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13931 questions

Question 13901Question

During soil profile development in humid regions, percolating water washes fine clay particles, iron, and aluminum oxides downward from the upper topsoil layer. In which soil profile horizon do these translocated materials primarily accumulate?

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Answer: The B horizon, which acts as the zone of illuviation

Answer

The B horizon, which acts as the zone of illuviation
The correct answer correctly identifies the subsoil layer as the primary site of illuviation. In a mature soil profile, downward-percolating rainwater carries fine clay particles, iron compounds, and organic material out of the top layer (eluviation) and deposits them into the subsoil below (illuviation), creating a distinct layer enriched in minerals.

Step-by-Step Solution

1
Identify the pedogenic translocational process described in the stem.
Downward movement of fine clay and oxides by percolating water is known as leaching/eluviation from topsoil.
Water moving through topsoil dissolves and suspends fine particulates.
2
Determine the destination horizon where translocated materials settle.
The subsoil layer receives and stores these deposited compounds through illuviation.
The subsoil horizon positioned immediately beneath the topsoil captures downward-migrating minerals.
3
Match the zone of illuviation to standard soil profile nomenclature.
The B horizon is designated as the subsoil accumulation layer.
Standard pedological classification defines the B horizon as the illuvial layer.

Key Concept

Soil Profile Horizons and Illuviation
Question 13902Question

A body is weighed on a distant planet where the acceleration due to gravity is 4 m s24\text{ m s}^{-2}. A spring balance calibrated on Earth (g=10 m s2g = 10\text{ m s}^{-2}) gives a reading of 20 N20\text{ N} for the body on this planet. What is the mass of the body as measured by a beam balance on the planet?

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Answer: 5.0 kg5.0\text{ kg}

Answer

The mass of the body as measured by a beam balance is 5.0 kg5.0\text{ kg}.
The spring balance measures weight (W=mgplanetW = mg_{\text{planet}}). Substituting the given values (20 N=m×4 m s220\text{ N} = m \times 4\text{ m s}^{-2}) gives m=5.0 kgm = 5.0\text{ kg}. Since mass is constant everywhere and a beam balance measures mass independently of local gravity variation, the beam balance reads 5.0 kg5.0\text{ kg}.

Step-by-Step Solution

1
Calculate the mass of the body using the spring balance reading on the planet.
m=Wplanetgplanet=20 N4 m s2=5.0 kgm = \frac{W_{\text{planet}}}{g_{\text{planet}}} = \frac{20\text{ N}}{4\text{ m s}^{-2}} = 5.0\text{ kg}
Spring balances measure weight force (W=mgW = mg). The local weight divided by local gravity yields the mass.
2
Determine the reading on a beam balance.
The beam balance measures mass directly by comparison, yielding 5.0 kg5.0\text{ kg}.
Mass is an invariant property of matter and does not change with location or gravitational field strength.

Key Concept

Mass vs. Weight and Instrument Principles
Estimated Time:1m 0s
Question 13903Question

A cell with an electromotive force (e.m.f.) of 15.0 V15.0\text{ V} and an internal resistance of 2.0 Ω2.0\ \Omega is connected across a parallel combination of two resistors with resistances of 6.0 Ω6.0\ \Omega and 12.0 Ω12.0\ \Omega. What is the potential difference across the parallel resistor network?

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Answer: 10.0 V10.0\text{ V}

Answer

The potential difference across the parallel resistor network is 10.0 V10.0\text{ V}.
The parallel combination of 6.0 Ω6.0\ \Omega and 12.0 Ω12.0\ \Omega has an effective resistance of 4.0 Ω4.0\ \Omega. Adding the internal resistance of 2.0 Ω2.0\ \Omega gives a total circuit resistance of 6.0 Ω6.0\ \Omega. The circuit current is I=15.0 V6.0 Ω=2.5 AI = \frac{15.0\text{ V}}{6.0\ \Omega} = 2.5\text{ A}. The potential difference across the parallel load is therefore V=2.5 A×4.0 Ω=10.0 VV = 2.5\text{ A} \times 4.0\ \Omega = 10.0\text{ V}.

