Question 1

Consider the volumes of the following  1. A parallelopiped of length 5 cm, breadth 3 cm and height 4 cm 2. A cube of each side 4 cm 3. A cylinder of radius 3 cm and length 3 cm 4. A sphere of radius 3 cm  The volumes of these in the decreasing order…

Accepted Answer

Concept & Volume Calculations: Evaluate the volume of each fundamental 3D shape independently using their respective formulas, and then sort the numerical results. V_{cuboid} = l × b × h V_{cube} = a^3 V_{cylinder} = \pi r^2 h V_{sphere} = (4/3) \pi r^3 Step-by-Step Solution: - 1. Parallelopiped (Cuboid): l = 5, b = 3, h = 4. V_1 = 5 × 3 × 4 = 60 cm^3. - 2. Cube: side a = 4. V_2 = 4^3 = 4 × 4 × 4 = 64 cm^3. - 3. Cylinder: radius r = 3, length (height) h = 3. Use \pi \approx 3.14. V_3 = \pi × 3^2 × 3 = 27\pi \approx 27 × 3.1415 = 84.82 cm^3. - 4. Sphere: radius r = 3. V_4 = (4/3) \pi × 3^3 = (4/3) \pi × 27 = 36\pi \approx 36 × 3.1415 = 113.09 cm^3. - Comparing the values: V_4 (113) > V_3 (84.8) > V_2 (64) > V_1 (60). - The decreasing order is 4, 3, 2, 1. Exam Strategy & Shortcut: To quickly approximate without deep decimals, use \pi > 3. For cylinder: 27 × 3 = 81 (so it's > 81). For sphere: 36 × 3 = 108 (so it's > 108). This immediately establishes that 36\pi > 27\pi > 64 > 60, giving the correct sequence instantly. Common Pitfall: Using the diameter instead of the radius for the sphere or cylinder, or incorrectly calculating 4^3 as 16 instead of 64. Final Answer: Therefore, the correct answer is 4, 3, 2, 1.

Question 2

A hollow spherical metallic ball has an external diameter $6$ cm and is $\frac{1}{2}$ cm thick. The volume of metal used in the ball is

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Concept & Volume of a Hollow Sphere: The volume of material in a hollow sphere is the difference between the external volume and the internal volume. V = (4/3)\pi (R^3 - r^3) Where R is the external radius and r is the internal radius. Internal radius r = R - thickness. Step-by-Step Solution: • Given: External diameter = 6 cm \Rightarrow External radius R = 3 cm. • Thickness = 0.5 cm. • Internal radius r = 3 - 0.5 = 2.5 cm = (5/2) cm. • Volume of metal = (4/3)\pi (R^3 - r^3) • = (4/3) × (22/7) × (3^3 - ((5/2))^3) • = (4/3) × (22/7) × (27 - (125/8)) • = (4/3) × (22/7) × ((216 - 125/8)) • = (4/3) × (22/7) × (91/8) • Simplify the expression: 91 \div 7 = 13, and 4 \div 8 = (1/2). • Volume = (1/3) × 22 × 13 × (1/2) = (11 × 13/3) = (143/3) • Converting to a mixed fraction: 143 \div 3 = 47 with a remainder of 2. • Volume = 47 (2/3) cm^3. Exam Strategy & Shortcut: Keep terms in fraction format (like 5/2) rather than decimals (like 2.5) when substituting into volume equations. Cubing and subtracting fractions is structurally easier to simplify with factors like 22/7. Common Pitfall: Using the diameter (6 cm) directly in the formula instead of the radius (3 cm), or subtracting the thickness from the diameter instead of the radius to find the inner dimension. Final Answer: Therefore, the correct answer is 47 (2/3) cm^3.

Question 3

How many lead shots each $3$ mm in diameter can be made from a cuboid of dimensions $9 	ext{ cm} 	imes 11 	ext{ cm} 	imes 12 	ext{ cm}$?

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Concept & Conservation of Volume and Unit Conversion: To find the number of identical small objects formed from melting a larger object, divide the volume of the larger object by the volume of a single small object. All measurements must be converted to the same unit before calculation. n = \frac{Volume of Cuboid}{Volume of one Lead Shot} Step-by-Step Solution: • Given: Cuboid dimensions = 9 cm × 11 cm × 12 cm. • Volume of Cuboid = 9 × 11 × 12 = 1188 cm^3. • Diameter of lead shot = 3 mm. Convert to cm: 3 mm = 0.3 cm. • Radius of lead shot r = (0.3/2) = 0.15 cm (or (3/20) cm). • Volume of one spherical shot = (4/3) × \pi × r^3. • = (4/3) × (22/7) × ((3/20))^3 = (4/3) × (22/7) × (27/8000). • = (88/21) × (27/8000) = (11/7) × (9/1000) = (99/7000) cm^3. • Number of shots n = (1188/\frac{99){7000}} = 1188 × (7000/99). • Divide 1188 by 99: 1188 \div 99 = 12. • n = 12 × 7000 = 84000. Exam Strategy & Shortcut: Work entirely in fractions and do not calculate intermediate decimal values. Set up the grand fraction: N = (9 × 11 × 12/\frac{4){3} × (22/7) × (3/20) × (3/20) × (3/20)}. Notice how the 11 in the numerator beautifully cancels the 22 in the denominator, reducing the math burden drastically. Common Pitfall: Failing to convert millimeters to centimeters. Mixing units will result in an answer off by a massive magnitude (like 72 or 84 instead of 84000). Final Answer: Therefore, the correct answer is 84000.

