All practice questions

Question 12941Question →

At a specific location on Earth, the total magnetic field intensity is $40\ \mu\text{T}$ and the angle of dip is $60^\circ$ . What is the magnitude of the horizontal component of Earth's magnetic field at this location, in $\mu\text{T}$ ?

Show answer & explanation

Answer: 20

Answer

The magnitude of the horizontal component of Earth's magnetic field is

20\ \mu\text{T}

.

The horizontal component

B_h

of Earth's magnetic field is derived using

B_h = B \cos\theta

. Substituting

B = 40\ \mu\text{T}

and

\theta = 60^\circ

yields

B_h = 40 \times 0.5 = 20\ \mu\text{T}

.

Step-by-Step Solution

1

Recall the resolving formula for the horizontal component of Earth's magnetic field

B_h = B \cos\theta

The horizontal component is the vector projection of total field

B

onto the horizontal plane inclined at angle

\theta

.

2

Evaluate the cosine function for

60^\circ

\cos(60^\circ) = 0.5

Standard trigonometric value for

60^\circ

.

3

Multiply total magnetic field strength by

\cos(60^\circ)

B_h = 40 \times 0.5 = 20\ \mu\text{T}

Calculates the exact horizontal component magnitude.

Key Concept

Components of Earth's Magnetic Field

Question 12942Question →

An astronomical telescope operating in normal adjustment consists of an objective lens with a focal length of $80\text{ cm}$ and an eyepiece with a focal length of $4\text{ cm}$ . What is the magnitude of the angular magnification produced by the telescope?

Show answer & explanation

Answer: 20

Answer

The magnitude of the angular magnification produced by the telescope is 20.

The angular magnification

M

of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective lens

f_o

to the focal length of the eyepiece lens

f_e

, given by

M = \frac{f_o}{f_e}

. Substituting the given values

f_o = 80\text{ cm}

and

f_e = 4\text{ cm}

yields

M = \frac{80}{4} = 20

.

Step-by-Step Solution

1

Identify the given optical parameters and formula for angular magnification.

Objective focal length

f_o = 80\text{ cm}

, Eyepiece focal length

f_e = 4\text{ cm}

. Formula:

M = \frac{f_o}{f_e}

.

For an astronomical telescope in normal adjustment, the light rays emerge parallel, and the angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.

2

Calculate the angular magnification value.

M = \frac{80\text{ cm}}{4\text{ cm}} = 20

.

Dividing the focal length of the objective lens by the focal length of the eyepiece yields the dimensionless magnification ratio.

Key Concept

Angular Magnification of an Astronomical Telescope in Normal Adjustment

Question 12943Question →

A $600\text{ cm}^3$ sample of air contaminated with sulphur(IV) oxide ( $\text{SO}_2$ ) gas was passed through an excess aqueous solution of sodium hydroxide to absorb all the $\text{SO}_2$ , reducing the volume of the gas sample to $576\text{ cm}^3$ . The remaining gas mixture was then passed over excess heated copper turnings to remove oxygen gas, after which the unreacted gas volume measured $456\text{ cm}^3$ under the same conditions of temperature and pressure. What is the percentage by volume of sulphur(IV) oxide in the original air sample?

Show answer & explanation

Answer: 4

Answer

The percentage by volume of sulphur(IV) oxide in the original air sample is 4.0%.

Sodium hydroxide (

\text{NaOH}

) selectively absorbs acidic pollutant gases such as sulphur(IV) oxide (

\text{SO}_2

). The volume decrease from

600\text{ cm}^3

to

576\text{ cm}^3

indicates that

24\text{ cm}^3

of

\text{SO}_2

was absorbed. Dividing

24\text{ cm}^3

by the original sample volume of

600\text{ cm}^3

and multiplying by 100 gives

4.0\%

.

Step-by-Step Solution

1

Calculate the volume of sulphur(IV) oxide gas absorbed by the sodium hydroxide solution.

Volume of

\text{SO}_2 = 600\text{ cm}^3 - 576\text{ cm}^3 = 24\text{ cm}^3

.

Sodium hydroxide reacts with acidic oxide pollutants like

\text{SO}_2

, causing a reduction in gas volume equal to the volume of

\text{SO}_2

present.

2

Calculate the percentage composition by volume relative to the total initial sample.

\text{Percentage of } \text{SO}_2 = \left(\frac{24\text{ cm}^3}{600\text{ cm}^3}\right) \times 100\% = 4.0\%.

