Question 1

In comminution theory, which empirical law states that the crushing energy is proportional to the new surface area created during size reduction?

Accepted Answer

Introduction / Context:Energy–size reduction relationships help estimate power requirements and select suitable grinders. Three classic laws—Rittinger, Kick, and Bond—apply in different size ranges and fragmentation regimes.Given Data / Assumptions:The question asks specifically for energy proportional to new surface area.Applies best to fine grinding where surface area increase is large.Concept / Approach:Rittinger’s law: E ∝ ΔA (increase in surface area). Kick’s law: E ∝ ln(d1/d2). Bond’s law: E ∝ (1/√d2 − 1/√d1). Thus, only Rittinger directly ties energy to new surface created, matching systems where fracturing creates many fines.Step-by-Step Solution:Identify law with energy–surface proportionality.Match to Rittinger’s statement.Select Rittinger’s law.Verification / Alternative check:Fine grinding design commonly references Rittinger for micronization contexts, validating the selection.Why Other Options Are Wrong:Taggart’s rule: screening/misc. correlations, not energy law.Fick’s law: diffusion, unrelated to grinding energy.Kick/Bond: different functional forms, not directly to surface area.Common Pitfalls:Confusing Bond’s widely used work index with Rittinger’s fine-grinding emphasis.Final Answer:Rittinger’s law

Question 2

In industrial bag filters (fabric filters for dust collection), filter media are typically textiles or felts. Which of the following is never used as the filter fabric itself?

Accepted Answer

Introduction / Context:Baghouse filters remove particulates from gas streams using fabric bags. The choice of fabric affects temperature resistance, chemical compatibility, and collection efficiency.Given Data / Assumptions:Bag “fabric” refers to the actual filtering medium.Common fabrics include cotton, polyester, nylon, acrylic, aramid, PPS, etc.Concept / Approach:Bag filters require flexible, permeable fabrics. Metallic wire meshes are used as internal cages (support structures) but not as the fabric itself in a conventional baghouse. Textile felts and woven fabrics provide the dust cake support and filtration layer.Step-by-Step Solution:Differentiate between filter media (fabric) and cage/support.Identify polymers/natural fibres as valid media.Exclude metallic wire woven mesh as the fabric.Verification / Alternative check:Baghouse supplier catalogs list textile media; metal elements show up only as cages or sintered-metal cartridges (a different technology than cloth bags).Why Other Options Are Wrong:Polyester/cotton/nylon/acrylic: all are used for bag fabrics depending on conditions.Common Pitfalls:Confusing sintered-metal cartridge filters with fabric baghouses.Final Answer:Metallic wire woven mesh

Question 3

For separating isotopes (extremely small mass differences), which centrifuge design is generally employed in practice?

Accepted Answer

Introduction / Context:Isotope separation requires distinguishing species with minute density differences. Very high centrifugal fields and precise rotors are essential to achieve measurable fractionation.Given Data / Assumptions:Mass differences are tiny; separation factors per stage are small.High g-levels and small diffusion paths are required.Concept / Approach:Ultracentrifuges operate at extremely high rotational speeds, generating very large centrifugal accelerations. This enhances separation of isotopic species compared to disc-bowl machines, which are intended mainly for liquid–solid or liquid–liquid separations with larger density contrasts.Step-by-Step Solution:Identify required g-force for isotope fractionation.Match to ultracentrifuge capabilities.Select ultracentrifuge as the correct design.Verification / Alternative check:Isotope enrichment technologies historically leverage ultra-high-speed centrifugation and specialized rotor designs, consistent with this choice.Why Other Options Are Wrong:Disc-bowl: inadequate g-levels for isotope work.Batch basket: solid–liquid duty.Common Pitfalls:Assuming any centrifuge suits isotope separation; only ultra-high-g devices are effective.Final Answer:Ultracentrifuge

Question 4

For separating abrasive solids suspended in a corrosive liquid stream, which filtration device is generally the most suitable choice (consider material compatibility and wear)?