Question 4

A solid metallic sphere of radius $r$ is converted into a solid right circular cylinder of radius $R$. If the height of the cylinder is twice the radius of the sphere, then

Accepted Answer

Concept & Algebraic Volume Conservation: When reforming solids, equate their volume formulas. V_{sphere} = (4/3)\pi r^3 V_{cylinder} = \pi R^2h Step-by-Step Solution: 1. Given the sphere has radius r, its volume is (4/3)\pi r^3. 2. Given the cylinder has radius R and its height h = 2r. 3. The volume of the cylinder is \pi R^2(2r). 4. Equating the volumes: (4/3)\pi r^3 = \pi R^2 (2r). 5. Cancel \pi and r from both sides: (4/3)r^2 = 2R^2. 6. Divide by 2: R^2 = (2/3)r^2. 7. Take the square root of both sides: R = \sqrt{(2/3)r^2} = r\sqrt{(2/3)}. Exam Strategy & Shortcut: Substitute the height into the cylinder volume formula immediately and equate: 2\pi R^2 r = (4/3)\pi r^3. Canceling terms on sight yields R^2 = (2/3)r^2, skipping intermediate algebraic steps. Common Pitfall: Misinterpreting the wording "height of the cylinder is twice the radius of the sphere" and assigning h = 2R instead of h = 2r. Final Answer: Therefore, the correct answer is R = r\sqrt{(2/3)}.

Question 5

The volume 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

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Concept & Inscribed Geometries: To cut the "greatest sphere" from a cylinder, the sphere's diameter cannot exceed the smallest dimension of the cylinder. It is constrained by either the cylinder's diameter or its height. Volume_{sphere} = (4/3)\pi r^3 Step-by-Step Solution: 1. Identify the cylinder's constraints: Base radius = 1 cm (Diameter = 2 cm), Height = 5 cm. 2. The greatest sphere must fit entirely inside this cylinder. 3. If the sphere's diameter were equal to the cylinder's height (5 cm), it would not fit inside the cylinder's 2 cm diameter width. 4. Therefore, the maximum diameter of the sphere is limited by the cylinder's diameter, which is 2 cm. 5. The radius of this maximal sphere is 1 cm. 6. Calculate the volume of this sphere: V = (4/3)\pi (1)^3 = (4/3)\pi. Exam Strategy & Shortcut: Always look for the smallest limiting dimension when fitting shapes inside one another. The limiting factor here is the diameter of 2 cm. The height of 5 cm is extra space that cannot be utilized by a perfect sphere. Common Pitfall: Assuming the maximum sphere volume involves averaging the dimensions or attempting to equate volumes, instead of recognizing it as a geometric constraint problem limited by the smallest internal width. Final Answer: Therefore, the correct answer is (4/3)\pi.

Question 6

The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is

Accepted Answer

Concept & Circumscribed Geometry: When a cylinder circumscribes a sphere, it perfectly encloses it. This means the cylinder's radius is equal to the sphere's radius, and the cylinder's height is equal to the sphere's diameter. Surface Area_{sphere} = 4\pi r^2 Curved Surface Area_{cylinder} = 2\pi rh Step-by-Step Solution: 1. Let the radius of the sphere be r. The surface area of the sphere is 4\pi r^2. 2. For the circumscribing cylinder, the base radius R = r. 3. The height of the circumscribing cylinder h is exactly the diameter of the enclosed sphere, so h = 2r. 4. Calculate the curved surface area (CSA) of the cylinder: CSA = 2\pi R h = 2\pi (r) (2r) = 4\pi r^2. 5. Find the required ratio: \frac{Surface Area_{sphere}}{Curved Surface Area_{cylinder}} = (4\pi r^2/4\pi r^2) = (1/1). 6. The ratio is 1 : 1. Exam Strategy & Shortcut: This is a standard geometric theorem formulated by Archimedes. Memorize the fact that a sphere and its perfectly circumscribing cylinder have exactly the same surface area (ratio 1:1) and their volumes are in a 2:3 ratio. Common Pitfall: Confusing the "curved surface area" with the "total surface area" of the cylinder. If total surface area were used (2\pi r(r+h) = 6\pi r^2), the ratio would be 4:6 or 2:3. Final Answer: Therefore, the correct answer is 1 : 1.