Step-by-Step Solution

1
Calculate the equivalent resistance of the two parallel resistors (R1=6.0 ΩR_1 = 6.0\ \Omega and R2=12.0 ΩR_2 = 12.0\ \Omega).
Rp=R1R2R1+R2=6.0×12.06.0+12.0=72.018.0=4.0 ΩR_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\ \Omega
Resistors connected in parallel combine according to the reciprocal formula.
2
Determine the total resistance of the entire circuit including the internal resistance (r=2.0 Ωr = 2.0\ \Omega).
Rtotal=Rp+r=4.0 Ω+2.0 Ω=6.0 ΩR_{\text{total}} = R_p + r = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega
The cell's internal resistance is in series with the external equivalent load resistance.
3
Calculate the total current supplied by the cell using Ohm's law for a complete circuit.
I=ERtotal=15.0 V6.0 Ω=2.5 AI = \frac{E}{R_{\text{total}}} = \frac{15.0\text{ V}}{6.0\ \Omega} = 2.5\text{ A}
The current depends on the total e.m.f. divided by the total circuit resistance.
4
Calculate the potential difference across the parallel combination (terminal potential difference).
V=IRp=2.5 A×4.0 Ω=10.0 VV = I R_p = 2.5\text{ A} \times 4.0\ \Omega = 10.0\text{ V}
The potential difference across the parallel network is the product of the total current flowing into the combination and its equivalent resistance.

Key Concept

Terminal Potential Difference and Internal Resistance
Estimated Time:1m 30s
Question 13904Question

An agricultural survey conducted in a tropical region identified a commercial farming system characterized by large land holdings, monoculture production of cash crops, high capital investment, reliance on hired labor, and direct integration with export processing facilities. Which agricultural system is best described by these operational features?

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Answer: Plantation agriculture

Answer

Plantation agriculture is the commercial farming system defined by large-scale monoculture, heavy capital investment, and export-oriented production.
Plantation agriculture is defined by large-scale land holdings dedicated to the monoculture of crops such as rubber, oil palm, cocoa, or sugarcane, requiring significant financial capital, modern management, and processing infrastructure geared toward international trade.

Step-by-Step Solution

1
Analyze the operational parameters given in the stem.
Identified key features: large estate size, monoculture, high capital input, hired labor force, and processing for international export markets.
Systematic evaluation of inputs, labor organization, and market destination determines the farming classification.
2
Compare identified parameters against standard agricultural system typologies.
Plantation agriculture uniquely satisfies all five criteria, whereas subsistence and pastoral systems lack commercial infrastructure and monoculture estates.
Differentiating between commercial tree/cash crop estates and traditional subsistence or pastoral farming models.

Key Concept

Characteristics of Plantation Agriculture
Question 13905Question

At a weather station, a meteorological observer inspects an instrument equipped with a flexible, partially evacuated corrugated metal capsule that expands and contracts in response to changes in atmospheric conditions. Which weather element does this instrument measure, and what is its standard unit of measurement?

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Answer: Atmospheric pressure, measured in millibars

Answer

Atmospheric pressure, measured in millibars
The correct answer identifies atmospheric pressure measured in millibars because an aneroid barometer operates using an evacuated, sealed corrugated metal cell that flexes under atmospheric pressure changes, moving an indicator needle across a scale calibrated in millibars.

Step-by-Step Solution

1
Identify the operating mechanism described in the stem
The mechanism featuring a flexible, partially evacuated corrugated metallic cell (aneroid capsule) is characteristic of an aneroid barometer.
The aneroid cell reacts mechanically to external pressure differences without using liquid columns like mercury.
2
Determine the weather element measured by an aneroid barometer
Atmospheric pressure.
Barometers are designed specifically to record atmospheric force per unit area.
3
Identify the standard meteorological unit for atmospheric pressure
Millibars (mb) or hectopascals (hPa).
Standard meteorological convention uses millibars or hectopascals for pressure readings.

Key Concept

Operating principles of weather instruments and their parameters
Estimated Time:1m 0s
Question 13906Question

A Vernier caliper has 2020 divisions on its Vernier scale that coincide with 1919 main scale divisions of 1 mm1\text{ mm} each. When the jaws of the instrument are brought together without any object, the zero of the Vernier scale lies to the right of the main scale zero mark, and the 3rd3\text{rd} Vernier division coincides with a main scale mark. When used to measure the thickness of a wooden block, the main scale reads 3.5 cm3.5\text{ cm} and the 12th12\text{th} Vernier division coincides with a main scale line. What is the actual thickness of the wooden block?

