All practice questions

Question 12961Question →

Match each human regulatory hormone listed on the left with its corresponding physiological action or target mechanism on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Parathormone matches with increasing plasma calcium levels via bone resorption; Aldosterone matches with stimulating sodium ion reabsorption in kidney tubules; Oxytocin matches with inducing uterine contraction and milk ejection; Calcitonin matches with promoting calcium deposition into bone matrix.

Each hormone is correctly matched according to its endocrine target and action: Parathormone elevates plasma calcium concentration, Aldosterone increases renal sodium reabsorption, Oxytocin induces uterine contractions and milk let-down, and Calcitonin lowers plasma calcium concentration by depositing calcium into bone.

Step-by-Step Solution

1

Determine the physiological action of Parathormone (PTH)

PTH raises plasma

Ca^{2+}

levels by activating bone breakdown (resorption) and decreasing calcium excretion.

Parathyroid hormone regulates calcium homeostasis by preventing hypocalcemia.

2

Determine the function of Aldosterone

Aldosterone targets distal convoluted tubules and collecting ducts to reabsorb

Na^+

.

As a principal mineralocorticoid, it regulates electrolyte balance and blood pressure.

3

Analyze the action of Oxytocin

Oxytocin acts on target reproductive smooth muscle tissues during parturition and lactation.

It directly triggers uterine contractions and the milk ejection reflex.

4

Identify the function of Calcitonin

Calcitonin works antagonistically to PTH by depositing excess blood

Ca^{2+}

into bone tissue.

Thyroid C-cells release calcitonin to prevent hypercalcemia.

Key Concept

Endocrine regulation and physiological mechanisms of human hormones

Question 12962Question →

The standard enthalpies of combustion of carbon, hydrogen, and propane are given below:

\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -393.5\text{ kJ mol}^{-1}

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -285.8\text{ kJ mol}^{-1}

\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \quad \Delta H^\circ = -2220.0\text{ kJ mol}^{-1}

What is the standard enthalpy of formation of propane, $\text{C}_3\text{H}_8(g)$ ?

-103.7\text{ kJ mol}^{-1}

+103.7\text{ kJ mol}^{-1}

-4543.7\text{ kJ mol}^{-1}

+4543.7\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $-103.7\text{ kJ mol}^{-1}$

Answer

-103.7\text{ kJ mol}^{-1}

To calculate the standard enthalpy of formation of propane, combine the enthalpy changes for burning 3 moles of carbon and 4 moles of hydrogen, and subtract the enthalpy change for burning 1 mole of propane:

3(-393.5) + 4(-285.8) - (-2220.0) = -1180.5 - 1143.2 + 2220.0 = -103.7\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Write the target equation for the formation of propane from its elements in standard states.

3\text{C}(s) + 4\text{H}_2(g) \rightarrow \text{C}_3\text{H}_8(g)

The enthalpy of formation definition requires forming 1 mole of propane from carbon and hydrogen gas.

2

Scale the given combustion reactions to match the stoichiometric coefficients of the target equation.

3\text{C}(s) + 3\text{O}_2(g) \rightarrow 3\text{CO}_2(g) \quad \Delta H_1^\circ = 3 \times (-393.5) = -1180.5\text{ kJ mol}^{-1}

4\text{H}_2(g) + 2\text{O}_2(g) \rightarrow 4\text{H}_2\text{O}(l) \quad \Delta H_2^\circ = 4 \times (-285.8) = -1143.2\text{ kJ mol}^{-1}

3 moles of C and 4 moles of H2 are needed on the reactant side.

3

Reverse the propane combustion equation so propane appears on the product side, changing the sign of

\Delta H^\circ

.

3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \rightarrow \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \quad \Delta H_3^\circ = +2220.0\text{ kJ mol}^{-1}

Reversing a reaction step changes the sign of its enthalpy change according to Hess's Law.

4

Sum the modified enthalpy values.

\Delta H_f^\circ = -1180.5 + (-1143.2) + 2220.0 = -103.7\text{ kJ mol}^{-1}

According to Hess's Law, the total enthalpy change is independent of the pathway.

Key Concept

Hess's Law and Calculation of Enthalpy of Formation from Enthalpies of Combustion

Estimated Time:1m 30s

Question 12963Question →

A radioactive sample has a half-life of $5\text{ days}$ . If the mass of the sample that has decayed after $20\text{ days}$ is $45\text{ g}$ , what was the initial mass of the sample?

