Question 1

If the radius of a sphere is increased by 3%, by what percentage does its total surface area increase?

Accepted Answer

Introduction / Context: For similar shapes, areas scale with the square of the linear scale factor. A small percentage change in radius produces approximately double that percentage for area, but the exact value uses the square of the scale factor. Given Data / Assumptions: Initial radius r. New radius r′ = 1.03 r (3% increase). Surface area of a sphere: S = 4πr^2. Concept / Approach: Compute ratio S′/S = (4πr′^2)/(4πr^2) = (r′/r)^2 = (1.03)^2. Step-by-Step Solution: S′/S = (1.03)^2 = 1.0609Percentage increase = (1.0609 − 1) * 100% = 6.09% Verification / Alternative check: Binomial: (1 + 0.03)^2 = 1 + 2(0.03) + (0.03)^2 = 1 + 0.06 + 0.0009 = 1.0609 → 6.09%. Why Other Options Are Wrong: 7%, 9%: Do not match exact square scaling. 5.06%: This would be for (1.025)^2 − 1, not 3%. Common Pitfalls: Doubling the percent to 6% and ignoring the small square term 0.09%. Final Answer: 6.09%

Question 2

If the radius of a sphere decreases by 24%, by what percentage does its total surface area decrease?

Accepted Answer

Introduction / Context: Area scales as the square of a linear dimension. A decrease in radius reduces surface area by the square of the scale factor. Given Data / Assumptions: Initial radius r; decreased radius r′ = 0.76 r (since 24% decrease). Surface area S = 4πr^2 and S′ = 4πr′^2. Concept / Approach: Compute ratio S′/S = (r′/r)^2 = (0.76)^2 and convert to percent decrease. Step-by-Step Solution: S′/S = 0.76^2 = 0.5776Percent decrease = (1 − 0.5776) * 100% = 42.24% Verification / Alternative check: Square 76/100 precisely to avoid rounding errors; 76^2 = 5776; move decimal: 0.5776. Why Other Options Are Wrong: 44%, 49%, 46.2%: Not equal to the exact square-based reduction. Common Pitfalls: Subtracting 24% twice or assuming area decreases by exactly double (48%). Final Answer: 42.24%

Question 3

If the radius of a sphere is increased by 100% (i.e., doubled), by what percentage does its volume increase?

Accepted Answer

Introduction / Context: Volumes scale with the cube of a linear dimension. Doubling a radius multiplies the volume by 2^3 = 8 times the original volume. Given Data / Assumptions: Initial radius r, new radius r′ = 2r. Sphere volume V = (4/3)πr^3. Concept / Approach: Compute V′/V = (r′/r)^3 = 2^3 = 8, then convert to a percentage increase over the original (i.e., compared to 1×). Step-by-Step Solution: V′ = 8VIncrease factor = 8 − 1 = 7Percentage increase = 7 * 100% = 700% Verification / Alternative check: Plug r = 1 into V = (4/3)π; doubling to r = 2 gives V′ = (4/3)π * 8 = 8V → +700%. Why Other Options Are Wrong: 300%, 500%, 900%: Do not match the cubic scaling for doubling. Common Pitfalls: Confusing area scaling (square) with volume scaling (cube). Final Answer: 700%

Question 4

A solid metallic sphere of radius 4 cm weighs 4 kg. A hollow spherical shell of the same metal has outer diameter 16 cm and inner diameter 12 cm. Find the weight (in kg) of the hollow sphere.

Accepted Answer

Introduction / Context: For the same material, weight is proportional to volume (same density). We can find the density from the solid sphere, then multiply by the shell volume to get the new weight. Given Data / Assumptions: Solid sphere: radius 4 cm, weight 4 kg. Hollow sphere: outer radius R = 8 cm, inner radius r = 6 cm. Sphere volume: V = (4/3)πr^3. Concept / Approach: Density k = weight / volume for the first sphere. Then shell volume V_shell = (4/3)π(R^3 − r^3). Multiply k * V_shell for the new weight. Step-by-Step Solution: V_solid = (4/3)π * 4^3 = (4/3)π * 64 = (256/3)πk = 4 / [(256/3)π] = 12/(256π)R^3 − r^3 = 8^3 − 6^3 = 512 − 216 = 296V_shell = (4/3)π * 296 = (1184/3)πWeight = k * V_shell = [12/(256π)] * [(1184/3)π] = 37/2 = 18.5 kg Verification / Alternative check: Units cancel; π cancels; exact fractional arithmetic yields 18.5 kg. Why Other Options Are Wrong: 20.5, 16.5, 15.5 kg: Do not match the exact shell volume times density. Common Pitfalls: Using surface area instead of volume; forgetting to convert diameters to radii; arithmetic slips in cube values. Final Answer: 18.5 kg