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Answer: 3.545 cm3.545\text{ cm}

Answer

The actual thickness of the wooden block is 3.545 cm3.545\text{ cm}.
The least count of a 20-division Vernier caliper with 1 mm1\text{ mm} main scale divisions is 0.1 cm20=0.005 cm\frac{0.1\text{ cm}}{20} = 0.005\text{ cm}. Since the Vernier zero lies to the right of the main scale zero when closed, it has a positive zero error of +(3×0.005 cm)=+0.015 cm+ (3 \times 0.005\text{ cm}) = +0.015\text{ cm}. The observed measurement is 3.5 cm+(12×0.005 cm)=3.560 cm3.5\text{ cm} + (12 \times 0.005\text{ cm}) = 3.560\text{ cm}. Subtracting the positive zero error gives the actual thickness: 3.560 cm0.015 cm=3.545 cm3.560\text{ cm} - 0.015\text{ cm} = 3.545\text{ cm}.

Step-by-Step Solution

1
Determine the least count (precision) of the Vernier caliper.
Least Count (LC)=1 Main Scale Division (MSD)Number of Vernier Divisions=1 mm20=0.05 mm=0.005 cm\text{Least Count (LC)} = \frac{1\text{ Main Scale Division (MSD)}}{\text{Number of Vernier Divisions}} = \frac{1\text{ mm}}{20} = 0.05\text{ mm} = 0.005\text{ cm}.
The least count is the smallest value that can be measured directly by the instrument.
2
Calculate the zero error of the instrument.
Zero Error=+(3×0.005 cm)=+0.015 cm\text{Zero Error} = +(3 \times 0.005\text{ cm}) = +0.015\text{ cm}.
Because the Vernier zero lies to the right of the main scale zero mark when closed, the instrument has a positive zero error.
3
Calculate the total observed reading from the scales.
Observed Reading=3.5 cm+(12×0.005 cm)=3.5 cm+0.060 cm=3.560 cm\text{Observed Reading} = 3.5\text{ cm} + (12 \times 0.005\text{ cm}) = 3.5\text{ cm} + 0.060\text{ cm} = 3.560\text{ cm}.
The observed reading combines the main scale reading and the coincidental Vernier scale division multiplied by the least count.
4
Apply zero error correction to obtain the actual reading.
Actual Reading=Observed ReadingZero Error=3.560 cm(+0.015 cm)=3.545 cm\text{Actual Reading} = \text{Observed Reading} - \text{Zero Error} = 3.560\text{ cm} - (+0.015\text{ cm}) = 3.545\text{ cm}.
Zero error correction always requires subtracting the zero error (with its sign) from the observed measurement.

Key Concept

Measurement of length using Vernier calipers and zero error correction
Estimated Time:1m 30s
Question 13907Question

In Nigeria's domestic commercial trade network, agricultural commodities move between distinct ecological zones to satisfy regional demand. While timber, palm oil, and kola nuts are shipped northward from the humid southern rainforests, which of the following commodity groups forms the major reverse trade flow transported from the northern savanna regions to southern urban markets?

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Answer: Cattle, onions, and groundnuts

Answer

Cattle, onions, and groundnuts form the primary northern-to-southern agricultural trade flow in Nigeria.
Nigeria's internal commercial trade relies on ecological complementarity. The open savanna of the north provides ideal conditions for livestock (cattle) and crops adapted to drier environments (onions, groundnuts, beans). These items are transported southward to meet dense urban consumer demand, balancing the northward flow of forest products like palm oil, timber, and kola nuts.

Step-by-Step Solution

1
Analyze regional ecological specialization in Nigeria
Southern Nigeria is characterized by high rainfall and rainforest vegetation suited for tree crops (cocoa, oil palm, rubber, timber), while Northern Nigeria is dominated by savanna grassland and lower rainfall.
Agricultural production and trade flows are directly tied to climatic and vegetation belts.
2
Identify northern agricultural specializations
Northern savanna conditions favor pastoral cattle rearing as well as crops like groundnuts, onions, beans, sorghum, and tomatoes.
The absence of tsetse flies in open savanna facilitates cattle rearing, and drier soil conditions support grain and bulb crop cultivation.
3
Determine the direction of inter-regional trade flow
High urban population density in southern cities creates high demand for meat and northern crops, producing a steady South-bound trade flow of cattle, onions, and groundnuts.
Complementary ecological zones drive domestic trade exchange between northern and southern Nigeria.

