All practice questions

Match each mixture or suspension to the physical separation technique best suited for isolating its components.

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Answer

Ammonium chloride and sodium chloride are separated by sublimation; sulfur and iron filings are separated by magnetization; red blood cells in plasma are separated by centrifugation.

Each mixture matches its technique based on fundamental physical characteristics: ammonium chloride sublimes readily upon heating; iron is attracted to magnets; suspended blood cells require high-speed centrifugal acceleration to separate from plasma.

Step-by-Step Solution

1

Identify the key physical property difference in the solid mixture of ammonium chloride and sodium chloride.

Ammonium chloride is a sublimable solid, whereas sodium chloride is thermally stable and non-volatile at gentle heating temperatures.

Sublimation vaporizes ammonium chloride directly into gas, which condenses on a cool surface while sodium chloride remains behind.

2

Analyze the magnetic properties of sulfur and iron filings.

Iron is strongly magnetic (ferromagnetic) while sulfur is non-magnetic.

Applying a magnetic field pulls the iron particles out of the mixture cleanly.

3

Determine the appropriate method for separating solid blood cells from liquid blood plasma.

Blood cells exist as a fine suspension of particles with subtle density differences relative to plasma, which settle too slowly under normal gravity.

Centrifugation applies high rotational speed to force dense suspended cells to the bottom of the container rapidly.

Key Concept

Selecting separation techniques based on physical properties (sublimability, magnetism, and particle density in suspension)

Question 1222Question →

Match each ecological measuring instrument listed in Column A with the corresponding abiotic parameter it measures in Column B.

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Answer

Wind vane matches with direction of wind movement; Lux meter matches with intensity of light in a habitat; Rain gauge matches with amount of precipitation over a period; Maximum-minimum thermometer matches with daily temperature range and extremes.

Each instrument correctly corresponds to its specific environmental measurement: wind vanes indicate wind direction, lux meters measure light intensity, rain gauges record precipitation amount, and maximum-minimum thermometers capture daily temperature extremes.

Step-by-Step Solution

1

Identify the primary function of each ecological instrument.

Wind vane indicates wind direction; Lux meter quantifies light intensity; Rain gauge collects rainfall; Maximum-minimum thermometer measures extreme temperatures.

Abiotic environmental factors are quantified using specific instruments designed for physical measurements.

2

Pair each instrument from Column A to its matching abiotic factor in Column B.

All instruments are mapped correctly to their respective environmental measurement parameters.

Direct one-to-one mapping links each equipment item with its intended ecological variable.

Key Concept

Measurement of Abiotic Ecological Factors

Question 1223Question →

Match each plant group on the left with its corresponding structural feature on the right.

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Answer

Thallophytes correspond to an undifferentiated thallus lacking roots, stems, leaves, and vascular tissue; Bryophytes correspond to non-vascular plants anchored by rhizoids with gametophyte dominance; Pteridophytes correspond to seedless vascular plants with true roots, stems, leaves, and sporophyte dominance.

Thallophytes feature a simple thallus without specialized conducting tissue or organs. Bryophytes are non-vascular plants with rhizoids and gametophyte dominance. Pteridophytes are true vascular seedless plants with true roots, stems, leaves, and sporophyte dominance.

Step-by-Step Solution

1

Identify the characteristic features of Thallophytes.

Thallophytes represent the simplest algae/plant forms with an unspecialized thallus body devoid of vascular tissues.

They have not evolved vascular organs or tissue differentiation.

2

Identify the characteristic features of Bryophytes.

Bryophytes are non-vascular land plants (mosses, liverworts) anchored by root-like rhizoids where the gametophyte phase is dominant.

They lack lignified xylem and true roots.

3

Identify the characteristic features of Pteridophytes.

Pteridophytes are vascular seedless plants (ferns) possessing true vegetative structures (roots, stems, leaves) and a dominant sporophyte generation.

They are the first plant group to evolve conducting tissues (xylem and phloem).

