Question 1

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

Accepted Answer

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 2

The volume of the largest sphere which can be carved out of a cube of side $6$ cm is

Accepted Answer

Concept & Inscribed Sphere: The largest sphere that can be carved from a cube perfectly touches the centers of the cube's six faces. Therefore, the diameter of this largest sphere is exactly equal to the side length of the cube. d_{sphere} = a_{cube} Volume of a sphere = (4/3)\pi r^3. Step-by-Step Solution: • Given: Side of the cube a = 6 cm. • Diameter of the largest possible sphere d = 6 cm. • Radius of the sphere r = (6/2) = 3 cm. • Volume of the sphere = (4/3) × \pi × r^3. • Volume = (4/3) × (22/7) × 3^3 = (4/3) × (22/7) × 27. • Volume = 4 × (22/7) × 9 = (792/7). • Convert to decimal: 792 \div 7 = 113.1428... cm^3. Exam Strategy & Shortcut: Recognize that the volume of a sphere inscribed in a cube is about 52.4% of the cube's volume ((\pi/6)). The cube's volume is 6^3 = 216. Roughly half of 216 is 108. The only option close to this estimation is 113.14. Common Pitfall: Using the cube's diagonal to find the sphere's radius. The diagonal determines the *circumscribed* sphere (outside the cube), but a *carved* sphere is inscribed (inside), limited by the flat faces, not the corners. Final Answer: Therefore, the correct answer is 113.14 cm^3.

Question 3

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

Accepted Answer

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

The volume of the largest possible cube that can be inscribed in a hollow spherical ball of radius $r$ cm is

Accepted Answer

Concept & Cube Inscribed in a Sphere: When the largest possible cube is inscribed inside a sphere, the longest interior distance of the cube (its main body diagonal) perfectly spans the maximum width of the sphere (its diameter). Body Diagonal of Cube = Diameter of Sphere \sqrt{3}a = 2r Where a is the side of the cube and r is the radius of the sphere. Step-by-Step Solution: • Given: Radius of the sphere is r. Let the side of the inscribed cube be a. • Equate the body diagonal of the cube to the sphere's diameter: \sqrt{3}a = 2r. • Isolate the side length of the cube, a: a = (2r/\sqrt{3)}. • The volume of the cube V = a^3. • Substitute the value of a into the volume formula: V = ((2r/\sqrt{3)})^3 • Cube the numerator and the denominator separately: Numerator = (2r)^3 = 8r^3. Denominator = (\sqrt{3})^3 = \sqrt{3} × \sqrt{3} × \sqrt{3} = 3\sqrt{3}. • Resulting volume V = (8r^3/3\sqrt{3)}. • This matches the format (8/3\sqrt{3)}r^3. Exam Strategy & Shortcut: Use dimensional analysis. Volume must be a cubic unit (r^3). Options (a) and (b) are squared units (r^2), which represent area, not volume. They can be instantly eliminated. Between the remaining options, knowing the diagonal relates via \sqrt{3} directly leads to 3\sqrt{3} in the denominator when cubed. Common Pitfall: Confusing an inscribed cube with an inscribed sphere. If the sphere is inscribed, a=2r. If the cube is inscribed, \sqrt{3}a=2r. Mixing these up leads to entirely incorrect structural equations. Final Answer: Therefore, the correct answer is (8/3\sqrt{3)}r^3.

Question 5

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 6

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

Accepted Answer

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 7

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 8

A spherical iron ball is dropped into a cylindrical vessel of base diameter 14 cm containing water. The water level is increased by $9\frac{1}{3}$ cm. What is radius of the ball?