Question 7

In what ratio are the volumes of a cylinder, a cone and a sphere, if each has the same diameter and the same height?

Accepted Answer

Concept & Proportional Volumes: To compare the volumes of different geometric shapes with shared dimensions, express all dimensions in terms of a single variable, compute the standard formulas, and evaluate the final ratio. Step-by-Step Solution: 1. Identify Common Variables: Let diameter = d. Then radius r = (d/2) for all shapes. For a sphere, the height is its diameter. Thus, height h = d = 2r. Since all shapes have the same height, h = 2r for the cylinder and cone as well. 2. Express Volumes in terms of r: - Cylinder: V_{cyl} = \pi r^2h = \pi r^2(2r) = 2\pi r^3 - Cone: V_{cone} = (1/3)\pi r^2h = (1/3)\pi r^2(2r) = (2/3)\pi r^3 - Sphere: V_{sph} = (4/3)\pi r^3 3. Calculate the Ratio: V_{cyl} : V_{cone} : V_{sph} = 2\pi r^3 : (2/3)\pi r^3 : (4/3)\pi r^3 Divide by 2\pi r^3: = 1 : (1/3) : (2/3) Multiply by 3 to clear the fractions: = 3 : 1 : 2 Exam Strategy & Shortcut: If you recall Archimedes' principle regarding the cylinder and inscribed sphere, a sphere is (2/3) the volume of its circumscribed cylinder, and a cone is (1/3) of it. Thus the ratio of Cylinder:Cone:Sphere is 1 : (1/3) : (2/3), which scales immediately to 3 : 1 : 2. Common Pitfall: Misidentifying the "height" of the sphere. The height of a sphere is equal to its diameter (2r), not its radius (r). Final Answer: Therefore, the correct answer is 3 : 1 : 2.

Question 8

A right pyramid has an equilateral triangular base of side $4	ext{ units}$. If the number of square units of its whole surface area is three times the number of cubic units of its volume, find its height.

Accepted Answer

Concept & Formula: The problem asks us to equate the numeric value of the total surface area to three times the volume. Total Surface Area = Base Area + Lateral Area Volume = (1/3) × Base Area × Height *(Note: There is a known typographical error in standard textbook printings of this question where none of the options perfectly match the mathematical derivation of 8 units. We illustrate the structural logic below.)* Step-by-Step Solution: 1. Base Parameters: - Side a = 4 units. - Base Area (B) = \frac{\sqrt{3}}{4} × 4^2 = 4\sqrt{3}. - Inradius r (distance from center to side) = (a/2\sqrt{3)} = (4/2\sqrt{3)} = (2/\sqrt{3)}. 2. Volume (V): - V = (1/3) × B × h = (1/3) × 4\sqrt{3} × h = \frac{4\sqrt{3}}{3}h. 3. Surface Area (S): - Slant height l = \sqrt{h^2 + r^2} = \sqrt{h^2 + (4/3)}. - Perimeter P = 3 × 4 = 12. - Lateral Area (L) = (1/2) × P × l = 6\sqrt{h^2 + (4/3)}. - S = 4\sqrt{3} + 6\sqrt{h^2 + (4/3)}. 4. Equating S = 3V: - 4\sqrt{3} + 6\sqrt{h^2 + (4/3)} = 3 × (\frac{4\sqrt{3}}{3}h) - 4\sqrt{3} + 6\sqrt{h^2 + (4/3)} = 4\sqrt{3}h - Dividing by 2\sqrt{3}: 2 + 3\sqrt{(3h^2+4/9)} = 2h \implies \sqrt{3h^2+4} = 2h - 2 - Squaring both sides: 3h^2 + 4 = 4h^2 - 8h + 4 - h^2 - 8h = 0 \implies h = 8. - Mathematically, the height is 8 units. In many competitive exams, typographical errors occur in the options. Exam Strategy & Shortcut: When setting up algebraic equations equating area to volume, isolate the radical before squaring to prevent messy polynomial expansions. If your calculated answer (8) isn't in the options, re-read carefully to ensure no misinterpretations (like confusing total vs lateral area). Common Pitfall: Forgetting the base area when calculating the "whole" surface area. If the problem meant "lateral surface area is three times the volume", 6\sqrt{h^2 + 4/3} = 4\sqrt{3}h \implies h=2, which is also not an option. Final Answer: Therefore, mathematically the height is 8 units (none of the options accurately reflect this standard textbook misprint).

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