Key Concept

Inter-Regional Agricultural Trade Complementarity in Nigeria
Estimated Time:1m 0s
Question 13908Question

Arrange the following sequential stages of a satellite remote sensing and GIS land-use mapping workflow in the correct order from initial energy capture to final spatial analysis.

Drag items to arrange them in the correct order

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Answer

The correct sequence is: (1) Detection and digital recording of reflected electromagnetic radiation by the satellite sensor, (2) Application of radiometric and geometric corrections to eliminate atmospheric and orbital distortions, (3) Supervised spectral classification to categorize land surface pixels into distinct land-use cover classes, and (4) Conversion of classified raster data into vector layers for overlay and spatial query in a GIS.
The workflow follows a standard sequence from physical signal capture to analytical integration: electromagnetic radiation is first recorded by satellite sensors, pre-processed to remove radiometric and atmospheric distortions, classified spectrally into thematic land-use categories, and finally converted into vector layers within a GIS database for spatial decision-making.

Step-by-Step Solution

1
Identify the initial physical process of data collection
Recording reflected electromagnetic energy at the sensor stage
Remote sensing workflows begin when sensors collect radiant energy from target surface features.
2
Determine the necessary data preparation step
Radiometric and geometric image pre-processing
Raw satellite imagery contains atmospheric noise and sensor distortions that must be rectified before extraction of analytical information.
3
Identify the thematic information extraction phase
Supervised spectral image classification
Corrected pixels are grouped into meaningful environmental or land-use categories using spectral response patterns.
4
Determine the final GIS integration and mapping stage
Vector conversion and GIS database overlay
Classified thematic layers are imported into GIS vector models for overlay analysis, spatial querying, and decision support.

Key Concept

Remote Sensing Data Processing and GIS Integration Workflow
Question 13909Question

X-rays undergo diffraction when passed through crystalline solids because the interplanar atomic spacing of a crystal lattice is comparable to the wavelength of X-rays.

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Answer: True

Answer

The statement is true. X-rays have wavelengths of approximately 1010 m10^{-10}\text{ m}, which is of the same order of magnitude as the spacing between adjacent atomic planes in crystal lattices, allowing crystals to act as natural diffraction gratings.
The statement is true because wave diffraction requires the grating aperture or obstacle spacing to be of the same order of magnitude as the incident wavelength. Since X-ray wavelengths (approx. 1010 m10^{-10}\text{ m}) match the interatomic spacing of crystals, crystalline solids act as effective diffraction gratings.

Step-by-Step Solution

1
Identify the characteristic wavelength range of X-rays.
X-rays are high-energy electromagnetic waves with wavelengths typically ranging from 1011 m10^{-11}\text{ m} to 108 m10^{-8}\text{ m} (around 0.1 nm0.1\text{ nm}).
Wave diffraction requires the width of the aperture or obstacle spacing to be comparable in size to the wavelength of the incident wave.
2
Determine the interatomic spacing of crystalline solids.
The distance between neighboring atomic planes in a typical crystal lattice is on the order of 1010 m10^{-10}\text{ m} (0.1 nm0.1\text{ nm} to 0.3 nm0.3\text{ nm}).
Comparing these dimensions shows that crystal lattice spacing matches X-ray wavelengths.
3
Evaluate the condition for wave diffraction.
Because the interatomic spacing matches the X-ray wavelength, constructive and destructive interference occurs, forming diffraction patterns.
This confirms that the given statement is true.

Key Concept

X-ray Diffraction by Crystal Lattices
Estimated Time:1m 0s
Question 13910Question

Match each climatic control on the left with its primary physical impact on global atmospheric and weather patterns on the right.

Click a left item, then click its matching right item

Items

Continentality
Cold offshore ocean currents
Latitude
Seasonal migration of planetary wind belts

Matches

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Answer

Continentality matches with large annual temperature ranges; Cold offshore ocean currents match with hyper-arid coastal deserts with advection fog; Latitude matches with primary determination of solar radiation intensity; Seasonal migration of planetary wind belts matches with alternation of distinct wet and dry seasons.
Each listed climatic factor directly determines specific environmental conditions: continentality leads to wide temperature swings inland; cold currents stabilize coastal air and cause aridity with fog; latitude governs global solar radiation reception; and wind belt shifts drive seasonal wet-dry dynamics in tropical regions.