Question 5

A right triangular prism has a right-angled triangular base with legs 10 cm and 12 cm, and height 20 cm. If the material density is 6 g/cm^3, find the weight (in kg) of the prism.

Accepted Answer

Introduction / Context: The mass of a solid is density times volume. For a right prism, volume equals the area of its base multiplied by its height (length). Given Data / Assumptions: Right triangle legs: 10 cm and 12 cm. Prism height (length): 20 cm. Density: 6 g/cm^3. Concept / Approach: Compute base area A = (1/2)*a*b, then volume V = A*height. Convert grams to kilograms at the end (1000 g = 1 kg). Step-by-Step Solution: A = (1/2) * 10 * 12 = 60 cm^2V = 60 * 20 = 1200 cm^3Mass = density * volume = 6 * 1200 = 7200 g = 7.2 kg Verification / Alternative check: Units: (g/cm^3)*(cm^3) = g → convert to kg by dividing by 1000. Why Other Options Are Wrong: 6.4, 4.8, 3.4 kg are off the exact density-volume product. Common Pitfalls: Using hypotenuse instead of legs; forgetting the 1/2 in the triangular area; missing unit conversion to kg. Final Answer: 7.2 kg

Question 6

The perimeter of the triangular base of a right prism is 60 cm. The sides of the base are in the ratio 5 : 12 : 13. If the prism height is 60 cm, find the volume of the prism (in cm^3).

Accepted Answer

Introduction / Context: For a right prism, the volume equals the area of the base times the prism height. The 5:12:13 ratio indicates a Pythagorean triple (right triangle), simplifying the base area calculation. Given Data / Assumptions: Perimeter = 60 cm; ratio 5:12:13 → sides 10 cm, 24 cm, 26 cm. Right triangle legs: 10 cm and 24 cm; hypotenuse 26 cm. Prism height H = 60 cm. Concept / Approach: Base area A = (1/2) * (product of legs). Then volume V = A * H. Step-by-Step Solution: Scale: 5k + 12k + 13k = 30k = 60 ⇒ k = 2Sides: 10, 24, 26 (right triangle)A = (1/2) * 10 * 24 = 120 cm^2V = 120 * 60 = 7200 cm^3 Verification / Alternative check: 5-12-13 confirms right angle; area via Heron also yields 120 cm^2, validating result. Why Other Options Are Wrong: 6000, 6600, 5400 cm^3 do not match the exact base area times height. Common Pitfalls: Using hypotenuse in the area formula; mis-scaling side lengths from the ratio; mistaking perimeter for semi-perimeter in Heron’s formula. Final Answer: 7200 cm3

Question 7

A prism and a pyramid have the same base area and the same height. What is the ratio of the volume of the prism to that of the pyramid?

Accepted Answer

Introduction / Context: Comparing volumes of solids with the same base and height is a classic geometry result. Pyramids (including cones) have one-third the volume of the corresponding prism (or cylinder) with equal base and height. Given Data / Assumptions: Same base area B and same height H. Concept / Approach: Volume formulas: V_prism = B * H; V_pyramid = (1/3) * B * H. Take their ratio. Step-by-Step Solution: V_prism / V_pyramid = (B*H) / [(1/3)B*H] = 3Therefore, ratio = 3 : 1 Verification / Alternative check: Analogy: Cylinder vs cone with same base and height also yields 3:1 for volumes. Why Other Options Are Wrong: 1:1 and 1:3 contradict the standard formula. Cannot be determined: Sufficient information is provided via general formulas. Common Pitfalls: Mixing up the factor 1/3 for pyramids and cones; thinking perimeter rather than area influences the ratio. Final Answer: 3 : 1

Question 8

A right circular cylinder and a right circular cone have the same base radius 6 cm and the same height 8 cm. Find the ratio of the curved surface area of the cylinder to that of the cone.