Key Concept

Structural organization and vascular tissue presence across lower plant divisions

Question 1224Question →

Match each plant growth regulator on the left with its primary physiological action on the right.

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Answer

Auxin pairs with cell elongation and apical dominance; Ethylene pairs with fruit ripening and leaf abscission; Abscisic acid pairs with stomatal closure during drought stress; Gibberellin pairs with seed germination and stem elongation.

Each plant growth regulator is matched directly to its physiological function: Auxin promotes cell elongation and apical dominance, Ethylene stimulates fruit ripening and abscission, Abscisic acid triggers stomatal closure during drought, and Gibberellin initiates seed germination.

Step-by-Step Solution

1

Identify the primary function of Auxin.

Auxin stimulates cell elongation and apical dominance.

Auxin is synthesized in apical meristems and controls longitudinal growth.

2

Identify the primary function of Ethylene.

Ethylene promotes fruit ripening and leaf abscission.

Ethylene acts as a gaseous growth regulator during organ maturation.

3

Identify the primary function of Abscisic acid.

Abscisic acid induces stomatal closure under water deficit.

Abscisic acid functions as an inhibitory stress hormone during drought conditions.

4

Identify the primary function of Gibberellin.

Gibberellin mobilizes food reserves to promote seed germination.

Gibberellins stimulate the synthesis of hydrolytic enzymes during seed germination.

Key Concept

Physiological Roles of Plant Hormones

Question 1225Question →

Match each commercial banking concept or credit creation term in Column A with its correct definition or description in Column B.

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Answer

Primary Functions match with accepting customer deposits and granting loans; Secondary Functions match with providing agency services such as clearing cheques; Credit Multiplier matches with the factor determining overall money expansion (1/CRR); Cash Reserve Ratio matches with the legally mandatory percentage of deposits retained as cash reserves.

Each concept correctly aligns with its precise economic role: Primary Functions represent core deposit-taking and lending; Secondary Functions cover agency services like clearing cheques; Credit Multiplier represents the inverse of the reserve ratio (1/CRR); and Cash Reserve Ratio represents the statutory percentage of deposits kept as reserves.

Step-by-Step Solution

1

Classify the core operations of commercial banking

Primary functions represent direct financial intermediation (taking deposits and advancing loans), while secondary functions represent ancillary agency and general utility services.

Commercial banking activities are categorized based on whether they form core intermediation or supplementary services.

2

Analyze quantitative parameters of credit creation

The Cash Reserve Ratio (CRR) sets the required reserve fraction, and the Credit Multiplier is derived as

\frac{1}{\text{CRR}}

.

The reserve requirement dictates how much of each deposit can be converted into new derivative loans across the commercial banking system.

Key Concept

Commercial Bank Functional Taxonomy and Credit Expansion Parameters

Estimated Time:1m 30s

Question 1226Question →

Match each type of plant meristematic tissue on the left with its correct growth function on the right.

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Answer

Apical meristem pairs with responsible for primary elongation at the apex of stems and roots; Lateral meristem pairs with facilitates increase in girth or secondary growth in stems and roots; Intercalary meristem pairs with promotes internodal elongation located at the base of leaf blades or nodes.

Apical meristems cause lengthening at shoot and root apices. Lateral meristems increase stem and root diameter. Intercalary meristems enable internodal lengthening in monocot stems.

Step-by-Step Solution

1

Analyze the primary role of apical meristems in plant growth.

Apical meristems produce primary growth leading to increased length at tips.

These meristems exist at root and shoot apices where active cell division extends the plant bodies longitudinally.

2

Analyze the role of lateral meristems.

Lateral meristems produce secondary growth increasing thickness.

Located parallel to the long axis, lateral meristems add vascular and cork layers outward and inward.

3

Analyze the function of intercalary meristems.

Intercalary meristems drive internodal extension in monocots.

They remain active at leaf bases and stem nodes, allowing stems to elongate quickly.