48\text{ g}

720\text{ g}

60\text{ g}

180\text{ g}

Show answer & explanation

Answer: $48\text{ g}$

Answer

The initial mass of the sample was

48\text{ g}

.

The elapsed time corresponds to 4 half-lives (

20 / 5 = 4

). The remaining mass fraction is

(1/2)^4 = 1/16

, which means

15/16

of the original mass has decayed. Setting

15/16

of the initial mass equal to

45\text{ g}

gives an initial mass of

48\text{ g}

.

Step-by-Step Solution

1

Calculate the number of half-lives elapsed

n = \frac{t}{T_{1/2}} = \frac{20\text{ days}}{5\text{ days}} = 4\text{ half-lives}

Determining how many half-life cycles occurred within the total elapsed time.

2

Determine the fraction of sample decayed

Fraction remaining =

(1/2)^4 = 1/16

; Fraction decayed =

1 - 1/16 = 15/16

The decayed fraction is the complement of the remaining fraction.

3

Calculate the initial mass

M_{\text{decayed}} = \frac{15}{16} M_0 = 45\text{ g} \implies M_0 = 45 \times \frac{16}{15} = 48\text{ g}

Relating the given decayed mass to the total initial mass.

Key Concept

Radioactive Decay Law and Half-life

Question 12964Question →

A sustained boom in Nigeria's crude oil revenue leads to massive foreign exchange inflows, causing the domestic currency to appreciate and rendering traditional export sectors like agriculture internationally uncompetitive. Which economic concept describes this structural distortion?

Import substitution

Dutch disease

Economic growth

Resource allocation

Show answer & explanation

Answer: Dutch disease

Answer

Dutch disease

The correct answer is Dutch disease. When a country experiences a natural resource export boom (such as crude oil in Nigeria), the heavy influx of foreign currency causes the domestic currency to appreciate. This real exchange rate appreciation makes non-oil exports, such as agriculture and manufacturing, less competitive in global markets, leading to structural decline in those sectors.

Step-by-Step Solution

1

Identify the economic trigger and mechanism in the scenario

Crude oil exports generate large inflows of foreign currency, causing domestic currency appreciation.

Increased demand for domestic currency or massive foreign exchange revenue strengthens the exchange rate.

2

Analyze the impact on non-oil export sectors

Agricultural goods become relatively more expensive abroad and less competitive, shrinking non-oil export earnings.

A stronger local currency makes domestic products costlier for foreign buyers.

3

Match the observed macroeconomic condition with the correct economic term

The phenomenon where a primary resource boom harms other tradeable sectors via currency appreciation is known as Dutch disease.

Dutch disease specifically describes this resource curse channel.

Key Concept

Dutch Disease in Nigeria's Petroleum Sector

Question 12965Question →

Phosphorus forms two distinct chlorides, X and Y. Analysis shows that $3.10\text{ g}$ of phosphorus combines with $10.65\text{ g}$ of chlorine to form compound X, whereas $3.10\text{ g}$ of phosphorus combines with $17.75\text{ g}$ of chlorine to form compound Y. Which law of chemical combination is illustrated by these experimental data, and what is the simple mass ratio of chlorine reacting with the fixed mass of phosphorus?

Law of Multiple Proportions with a ratio of 3 : 5

Law of Definite Proportions with a ratio of 3 : 5

Law of Multiple Proportions with a ratio of 2 : 3

Law of Conservation of Mass with a ratio of 1 : 2

Show answer & explanation

Answer: Law of Multiple Proportions with a ratio of 3 : 5

Answer

Law of Multiple Proportions with a ratio of 3 : 5

The option specifying the Law of Multiple Proportions with a ratio of 3 : 5 is correct because the mass of phosphorus is held constant at

3.10\text{ g}

, while the masses of chlorine in compounds X and Y are

10.65\text{ g}

and

17.75\text{ g}

respectively. Simplifying the ratio

\frac{10.65}{17.75}

yields

\frac{3}{5}

(

3 : 5

). Because two elements form different compounds whose mass ratios reduce to simple integers, the observation demonstrates John Dalton's Law of Multiple Proportions.

Step-by-Step Solution

1

Identify the fixed mass and variable mass values from the given data.

Fixed mass of phosphorus =

3.10\text{ g}

in both compounds. Mass of chlorine in compound X =

10.65\text{ g}

. Mass of chlorine in compound Y =

17.75\text{ g}

.

The Law of Multiple Proportions compares the varying masses of one element combining with a constant mass of another.

