \( 2^2=4 \), not just 2.

Question 8. The total surface area of a hemisphere is how many times the square of its radius.
(a) \( \pi \)
(b) \( 4\pi \)
(c) \( 3\pi \)
(d) \( 2\pi \)
Answer: (c) \( 3\pi \)
In simple words: A hemisphere is half of a sphere. Its total surface area includes the curved surface and its flat circular base. The formula for the total surface area of a hemisphere is \( 3\pi r^2 \), which means it is \( 3\pi \) times the square of its radius.

🎯 Exam Tip: Differentiate between the curved surface area of a hemisphere (\( 2\pi r^2 \)) and its total surface area (\( 3\pi r^2 \)), which includes the flat base.

Question 9. A solid sphere of radius x cm is melted and cast into a shape of a solid cone of the same radius. The height of the cone is
(a) 3x cm
(b) x cm
(c) 4x cm
(d) 2x cm
Answer: (c) 4x cm
In simple words: When a solid is melted and reshaped, its volume stays the same. So, the volume of the sphere will be equal to the volume of the cone. By setting their formulas equal and using the given radius, we can find the height of the cone.

🎯 Exam Tip: In melting and recasting problems, the key is always the conservation of volume. Equate the volumes of the initial and final shapes.

Question 10. A frustum of a right circular cone is of height 16cm with radii of its ends as 8cm and 20cm. Then, the volume of the frustum is
(a) \( 3328\pi \text{ cm}^3 \)
(b) \( 3228\pi \text{ cm}^3 \)
(c) \( 3240\pi \text{ cm}^3 \)
(d) \( 3340\pi \text{ cm}^3 \)
Answer: (a) \( 3328\pi \text{ cm}^3 \)
In simple words: A frustum is the part of a cone that remains when the top part is cut off by a plane parallel to the base. Its volume formula uses its height and the radii of both its circular ends. We simply plug in the given values into the formula to get the answer.

🎯 Exam Tip: Be careful with the frustum volume formula: \( V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \). Ensure you use the capital R for the larger radius and small r for the smaller radius correctly.

Question 11. A shuttlecock used for playing badminton has the shape of the combination of
(a) a cylinder and a sphere
(b) a hemisphere and a cone
(c) a sphere and a cone
(d) frustum of a cone and a hemisphere
Answer: (d) frustum of a cone and a hemisphere
In simple words: If you look at a shuttlecock, its top part (the cork) is shaped like a hemisphere. The feathery part resembles a frustum, which is a cone with its tip cut off. So, a shuttlecock is a combination of these two specific shapes.

🎯 Exam Tip: Practice identifying common 3D shapes within everyday objects to easily answer combination questions.

Question 12. A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is
(a) 2:1
(b) 1:2
(c) 4:1
(d) 1:4
Answer: (a) 2:1
In simple words: When one large sphere is melted to make several smaller spheres, the total amount of material (volume) remains constant. So, the volume of the big sphere equals the sum of the volumes of all the small spheres. By comparing these volumes, we can find the ratio of their radii.

🎯 Exam Tip: Always remember that melting and recasting preserves volume. For spheres, this leads to a relationship involving the cube of the radii, so you'll often need to find cube roots.

Question 13. The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is
(a) \( \frac{4}{3} \pi \)
(b) \( \frac{10}{3} \pi \)
(c) \( 5\pi \)
(d) \( \frac{20}{3} \pi \)
Answer: (a) \( \frac{4}{3} \pi \)
In simple words: The largest sphere that can be cut from a cylinder will have a diameter equal to the smallest dimension of the cylinder. In this case, it's the diameter of the cylinder's base. The height of the cylinder must also be at least the diameter of the sphere. Then, we use the sphere's volume formula.

🎯 Exam Tip: To find the maximum size of a shape nested inside another, always check which dimension (radius or height) limits the size of the inner shape. For a sphere in a cylinder, the sphere's diameter is limited by the cylinder's diameter and its height.

Question 14. The height and radius of the cone of which the frustum is a part are h₁ units and r₁ units respectively. Height of the frustum is h₂ units and the radius of the smaller base is r₂ units. If h₂ : h₁ = 1 : 2 then r₂ : r₁ is
(a) 1:3
(b) 1:2
(c) 2:1<
(d) 3:1
Answer: (b) 1:2
In simple words: When a frustum is formed by cutting a smaller cone from the top of a larger cone, the remaining frustum and the cut-off small cone are similar in shape to the original large cone. For similar cones, the ratio of their heights is equal to the ratio of their corresponding radii. We use this property to find the required ratio.

🎯 Exam Tip: For similar figures (like cones derived from each other), the ratio of corresponding linear dimensions (radii, heights) is always constant. This is a powerful property for many geometry problems.

Question 15. The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is
(a) 1:2:3
(b) 2:1:3
(c) 1:3:2
(d) 3:1:2
Answer: (d) 3:1:2
In simple words: We compare the volume formulas for a cylinder, cone, and sphere. If they all have the same radius and the height of the cylinder and cone is equal to the sphere's diameter (which is twice its radius), a clear ratio emerges. We divide each volume by the smallest common factor to get the simplest ratio. This relationship is a fundamental concept in solid geometry.

🎯 Exam Tip: This is a standard result to remember for quick problem-solving. Make sure to understand why the height of the sphere is taken as its diameter (2r) when comparing to the height of the cylinder and cone.