Key Concept

Plant Meristematic Tissues and Primary vs Secondary Growth

Estimated Time:45s

Question 1227Question →

Match each pair of organic compounds on the left with its corresponding type of isomerism on the right.

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Answer

Hexan-2-one and Hexan-3-one exhibit Positional isomerism; Pentane and 2,2-Dimethylpropane exhibit Chain isomerism; Ethoxyethane and Butan-1-ol exhibit Functional group isomerism; (+)-Lactic acid and (-)-Lactic acid exhibit Optical isomerism.

Hexan-2-one and Hexan-3-one differ only in the locant of the carbonyl group along an unchanged six-carbon backbone (positional isomerism). Pentane and 2,2-dimethylpropane differ in the branching of their carbon skeletons (chain isomerism). Ethoxyethane and Butan-1-ol share the formula C4H10O but contain different functional groups (functional group isomerism). (+)-Lactic acid and (-)-Lactic acid are optical enantiomers due to an asymmetric chiral carbon center.

Step-by-Step Solution

1

Examine Hexan-2-one and Hexan-3-one

Both share the molecular formula

C_6H_{12}O

and contain the carbonyl (

C=O

) functional group. In Hexan-2-one, the carbonyl carbon is at C-2, whereas in Hexan-3-one, it is at C-3.

Molecules with identical functional groups located at different positions on the carbon chain are positional isomers.

2

Examine Pentane and 2,2-Dimethylpropane

Both share the formula

C_5H_{12}

. Pentane is a straight 5-carbon chain (

CH_3-CH_2-CH_2-CH_2-CH_3

), whereas 2,2-Dimethylpropane consists of a 3-carbon chain with two methyl branches,

C(CH_3)_4

.

Molecules with the same molecular formula but different carbon chain structures are chain isomers.

3

Examine Ethoxyethane and Butan-1-ol

Both share the molecular formula

C_4H_{10}O

. Ethoxyethane (

C_2H_5-O-C_2H_5

) is an ether, while Butan-1-ol (

C_4H_9OH

) is a primary alkanol.

Molecules possessing the same molecular formula but belonging to different homologous series with distinct functional groups are functional group isomers.

4

Examine (+)-Lactic acid and (-)-Lactic acid

Lactic acid (

2

-hydroxypropanoic acid) features a central carbon atom bonded to four distinct groups:

-H

,

-OH

,

-CH_3

, and

-COOH

. This chiral center generates two non-superimposable mirror-image forms.

Stereoisomers that rotate plane-polarized light in opposite directions due to molecular chirality are optical isomers.

Key Concept

Types of Structural Isomerism and Stereoisomerism

Question 1228Question →

Match each experimental observation of cathode rays in a discharge tube with the corresponding physical property or characteristic it demonstrates.

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Answer

1. Formation of a sharp shadow of a Maltese cross matches with Cathode rays travel in straight lines. 2. Rotation of a paddle wheel matches with Cathode rays possess particle mass and mechanical momentum. 3. Deflection toward a positively charged plate matches with Cathode rays carry a negative electrical charge. 4. Deflection perpendicular to a magnetic field matches with Cathode rays act as a current of moving charged particles obeying magnetic force laws.

Each experimental setup provides specific proof of a cathode ray property: sharp shadow formation proves straight-line motion; turning a paddle wheel demonstrates particle mass and momentum; attraction to a positive plate confirms negative charge; and deflection in a magnetic field confirms that the beam acts as moving electrical charges.

Step-by-Step Solution

1

Analyze the Maltese cross shadow experiment

Sharp shadows indicate straight-line propagation of rays from the cathode surface.

Light and particle beams traveling in straight lines produce sharp geometrical shadows of opaque obstructions.

2

Analyze the paddle wheel experiment

The paddle wheel rotates when struck by cathode rays, demonstrating kinetic energy and momentum transfer.

Mechanical rotation requires a force resulting from the momentum transfer of moving material particles.

3

Analyze electric field deflection

The ray path bends toward the positive anode plate.

Electrostatic attraction pulls negatively charged entities toward positive electric potentials.