Accepted Answer

Concept & Volume Displacement: When a solid object is submerged in a fluid inside a uniform container, the volume of the displaced fluid equals the volume of the object. Volume\ of\ Object = Area\ of\ Base × Rise\ in\ Height Step-by-Step Solution: 1. Identify the container (cylinder) dimensions: Base diameter = 14 cm \implies radius r = 7 cm. 2. Identify the rise in water level: h = 9(1/3) cm = (28/3) cm. 3. Calculate the volume of the displaced water: V_{water} = \pi r^2 h = \pi (7)^2 ((28/3)) = 49\pi ((28/3)) = (1372\pi/3) cm^3. 4. This displaced volume equals the volume of the spherical iron ball. Let the ball's radius be R. 5. V_{sphere} = (4/3)\pi R^3. 6. Equate the volumes: (4/3)\pi R^3 = (1372\pi/3). 7. Cancel (\pi/3) from both sides: 4R^3 = 1372. 8. Divide by 4: R^3 = (1372/4) = 343. 9. Find the cube root: R = \sqrt[3]{343} = 7 cm. Exam Strategy & Shortcut: Set up the displacement equation directly: (4/3)\pi R^3 = \pi (7^2)((28/3)). The \pi and the denominator 3 cancel out instantly. 4R^3 = 49 × 28. Divide 28 by 4 to get R^3 = 49 × 7 = 343. Thus, R = 7. Common Pitfall: Misconverting the mixed fraction 9(1/3) into an improper fraction. Always multiply the whole number by the denominator and add the numerator: ((9 × 3) + 1/3) = (28/3). Final Answer: Therefore, the correct answer is 7 cm.

Question 9

A metallic sphere of radius 10.5 cm is melted and recast into small right circular cones, each of base radius 3.5 cm and height 3 cm. The number of cones so formed is

Accepted Answer

Concept & Recasting Objects: When a single large object is recast into multiple identical smaller objects, the total volume is conserved. The number of smaller objects is the total initial volume divided by the volume of one smaller object. Step-by-Step Solution: 1. Given: Sphere radius (R) = 10.5 cm. Cone radius (r) = 3.5 cm, height (h) = 3 cm. 2. Set up the Equation: Let n be the number of cones. V_{sphere} = n × V_{cone} (4/3)\pi R^3 = n × ((1/3)\pi r^2h) 3. Substitute and Simplify: Cancel out (\pi/3) from both sides: 4 × (10.5)^3 = n × (3.5)^2 × 3 n = (4 × 10.5 × 10.5 × 10.5/3.5 × 3.5 × 3) 4. Calculate: Notice that 10.5 / 3.5 = 3. n = (4 × (3 × 3.5) × (3 × 3.5) × 10.5/3.5 × 3.5 × 3) n = (4 × 3 × 3 × 10.5/3) = 4 × 3 × 10.5 n = 12 × 10.5 = 126 Exam Strategy & Shortcut: Recognize ratio multiples immediately. 10.5 is exactly 3 times 3.5. The equation for n simplifies to: n = 4 × ((R/r))^2 × ((R/h)) = 4 × (3)^2 × ((10.5/3)) = 36 × 3.5 = 126. Common Pitfall: Converting decimal numbers to fractions (e.g., 21/2 and 7/2) works but can sometimes increase the chance of flipping numerator/denominator incorrectly. Rely on direct division if the numbers form clean integer ratios like 10.5 / 3.5 = 3. Final Answer: Therefore, the correct answer is 126.

Question 10

A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each. How many spheres are there?

Accepted Answer

Concept & Volume Equivalence in Recasting: When a large shape is melted down to create multiple smaller shapes, the total volume of the original shape is divided by the volume of one smaller shape to determine the total quantity produced. n = \frac{V_{cone}}{V_{sphere}} Step-by-Step Solution: 1. Volume of the Original Cone: r = 12 cm, h = 24 cm V_{cone} = (1/3)\pi r^2h = (1/3)\pi (12)^2(24) = 8\pi × 144 = 1152\pi cm^3 2. Volume of One Small Sphere: R = 2 cm V_{sphere} = (4/3)\pi R^3 = (4/3)\pi (2)^3 = (32/3)\pi cm^3 3. Calculate Number of Spheres (n): n = (1152\pi/\frac{32){3}\pi} = 1152 × (3/32) n = 36 × 3 = 108 Exam Strategy & Shortcut: Set up the fraction before fully expanding the multiplications: n = (\frac{1/3)\pi(12 × 12 × 24)}{(4/3)\pi(2 × 2 × 2)} Cancel the (\pi/3): n = (12 × 12 × 24/4 × 8) = (12 × 12 × 24/32) Simplify: n = (144 × 24/32) = (144 × 3/4) = 36 × 3 = 108. Common Pitfall: A standard calculation error occurs when dividing by a fraction; always remember to multiply by the reciprocal (e.g., multiply by 3/32 instead of dividing by 32/3). Final Answer: Therefore, the correct answer is 108.