2

Calculate the simple whole-number ratio of the masses of chlorine.

\frac{10.65}{17.75} = \frac{3}{5} \implies 3 : 5

Dividing both chlorine masses by their greatest common factor (

3.55

) gives the integer ratio

3 : 5

.

3

Determine which law of chemical combination corresponds to this relationship.

Law of Multiple Proportions

When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a ratio of small whole numbers.

Key Concept

Law of Multiple Proportions

Question 12966Question →

A textiles trader in Kano has a budget of $\text{₦}200,000$ and wants to buy both a modern sewing machine and additional fabric stock. However, because the total cost of both items is $\text{₦}350,000$ , she chooses to purchase only the fabric stock after ranking her pressing needs. Which fundamental aspect of the definition and scope of Economics does this decision best illustrate?

Economics studies human behavior as a relationship between unlimited ends and scarce means that have alternative uses.

Economics is primarily concerned with calculating the total money cost and financial expenditure incurred in business operations.

Economics focuses strictly on how government interventions cause an outward shift in a nation's production possibility curve.

Economics is defined exclusively as the scientific inquiry into the accumulation of physical wealth and material commodities.

Show answer & explanation

Answer: Economics studies human behavior as a relationship between unlimited ends and scarce means that have alternative uses.

Answer

Economics studies human behavior as a relationship between unlimited ends and scarce means that have alternative uses.

The correct answer accurately states the widely accepted modern definition of Economics formulated by Lionel Robbins. The trader's situation highlights how human beings must choose between competing wants (ends) using limited funds (scarce means) that could be used for other purposes.

Step-by-Step Solution

1

Identify the key elements in the scenario

The trader has limited financial resources (

\text{₦}200,000

) relative to her total desires (

\text{₦}350,000

), forcing her to prioritize and select one item.

Recognizing scarcity and choice is the foundation of economic definitions.

2

Relate the scenario to formal economic definitions

Lord Lionel Robbins defined Economics in 1932 as 'the science which studies human behavior as a relationship between ends and scarce means which have alternative uses.'

The trader's funds represent scarce means with alternative uses, while the desired inventory items represent ends.

3

Evaluate the correct statement

The choice emphasizing human behavior, unlimited ends, and scarce means directly describes the core definition and scope of Economics.

It captures both the micro-level choice context and the universal scope of economic inquiry.

Key Concept

Definition and Scope of Economics (Robbins Definition)

Estimated Time:1m 0s

Question 12967Question →

Members of Kingdom Fungi display diverse structural forms ranging from microscopic unicellular yeasts to filamentous moulds. Which pair of characteristics is shared by both unicellular yeasts (*Saccharomyces*) and multicellular moulds (*Rhizopus*)?

Cell walls composed of chitin and carbohydrate storage in the form of glycogen

Cell walls composed of cellulose and carbohydrate storage in the form of starch

Presence of photosynthetic pigments and autotrophic synthesis of organic food

Holozoic ingestion of solid organic particles followed by intracellular digestion

Show answer & explanation

Answer: Cell walls composed of chitin and carbohydrate storage in the form of glycogen

Answer

Cell walls composed of chitin and carbohydrate storage in the form of glycogen

All members of Kingdom Fungi, regardless of whether they are unicellular (yeasts) or multicellular filamentous structures (moulds and mushrooms), share fundamental cellular and biochemical traits: their cell walls are constructed primarily of chitin (a polymer of N-acetylglucosamine) and they store surplus carbohydrates in the form of glycogen.

Step-by-Step Solution

1

Identify the key biochemical and structural features defining Kingdom Fungi.

Fungi possess cell walls containing chitin and store energy as glycogen, distinct from plants (cellulose/starch).

These biochemical markers are universal across both unicellular (yeasts) and multicellular (moulds, mushrooms) fungal groups.

2

Evaluate the nutritional mode of fungi.

Fungi are heterotrophs performing extracellular saprophytic digestion rather than photosynthesis or holozoic ingestion.

Eliminating options proposing autotrophic nutrition or intracellular digestion.

Key Concept

Shared structural and biochemical characteristics of Kingdom Fungi

Estimated Time:1m 0s

Question 12968Question →

A solid sample of lead with a mass of $0.80\text{ kg}$ is kept at its melting point of $327^\circ\text{C}$ . If $15,000\text{ J}$ of thermal energy is supplied to the sample, calculate the mass of lead, in kilograms, that remains in the solid state. (Take the specific latent heat of fusion of lead as $2.5 \times 10^4\text{ J/kg}$ ).