4

Analyze magnetic field deflection

The beam bends laterally perpendicular to both the trajectory and the magnetic field vector.

Moving electrical charges experience a magnetic Lorentz force given by

F = qvB\sin\theta

.

Key Concept

Experimental evidence establishing the properties of cathode rays

Estimated Time:1m 30s

Question 1229Question →

Match each physical quantity on the left with its correct fundamental SI base unit expression or fundamental status on the right.

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Answer

Thermodynamic temperature matches with 'Fundamental physical quantity measured in kelvin (K)'; Electric charge matches with 'Derived physical quantity expressed in base SI units as A·s'; Linear momentum matches with 'Derived physical quantity expressed in base SI units as kg·m·s⁻¹'; Power matches with 'Derived physical quantity expressed in base SI units as kg·m²·s⁻³'.

Each physical quantity is correctly paired with either its fundamental status or its base SI unit breakdown derived from core physics definitions.

Step-by-Step Solution

1

Identify fundamental quantities versus derived quantities.

Thermodynamic temperature is a basic fundamental quantity (unit:

\text{K}

). Electric current is fundamental (unit:

\text{A}

), but electric charge is derived (

Q = I t

).

Fundamental quantities cannot be defined in terms of other physical quantities.

2

Decompose Electric Charge into base units.

Since

Q = I \cdot t

, its unit is

\text{A}\cdot\text{s}

.

Current is measured in amperes and time in seconds.

3

Decompose Linear Momentum into base units.

p = m \cdot v \Rightarrow \text{unit} = \text{kg} \cdot \text{m}\cdot\text{s}^{-1}

.

Mass is in kilograms and velocity is in meters per second.

4

Decompose Power into base units.

P = \frac{W}{t} = \frac{F \cdot d}{t} = \frac{(m \cdot a) \cdot d}{t} \Rightarrow \frac{\text{kg} \cdot (\text{m}\cdot\text{s}^{-2}) \cdot \text{m}}{\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

.

Power is work done per unit time.

Key Concept

Fundamental and Derived Quantities

Question 1230Question →

Match each kinetic theory concept on the left with its correct physical description on the right.

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Answer

Temperature matches with the measure of average translational kinetic energy; Pressure matches with the average force per unit area exerted by colliding gas molecules on container walls; Root-mean-square speed matches with the square root of the mean of squared speeds.

Temperature measures average translational kinetic energy per particle. Pressure originates from force per unit area due to elastic wall collisions. Root-mean-square speed is the square root of the mean of squared molecular speeds.

Step-by-Step Solution

1

Identify the kinetic theory definition of Temperature

Temperature is directly proportional to the mean translational kinetic energy of the gas particles (

E_k \propto T

).

Absolute temperature reflects the average kinetic energy of molecular motion.

2

Identify the microscopic origin of Gas Pressure

Pressure is caused by molecular collisions with the container walls, transferring momentum and creating force per unit area.

Frequent elastic collisions of particles on container walls produce measurable pressure.

3

Identify the mathematical definition of Root-Mean-Square Speed

v_{rms} = \sqrt{\overline{v^2}}

, representing the square root of the average of squared molecular velocities.

This parameter represents the effective speed of gas particles relevant to thermal kinetic energy.

Key Concept

Kinetic Theory Interpretation of Gas Properties

Question 1231Question →

Match each redox description on the left with its corresponding classical or modern definition concept on the right.

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Answer

Addition of oxygen matches Classical oxidation; Loss of electrons matches Modern oxidation (electron transfer); Decrease in oxidation state matches Modern reduction (oxidation state change); Removal of oxygen matches Classical reduction.

Classical oxidation involves gaining oxygen, while classical reduction involves losing oxygen. In modern electronic terms, oxidation is the loss of electrons (OIL), and reduction is a decrease in oxidation state (reduction of oxidation number).

Step-by-Step Solution

1

Identify classical redox concepts based on oxygen transfer.