The volumetric percentage is determined by expressing the volume of the target gas component over the total volume of the original air sample.

Key Concept

Volumetric determination of air composition and gaseous pollutants.

Question 12944Question →

In an economy, the Central Bank issues a total of $\text{N}450\text{ billion}$ in currency. Out of this amount, commercial banks hold $\text{N}50\text{ billion}$ as vault cash in their tills. If the demand deposits held by the public in commercial banks total $\text{N}800\text{ billion}$ , what is the value of the narrow money supply ( $M_1$ ) in billions of Naira?

Show answer & explanation

Answer: 1200

Answer

The value of the narrow money supply (

M_1

) is 1200 billion Naira.

The narrow money supply (

M_1

) consists of currency in circulation outside commercial banks plus demand deposits. Currency in circulation is found by subtracting vault cash from total currency issued by the central bank (

\text{N}450\text{ billion} - \text{N}50\text{ billion} = \text{N}400\text{ billion}

). Adding demand deposits (

\text{N}800\text{ billion}

) yields a total narrow money supply of

\text{N}1200\text{ billion}

.

Step-by-Step Solution

1

Calculate currency in circulation outside commercial banks

400 billion Naira

Vault cash held inside commercial bank vaults is excluded from currency in circulation outside the banking system.

2

Calculate narrow money supply (

M_1

)

1200 billion Naira

Narrow money supply (

M_1

) is the sum of currency in circulation outside commercial banks and demand deposits.

Key Concept

Components and Calculation of Narrow Money Supply (

M_1

)

Question 12945Question →

Arrange the following electromagnetic radiation applications in order of increasing photon energy, starting from the radiation with the lowest energy to the one with the highest energy.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from lowest to highest photon energy is: Radar waves (microwaves), radiant heat (infrared), sterilizing radiation (ultraviolet), and nuclear gamma emissions.

The photon energy of electromagnetic radiation is directly proportional to its frequency (

E = hf

). Microwaves have the lowest frequency among the listed types, followed by infrared radiation, then ultraviolet radiation, with gamma rays having the highest frequency and thus the greatest photon energy per photon.

Step-by-Step Solution

1

Identify the region of the electromagnetic spectrum for each listed application.

Radar waves belong to microwaves; radiant heat corresponds to infrared radiation; sterilization uses ultraviolet light; nuclear emissions are gamma rays.

Connecting applications to their respective spectral regions is required to compare physical properties.

2

Recall the relationship between frequency and photon energy in the electromagnetic spectrum using

E = h f

.

Photon energy is directly proportional to frequency (

E \propto f

), meaning higher frequency waves carry greater energy per photon.

Understanding quantum photon energy helps determine the correct energy ranking.

3

Order the spectral regions from lowest frequency to highest frequency.

The order of increasing frequency (and energy) is: Microwaves < Infrared < Ultraviolet < Gamma rays.

This matches the physical ordering of the electromagnetic spectrum by increasing frequency.

Key Concept

Photon energy across the electromagnetic spectrum increases with increasing frequency (

E = h f

).

Question 12946Question →

Match each chemical process on the left with the correct classical or modern redox concept on the right that specifically describes the change taking place.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Removal of oxygen from zinc oxide corresponds to classical reduction; addition of oxygen to carbon corresponds to classical oxidation; loss of electrons by sodium corresponds to modern electron-transfer oxidation; and the decrease in oxidation number of nitrogen corresponds to modern oxidation number reduction.

Each chemical transformation is matched according to whether it describes classical oxygen transfer or modern electron/oxidation state changes. Removal of oxygen represents classical reduction; addition of oxygen represents classical oxidation; loss of electrons represents modern electronic oxidation; and a decrease in oxidation number represents modern reduction.

Step-by-Step Solution

1

Analyze classical definitions of redox

Classical oxidation involves the addition of oxygen or removal of hydrogen, while classical reduction involves the removal of oxygen or addition of hydrogen.

This establishes the ground rules for the first two pairs.

2

Analyze modern electronic and oxidation number definitions

Modern oxidation is the loss of electrons (OIL) or an increase in oxidation state. Modern reduction is the gain of electrons (RIG) or a decrease in oxidation state.

This establishes the ground rules for the remaining two pairs.