Show answer & explanation

Answer: 0.2

Answer

The mass of lead remaining in the solid state is

0.20\text{ kg}

.

Thermal energy

Q = 15,000\text{ J}

supplied to lead at its melting point melts a portion calculated by

m_{\text{melted}} = \frac{Q}{L_f} = \frac{15,000}{25,000} = 0.60\text{ kg}

. Subtracting this melted mass from the original

0.80\text{ kg}

yields

0.20\text{ kg}

of remaining solid lead.

Step-by-Step Solution

1

Calculate the mass of lead that melts.

Melted mass

m_{\text{melted}} = 0.60\text{ kg}

.

At the melting point, thermal energy supplied goes entirely into phase change without changing temperature:

Q = m_{\text{melted}} L_f

.

2

Determine the remaining mass of solid lead.

Remaining solid mass

m_{\text{solid}} = 0.20\text{ kg}

.

The un-melted portion equals the initial total mass minus the mass that has melted (

m_{\text{solid}} = m_{\text{total}} - m_{\text{melted}}

).

Key Concept

Latent Heat of Fusion and Phase Change

Question 12969Question →

Two closely related species of field crickets inhabit the same meadow and are capable of producing viable hybrids in laboratory settings. However, in nature, interbreeding never occurs because one species actively calls and mates at dawn while the other mates exclusively at dusk. Which type of reproductive isolating mechanism is maintaining speciation between these two populations?

Temporal isolation

Geographical isolation

Mechanical isolation

Hybrid breakdown

Show answer & explanation

Answer: Temporal isolation

Answer

Temporal isolation prevents gene flow between the two cricket populations because their reproductive activities occur at different times of the day (dawn versus dusk) despite sharing the same geographic habitat.

Temporal isolation is a pre-zygotic reproductive barrier that prevents interbreeding between sympatric species because they breed at different times of the day, seasons, or years. In this scenario, mating at dawn versus dusk ensures that mature gametes are never exchanged in nature.

Step-by-Step Solution

1

Analyze the environmental context and spatial distribution of the two populations.

Both species inhabit the exact same meadow, eliminating physical spatial barriers (geographical isolation).

Sympatric populations occupy overlapping geographic ranges.

2

Identify the precise barrier preventing interbreeding.

Mating activity is separated by time of day (dawn vs. dusk).

Different mating schedules act as a pre-zygotic barrier preventing gametes from meeting.

3

Classify the reproductive isolation mechanism.

Separation based on timing of reproductive behavior is defined as temporal isolation.

Temporal isolation prevents gene flow between populations due to differences in reproductive timing.

Key Concept

Pre-zygotic Reproductive Isolation Mechanisms (Temporal Isolation)

Question 12970Question →

A spring with a stiffness constant of $200\text{ N m}^{-1}$ is compressed by $0.3\text{ m}$ on a frictionless horizontal table. A block of mass $0.5\text{ kg}$ is placed against the compressed spring. When the system is released from rest, all the stored elastic potential energy of the spring is transferred to the block. What is the speed of the block as it leaves the spring?

6.0\text{ m s}^{-1}

36.0\text{ m s}^{-1}

18.0\text{ m s}^{-1}

3.0\text{ m s}^{-1}

Show answer & explanation

Answer: $6.0\text{ m s}^{-1}$

Answer

The speed of the block as it leaves the spring is

6.0\text{ m s}^{-1}

.

By the law of conservation of energy, the elastic potential energy stored in the spring when compressed by

x = 0.3\text{ m}

is

E_p = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.3)^2 = 9\text{ J}

. Upon release, this energy converts fully into the block's kinetic energy

E_k = \frac{1}{2}mv^2 = 9\text{ J}

. Substituting

m = 0.5\text{ kg}

gives

0.25 v^2 = 9

, so

v^2 = 36

and

v = 6.0\text{ m s}^{-1}

.

Step-by-Step Solution

1

Calculate the elastic potential energy (

E_p

) stored in the compressed spring.

E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.3)^2 = 100 \times 0.09 = 9.0\text{ J}

Energy stored in a compressed ideal spring is given by Hooke's law energy formula.

2

Apply the law of conservation of mechanical energy to find the kinetic energy (

E_k

) of the block.

E_k = E_p = 9.0\text{ J}

On a frictionless surface, all elastic potential energy converts entirely into translational kinetic energy.