Step-by-Step Solution

1
Analyze the climatic impact of land-water distribution (continentality)
Lacking ocean thermal buffering, inland regions experience severe seasonal temperature extremes.
Land has a lower specific heat capacity than water.
2
Evaluate atmospheric dynamics over cold oceanic currents
Lower-layer air cooling prevents convection and rainfall, leading to coastal aridity accompanied by fog.
Temperature inversion over cold water stabilizes the lower atmosphere.
3
Assess how geographic latitude controls temperature distribution
Higher latitudes receive lower solar intensity due to oblique rays, establishing global temperature zones.
Insolation per unit surface area decreases from the equator toward the poles.
4
Examine the effect of shifting pressure and wind belts
Seasonal movement of pressure systems yields distinct rainfall peaks and dry spells in intermediate tropical zones.
Planetary winds move north and south following the overhead sun.

Key Concept

Climatic Controls and World Climate Types
Question 13911Question

In an isolated system of colliding bodies where no net external force acts, total linear momentum is conserved only if the collision is perfectly elastic.

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Answer: False

Answer

The statement is False. Total linear momentum is conserved in all isolated collisions regardless of whether the collision is elastic or inelastic.
Linear momentum conservation depends strictly on the absence of net external forces. Internal forces, such as those causing deformation or heat release in inelastic collisions, cancel out in equal and opposite pairs according to Newton's Third Law and therefore do not change total linear momentum.

Step-by-Step Solution

1
Analyze the condition for linear momentum conservation.
Linear momentum is conserved whenever the net external force acting on the system is zero (Fext=0∑ F_{\text{ext}} = 0).
By Newton's Second Law written in terms of momentum (Fnet=ΔpΔtF_{\text{net}} = \frac{\Delta p}{\Delta t}), if Fnet=0F_{\text{net}} = 0, then Δp=0\Delta p = 0, meaning initial total momentum equals final total momentum.
2
Differentiate between momentum conservation and kinetic energy conservation.
Total linear momentum is conserved in both elastic and inelastic collisions, whereas total kinetic energy is conserved only in elastic collisions.
In inelastic collisions, internal forces convert kinetic energy into heat, sound, or mechanical deformation, but internal forces cannot alter the net momentum of the system.

Key Concept

Conservation of Linear Momentum in Collisions
Question 13912Question

In agricultural geography, land-use intensity increases as fallow periods shorten and labor inputs per unit area rise. Arrange the following tropical agricultural land-use practices in sequence from the lowest land-use intensity (longest fallow duration) to the highest land-use intensity (continuous cropping).

Drag items to arrange them in the correct order

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Answer

The correct sequence from lowest to highest land-use intensity is Forest fallow (shifting cultivation), Bush fallow cultivation, Short fallow system, and Continuous multi-cropping.
Forest fallow allows vegetation to regenerate over 20 to 25 years, representing the lowest land-use intensity. Under demographic pressure, fallow periods shorten to bush fallow (6–10 years), then short fallow (1–2 years), and ultimately continuous multi-cropping, where plots produce multiple harvests per year without any fallow period.

Step-by-Step Solution

1
Determine the criterion for land-use intensity in agricultural systems.
Land-use intensity is inversely related to fallow length: systems with long rest periods have low intensity, while systems with zero rest period have high intensity.
Theoretical frameworks of agricultural intensification (such as Boserup's model) rank farming systems by cropping frequency and labor density.
2
Evaluate the fallow duration of each given practice.
Forest fallow requires 20–25 years of rest; Bush fallow requires 6–10 years; Short fallow requires 1–2 years; Continuous multi-cropping requires 0 rest years.
Systematic population pressure and land scarcity force farmers to reduce fallow intervals and increase labor inputs.
3
Sequence the items from longest fallow duration to zero fallow duration.
Forest fallow → Bush fallow → Short fallow → Continuous multi-cropping.
This progression reflects increasing land-use frequency and intensity.

Key Concept

Agricultural Intensification and Fallow Sequences
Estimated Time:1m 30s
Question 13913Question

A meteorological technician at an airfield weather station needs to measure atmospheric pressure to assist aircraft pilots with altimeter settings. Which measuring instrument must be used, and in what standard unit is this atmospheric parameter recorded?