3

Match each process to its corresponding definition

Zinc oxide losing oxygen is classical reduction. Carbon gaining oxygen is classical oxidation. Sodium losing electrons is modern electronic oxidation. Nitrogen decreasing in oxidation number from 0 to -3 is modern oxidation number reduction.

Applying the concepts directly aligns each process with its primary redox definition.

Key Concept

Classical vs Modern Concepts of Redox Reactions

Question 12947Question →

An oxygen nucleus $^{16}_{8}\text{O}$ has a measured nuclear mass of $15.9906\text{ u}$ . Given that the mass of a proton is $1.0078\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , calculate the total binding energy of the nucleus in $\text{MeV}$ . (Take $1\text{ u} = 931.5\text{ MeV}$ )

Show answer & explanation

Answer: 131.71

Answer

The total binding energy of the oxygen nucleus is

131.71\text{ MeV}

.

Summing the masses of 8 individual protons and 8 individual neutrons gives

16.1320\text{ u}

. Subtracting the actual nuclear mass of

15.9906\text{ u}

leaves a mass defect of

0.1414\text{ u}

. Multiplying this mass defect by the equivalence constant

931.5\text{ MeV/u}

yields a total binding energy of

131.71\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total combined mass of the free constituent nucleons

Mass of 8 protons + 8 neutrons =

8(1.0078\text{ u}) + 8(1.0087\text{ u}) = 8.0624\text{ u} + 8.0696\text{ u} = 16.1320\text{ u}

Oxygen-16 has

Z = 8

protons and

A - Z = 16 - 8 = 8

neutrons.

2

Calculate the mass defect (

\Delta m

)

\Delta m = 16.1320\text{ u} - 15.9906\text{ u} = 0.1414\text{ u}

Mass defect is the difference between total constituent mass and actual nuclear mass.

3

Convert mass defect into total binding energy (

E_b

)

E_b = 0.1414\text{ u} \times 931.5\text{ MeV/u} = 131.7141\text{ MeV} \approx 131.71\text{ MeV}

Applying the energy equivalent factor of

931.5\text{ MeV}

per atomic mass unit.

Key Concept

Mass defect and nuclear binding energy equivalence

Question 12948Question →

In medical practice, understanding the genetic basis of the ABO blood group system is essential for safe blood transfusions. If an individual with blood group AB ( $I^A I^B$ ) requires a red blood cell transfusion, which of the following statements correctly explains their biological donor compatibility?

They can safely receive red blood cells from any ABO blood group donor because their blood plasma contains neither anti-A nor anti-B antibodies.

They can only receive blood from group O individuals because allele

i

exhibits complete dominance over both alleles

I^A

and

I^B

.

They exhibit a continuous spectrum of blood phenotypes that changes depending on environmental temperature and nutrition.

They can donate red blood cells to any ABO recipient because acquired plasma antibodies are inherited as somatic adaptations.

Show answer & explanation

Answer: They can safely receive red blood cells from any ABO blood group donor because their blood plasma contains neither anti-A nor anti-B antibodies.

Answer

Individuals with blood group AB (

I^A I^B

) can safely receive red blood cells from any ABO blood group donor because their plasma contains neither anti-A nor anti-B antibodies.

Individuals with blood group AB (

I^A I^B

) express both A and B surface antigens on their erythrocytes due to codominance between the

I^A

and

I^B

alleles. Because their immune system recognizes both antigens as self-antigens, their plasma contains neither anti-A nor anti-B antibodies. Consequently, receiving red blood cells from group A, B, AB, or O donors will not induce antibody-mediated agglutination, making them universal red blood cell recipients.

Step-by-Step Solution

1

Identify the genotype and phenotypic antigen expression of blood group AB.

Genotype

I^A I^B

results in the codominant expression of both A and B antigens on the surface of red blood cells.

Both alleles

I^A

and

I^B

are fully expressed simultaneously.

2

Determine the corresponding antibody composition in the blood plasma.

Because both A and B antigens are recognized as self-antigens, the plasma produces neither anti-A nor anti-B antibodies.

Producing antibodies against self-antigens would trigger autoimmune destruction of erythrocytes.

3

Assess transfusion compatibility for red blood cell donation.

The absence of anti-A and anti-B antibodies prevents immune reaction against donated red blood cells from group A, B, AB, or O donors.

Lacking these antibodies makes group AB individuals universal recipients of red blood cells.