Accepted Answer

Introduction / Context:Abrasive solids erode internals, while corrosive liquids attack metals and elastomers. The chosen filter must allow corrosion-resistant contact surfaces and withstand abrasion without rapid failure.Given Data / Assumptions:Abrasive solids + corrosive liquid.Need robust, maintainable, pressure-driven separation.Flexibility in media and plate materials is advantageous.Concept / Approach:Plate and frame presses accommodate a wide range of chemically resistant plates, gaskets, and fabrics (e.g., polypropylene, PTFE-lined components). The batch operation and high-pressure driving force make them suitable for difficult slurries, while replaceable cloths mitigate abrasion wear. Sand beds and vacuum filters struggle with severe abrasion and material compatibility. Basket centrifuges can handle abrasives but are less favored when corrosion resistance and flexible media selection dominate concerns.Step-by-Step Solution:Prioritize corrosion-resistant construction and replaceable cloths.Evaluate abrasion handling and maintenance practicality.Select plate and frame press.Verification / Alternative check:Industry practice frequently deploys polypropylene plate presses with appropriate fabrics for corrosive, abrasive slurries.Why Other Options Are Wrong:Sand bed: limited solids handling, difficult backwash with abrasives.Vacuum filter: larger wear surfaces; material compatibility may be limiting.Basket centrifuge: viable for some cases, but material-of-construction and cloth options are more versatile in plate and frame systems.Common Pitfalls:Overlooking fabric and plate material selection to combat corrosion.Final Answer:Plate and frame filter press

Question 5

Which of the following ultrafine grinders primarily employ attrition (particle–particle and particle–wall friction) as the main mechanism for achieving micron/submicron sizes?

Accepted Answer

Introduction / Context:Ultrafine grinding targets d50 in microns or below. Fluid-energy devices accelerate particles in gas streams; size reduction occurs by high-velocity collisions and attrition rather than bulk compression.Given Data / Assumptions:Devices listed are gas-entrained mills.Focus is on dominant breakage mode.Concept / Approach:Jet mills, fluid energy mills, and branded “micronizers” create intense particle–particle impacts and rubbing in turbulent jets. The absence of mechanical media avoids contamination, and attrition plus impact between like particles achieves very fine sizes.Step-by-Step Solution:Identify each device as a fluid-energy type.Note attrition/impact dominance without grinding media.Conclude that all three employ attrition significantly.Verification / Alternative check:Manufacturer data cite gas energy and inter-particle collisions as the core mechanism for these mills.Why Other Options Are Wrong:High-pressure rolls: compressive bed grinding, not attrition-dominant ultrafine gas milling.Common Pitfalls:Assuming mechanical media are required for ultrafines; fluid energy suffices.Final Answer:All (a), (b) and (c)

Question 6

Which grinding mill uses horizontally arranged rods as the grinding media and is known to deliver a more uniform granular product with minimal fines generation?

Accepted Answer

Introduction / Context:Different tumbling mills use different media: balls, rods, pebbles, or ore itself. Media geometry affects breakage mode and product size distribution.Given Data / Assumptions:Goal is uniform product with fewer fines.Grinding elements are horizontally arranged rods.Concept / Approach:Rod mills grind by line contact rather than point contact of balls. This reduces overgrinding and limits fines production, making rod mills suitable for feed to ball mills or for specific sizing where slimes are undesirable.Step-by-Step Solution:Identify mill with rods: rod mill.Relate line-contact to minimized fines.Select rod mill accordingly.Verification / Alternative check:Flowsheets often place rod mills ahead of ball mills to control fines, validating the choice.Why Other Options Are Wrong:Compartment/Tube mills: ball media.Pebble: ceramic/natural pebbles; different duty.Common Pitfalls:Assuming all tumbling mills behave the same; media geometry matters.Final Answer:Rod mill

Question 7

Among the following minerals, which one has the minimum Rittinger’s number (i.e., requires the least energy per unit new surface area created) under comparable conditions?

Accepted Answer

Introduction / Context:Rittinger’s number reflects the energy needed to create unit new surface. Softer, cleavage-prone minerals typically show lower values than hard, tough minerals.Given Data / Assumptions:Minerals listed: calcite, pyrite, quartz, galena.Measurements compared at similar size ranges and conditions.Concept / Approach:Quartz is very hard and tough, implying a high Rittinger’s number. Pyrite and calcite are moderate; galena (lead sulfide) is very soft and exhibits perfect cleavage, making surface generation easier and energy per area lower, hence a smaller Rittinger’s number.Step-by-Step Solution:Rank by hardness/toughness: quartz > pyrite ≈ calcite > galena.Associate lower hardness/cleavage with lower energy per area.Select galena as minimum.Verification / Alternative check:Grinding practice shows galena liberates readily with less power compared with quartz-bearing ores.Why Other Options Are Wrong:Quartz: very high energy per surface created.Pyrite/Calcite: higher than galena though lower than quartz.Common Pitfalls:Equating Mohs hardness linearly with grindability; cleavage also matters.Final Answer:Galena

Question 8

Mechanical conveyors that push material along a trough or through a tube are often called “scrapers.” Which of the following equipment types fall under this pushing/scraper category?