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Answer: Barometer, measured in millibars

Answer

The barometer is the correct instrument used to measure atmospheric pressure, which is standardly reported in millibars (mb) or hectopascals (hPa).
Atmospheric pressure is the force exerted per unit area by the weight of the air column above a location. It is measured using a barometer (mercury or aneroid barometer) and expressed standardly in millibars (mb) or hectopascals (hPa).

Step-by-Step Solution

1
Identify the target meteorological parameter required
The target parameter is atmospheric pressure.
Airfield weather stations report pressure settings to ensure accurate aircraft altimeter calibration.
2
Match the target parameter to its correct measuring instrument and standard unit
Atmospheric pressure is measured using a barometer (either mercury or aneroid type) and recorded in millibars (1 mb=100 Pa1\text{ mb} = 100\text{ Pa}) or hectopascals (hPa).
Each weather element requires a specific instrument and SI/meteorological unit.

Key Concept

Atmospheric pressure measurement and meteorological instruments
Question 13914Question

Consider the following West African nations: Ghana, Senegal, Côte d'Ivoire, and Nigeria. To which regional economic grouping do these countries belong as core member states seeking to foster sub-regional integration and free movement of goods and people?

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Answer: Economic Community of West African States (ECOWAS)

Answer

Economic Community of West African States (ECOWAS)
The Economic Community of West African States (ECOWAS) is the regional organization established in 1975 specifically for West African nations, including Nigeria, Ghana, Senegal, and Côte d'Ivoire, to facilitate free trade, economic development, and borderless movement within the sub-region.

Step-by-Step Solution

1
Identify the geographical sub-region of the listed countries
Ghana, Senegal, Côte d'Ivoire, and Nigeria are all located in West Africa.
Determining the geographical location narrows down the relevant sub-regional trade community.
2
Match the sub-region with its corresponding Regional Economic Community (REC)
The Economic Community of West African States (ECOWAS) is the trade bloc dedicated to West Africa.
ECOWAS was formed in 1975 specifically to drive economic integration across West African states.

Key Concept

Regional Economic Communities (RECs) in Africa
Question 13915Question

A voltmeter having an internal resistance of 900 Ω900\ \Omega is connected across the terminals of a cell with an electromotive force (e.m.f.) of 1.50 V1.50\text{ V} and an internal resistance of 100 Ω100\ \Omega. What is the reading on the voltmeter?

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Answer: 1.35 V1.35\text{ V}

Answer

The reading on the voltmeter is 1.35 V1.35\text{ V}.
When connected across the cell, the voltmeter's resistance forms a series circuit with the cell's internal resistance. The total resistance is 900 Ω+100 Ω=1000 Ω900\ \Omega + 100\ \Omega = 1000\ \Omega. The current drawn from the cell is I=1.50 V1000 Ω=0.0015 AI = \frac{1.50\text{ V}}{1000\ \Omega} = 0.0015\text{ A}. The voltage indicated by the meter is the potential difference across its terminals, V=0.0015 A×900 Ω=1.35 VV = 0.0015\text{ A} \times 900\ \Omega = 1.35\text{ V} (or EIr=1.500.15=1.35 VE - Ir = 1.50 - 0.15 = 1.35\text{ V}).

Step-by-Step Solution

1
Calculate the total resistance of the circuit.
Rtotal=Rv+r=900 Ω+100 Ω=1000 ΩR_{\text{total}} = R_v + r = 900\ \Omega + 100\ \Omega = 1000\ \Omega
The voltmeter resistance and the internal resistance of the cell are connected in series.
2
Calculate the current flowing in the circuit using Ohm's law.
I=ERtotal=1.50 V1000 Ω=0.0015 AI = \frac{E}{R_{\text{total}}} = \frac{1.50\text{ V}}{1000\ \Omega} = 0.0015\text{ A}
The electromotive force drives current through the total circuit resistance.
3
Calculate the potential difference across the voltmeter (terminal potential difference).
V=I×Rv=0.0015 A×900 Ω=1.35 VV = I \times R_v = 0.0015\text{ A} \times 900\ \Omega = 1.35\text{ V}
The voltmeter measures the potential drop across its own internal resistance.

Key Concept

Terminal Potential Difference and Voltmeter Loading Effect
Estimated Time:1m 30s
Question 13916Question

In recent decades, rapid urbanization across East African nations such as Kenya and Tanzania has led to significant demographic shifts. When analyzing population movements from rural agricultural districts to major urban metropolitan areas like Nairobi, which of the following represents a primary economic pull factor drawing migrants to these urban destinations?