Key Concept

ABO blood group codominance, antibody presence, and transfusion compatibility in medical genetics

Estimated Time:1m 0s

Question 12949Question →

A student measures the time taken for $40$ complete oscillations of a simple pendulum using a digital stopwatch that has a negative zero error of $-0.40\text{ s}$ . If the stopwatch displays a reading of $47.60\text{ s}$ for the oscillations, what will be the correct period of oscillation when the pendulum's length is reduced to one-fourth ( $1/4$ ) of its initial length?

0.60\text{ s}

0.59\text{ s}

0.30\text{ s}

2.40\text{ s}

Show answer & explanation

Answer: $0.60\text{ s}$

Answer

The correct period of oscillation for the shortened pendulum is

0.60\text{ s}

.

The correct answer is obtained by first adjusting the measured time for negative zero error (

47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}

), calculating the initial period per oscillation (

48.00 / 40 = 1.20\text{ s}

), and then taking into account that period varies with the square root of pendulum length. Reducing length to

1/4

reduces period by a factor of

\sqrt{4} = 2

, yielding

0.60\text{ s}

.

Step-by-Step Solution

1

Correct the recorded total time for instrument zero error

True total time

t = \text{Uncorrected Time} - \text{Zero Error} = 47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}

Negative zero error means the timer reads less than the actual elapsed time, so the absolute value of the error must be added back.

2

Calculate the initial period of oscillation

Initial period

T_1 = \frac{t}{N} = \frac{48.00\text{ s}}{40} = 1.20\text{ s}

Period is defined as the time per single oscillation.

3

Apply the pendulum scaling relationship for length and period

New period

T_2 = T_1 \times \sqrt{\frac{L_2}{L_1}} = 1.20\text{ s} \times \sqrt{\frac{1}{4}} = 1.20\text{ s} \times 0.5 = 0.60\text{ s}

The period of a simple pendulum is proportional to

\sqrt{L}

, so quartering the length reduces the period to half of its original value.

Key Concept

Measurement of time using a stopwatch with zero error correction combined with simple pendulum period dependence on length.

Estimated Time:2m 0s

Question 12950Question →

Read the extract below from Frank Ogodo Ogbeche's prescribed African drama, Harvest of Corruption:

"My hands are clean. I am an honorable minister of the state, serving my nation with selflessness. Nobody can link me to any shady deal in this ministry."

Given the audience's prior knowledge of Chief Haladu Ade-Amaka's corrupt practices and involvement in drug trafficking, which dramatic device is predominantly demonstrated in his utterance above?

Dramatic irony, because the audience is fully aware of Chief's criminal involvement while he publicly asserts his innocence.

Soliloquy, because Chief is expressing his genuine moral remorse privately to the audience without other characters present.

Comic relief, because his declaration creates a lighthearted mood intended to entertain the audience during a tense scene.

Tragic flaw, because it illustrates Chief's noble heroic character being ruined by uncontrollable external circumstances.

Show answer & explanation

Answer: Dramatic irony, because the audience is fully aware of Chief's criminal involvement while he publicly asserts his innocence.

Answer

Dramatic irony, because the audience is fully aware of Chief's criminal involvement while he publicly asserts his innocence.

Dramatic irony is present because the audience possesses full knowledge of Chief Haladu Ade-Amaka's illicit drug operations and embezzlement in Harvest of Corruption, making his bold claim of having clean hands sharply ironic to the viewer.

Step-by-Step Solution

1

Analyze the context of the extract within Frank Ogodo Ogbeche's play Harvest of Corruption.

Chief Haladu Ade-Amaka claims that his hands are clean and that no one can connect him to corruption.

Understanding the character's statement is the first step in contextual text analysis.

2

Compare the character's statement with the audience's knowledge of the plot events.

The audience already knows Chief is deeply involved in embezzlement, bribery, and drug trafficking through Ochuole and Aloho.

Identifying the gap between character perception/speech and audience awareness determines the dramatic technique.

3

Identify the dramatic device that matches this contradiction.

The contradiction between Chief's proclamation of innocence and the audience's knowledge of his guilt defines dramatic irony.

Dramatic irony occurs precisely when the audience knows important facts that contradict a character's words or beliefs.

Key Concept

Dramatic Irony in Prescribed African Drama

Question 12951Question →

In a resonance tube experiment using a tuning fork of frequency $340\text{ Hz}$ , the first two consecutive resonance lengths of the air column closed at one end are $22\text{ cm}$ and $72\text{ cm}$ . If a pipe open at both ends with a physical length of $28\text{ cm}$ is operated in the same environment, what is its fundamental frequency when end corrections at both open ends are taken into account?