Accepted Answer

Introduction / Context:Conveyors may carry material on belts/buckets or push it along using blades/flights or a rotating screw. The latter operate by displacement within a trough or tube and are grouped as “scraper” types in many texts.Given Data / Assumptions:Focus on push/drag action rather than carrying on a surface.Common industrial conveyor types listed.Concept / Approach:Flight conveyors use moving flights to drag material along a trough. Screw conveyors use a rotating helical flight to push bulk solids through a U-trough or pipe. Both displace/push material rather than carry it externally like belts or buckets.Step-by-Step Solution:Identify push-type conveyors among options.Exclude carry-type conveyors (belt/bucket).Select flight and screw conveyors together.Verification / Alternative check:Plant handling references classify flight/drag and screw conveyors as displacement/pusher types.Why Other Options Are Wrong:Bucket/Belt: carry material on containers/surfaces; not “scrapers.”Common Pitfalls:Using “scraper” only for chain-drag flights and overlooking screws, which also push.Final Answer:Both (b) and (c)

Question 9

State Rittinger’s crushing law in terms of energy–size reduction: which statement correctly represents its core idea?

Accepted Answer

Introduction / Context:Among the classical size-reduction laws, Rittinger’s law best fits fine grinding where surface area change dominates. It is fundamental for estimating energy when large increases in surface area occur.Given Data / Assumptions:We compare verbal statements of different laws.We want the one proportional to new surface area.Concept / Approach:Rittinger’s law: E ∝ ΔA. Kick’s law: E ∝ ln(d1/d2). Bond’s law: E ∝ (1/√d2 − 1/√d1). Only Rittinger ties energy directly to newly created surface, which is especially relevant as particle size approaches fine ranges.Step-by-Step Solution:Identify the statement matching Rittinger’s proportionality to surface area.Select the option expressing “energy ∝ new surface.”Verification / Alternative check:Fine milling design calculations and literature consistently pair Rittinger’s law with surface area creation.Why Other Options Are Wrong:Options (a), (b), (d), (e) do not reflect Rittinger’s formulation.Common Pitfalls:Confusing Bond and Kick with Rittinger due to overlapping usage domains.Final Answer:Energy required is proportional to the new surface created

Question 10

For a revolving (tumbling) mill, how does wet grinding compare with dry grinding in terms of practical operation and downstream handling/classification?

Accepted Answer

Introduction / Context:Choice between wet and dry grinding impacts energy, capacity, dust control, and downstream classification. Understanding trade-offs guides circuit design.Given Data / Assumptions:Tumbling mill operation.Industry practice: wet grinding often improves breakage kinetics and capacity.However, slurry handling adds complexity.Concept / Approach:Wet grinding typically reduces power per ton and can increase capacity due to improved transport and cushioning effects. The downside is the need for slurry pumps, hydrocyclones, thickeners, and drying if a dry product is required—complicating handling and classification.Step-by-Step Solution:Contrast power/capacity trends (wet often favorable).Identify operational complexity of slurry circuits.Conclude handling/classification are more complicated.Verification / Alternative check:Typical plant flowsheets include hydrocyclones after wet mills, confirming added classification complexity.Why Other Options Are Wrong:More energy/less capacity: not generally true; often the opposite.“None”: incorrect because there is a clear operational difference.Common Pitfalls:Ignoring end-product moisture specifications that may require subsequent drying.Final Answer:Wet grinding complicates handling and classification of the product

Question 11

Grizzlies (stationary bar screens) are used in screening operations primarily to separate which size class of solids before downstream crushing or screening?