3

Solve for the velocity (

v

) using the kinetic energy formula

E_k = \frac{1}{2} m v^2

.

9.0 = \frac{1}{2} (0.5) v^2 \implies 0.25 v^2 = 9.0 \implies v^2 = 36 \implies v = 6.0\text{ m s}^{-1}

Isolating

v

requires dividing by half the mass and taking the principal square root.

Key Concept

Conservation of Mechanical Energy (Elastic Potential Energy to Kinetic Energy)

Question 12971Question →

Two naturally occurring species of chlorine are represented as $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ . Which of the following statements correctly describes the relationship between these two atoms?

They are isotopes because they have the same atomic number but different mass numbers due to differing neutron counts.

They are allotropes because they represent different structural forms of the same element in the same physical state.

They are isobars because they have identical mass numbers but different numbers of protons in their nuclei.

They are ions formed by gaining or losing electrons from the outermost valence shell.

Show answer & explanation

Answer: They are isotopes because they have the same atomic number but different mass numbers due to differing neutron counts.

Answer

The atoms are isotopes because they have the same atomic number (17 protons) but different mass numbers (35 and 37) resulting from different numbers of neutrons.

The statement identifying them as isotopes is correct because both species belong to chlorine with an atomic number of 17 (17 protons), but differ in mass numbers (35 and 37) due to having 18 and 20 neutrons respectively.

Step-by-Step Solution

1

Analyze the nuclide symbols

^{35}_{17}\text{Cl}

and

^{37}_{17}\text{Cl}

for atomic and mass numbers.

Both species have an atomic number

Z = 17

(protons = 17).

^{35}_{17}\text{Cl}

has mass number

A = 35

, while

^{37}_{17}\text{Cl}

has mass number

A = 37

.

The lower subscript indicates atomic number (number of protons) and the upper superscript indicates mass number (protons + neutrons).

2

Calculate the neutron count for each nuclide.

Neutrons in

^{35}_{17}\text{Cl} = 35 - 17 = 18

. Neutrons in

^{37}_{17}\text{Cl} = 37 - 17 = 20

.

Neutron number is determined by subtracting atomic number from mass number (

N = A - Z

).

3

Classify the relationship based on chemical definitions.

Atoms of the same element (

Z = 17

) possessing different neutron counts and mass numbers are defined as isotopes.

Isotopy is defined specifically by identical atomic number paired with different mass numbers.

Key Concept

Isotopes are atoms of the same element with identical atomic numbers (protons) but different mass numbers (neutrons).

Question 12972Question →

Over geological time, animal respiratory systems evolved structural adaptations to meet increasing metabolic demands and facilitate the transition from aquatic to terrestrial environments. Arrange the following respiratory mechanisms in order of their evolutionary complexity, starting from the most primitive method to the most advanced terrestrial adaptation.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct evolutionary sequence of respiratory mechanisms from most primitive to most advanced is: Direct cutaneous diffusion across a moist body surface → Filamentous vascularized gills → Simple sac-like lungs supplemented by skin → Highly compartmentalized lungs containing dense alveolar networks.

The evolutionary trend in animal respiratory systems progresses from simple diffusion across unspecialized body surfaces (in primitive aquatic organisms) to specialized external/internal gills (in aquatic invertebrates and fish), followed by simple sac-like lungs requiring cutaneous assistance (in amphibians transitioning to land), and culminates in highly compartmentalized alveolar lungs (in advanced terrestrial homoiotherms).

Step-by-Step Solution

1

Identify the baseline primitive state of gas exchange in animals.

Direct cutaneous diffusion across the cell membrane or general body surface requires no specialized tissue and is characteristic of lower invertebrates.

Evolutionary trends begin with unspecialized body surfaces before true organ systems evolved.

2

Determine the specialized aquatic respiratory adaptation.

Filamentous gills evolved as specialized, highly folded vascular structures to extract dissolved oxygen in aquatic vertebrates and higher invertebrates.

Gills represent organ-level specialization in aquatic habitats prior to land colonization.

3

Identify the early primitive terrestrial respiratory adaptation.

Simple sac-like lungs represent an intermediate evolutionary stage seen in early land dwellers like amphibians, which still rely partly on cutaneous respiration.

Primitive lungs had minimal folding and surface area, necessitating supplementary cutaneous gas exchange.

4

Identify the most complex, specialized terrestrial respiratory adaptation.