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Answer: Greater concentration of formal employment opportunities and higher potential wages

Answer

Greater concentration of formal employment opportunities and higher potential wages
The correct answer identifies expanded job availability and superior wage prospects as positive urban attributes. In African demographic studies, the perception of better economic livelihoods in cities is the primary economic pull factor encouraging rural-urban drift.

Step-by-Step Solution

1
Distinguish between push factors and pull factors in demographic migration theory.
Push factors are negative attributes of the origin area driving people away, whereas pull factors are positive attributes of the destination attracting migrants.
Correct identification of migration drivers requires separating origin forces from destination attractions.
2
Evaluate the urban economic incentives in the options.
Higher wages and diverse job markets in cities represent desirable destination conditions.
Urban industrial and service sectors concentrate economic capital, serving as the main economic pull factor for rural labor.

Key Concept

Push and Pull Factors of Migration in Africa
Estimated Time:1m 0s
Question 13917Question

A spring balance is calibrated in kilograms at sea level where g=10.0 m s2g = 10.0\text{ m s}^{-2}. An object measured with an equal-arm beam balance at sea level is found to have a mass of 36.0 kg36.0\text{ kg}. If this object is taken to a high altitude where g=9.0 m s2g = 9.0\text{ m s}^{-2} and suspended from the same spring balance, what reading will the scale of the spring balance display?

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Answer: 32.4 kg32.4\text{ kg}

Answer

The spring balance scale will display a reading of 32.4 kg32.4\text{ kg}.
An equal-arm beam balance measures true invariant mass (36.0 kg36.0\text{ kg}) by comparing gravitational moments on two arms. A spring balance measures force (weight). At high altitude, the gravitational force acting on the mass is W=36.0 kg×9.0 m s2=324 NW = 36.0\text{ kg} \times 9.0\text{ m s}^{-2} = 324\text{ N}. Because the spring balance scale was calibrated assuming sea-level gravity (10.0 m s210.0\text{ m s}^{-2}), it converts force to indicated mass as 324 N/10.0 m s2=32.4 kg324\text{ N} / 10.0\text{ m s}^{-2} = 32.4\text{ kg}.

Step-by-Step Solution

1
Calculate the true weight of the object at the new altitude
W=m×galtitude=36.0 kg×9.0 m s2=324 NW = m \times g_{\text{altitude}} = 36.0\text{ kg} \times 9.0\text{ m s}^{-2} = 324\text{ N}
Weight is the force of gravity acting on a mass at a specific location.
2
Determine the indicated mass reading on the calibrated spring balance scale
\text{Reading} = \frac{W}{g_{\text{calibration}}} = \frac{324\text{ N}}{10.0\text{ m s}^{-2}} = 32.4\text{ kg}
The spring balance measures tension force but its scale was calibrated using sea-level gravity (10.0 m s210.0\text{ m s}^{-2}) to indicate mass.

Key Concept

Mass is an intrinsic property measured independently of gravity by a beam balance, whereas weight is a force measured by a spring balance, causing mass-calibrated spring balances to give different readings under varying local gravitational accelerations.
Estimated Time:1m 0s
Question 13918Question

A cell with an electromotive force (e.m.f.) of 12.0 V12.0\text{ V} and an internal resistance of 1.5 Ω1.5\ \Omega is connected in series to an external resistor of 8.5 Ω8.5\ \Omega. What is the terminal potential difference across the cell?

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Answer: 10.2

Answer

The terminal potential difference across the cell is 10.2 V10.2\text{ V}.
The terminal potential difference VV across a real cell supplying current is less than its electromotive force EE due to the internal voltage drop IrIr. By first determining the circuit current I=ER+r=12.08.5+1.5=1.2 AI = \frac{E}{R + r} = \frac{12.0}{8.5 + 1.5} = 1.2\text{ A}, the terminal potential difference is calculated as V=I×R=1.2×8.5=10.2 VV = I \times R = 1.2 \times 8.5 = 10.2\text{ V} (or equivalently V=EIr=12.0(1.2×1.5)=10.2 VV = E - I r = 12.0 - (1.2 \times 1.5) = 10.2\text{ V}).