Accepted Answer

Introduction / Context:Grizzlies are robust bar screens installed ahead of crushers to scalp oversize lumps and protect downstream equipment.Given Data / Assumptions:Stationary, widely spaced bars.Used at the front end of a crushing plant.Concept / Approach:Because of their large bar spacing and rugged design, grizzlies remove very coarse rocks and tramp material. Fines pass through to subsequent screens or crushers, improving throughput and reducing wear.Step-by-Step Solution:Identify application: scalping.Associate with large opening sizes.Select “coarse.”Verification / Alternative check:Plant layouts show grizzlies directly above hoppers or primary crushers for oversize removal.Why Other Options Are Wrong:Fine/ultrafine: grizzlies lack the small apertures required.Any size/none: inaccurate.Common Pitfalls:Confusing grizzlies with vibrating decks used for finer separations.Final Answer:Coarse

Question 12

In the beneficiation of sulphide ores (e.g., copper, lead, zinc), which process has virtually no alternative for effective separation under most industrial conditions?

Accepted Answer

Introduction / Context:Sulphide minerals possess surface chemistry amenable to selective attachment of air bubbles using collectors, frothers, and modifiers—making flotation the primary enrichment route worldwide.Given Data / Assumptions:Sulphide ore feed with fine liberation sizes.Need for selective separation among multiple sulphide minerals.Concept / Approach:Froth flotation exploits differential hydrophobicity, allowing selective recovery by tuning reagents and pH. Gravity methods (tabling, jigging, heavy media) are ineffective at fine sizes and small density contrasts common in sulphides. Classification is only a sizing step, not an upgrading process.Step-by-Step Solution:Identify sulphide system requiring surface-selective separation.Select froth flotation as the only broadly viable method.Verification / Alternative check:Industrial concentrators universally use flotation circuits for sulphide ores, confirming this choice.Why Other Options Are Wrong:Gravity and classification lack selectivity/efficiency at fine sizes.Common Pitfalls:Overestimating gravity methods for fine, complex sulphides.Final Answer:Froth flotation

Question 13

Based on Mohs hardness, which of the following materials is the hardest: calcite, quartz, corundum, or gypsum?

Accepted Answer

Introduction / Context:Hardness influences wear, grinding energy, and equipment selection. Mohs scale provides a simple ranking used in mineral processing and materials engineering.Given Data / Assumptions:Compare typical Mohs values.Materials: gypsum 2, calcite 3, quartz 7, corundum 9.Concept / Approach:Corundum (aluminum oxide) at Mohs 9 is harder than quartz (7), calcite (3), and gypsum (2). Harder materials raise liner wear and energy needs during grinding.Step-by-Step Solution:Recall Mohs ranking.Pick the maximum value.Select corundum.Verification / Alternative check:Reference charts list corundum just below diamond (10).Why Other Options Are Wrong:Gypsum/Calcite: much softer.Quartz: hard but below corundum.Common Pitfalls:Confusing corundum with quartz due to both being common abrasives.Final Answer:Corundum

Question 14

Grinding aids (surface-active additives) are primarily employed in which grinding mode to improve flowability, reduce agglomeration, and enhance mill throughput?

Accepted Answer

Introduction / Context:Grinding aids reduce surface energy effects that cause powder agglomeration, especially in dry circuits where moisture is insufficient to disperse fines.Given Data / Assumptions:Additives are low-dosage organic agents.Target is to improve mill efficiency and transport.Concept / Approach:In dry grinding, sticky fine particles can coat media and liners, increasing circulating loads. Grinding aids adsorb on particle surfaces, lowering adhesion and improving flow, which raises throughput and reduces power per ton.Step-by-Step Solution:Identify agglomeration issues in dry circuits.Apply grinding aids to mitigate coating and improve classification.Select “dry grinding.”Verification / Alternative check:Cement plants widely use amine- or glycol-based aids in dry finish mills, confirming the practice.Why Other Options Are Wrong:Wet grinding: slurry already disperses fines; aids less critical.Ultrafine/intermediate: not modes; aids relate to dryness, not fineness alone.Common Pitfalls:Assuming aids are only for ultrafines; they are standard in many dry circuits.Final Answer:Dry grinding

Question 15

Critical speed Nc (rpm) of a ball mill, in terms of mill radius R1 and ball radius R2 (both in metres), is correctly given by which expression?