Compartmentalized lungs filled with microscopic alveoli provide vast internal surface area for high metabolic demands in advanced endothermic land vertebrates.

Alveoli represent the apex of respiratory structural evolution, allowing efficient gas exchange while preventing body desiccation.

Key Concept

Evolutionary progression of respiratory systems in animals from simple diffusion to specialized internal alveolar structures.

Question 12973Question →

In a nuclear fission reaction, a Uranium-235 nucleus absorbs a neutron and splits into Tellurium-137 and Zirconium-97, releasing energy. If the total mass of the reactants is $236.053\text{ u}$ and the total mass of the products is $235.853\text{ u}$ , what is the energy released in $\text{MeV}$ ? (Take $1\text{ u} = 931\text{ MeV}$ )

18.62\text{ MeV}

186.20\text{ MeV}

0.000215\text{ MeV}

186.20\text{ J}

Show answer & explanation

Answer: $186.20\text{ MeV}$

Answer

The total energy released in the fission reaction is

186.20\text{ MeV}

.

The energy released during nuclear fission is calculated from the mass defect (

\Delta m

). Subtracting product mass from reactant mass gives

\Delta m = 236.053\text{ u} - 235.853\text{ u} = 0.200\text{ u}

. Multiplying this mass defect by the standard conversion factor (

1\text{ u} = 931\text{ MeV}

) yields

0.200 \times 931 = 186.20\text{ MeV}

.

Step-by-Step Solution

1

Calculate the mass defect (

\Delta m

)

\Delta m = 236.053\text{ u} - 235.853\text{ u} = 0.200\text{ u}

Mass defect is the difference between the total mass of the reactants and the total mass of the fission products.

2

Convert the mass defect into energy in

\text{MeV}

E = 0.200\text{ u} \times 931\text{ MeV/u} = 186.20\text{ MeV}

According to mass-energy equivalence, each atomic mass unit (

\text{u}

) liberates

931\text{ MeV}

of energy.

Key Concept

Mass Defect and Energy Release in Nuclear Fission

Question 12974Question →

A short bar magnet with magnetic dipole moment $1.6\text{ A}\cdot\text{m}^2$ is placed along the magnetic meridian with its north pole pointing towards the Earth's magnetic south pole. A neutral point is located on the axial line of the magnet at a distance of $0.2\text{ m}$ from its center. What is the magnitude of the horizontal component of the Earth's magnetic field at this location, in microtesla ( $\mu\text{T}$ )? (Take $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$ )

Show answer & explanation

Answer: 40

Answer

The magnitude of the horizontal component of Earth's magnetic field at this location is 40 μT.

At a neutral point, the horizontal component of Earth's magnetic field is equal in magnitude and opposite in direction to the magnetic field generated by the bar magnet. For a magnet aligned with its north pole pointing south, neutral points lie on its axial line at distance

d

. Using

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

with

M = 1.6\text{ A}\cdot\text{m}^2

and

d = 0.2\text{ m}

yields

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

.

Step-by-Step Solution

1

Determine the condition for the neutral point

B_{\text{axial}} = B_h

When a magnet's north pole points south, its axial magnetic field opposes Earth's horizontal field, creating neutral points along the axis where the magnetic fields cancel out completely.

2

Apply the short bar magnet formula for field along the axial line

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

The magnetic field produced at an axial point at distance

d

from the center of a short bar magnet of magnetic moment

M

is given by this formula.

3

Substitute the given numerical parameters

B_h = 10^{-7} \times \frac{2 \times 1.6}{(0.2)^3}

Substituting

M = 1.6\text{ A}\cdot\text{m}^2

,

d = 0.2\text{ m}

, and

\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}

into the field equation.

4

Calculate the magnitude of the horizontal field component in microtesla

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

Dividing

3.2 \times 10^{-7}

by

8 \times 10^{-3}

gives

4 \times 10^{-5}\text{ T}

, which converts to

40\ \mu\text{T}

.

Key Concept

Neutral points created by a bar magnet aligned with Earth's magnetic meridian

Question 12975Question →

Hydrogen chloride gas ( $HCl$ ) dissolved in anhydrous methylbenzene turns blue litmus paper red and conducts an electric current. Is this statement true or false?

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Hydrogen chloride dissolved in anhydrous methylbenzene does not ionize and therefore does not show acidic properties or conduct electricity.