Addition of oxygen corresponds to classical oxidation, whereas removal of oxygen corresponds to classical reduction.

Classical definitions focused on the transfer of oxygen and hydrogen atoms.

2

Identify modern redox concepts based on electron transfer and oxidation numbers.

Loss of electrons corresponds to modern oxidation, and a decrease in oxidation state corresponds to modern reduction.

Modern concepts expand redox beyond oxygen/hydrogen to include electron movement and formal charge changes.

Key Concept

Distinguishing between classical (oxygen/hydrogen transfer) and modern (electron transfer and oxidation number) definitions of oxidation and reduction.

Estimated Time:1m 0s

Question 1232Question →

Match each feature observed during electrical discharge through a gas with its corresponding pressure stage or physical characteristic.

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Answer

Cathode Glow matches 'Luminous glow appearing right next to the cathode at around 10 mmHg pressure', Striations match 'Alternating bright and dark bands in the positive column at around 1 mmHg pressure', and Crookes Dark Space matches 'Dark region extending to fill most of the tube at very low pressure around 0.01 mmHg'.

Each feature of gas discharge corresponds to a specific pressure regime inside the discharge tube: Cathode Glow occurs at ~10 mmHg, Striations form at ~1 mmHg in the positive column, and Crookes Dark Space expands to cover most of the tube at ~0.01 mmHg where cathode rays are freely emitted.

Step-by-Step Solution

1

Identify the discharge stage for Cathode Glow.

Cathode glow occurs at moderate pressure (~10 mmHg) directly adjacent to the negative electrode.

Initial gas ionization near the cathode causes luminescence at this pressure stage.

2

Identify the discharge stage for Striations.

Striations represent light and dark divisions of the positive column at ~1 mmHg.

Periodic ionization and recombination along the tube produce alternating luminous discs.

3

Identify the discharge stage for Crookes Dark Space.

Crookes dark space expands as pressure drops to ~0.01 mmHg.

At very low pressures, electrons travel longer distances without colliding, causing the dark space to fill most of the discharge tube.

Key Concept

Stages of electric conduction through gases at reduced pressure.

Question 1233Question →

Match each physical wave propagation scenario on the left with its correct classification and particle vibration characteristics on the right.

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Answer

Oscillating electric charge in a vacuum matches non-mechanical wave with perpendicular field oscillations; pressure pulse in argon gas matches mechanical longitudinal wave with parallel oscillations; shear pulse on a stretched string matches mechanical transverse wave with perpendicular particle oscillations; water ripple on deep water matches mechanical surface wave with combined circular motion.

The correct pairings align each physical wave scenario with its underlying propagation mechanics. The oscillating charge producing field variations in a vacuum represents a non-mechanical transverse electromagnetic wave. The pressure disturbance in argon represents a longitudinal mechanical wave because fluids transmit energy through density compressions. The pulse on a stretched string is a transverse mechanical wave enabled by string tension. Surface water ripples are two-dimensional interface waves exhibiting combined longitudinal and transverse particle movement.

Step-by-Step Solution

1

Classify the disturbance in a vacuum (oscillating charge).

Identified as electromagnetic radiation, which is non-mechanical and transverse.

Electromagnetic waves do not require a material medium and involve mutually perpendicular field vectors perpendicular to propagation.

2

Analyze wave propagation through a gaseous medium (argon gas).

Identified as a purely longitudinal mechanical wave.

Gases lack shear strength and can only transmit mechanical energy through volume variations (compressions and rarefactions) parallel to the propagation vector.

3

Evaluate wave propagation along a stretched 1D solid string.

Identified as a transverse mechanical wave.

Tension in the solid string restores lateral shear displacements, causing medium elements to vibrate perpendicular to the string length.

4

Determine the nature of surface waves on liquid interfaces.

Identified as a mechanical surface wave with orbital motion.

Interface boundaries support gravity and surface tension restoring forces, producing elliptical/circular particle trajectories combining transverse and longitudinal components.