Question 11

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 12

If the diameter of a sphere is 6 m, its hemisphere will have a volume of

Accepted Answer

Concept & Hemisphere Volume: A hemisphere is exactly half of a sphere. Its volume is derived by halving the standard volume formula of a full sphere. V_{hemisphere} = (1/2) × ( (4/3)\pi r^3 ) = (2/3)\pi r^3 Step-by-Step Solution: 1. Find Radius: Given diameter d = 6 m. Radius r = (d/2) = (6/2) = 3 m. 2. Calculate Hemisphere Volume: V = (2/3)\pi r^3 V = (2/3)\pi (3)^3 V = (2/3)\pi (27) V = 2 × 9\pi = 18\pi m^3 Exam Strategy & Shortcut: Calculate the sphere's volume (4/3)\pi (3^3) = 36\pi, and immediately cut it in half to get 18\pi. This prevents over-memorization of hemisphere-specific formulas. Common Pitfall: Using the given diameter (6 m) directly in the formula instead of converting it to the radius (3 m), which would incorrectly lead to 144\pi. Final Answer: Therefore, the correct answer is 18\pi.

Question 13

The total surface area of a solid hemisphere of diameter 14 cm, is

Accepted Answer

Concept & Total Surface Area of Hemisphere: The total surface area of a solid hemisphere includes both its curved surface area and its flat circular base. TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2 Step-by-Step Solution: 1. Given: Diameter (d) = 14 cm. 2. Find Radius: Radius (r) = (d/2) = (14/2) = 7 cm. 3. Calculate Total Surface Area: TSA = 3\pi r^2 TSA = 3 × (22/7) × (7)^2 TSA = 3 × (22/7) × 49 TSA = 3 × 22 × 7 TSA = 66 × 7 = 462 cm^2 Exam Strategy & Shortcut: When the radius is 7, 14, 21, etc., use \pi = (22/7) for quick cancellation. Memorizing that \pi r^2 for r=7 is 154 allows you to jump straight to 3 × 154 = 462 instantly. Common Pitfall: A common mistake is calculating only the curved surface area (2\pi r^2 = 308 cm^2) and forgetting to add the flat circular base area (\pi r^2) for a *solid* hemisphere. Final Answer: Therefore, the correct answer is 462 cm^2.

Question 14

A water tank open at the top is hemispherical at the bottom and cylindrical above it. If radius of the hemisphere is 12 m and the total capacity of the tank is $3312\pi$ m$^3$, then the ratio of the surface areas of the hemispherical and the cylindr…

Accepted Answer

Concept & Composite Volumes: The total volume of the tank is the sum of the volume of the hemisphere and the cylinder. Once the height of the cylinder is found, the surface areas can be compared. Volume of hemisphere: (2/3)\pi r^3 Volume of cylinder: \pi r^2 h Curved Surface Area of hemisphere: 2\pi r^2 Curved Surface Area of cylinder: 2\pi r h Step-by-Step Solution: 1. The radius of both the hemisphere and cylinder is r = 12 m. 2. The total volume V = (2/3)\pi r^3 + \pi r^2 h = 3312\pi. 3. Substitute r = 12: (2/3)\pi (12)^3 + \pi (12)^2 h = 3312\pi (2/3)(1728) + 144h = 3312 1152 + 144h = 3312 4. Solve for h: 144h = 3312 - 1152 = 2160 h = (2160/144) = 15 m. 5. Surface area of hemispherical portion (bottom) = 2\pi r^2 = 2\pi(12)^2 = 288\pi. 6. Surface area of cylindrical portion (curved) = 2\pi r h = 2\pi(12)(15) = 360\pi. 7. Find the ratio: (288\pi/360\pi) = (288/360). Dividing by 72 gives (4/5). Exam Strategy & Shortcut: Instead of calculating full surface areas, use the ratio formula directly: (2\pi r^2/2\pi rh) = (r/h). Once you find h = 15 and r = 12, the ratio is simply (12/15) = (4/5). Common Pitfall: A frequent error is including the flat circular top in the surface area of the cylinder, but the problem states the tank is "open at the top". Final Answer: Therefore, the correct answer is 4 : 5.