Accepted Answer

Introduction / Context: Curved (lateral) surface area of a cylinder depends on radius and height; for a cone it depends on radius and slant height. With the same radius and height, we compute each and then form the ratio. Given Data / Assumptions: Radius r = 6 cm; height h = 8 cm. CSA_cylinder = 2πrh. CSA_cone = πrl, where l = √(r^2 + h^2). Concept / Approach: Compute l, then both CSAs, then the ratio. Step-by-Step Solution: l = √(6^2 + 8^2) = √(36 + 64) = √100 = 10CSA_cylinder = 2π * 6 * 8 = 96πCSA_cone = π * 6 * 10 = 60πRatio = 96π : 60π = 8 : 5 Verification / Alternative check: Cancel π to simplify; ratio reduces cleanly to 8:5. Why Other Options Are Wrong: Other ratios do not match the computed CSA values with l = 10. Common Pitfalls: Using height instead of slant height in the cone’s CSA; forgetting to compute l by Pythagoras. Final Answer: 8 : 5

Question 9

A solid cylinder of radius 3 cm and height 5 cm is melted to form tiny right circular cones, each with base radius 1 mm and height 1 cm. How many such cones can be made?

Accepted Answer

Introduction / Context: Melting and recasting conserves volume. We compute the cylinder volume and divide by the volume of one small cone to find the count of cones produced. Given Data / Assumptions: Cylinder: r = 3 cm, h = 5 cm → V_cyl = πr^2h. Cone: base radius = 1 mm = 0.1 cm, height = 1 cm → V_cone = (1/3)πr^2h. Concept / Approach: Number of cones n = V_cyl / V_cone (volumes in the same units). Convert millimetres to centimetres before substitution. Step-by-Step Solution: V_cyl = π * 3^2 * 5 = 45π cm^3V_cone = (1/3)π * (0.1)^2 * 1 = (1/3)π * 0.01 = π/300 cm^3n = 45π / (π/300) = 45 * 300 = 13500 Verification / Alternative check: π cancels; ensure radius in cm (0.1 cm) to avoid 100× error. Why Other Options Are Wrong: They arise from unit mistakes (using 1 mm as 1 cm) or misplacing the 1/3 in cone volume. Common Pitfalls: Unit conversion errors; forgetting the 1/3 factor for cone volume; arithmetic slips during division. Final Answer: 13500

Question 10

How many spherical bullets of diameter 4 cm can be cast by melting a solid cube of edge 44 cm? (Use π = 22/7 where needed.)

Accepted Answer

Introduction / Context: Volume is conserved in melting/recasting. We divide the cube’s volume by the volume of one sphere (bullet) to get the count. The standard approximation π = 22/7 is commonly used in such problems for an exact integer result. Given Data / Assumptions: Cube edge a = 44 cm → V_cube = a^3. Bullet sphere diameter = 4 cm → radius r = 2 cm. Use V_sphere = (4/3)πr^3 with π = 22/7. Concept / Approach: n = V_cube / V_sphere; compute carefully to preserve integer results with π = 22/7. Step-by-Step Solution: V_cube = 44^3 = 85184 cm^3V_sphere = (4/3)π * 2^3 = (32/3)π = (32/3)*(22/7) = 704/21n = 85184 ÷ (704/21) = 85184 * (21/704) = 121 * 21 = 2541 Verification / Alternative check: Reduce 85184/704 = 121 exactly; integer multiplication by 21 yields 2541. Why Other Options Are Wrong: Near misses result from rounding π or arithmetic slips; only 2541 is exact with 22/7. Common Pitfalls: Using π ≈ 3.14 leads to a non-integer; forgetting radius is half the diameter. Final Answer: 2541

Question 11

The volumes of a sphere and a right circular cylinder (with the same radius) are equal. Find the ratio of the diameter of the sphere to the height of the cylinder.