Accepted Answer

Introduction / Context:Critical speed defines the rotational speed at which grinding media just cling to the mill shell. Above this speed, cataracting ceases and grinding action diminishes.Given Data / Assumptions:R1 = mill radius, R2 = ball radius (m).g ≈ 9.81 m/s^2; unit conversion to rpm included.Concept / Approach:At the point of detachment, centripetal acceleration equals gravitational acceleration at the media center: ω^2 (R1 − R2) = g. Hence ω = sqrt(g/(R1 − R2)). Converting to rpm gives Nc = (60 / 2π) * sqrt(g/(R1 − R2)) ≈ 42.3 / sqrt(R1 − R2).Step-by-Step Solution:Set ω^2 (R1 − R2) = g.Solve ω = sqrt(g/(R1 − R2)).Convert to rpm: Nc = 60 ω / (2π) ≈ 42.3 / sqrt(R1 − R2).Verification / Alternative check:With D and d as diameters, the familiar form is Nc ≈ 42.3 / sqrt(D − d) (D,d in metres), consistent with the derived expression using radii.Why Other Options Are Wrong:Other constants or using (R1 + R2) are dimensionally/physically inconsistent.Common Pitfalls:Forgetting to subtract ball radius; the orbit center is at (R1 − R2).Final Answer:Nc = 42.3 / sqrt(R1 - R2)

Question 16

For mixing in the laminar regime (very low Reynolds number), how does the power number Po scale with the Reynolds number Re for a stirred tank (Po = P / (ρ N^3 D^5), Re = ρ N D^2 / μ)?

Accepted Answer

Introduction / Context:In mixing, dimensional analysis yields power number–Reynolds number correlations. In laminar flow, viscous forces dominate and inertial effects vanish from the power requirement.Given Data / Assumptions:Definitions: Po = P/(ρ N^3 D^5), Re = ρ N D^2/μ.Low Re (creeping flow) limit.Concept / Approach:For laminar mixing, P ∝ μ N^2 D^3, so substituting into Po gives Po ∝ (μ N^2 D^3) / (ρ N^3 D^5) = (μ / ρ N D^2) = 1/Re. Therefore, Po scales inversely with Re in the viscous regime.Step-by-Step Solution:Use laminar mixing power correlation P ∝ μ N^2 D^3.Form Po and simplify.Obtain Po ∝ Re^-1.Verification / Alternative check:Standard impeller correlations display a Po·Re = constant plateau in laminar flow, equivalent to Po ∝ 1/Re.Why Other Options Are Wrong:Zero/positive exponents imply inertia influence, contrary to laminar limit.Common Pitfalls:Extending turbulent correlations into laminar ranges.Final Answer:Po ∝ Re^-1.0

Question 17

In power requirement calculations for mixing in stirred tanks, which dimensionless group(s) are fundamentally important for correlating power number and operating regime?

Accepted Answer

Introduction / Context:Power correlations in mixing use dimensionless groups to capture viscous, inertial, and gravitational effects that influence flow patterns and power draw.Given Data / Assumptions:Re = ρ N D^2 / μ (viscous vs inertial).Fr = N^2 D / g (inertial vs gravity), important for vortex formation/free-surface effects.Concept / Approach:Reynolds number classifies laminar, transitional, and turbulent mixing, influencing Po. Froude number becomes important when free-surface vortices or gas–liquid interfaces are present, altering power and flow. Thus both groups are key in many mixing problems.Step-by-Step Solution:Identify roles of Re (viscosity) and Fr (free surface/rotation).Note their appearance in empirical power charts.Conclude both are important.Verification / Alternative check:Power number charts often plot Po vs Re with Fr (or vortex depth) as a parameter in free-surface systems.Why Other Options Are Wrong:Single-parameter views miss regimes where gravity matters.Common Pitfalls:Ignoring Fr in open tanks, leading to unexpected vortexing and power draw changes.Final Answer:Both (a) and (b)

Question 18

Regarding particle shape, what is the typical sphericity value for pulverised coal particles used in combustion systems?