Question 15

If the radius of the base and height of a cylinder and cone are each equal to $r$, and the radius of a hemisphere is also equal to $r$, then the volumes of the cone, cylinder and hemisphere are in the ratio

Accepted Answer

Concept & Volume Comparison: By standardizing all dimensions to a single variable r, we can directly compare the volumetric formulas of these three basic 3D shapes. Volume of a cone: V = (1/3)\pi r^2 h Volume of a cylinder: V = \pi r^2 h Volume of a hemisphere: V = (2/3)\pi r^3 Step-by-Step Solution: 1. Express the volume of the cone, substituting h = r: V_{cone} = (1/3)\pi r^2 (r) = (1/3)\pi r^3. 2. Express the volume of the cylinder, substituting h = r: V_{cyl} = \pi r^2 (r) = \pi r^3. 3. Express the volume of the hemisphere, which already uses only r: V_{hemi} = (2/3)\pi r^3. 4. Form the ratio V_{cone} : V_{cyl} : V_{hemi}: (1/3)\pi r^3 : \pi r^3 : (2/3)\pi r^3. 5. Divide all terms by the common factor \pi r^3: (1/3) : 1 : (2/3). 6. Multiply the entire ratio by 3 to remove fractions: 1 : 3 : 2. Exam Strategy & Shortcut: This is a fundamental geometric relationship discovered by Archimedes. A cone, hemisphere, and cylinder with the same radius and height always have a volume ratio of 1 : 2 : 3 respectively. Be careful to check the order asked in the question (cone, cylinder, hemisphere), which reorders it to 1 : 3 : 2. Common Pitfall: Misreading the order requested in the prompt. The problem asks for "cone, cylinder, hemisphere", not the traditional "cone, hemisphere, cylinder" sequence which yields 1:2:3. Final Answer: Therefore, the correct answer is 1 : 3 : 2.

Question 16

A sphere of maximum volume is cut out from a solid hemisphere of radius $r$. The ratio of the volume of the hemisphere to that of the cut out sphere is :

Accepted Answer

Concept & Inscribed Sphere in Hemisphere: To fit a sphere of maximum volume inside a hemisphere of radius r, the sphere must sit perfectly within the height and width constraints. The maximum possible diameter of this sphere equals the radius of the hemisphere r. Therefore, the radius of the new sphere is r/2. Volume of hemisphere: V = (2/3)\pi r^3 Volume of sphere: V = (4/3)\pi R^3 Step-by-Step Solution: 1. The original solid is a hemisphere with radius r. Its volume is V_{hemi} = (2/3)\pi r^3. 2. The largest sphere that can be carved out will have its diameter equal to the hemisphere's height (which is r). So, the radius of this inscribed sphere is R = (r/2). 3. Calculate the volume of this new sphere: V_{sphere} = (4/3)\pi ((r/2))^3 = (4/3)\pi ((r^3/8)) = (1/6)\pi r^3. 4. Find the required ratio of the hemisphere's volume to the cut-out sphere's volume: \frac{V_{hemi}}{V_{sphere}} = (\frac{2/3)\pi r^3}{(1/6)\pi r^3}. 5. Cancel out \pi r^3 and simplify the fraction: (2/3/1/6) = (2/3) × (6/1) = (12/3) = 4. 6. The ratio is 4 : 1. Exam Strategy & Shortcut: Visualize the dimensions. A sphere inscribed in a hemisphere has half the radius. Because volume scales with the cube of the radius, halving the radius means the sphere's volume is (1/8) of a *full* sphere of radius r. Since a hemisphere is half a full sphere, the ratio is (1/2/1/8) = (1/2) × 8 = 4. Common Pitfall: Students often guess that the inscribed sphere has a radius of r divided by \sqrt{2} or make an error assuming it shares the same base area. Recognizing that the height r strictly limits the diameter is key. Final Answer: Therefore, the correct answer is 4 : 1.

Question 17

What is the volume in cubic cm of a pyramid whose area of the base is 25 sq cm and height 9 cm?