Accepted Answer

Introduction / Context: Equate volumes of sphere and cylinder with common radius r. Express the cylinder height in terms of r, then form the ratio of sphere diameter to that height. Given Data / Assumptions: V_sphere = (4/3)πr^3. V_cyl = πr^2h. Equal volumes: V_sphere = V_cyl. Concept / Approach: Solve (4/3)πr^3 = πr^2h for h, then compute (diameter):(height). Step-by-Step Solution: (4/3)πr^3 = πr^2h ⇒ h = (4/3)rDiameter of sphere = 2rRatio = 2r : (4/3)r = 2 : 4/3 = 3 : 2 Verification / Alternative check: Cancel π and r^2 safely since r > 0; ratio reduces cleanly. Why Other Options Are Wrong: They do not reflect h = 4r/3 and D = 2r. Common Pitfalls: Using circumference or surface areas instead of volumes; forgetting diameter is 2r. Final Answer: 3 : 2

Question 12

A sphere and a right circular cylinder have the same radius and equal curved surface areas. Find the ratio of their volumes (sphere : cylinder).

Accepted Answer

Introduction / Context: Relate the curved surface areas (CSA) to tie the cylinder’s height to the common radius, then compute volumes and form the ratio sphere:cylinder. Given Data / Assumptions: CSA_sphere = 4πr^2. CSA_cylinder = 2πrh (lateral area). Equal CSAs and same radius r. Concept / Approach: Set 4πr^2 = 2πrh ⇒ h = 2r. Then V_sphere = (4/3)πr^3 and V_cyl = πr^2h = 2πr^3. Take the ratio. Step-by-Step Solution: h = 2r from CSA equalityV_sphere = (4/3)πr^3; V_cyl = 2πr^3Volume ratio = (4/3)πr^3 : 2πr^3 = 4/3 : 2 = 2 : 3 Verification / Alternative check: π and r^3 cancel; ratio simplifies exactly to 2:3. Why Other Options Are Wrong: They mismatch the derived height h = 2r and corresponding cylinder volume. Common Pitfalls: Using total surface area instead of curved area; confusing diameter with radius in the height relation. Final Answer: 2 : 3

Question 13

A cuboid of dimensions 9 cm × 8 cm × 6 cm has twice the volume of a cube. Find the total surface area (in cm^2) of the cube.

Accepted Answer

Introduction / Context: Relate the cuboid’s volume to twice the cube’s volume, then solve for the cube’s edge and compute its total surface area (TSA = 6a^2). Given Data / Assumptions: Cuboid: 9 × 8 × 6 cm ⇒ V_cuboid = 432 cm^3. V_cuboid = 2 × V_cube. Concept / Approach: Find V_cube = 432/2 = 216 cm^3. Then a^3 = 216 ⇒ a = 6 cm. Compute TSA = 6a^2. Step-by-Step Solution: V_cube = 216 cm^3a = 216^(1/3) = 6 cmTSA = 6 * a^2 = 6 * 36 = 216 cm^2 Verification / Alternative check: Check: cube volume 6^3 = 216; twice equals 432 (the cuboid’s volume). Why Other Options Are Wrong: They are close numerically but not the exact TSA from a = 6 cm. Common Pitfalls: Computing lateral area instead of total; arithmetic slips in squaring or cube root. Final Answer: 216 sq cm

Question 14

Seven identical cubes, each of side 5 cm, are joined end-to-end to form a cuboid. Find the total surface area (in cm^2) of the resulting cuboid.

Accepted Answer

Introduction / Context: Joining seven cubes of side 5 cm end-to-end yields a cuboid whose dimensions can be inferred. Then use the total surface area formula for a cuboid: TSA = 2(lw + lh + wh). Given Data / Assumptions: Each cube side = 5 cm. Seven cubes in a line ⇒ length = 7 * 5 = 35 cm; width = 5 cm; height = 5 cm. Concept / Approach: Compute TSA with l = 35, w = 5, h = 5. Step-by-Step Solution: TSA = 2(lw + lh + wh)= 2(35*5 + 35*5 + 5*5) = 2(175 + 175 + 25) = 2 * 375 = 750 cm^2 Verification / Alternative check: Count exposed faces mentally: long faces (2), side faces (2), end faces (2) match TSA formula result. Why Other Options Are Wrong: They stem from forgetting one pair of faces or miscomputing one product. Common Pitfalls: Treating it as seven separate cubes and double-counting internal faces; arithmetic slips in multiplication. Final Answer: 750 sq cm

Question 15

In a shower, 10 cm of rainfall falls uniformly over 1 hectare of ground. Compute the volume of water (give the correct value in m^3).