The statement is false because hydrogen chloride (

HCl

) exists purely as un-ionized covalent molecules in non-polar solvents like methylbenzene. Water or another polar solvent is strictly required for

HCl

to dissociate into hydroxonium ions (

H_3O^+

) and chloride ions (

Cl^-

), which are responsible for changing indicator colors and conducting electricity.

Step-by-Step Solution

1

Analyze the nature of the solvent and solute interaction.

Methylbenzene is a non-polar organic solvent, while

HCl

is a covalent compound.

Acids display acidic behavior only when ionized into hydrogen ions (

H^+

/

H_3O^+

), which requires a polar solvent such as water.

2

Determine the state of

HCl

in non-polar methylbenzene.

HCl

remains as un-ionized covalent molecules in methylbenzene.

Non-polar solvents cannot stabilize the separated

H^+

and

Cl^-

ions via solvation.

3

Evaluate the chemical and physical observations.

Without free

H^+

ions, blue litmus paper remains blue. Without mobile ions, the solution cannot conduct electricity.

Acidic indicators and electrical conductivity both depend directly on the presence of mobile ions in solution.

Key Concept

Role of Water in Acidic Properties and Ionization

Question 12976Question →

An economy experiences a persistent increase in real Gross Domestic Product (GDP) accompanied by a structural shift from low-productivity agriculture to high-value industrial production, improved literacy rates, and reduced poverty levels. How is this total economic transformation best classified?

Economic development, because it combines quantitative national income expansion with qualitative structural improvements in living standards.

Economic growth, because it involves an increase in real gross domestic product and total industrial output over time.

Economic growth only, since structural changes and improvements in literacy are automatic consequences of national income expansion.

Economic planning, because structural shifts across sectors occur exclusively through centralized government intervention.

Show answer & explanation

Answer: Economic development, because it combines quantitative national income expansion with qualitative structural improvements in living standards.

Answer

Economic development, because it combines quantitative national income expansion with qualitative structural improvements in living standards.

Economic development encompasses both quantitative expansion in real national income (growth) and qualitative structural transformations in society, such as sector modernization, poverty reduction, and improved literacy levels.

Step-by-Step Solution

1

Analyze the economic indicators presented in the stem.

The stem describes both quantitative output growth (increase in real GDP) and qualitative multidimensional progress (sectoral modernization, literacy improvements, and poverty reduction).

To classify the process, one must determine whether the changes are purely quantitative or encompass structural and qualitative dimensions.

2

Distinguish between Economic Growth and Economic Development.

Economic growth is narrow and quantitative (rising GDP/output). Economic development is broader and qualitative, involving growth alongside structural transformation, institutional changes, and enhanced quality of life.

Matching the broad socio-economic indicators in the scenario to economic definitions confirms that the process represents economic development.

Key Concept

Distinction Between Economic Growth and Economic Development

Estimated Time:1m 0s

Question 12977Question →

A student pays ₦15,000 cash to purchase a textbook, and the seller accepts the currency immediately in exchange for the book. Which primary function of money does this transaction illustrate?

Medium of exchange

Store of value

Standard of deferred payment

Unit of account

Show answer & explanation

Answer: Medium of exchange

Answer

Medium of exchange

The option stating 'Medium of exchange' is correct because money functions as a medium of exchange whenever it acts as an intermediary token during the buying and selling of goods and services, eliminating the inefficiencies of barter.

Step-by-Step Solution

1

Analyze the action taking place in the scenario.

A buyer hands over currency to receive a physical item in a direct trade.

Identifying the nature of the economic exchange helps determine the function of money involved.

2

Map the transaction to the corresponding function of money.

The money facilitates the trade of goods, serving as a medium of exchange.

When money is used to facilitate immediate buying and selling, it acts as a medium of exchange.

Key Concept

Functions of Money - Medium of Exchange

Estimated Time:45s

Question 12978Question →

Light of frequency $8.0 \times 10^{14}\text{ Hz}$ illuminates a photosensitive plate, causing photoelectrons to be emitted with a maximum kinetic energy of $1.2\text{ eV}$ . If the same plate is subsequently illuminated by light of frequency $1.2 \times 10^{15}\text{ Hz}$ , what is the stopping potential, in volts, needed to reduce the photoelectric current to zero? (Take $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 2.85

Answer

The stopping potential needed to reduce the photoelectric current to zero is 2.85 V.