Accepted Answer

Introduction / Context:Sphericity (ψ) compares the surface area of a sphere with the same volume to the actual particle surface area. Real fragmented particles are irregular, influencing drag, settling, and combustion behavior.Given Data / Assumptions:Pulverised coal results from mechanical grinding.Particles are angular/irregular rather than perfectly spherical.Concept / Approach:A perfect sphere has ψ = 1. Irregular particles have greater surface area than a sphere of equal volume, so ψ < 1. Pulverised coal commonly exhibits ψ appreciably below unity, affecting aerodynamic and combustion characteristics (e.g., higher drag, quicker burnout due to higher specific surface).Step-by-Step Solution:Define sphericity.Relate irregularity to increased surface area.Conclude ψ < 1 for pulverised coal.Verification / Alternative check:Measured shape factors for milled coals in literature typically lie below 1, consistent with fragmented solids.Why Other Options Are Wrong:=1 implies perfect spheres; >1 is impossible by definition; ∞/0 are nonphysical here.Common Pitfalls:Assuming atomized liquids (nearly spherical droplets) when dealing with crushed solids.Final Answer:<1

Question 19

For a sphere falling where the drag coefficient is essentially constant (Newton’s regime, high Reynolds number), the terminal settling velocity varies with diameter d according to which proportionality?

Accepted Answer

Introduction / Context:Settling behavior transitions across regimes. In the Newton (constant C_d) regime at high Reynolds numbers, drag force is quadratic in velocity and proportional to cross-sectional area, while weight–buoyancy scales with volume.Given Data / Assumptions:Constant drag coefficient C_d.Sphere diameter = d; area ∝ d^2; volume ∝ d^3.Concept / Approach:At terminal velocity v_t, drag balances net weight: (1/2) C_d ρ_f A v_t^2 = (ρ_p − ρ_f) g V. Since A ∝ d^2 and V ∝ d^3, rearranging gives v_t ∝ sqrt(d), i.e., v_t ∝ d^0.5.Step-by-Step Solution:Write force balance: drag = net weight.Substitute A and V scalings with d.Solve for v_t ∝ (d^3 / d^2)^0.5 = d^0.5.Verification / Alternative check:In contrast, in Stokes (laminar) regime, v_t ∝ d^2; the constant-C_d regime clearly yields the square-root dependence, supporting the selected option.Why Other Options Are Wrong:d or d^2 correspond to other regimes/assumptions; 1/d and d^3 are inconsistent with Newton-law drag.Common Pitfalls:Using Stokes-law dependence outside its low-Re validity range.Final Answer:v_t ∝ d^0.5

Question 20

Cyclone separator design calculation: For a cyclone with body diameter 0.2 m and a measured tangential gas velocity of 15 m/s at the wall, estimate the dimensionless separation factor (also called the centrifugal acceleration ratio) using the standa…

Accepted Answer

Introduction / Context:In gas–solid separation, the cyclone's ability to drive particles toward the wall is often expressed using a separation factor (also termed the centrifugal acceleration ratio). It compares the centrifugal acceleration acting on particles at a given radius to gravitational acceleration. This dimensionless number helps relate operating conditions to expected cut size and collection efficiency for a given particle density and size range.Given Data / Assumptions:Cyclone body diameter, D = 0.2 m, so local wall radius, r ≈ D/2 = 0.1 m.Tangential gas velocity near the wall, v_t = 15 m/s.Gravity, g = 9.81 m/s^2.Use the standard relation: Separation factor, S = v_t^2 / (r * g).Concept / Approach:Centrifugal acceleration at radius r is a_c = v_t^2 / r. The separation factor S compares a_c to g: S = a_c / g = v_t^2 / (r * g). Larger S generally implies stronger outward particle forcing and potentially smaller achievable cut size, all else equal (geometry, inlet design, and dust properties).Step-by-Step Solution:Compute radius: r = 0.2 / 2 = 0.1 m.Compute v_t^2: 15^2 = 225 m^2/s^2.Compute denominator: r * g = 0.1 * 9.81 = 0.981.Evaluate S = 225 / 0.981 ≈ 229.46 ≈ 230 (rounded).Verification / Alternative check:The order of magnitude is typical for small industrial cyclones at moderate velocities (hundreds of g's). Using a representative wall radius is standard practice for quick estimates; more detailed designs consider inlet configuration, vortex finder, and actual velocity profiles.Why Other Options Are Wrong:2250, 1125, 460: These values are one or more orders too large for the given v_t and diameter. They would imply unrealistically high accelerations for the stated conditions.Common Pitfalls:Using cyclone body diameter instead of radius in the denominator; forgetting to square the velocity; or mixing units (e.g., cm vs m) can inflate S by factors of 10–100 erroneously.Final Answer:230

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