Accepted Answer

Introduction / Context: Uniform rainfall depth over an area yields a water “prism” of that depth. Volume is simply area multiplied by depth, with careful unit conversion to cubic metres (m^3). Given Data / Assumptions: Rain depth = 10 cm = 0.10 m. Area = 1 hectare = 10,000 m^2. Concept / Approach: Volume V = Area * Depth, ensuring both are in metres to produce m^3 directly. Step-by-Step Solution: V = 10,000 m^2 * 0.10 m = 1,000 m^3 Verification / Alternative check: Convert to cm^3: 1,000 m^3 = 1,000 × 10^6 cm^3 = 10^9 cm^3 (useful cross-check). Why Other Options Are Wrong: Values like 750 m^3 or 900 m^3 result from misusing hectare or depth units. Common Pitfalls: Confusing m^2 and cm^2; forgetting that 1 hectare = 10,000 m^2; converting depth incorrectly. Final Answer: 1000 m3

Question 16

Water flows through a cylindrical pipe of diameter 5 mm at a speed of 10 m per minute. How long will it take to fill a conical vessel of base diameter 40 cm and depth 24 cm?

Accepted Answer

Introduction / Context: Time = (volume to fill) / (flow rate). Compute the cone volume and the pipe’s volumetric flow rate from its cross-sectional area and linear speed. Given Data / Assumptions: Pipe diameter = 5 mm ⇒ radius = 2.5 mm = 0.0025 m. Flow speed = 10 m/min. Cone: base diameter 40 cm ⇒ radius 0.20 m; height 0.24 m. Concept / Approach: Flow rate Q = area * speed = πr^2 * v. Cone volume V = (1/3)πR^2h. Time t = V / Q. Step-by-Step Solution: Q = π * (0.0025)^2 * 10 = 6.25×10^-5 π m^3/minV = (1/3)π * (0.20)^2 * 0.24 = 0.0032 π m^3t = V/Q = (0.0032 π) / (6.25×10^-5 π) = 51.2 min = 51 min 12 s Verification / Alternative check: π cancels; ensure consistent SI units (metres). 0.2^2 = 0.04; 0.04*0.24/3 = 0.0032. Why Other Options Are Wrong: Small deviations come from rounding or mm–cm–m conversion errors. Common Pitfalls: Using radius as 5 mm (not diameter); forgetting 1/3 in cone volume; mixing cm and m. Final Answer: 51 min 12 s

Question 17

A cylindrical jar with base radius 15 cm is filled with water to a height of 20 cm. A solid iron sphere of radius 10 cm is fully submerged in the jar. By how many centimetres does the water level rise?

Accepted Answer

Introduction / Context: Submerging a solid displaces an equal volume of water. The rise in water height equals displaced volume divided by the jar’s base area. The initial 20 cm level is irrelevant to the rise amount as long as there is capacity. Given Data / Assumptions: Jar radius R = 15 cm. Sphere radius r = 10 cm. Sphere volume V_s = (4/3)πr^3. Jar base area A = πR^2. Concept / Approach: Rise Δh = V_displaced / A = V_s / (πR^2) since the sphere is fully submerged. Step-by-Step Solution: V_s = (4/3)π * 10^3 = (4000/3)π cm^3A = π * 15^2 = 225π cm^2Δh = [(4000/3)π] / (225π) = 4000 / 675 = 160/27 cm ≈ 5.926 cm Verification / Alternative check: Cancel π; reduce the fraction 4000/675 by 25 to get 160/27 exactly. Why Other Options Are Wrong: Other fractions correspond to incorrect radius usage or arithmetic; 160/27 is the exact simplified value. Common Pitfalls: Using diameter instead of radius; forgetting to square R in the base area; including the initial 20 cm in the rise computation. Final Answer: 160/27 cm

Question 18

A conical flask of base radius r and height h is full of water. The water is poured into a cylindrical flask whose base radius is m·r. What is the height of water in the cylinder?