Using Einstein's photoelectric equation

E = W_0 + K_{\text{max}}

, the initial photon energy is

E_1 = h f_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

. Given

K_1 = 1.2\text{ eV}

, the work function of the metal is

W_0 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}

. For the second frequency

f_2 = 1.2 \times 10^{15}\text{ Hz}

, the photon energy is

E_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}

. The new maximum kinetic energy is

K_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}

. Since

e V_s = K_{\text{max}}

, the stopping potential required to reduce the current to zero is

2.85\text{ V}

.

Step-by-Step Solution

1

Calculate the photon energy

E_1

of the initial light in electron-volts

E_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

Photon energy is related to frequency by

E = h f

.

2

Determine the work function

W_0

of the photosensitive plate

W_0 = E_1 - K_1 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}

By Einstein's photoelectric equation,

K_{\text{max}} = E - W_0

.

3

Calculate the photon energy

E_2

for the second light frequency

E_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}

The energy of the second photon is calculated using

f_2 = 1.2 \times 10^{15}\text{ Hz}

.

4

Find the maximum kinetic energy

K_2

and corresponding stopping potential

V_s

K_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}

, giving

V_s = 2.85\text{ V}

The stopping potential in volts is numerical equal to the maximum kinetic energy expressed in electron-volts (

e V_s = K_{\text{max}}

).

Key Concept

Einstein's Photoelectric Equation and Stopping Potential

Estimated Time:2m 0s

Question 12979Question →

An agricultural biologist examines the floral morphology of a newly cultivated crop species to determine its primary pollination mechanism. The flowers are small, lack petals and nectar, possess long pendulous filaments bearing abundant light, smooth pollen grains, and feature feathery, exposed stigmas. Based on these observed structural adaptations, what is the primary mode of pollination for this species?

Anemophily, because lightweight pollen grains and feathery stigmas facilitate wind transport and pollen capture

Entomophily, because feathery stigmas and smooth pollen grains are specialized adaptations to adhere to foraging insect legs

Hydrophily, because exposed filaments and dull bracts function exclusively for pollen dispersion in aquatic currents

Ornithophily, because pendulous filaments attract perching birds to transfer pollen between floral bracts

Show answer & explanation

Answer: Anemophily, because lightweight pollen grains and feathery stigmas facilitate wind transport and pollen capture

Answer

Anemophily, because lightweight pollen grains and feathery stigmas facilitate wind transport and pollen capture

The correct answer identifies wind pollination (anemophily). Wind-pollinated flowers lack colorful petals, scent, and nectar to conserve energy, producing large quantities of smooth, lightweight pollen that float easily in air currents, while exposed feathery stigmas maximize the surface area for trapping floating pollen.

Step-by-Step Solution

1

Analyze the floral morphological features described in the scenario

Identified small petal-less flowers, abundant light smooth pollen grains, long pendulous filaments, and feathery exposed stigmas.

Floral structures directly reflect the specific pollination vector to which the plant is adapted.

2

Correlate the features with mechanisms of pollen transfer

Lightweight, smooth pollen grains easily float in air currents without clumping, and feathery stigmas effectively trap drifting pollen grains out of the air.

These combined structural features are classic adaptations for wind pollination (anemophily).

Key Concept

Structural adaptations of anemophilous (wind-pollinated) flowers

Question 12980Question →

Which of the following measures is a primary feature of an Export Promotion Industrialization strategy?

Providing export subsidies and tax incentives to domestic manufacturers producing for international markets

Imposing heavy protective tariffs and quotas on imported goods to encourage local substitute production

Restricting all foreign investment to ensure total government ownership of manufacturing firms

Prohibiting the importation of essential capital equipment and industrial raw materials

Show answer & explanation

Answer: Providing export subsidies and tax incentives to domestic manufacturers producing for international markets

Answer

Providing export subsidies and tax incentives to domestic manufacturers producing for international markets

Export Promotion Industrialization is an outward-looking development strategy designed to boost industrial growth by producing manufactured goods for foreign markets. Providing export subsidies, financial support, and tax relief helps domestic firms compete effectively in international trade.

Step-by-Step Solution

1

Identify the core goal of Export Promotion Industrialization (EPI).

EPI aims to expand manufacturing production aimed primarily at foreign buyers to earn foreign exchange.

Understanding the policy intent distinguishes export promotion from trade restriction or domestic replacement strategies.

2

Evaluate the policy instruments associated with export promotion.

Governments support export-oriented domestic producers through financial incentives, tax holidays, export subsidies, and trade facilitation.

Subsidies and incentives lower export costs, making domestic goods competitive in foreign markets.

Key Concept

Export Promotion Industrialization (EPI)

Answer

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