Key Concept

Classification of waves by medium requirement (mechanical vs. electromagnetic) and directional displacement mode (transverse, longitudinal, and surface orbital motion).

Estimated Time:2m 0s

Question 1234Question →

Match each ecological pyramid phenomenon or energy flow concept on the left with its correct metabolic or thermodynamic explanation on the right. Which pairing correctly matches each concept to its underlying biological cause?

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Answer

The correct pairings match: (1) Inverted biomass pyramid in aquatic systems with producer turnover rate and rapid reproduction; (2) Invariably upright energy pyramid with Second Law of Thermodynamics heat loss; (3) Stepwise reduction in available energy with metabolic and waste loss of approximately 90% per level; (4) Biomagnification with non-biodegradable fat-soluble toxins concentrating in smaller higher-level biomasses.

Each ecological concept matches its precise biological mechanism: aquatic biomass inversion is caused by rapid turnover of primary producers; energy pyramids are strictly upright due to metabolic heat loss (Second Law of Thermodynamics); energy reduction across trophic levels stems from respiration and excretion losses (~90%); and biomagnification occurs because persistent toxins accumulate in fat tissue as biomass decreases at higher levels.

Step-by-Step Solution

1

Analyze standing crop vs. productivity in aquatic habitats

Recognize that phytoplankton productivity is high despite low standing biomass due to rapid population turnover, creating an inverted biomass pyramid.

Explains why biomass pyramids can be inverted while energy production rates remain normal.

2

Apply thermodynamic laws to ecological energy transfer

Establish that energy cannot be recycled and entropy increases via metabolic heat release during each conversion.

Demonstrates why energy pyramids are strictly upright in every natural ecosystem.

3

Examine trophic efficiency calculations

Relate energy loss across trophic levels to physiological processes like movement, excretion, and cellular respiration (~90% lost).

Calculates net energy available to secondary and tertiary consumers.

4

Trace toxic substance dynamics in trophic chains

Correlate fat-solubility and biological persistence with increased toxin concentration at apex trophic levels.

Defines the biological mechanism behind biomagnification.

Key Concept

Thermodynamic laws in energy flow, trophic efficiencies, ecological pyramid structures, and biological magnification.

Question 1235Question →

Match each statistical data representation term on the left with its corresponding definition or mathematical property on the right.

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Answer

Class Boundary matches with the value separating adjacent non-overlapping class intervals; Sector Angle matches with the central angle in a pie chart calculated as

\frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

; Frequency Density matches with the quotient of class frequency and class width; Ogive matches with a line graph produced by plotting cumulative frequencies against upper class boundaries.

Each data representation term directly corresponds to its core definition: class boundary closes gaps between discrete class intervals, sector angle measures central circle proportion in a pie chart, frequency density scales histogram height when class widths differ, and an ogive graphs cumulative frequency against upper boundaries.

Step-by-Step Solution

1

Define Class Boundary

Class boundary is the continuous point midway between adjacent class limits.

Class boundaries remove gaps in discrete grouped frequency distributions.

2

Define Sector Angle formula for a pie chart

Sector Angle

= \frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

.

The complete circle represents total frequency, so individual sectors scale proportionally with

360^\circ

.

3

Define Frequency Density for histograms

Frequency Density

= \frac{\text{Frequency}}{\text{Class Width}}

.

Histogram area equals frequency; when widths differ, height must represent frequency per unit width.

4

Define Ogive

An Ogive is a cumulative frequency curve plotted against upper boundaries.

Each point on an ogive shows the cumulative frequency up to that class's upper boundary.

Key Concept

Data Representation Terms and Formulas

Estimated Time:1m 30s

Question 1236Question →

The table below displays the distribution of scores obtained by $50$ candidates in a Mathematics assessment test:

Score Interval	Frequency ( $f$ )
$10 - 19$	$8$
$20 - 29$	$12$
$30 - 39$	$18$
$40 - 49$	$7$
$50 - 59$	$5$

Match each statistical feature of the charts representing this data on the left with its corresponding numerical value on the right.