Accepted Answer

Introduction / Context: Equate the cone’s volume to the cylinder’s volume to find the resulting water height. Work symbolically in terms of r, h, and m to derive a general formula. Given Data / Assumptions: V_cone = (1/3)πr^2h. V_cyl = π(mr)^2 * H = πm^2 r^2 H, where H is required height. Concept / Approach: Set (1/3)πr^2h = πm^2 r^2 H and solve for H; cancel common factors π and r^2 (r ≠ 0). Step-by-Step Solution: (1/3)πr^2h = πm^2 r^2 HH = [ (1/3) h ] / m^2 = h / (3 m^2) Verification / Alternative check: Dimensional check: H has units of h (length) divided by dimensionless m^2, so units are consistent. Why Other Options Are Wrong: They invert or misplace m or the factor 1/3 from cone volume. Common Pitfalls: Forgetting the 1/3 factor in the cone volume; not squaring m in the cylinder’s base area. Final Answer: h/(3 m^2)

Question 19

A hemispherical basin of diameter 150 cm holds 120 times the volume of a cylindrical tube. If the height of the tube is 15 cm, find the diameter (in cm) of the tube.

Accepted Answer

Introduction / Context: Relate the hemisphere’s volume to a multiple (120×) of a cylinder’s volume to solve for the cylinder’s radius, then double it to get the diameter. Given Data / Assumptions: Hemisphere radius R = 150/2 = 75 cm. Hemisphere volume V_h = (2/3)πR^3. Cylinder height h = 15 cm; cylinder radius = r (unknown). V_h = 120 × V_cylinder = 120 × πr^2h. Concept / Approach: Set (2/3)πR^3 = 120πr^2h, cancel π, solve for r, then diameter = 2r. Step-by-Step Solution: (2/3) * 75^3 = 120 * r^2 * 15Left side = 2/3 * 421875 = 281250Right side = 1800 r^2 ⇒ r^2 = 281250 / 1800 = 156.25r = 12.5 cm ⇒ diameter = 25 cm Verification / Alternative check: Square root of 156.25 is exactly 12.5; substitution confirms equality. Why Other Options Are Wrong: 24, 26, 27 cm are near but not exact; they fail the volume equality. Common Pitfalls: Using sphere instead of hemisphere volume; forgetting to double the radius to get diameter; arithmetic slips with large cubes. Final Answer: 25 cm

Question 20

A solid copper sphere has diameter 36 cm and is drawn into a uniform cylindrical wire of diameter 4 mm (0.4 cm). Assuming no loss of material, find the length of the wire (in metres).

Accepted Answer

Introduction / Context: When a solid is reshaped without loss, its volume remains constant. Here, a copper sphere is redrawn into a thin cylindrical wire; therefore, sphere volume = wire volume. The task is to compute the wire length from these equal volumes. Given Data / Assumptions: Sphere diameter = 36 cm ⇒ radius r_s = 18 cm. Wire diameter = 4 mm = 0.4 cm ⇒ radius r_w = 0.2 cm. No loss of material; volumes are equal. Answer required in metres. Concept / Approach: Volume of sphere: V_s = (4/3)*π*r_s^3. Volume of cylinder (wire): V_w = π*r_w^2*L. Set V_s = V_w and solve for length L, then convert cm → m. Step-by-Step Solution: V_s = (4/3)*π*(18)^3 cm^3V_w = π*(0.2)^2*L cm^3Equate: (4/3)*π*18^3 = π*0.2^2*LCancel π: (4/3)*18^3 = 0.04*LCompute 18^3 = 5832 ⇒ (4/3)*5832 = 77760.04*L = 7776 ⇒ L = 7776 / 0.04 = 194400 cmConvert to metres: L = 194400 / 100 = 1944 m Verification / Alternative check: Quick ratio check: Length ∝ r_s^3 / r_w^2. With r_s = 18 and r_w = 0.2, length scale is huge; a result in the order of 2 km is reasonable, matching 1944 m. Why Other Options Are Wrong: 2000 m: Rounds too loosely; exact computation gives 1944 m. 15000 m: Exaggerates by ~7.7×; inconsistent with volume equality. 1855 m: Underestimates the exact ratio. None of these: Incorrect since 1944 m is attainable. Common Pitfalls: Forgetting to convert 4 mm to 0.4 cm. Using diameter instead of radius in formulas. Failing to convert final length from cm to m. Final Answer: 1944 m

Volume and Surface Area Questions