Step-by-Step Solution

1
Calculate the total resistance of the circuit
Rtotal=10.0 ΩR_{\text{total}} = 10.0\ \Omega
The external load resistor and the cell's internal resistance are in series.
2
Determine the total current drawn from the cell
I=1.2 AI = 1.2\text{ A}
Using Ohm's law applied to the entire circuit, I=ER+rI = \frac{E}{R + r}.
3
Calculate the potential drop across the external load resistor
V=10.2 VV = 10.2\text{ V}
Terminal voltage equals potential difference across external load (V=IRV = I R) or V=EIrV = E - I r.

Key Concept

Terminal Potential Difference and Internal Resistance
Question 13919Question

Match each meteorological instrument on the left with its corresponding physical operational principle on the right.

Click a left item, then click its matching right item

Items

Six's thermometer
Wet-and-dry bulb psychrometer
Aneroid barometer
Campbell-Stokes recorder

Matches

Show answer & explanation

Answer

Six's thermometer matches with liquid differential expansion driving steel indexes; Wet-and-dry bulb psychrometer matches with evaporative cooling temperature depression; Aneroid barometer matches with flexure of an evacuated metallic cell under air pressure; Campbell-Stokes recorder matches with solar ray focal concentration burning a calibrated card.
Each weather recording instrument operates on a distinct physical property: Six's thermometer uses differential fluid expansion to record maximum/minimum temperatures, the psychrometer uses evaporative cooling depression to compute relative humidity, the aneroid barometer relies on non-liquid capsule flexing under atmospheric weight, and the Campbell-Stokes recorder uses optics to focus direct sunlight and scorch duration marks on paper.

Step-by-Step Solution

1
Analyze the functional mechanism of temperature extreme measurement
Identify Six's thermometer as utilizing alcohol and mercury expansion to move steel indicators to peak high and low values.
Extreme temperature recording requires dual fluids and physical position markers.
2
Analyze the operational principle for atmospheric humidity assessment
Identify the psychrometer as measuring moisture content via wet-bulb depression caused by evaporation.
Lower atmospheric humidity increases evaporation rates, producing a larger temperature difference between wet and dry bulbs.
3
Analyze pressure and sunshine duration mechanisms
Link the aneroid barometer to metallic capsule deformation under air mass weight, and the Campbell-Stokes recorder to glass sphere lens focusing of solar radiation.
Barometers rely on atmospheric weight changes on enclosed cell surfaces, whereas sunshine recorders rely on thermal scorching by concentrated light rays.

Key Concept

Meteorological Instrument Operational Mechanisms
Estimated Time:1m 30s
Question 13920Question

An electric immersion heater rated at 1.5kW1.5\,\text{kW} is operated on a 240V240\,\text{V} mains supply for 14minutes14\,\text{minutes}. Calculate the total electrical energy consumed by the heater during this period, in megajoules (MJ\text{MJ}).

Show answer & explanation

Answer: 1.26

Answer

The total electrical energy consumed by the heater is 1.26MJ1.26\,\text{MJ}.
Electrical energy consumed is obtained by multiplying electrical power by time (E=P×tE = P \times t). Converting power to watts (1.5kW=1500W1.5\,\text{kW} = 1500\,\text{W}) and time to seconds (14min=840s14\,\text{min} = 840\,\text{s}) yields E=1500×840=1,260,000J=1.26MJE = 1500 \times 840 = 1,260,000\,\text{J} = 1.26\,\text{MJ}.

Step-by-Step Solution

1
Convert power rating from kilowatts to watts
P=1.5kW=1500WP = 1.5\,\text{kW} = 1500\,\text{W}
The standard SI unit of power for energy calculation in Joules is Watts.
2
Convert time duration from minutes to seconds
t=14minutes×60seconds/minute=840secondst = 14\,\text{minutes} \times 60\,\text{seconds/minute} = 840\,\text{seconds}
The standard SI unit of time in Joule calculations is seconds.
3
Calculate energy consumed in Joules using E=P×tE = P \times t
E=1500W×840s=1,260,000JE = 1500\,\text{W} \times 840\,\text{s} = 1,260,000\,\text{J}
Electrical energy is the product of power in watts and time in seconds.
4
Convert energy from Joules to Megajoules
E=1,260,000106=1.26MJE = \frac{1,260,000}{10^6} = 1.26\,\text{MJ}
One Megajoule (1MJ1\,\text{MJ}) is equivalent to 106Joules10^6\,\text{Joules}.

Key Concept

Electrical Energy Consumption
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