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Answer

The correct pairings are: Sector angle for the modal class matches

129.6^\circ

; Frequency density of the

30 - 39

class interval matches

1.8

; Lower class boundary of the median class matches

29.5

; Cumulative frequency up to

29.5

matches

20

.

Each chart feature correctly aligns with its mathematically derived value: the modal class sector angle is

\frac{18}{50} \times 360^\circ = 129.6^\circ

, the frequency density is

\frac{18}{10} = 1.8

, the median class lower boundary is

29.5

, and the cumulative frequency up to

29.5

is

8 + 12 = 20

.

Step-by-Step Solution

1

Determine the total frequency and locate the modal and median classes.

Total frequency

N = 8 + 12 + 18 + 7 + 5 = 50

. The modal class is

30 - 39

(highest frequency =

18

). The median position is

\frac{50}{2} = 25^{\text{th}}

, which falls within the

30 - 39

class interval since cumulative frequency reaches

38

at the end of this class.

Identifying

N

, the modal class, and the median position is required for calculating chart parameters.

2

Calculate the pie chart sector angle for the modal class (

30 - 39

).

Sector angle

= \frac{\text{Frequency}}{N} \times 360^\circ = \frac{18}{50} \times 360^\circ = 129.6^\circ

.

Pie chart sectors represent relative frequencies scaled to

360^\circ

.

3

Compute the frequency density for the histogram bar of class

30 - 39

.

Class width

= 39.5 - 29.5 = 10

. Frequency density

= \frac{\text{Frequency}}{\text{Class width}} = \frac{18}{10} = 1.8

.

Histogram height represents frequency density, defined as frequency divided by class width.

4

Identify the lower class boundary of the median class (

30 - 39

) and cumulative frequency at upper boundary

29.5

.

Lower boundary of

30 - 39

is

30 - 0.5 = 29.5

. Cumulative frequency up to

29.5

is

8 + 12 = 20

.

Class boundaries eliminate gaps between intervals for continuous plots like ogives and histograms.

Key Concept

Calculating statistical chart parameters (pie chart sector angles, histogram frequency densities, class boundaries, and cumulative frequencies) from grouped frequency distributions.

Question 1237Question →

Match each physical quantity listed on the left with its corresponding SI classification and base unit representation on the right.

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Answer

Electric current matches Fundamental quantity measured in amperes; Thermodynamic temperature matches Fundamental quantity measured in kelvins; Electric potential difference matches Derived quantity expressed as kg·m²·s⁻³·A⁻¹; Specific heat capacity matches Derived quantity expressed as m²·s⁻²·K⁻¹.

Electric current and thermodynamic temperature are basic SI fundamental quantities. Electric potential difference and specific heat capacity are derived quantities whose fundamental unit decompositions follow directly from their governing formulas.

Step-by-Step Solution

1

Identify fundamental physical quantities and their base units

Electric current and thermodynamic temperature are fundamental SI quantities with base units ampere (

\text{A}

) and kelvin (

\text{K}

) respectively.

Fundamental physical quantities are defined independently and serve as the foundation for the SI system.

2

Decompose electric potential difference into SI base units

Electric potential difference

V = \frac{\text{Work}}{\text{Charge}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

Because it is expressed by combining fundamental quantities, it is a derived quantity.

3

Decompose specific heat capacity into SI base units

Specific heat capacity

c = \frac{\text{Energy}}{\text{Mass} \times \text{Temperature change}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

.

It is calculated from energy, mass, and temperature, making it a derived quantity.

Key Concept

Fundamental quantities are independent basic quantities, whereas derived quantities are formed through algebraic combination of fundamental quantities.

Estimated Time:1m 30s

Question 1238Question →

Match each historical atomic model on the left with its defining postulate, experimental outcome, or theoretical limitation on the right.