Accepted Answer

Concept & Volume of a Pyramid: The volume of any pyramid is one-third of the product of its base area and its perpendicular height, regardless of the shape of the base (square, triangular, etc.). Volume Formula: V = (1/3) × Base Area × Height Step-by-Step Solution: 1. Extract the given values from the problem: Base Area (A) = 25 sq cm Height (h) = 9 cm 2. Apply the volume formula for a pyramid: V = (1/3) × A × h. 3. Substitute the given numbers into the formula: V = (1/3) × 25 × 9. 4. Simplify the expression by multiplying (1/3) with 9 first: V = 25 × 3. 5. Calculate the final result: V = 75 cubic cm. Exam Strategy & Shortcut: You don't need to know what kind of polygon forms the base; the formula V = (Ah/3) applies universally. Always look to divide the height by 3 first to make mental math trivial (9/3 = 3, 25 × 3 = 75). Common Pitfall: A common mistake is treating the pyramid like a prism and forgetting to divide by 3 (which would incorrectly yield 225). Final Answer: Therefore, the correct answer is 75.

Question 18

A right pyramid is on a regular hexagonal base. Each side of the base is $10	ext{ m}$ and the height is $60	ext{ m}$. The volume of the pyramid is

Accepted Answer

Concept & Formula: The area of a regular hexagon can be found by treating it as six equilateral triangles. Base Area = 6 × \frac{\sqrt{3}}{4} × a^2 Volume = (1/3) × Base Area × h Step-by-Step Solution: 1. Given: - Side of hexagonal base, a = 10 m - Height, h = 60 m 2. Calculate Base Area: - Base Area = 6 × \frac{\sqrt{3}}{4} × (10)^2 - Base Area = 6 × \frac{\sqrt{3}}{4} × 100 = 6 × 25\sqrt{3} = 150\sqrt{3} m^2 3. Calculate Volume: - Volume = (1/3) × 150\sqrt{3} × 60 - Volume = 50\sqrt{3} × 60 = 3000\sqrt{3} 4. Convert to Decimal: - Since \sqrt{3} \approx 1.732 - Volume = 3000 × 1.732 = 5196 m^3 Exam Strategy & Shortcut: Instead of multiplying large numbers, group them logically: (1/3) × 60 = 20. Then 20 × 150\sqrt{3} = 3000\sqrt{3}. You can mentally shift the decimal place of 1.732 three places to the right to get 1732, and then multiply by 3 to get 5196. Common Pitfall: A common mistake is calculating the perimeter instead of the area of the hexagon, or forgetting the (1/3) in the pyramid volume formula. Final Answer: Therefore, the correct answer is 5196 m^3.

Question 19

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).

Question 20

The radius of a hemispherical bowls is $6	ext{cm}$. The capacity of the bowl is $\left(	ext{Take } \pi=\frac{22}{7}ight)$

Accepted Answer

Concept & Formula: The capacity of a bowl is synonymous with its internal volume. For a hemisphere, the volume is exactly half of a full sphere. Volume of Hemisphere = (2/3)\pi r^3 Step-by-Step Solution: 1. Given: - Radius r = 6cm - \pi = (22/7) 2. Apply the Formula: - Volume = (2/3) × (22/7) × (6)^3 - Volume = (2/3) × (22/7) × 216 3. Simplify and Calculate: - Divide 216 by 3 to get 72. - Volume = 2 × (22/7) × 72 - Volume = (44 × 72/7) = (3168/7) 4. Convert to Decimal: - 3168 \div 7 = 452.5714... - Rounding to two decimal places, we get 452.57 cm^3. Exam Strategy & Shortcut: When multiplying 44 × 72, you can split it as 44 × (70 + 2) = 3080 + 88 = 3168. When dividing by 7, notice that 7 × 400 = 2800 and 7 × 50 = 350, so the result must be slightly above 450. Option (b) is the only reasonable match without fully completing the long division. Common Pitfall: Using the formula for the volume of a full sphere ((4/3)\pi r^3) instead of a hemisphere, which would yield exactly double the correct answer. Final Answer: Therefore, the correct answer is 452.57cm^3.

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