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Answer

Thomson's model matches the diffuse positive sphere disproved by alpha particle backscattering; Rutherford's model matches the dense nucleus with classical radiation collapse limitations; Bohr's model matches quantized angular momentum in non-radiating orbits; Sommerfeld's extension matches elliptical sub-shells and relativistic adjustments for fine-structure splitting.

Each model directly maps to its defining theoretical contribution or failure mechanism: Thomson's diffuse charge sphere failed under

\alpha

-particle scattering; Rutherford's nuclear atom suffered from classical radiation instability; Bohr's model introduced angular momentum quantization

L = n\hbar

; and Sommerfeld's model extended orbits to ellipses with relativistic velocity corrections to account for fine structure.

Step-by-Step Solution

1

Analyze Thomson's Plum Pudding Model

Thomson proposed electrons embedded in a sea of positive charge. This continuous distribution could not account for

\alpha

-particles rebounding at angles greater than

90^\circ

.

Identify the historical assumption and experimental contradiction for Thomson's model.

2

Analyze Rutherford's Planetary Model

Rutherford deduced a concentrated positive core (nucleus). However, according to Maxwellian electrodynamics, orbiting electrons accelerate continuously, radiating energy until collapsing into the nucleus.

Identify the primary theoretical failure of classical planetary electron orbits.

3

Analyze Bohr's Quantized Model

Bohr introduced the non-classical postulate that electrons exist in stable stationary states with angular momentum

L = \frac{nh}{2\pi}

, accurately yielding the Rydberg formula for hydrogen.

Recognize the quantum postulate resolving Rutherford's radiation collapse.

4

Analyze Sommerfeld's Extension

To explain fine spectral line splitting not accounted for by circular Bohr orbits, Sommerfeld introduced elliptical orbits with azimuthal quantum numbers and relativistic mass variation at high electron velocities.

Connect fine-structure spectral features to relativistic elliptical orbital modifications.

Key Concept

Development and Limitations of Historical Atomic Models

Question 1239Question →

Match each physical quantity to its correct classification as either a fundamental or a derived physical quantity.

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Answer

Luminous intensity matches with fundamental quantity measuring light brightness; Mass matches with fundamental quantity measuring quantity of matter; Force matches with derived quantity defined as rate of change of linear momentum; Electric potential matches with derived quantity defined as work done per unit electric charge.

Luminous intensity and mass are two of the seven base SI quantities. Force and electric potential are derived quantities defined through mathematical combinations of base quantities.

Step-by-Step Solution

1

Identify the fundamental physical quantities

Luminous intensity and Mass are fundamental physical quantities.

Fundamental physical quantities are basic quantities that do not depend on any other physical quantity for their definition.

2

Identify the derived physical quantities

Force and Electric potential are derived physical quantities.

Derived physical quantities are obtained by combining fundamental physical quantities through mathematical relationships.

Key Concept

Fundamental quantities (length, mass, time, electric current, thermodynamic temperature, amount of substance, luminous intensity) are independent, whereas derived quantities are formed by combining fundamental quantities.

Question 1240Question →

Match each type of thermometer with its corresponding physical thermometric property.

Click a left item, then click its matching right item

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Answer

Liquid-in-glass thermometer matches change in length or volume of a liquid column; Constant-volume gas thermometer matches change in pressure of a gas; Resistance thermometer matches change in electrical resistance; Thermocouple matches electromotive force (e.m.f.) produced across junctions.

Each thermometer is accurately matched to the physical property that undergoes a measurable change as temperature varies.

Step-by-Step Solution

1

Identify the defining physical property that varies with temperature for each instrument.

Each thermometer operates on a distinct physical property that changes predictably when heated or cooled.

Thermometric properties must be reproducible and continuously measurable across a temperature range.

2

Pair each instrument with its specific thermometric property.

Liquid-in-glass pairs with liquid column expansion/length; constant-volume gas thermometer pairs with gas pressure; resistance thermometer pairs with electrical resistance; thermocouple pairs with thermoelectric e.m.f.

These pairs represent standard physical principles used in thermometry.

Key Concept

Thermometric properties and operating principles of thermometers

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