500\text{ Hz}

250\text{ Hz}

680\text{ Hz}

1000\text{ Hz}

Show answer & explanation

Answer: $500\text{ Hz}$

Answer

The fundamental frequency of the open pipe is

500\text{ Hz}

.

The difference between successive resonant lengths in the closed tube gives half a wavelength (

L_2 - L_1 = 0.50\text{ m} \implies \lambda = 1.00\text{ m}

). Using the tuning fork frequency

340\text{ Hz}

, the speed of sound is

340\text{ m/s}

. The end correction is

e = \lambda / 4 - L_1 = 0.25\text{ m} - 0.22\text{ m} = 0.03\text{ m}

. For a pipe open at both ends, end corrections apply at both openings, making the effective length

L_{\text{eff}} = 0.28\text{ m} + 2(0.03\text{ m}) = 0.34\text{ m}

. Its fundamental frequency is

f_0 = v / (2 L_{\text{eff}}) = 340 / (2 \times 0.34) = 500\text{ Hz}

.

Step-by-Step Solution

1

Determine the wavelength and speed of sound from the resonance tube data.

\lambda = 2(L_2 - L_1) = 2(0.72\text{ m} - 0.22\text{ m}) = 1.00\text{ m}

. Speed of sound

v = f \lambda = 340\text{ Hz} \times 1.00\text{ m} = 340\text{ m/s}

.

The distance between consecutive resonance positions in a closed pipe is equal to half a wavelength.

2

Calculate the end correction

e

of the tube.

L_1 + e = \frac{\lambda}{4} \implies 0.22\text{ m} + e = 0.25\text{ m} \implies e = 0.03\text{ m} = 3\text{ cm}

.

The first resonance of a pipe closed at one end occurs when the effective length equals one quarter of a wavelength.

3

Calculate the effective length

L_{\text{eff}}

of the open pipe.

L_{\text{eff}} = L + 2e = 28\text{ cm} + 2(3\text{ cm}) = 34\text{ cm} = 0.34\text{ m}

.

A pipe open at both ends requires an end-correction term added at each open boundary.

4

Calculate the fundamental frequency of the open pipe.

f_0 = \frac{v}{2 L_{\text{eff}}} = \frac{340\text{ m/s}}{2 \times 0.34\text{ m}} = 500\text{ Hz}

.

The fundamental wavelength of a pipe open at both ends is twice its effective length.

Key Concept

Resonance tube end correction and boundary conditions of open vs closed pipes

Estimated Time:2m 0s

Question 12952Question →

A light wave travels through medium A at a speed of $2.25 \times 10^8\text{ m s}^{-1}$ and enters medium B, where its speed decreases to $1.50 \times 10^8\text{ m s}^{-1}$ . What is the value of the sine of the critical angle for total internal reflection between these two media, and in which medium must the light ray originate?

\frac{2}{3}

, originating in medium B

\frac{2}{3}

, originating in medium A

\frac{3}{2}

, originating in medium B

\frac{3}{2}

, originating in medium A

Show answer & explanation

Answer: $\frac{2}{3}$ , originating in medium B

Answer

The sine of the critical angle is

\frac{2}{3}

, and the light ray must originate in medium B.

Total internal reflection occurs when light travels from an optically denser medium to an optically rarer medium. Since the speed of light is lower in medium B (

1.50 \times 10^8\text{ m s}^{-1}

) than in medium A (

2.25 \times 10^8\text{ m s}^{-1}

), medium B is the denser medium. The critical angle

C

satisfies

\sin C = \frac{v_{\text{dense}}}{v_{\text{rare}}} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}

. Therefore, the light must originate in medium B and the sine of the critical angle is

\frac{2}{3}

.

Step-by-Step Solution

1

Determine the relative optical densities of medium A and medium B from wave speed

Medium B has a lower light speed (

1.50 \times 10^8\text{ m s}^{-1}

) than medium A (

2.25 \times 10^8\text{ m s}^{-1}

), so medium B is optically denser than medium A.

Refractive index is inversely proportional to wave speed (

n \propto \frac{1}{v}

).

2

Identify the required direction of light propagation for total internal reflection

The light ray must originate in medium B and travel toward medium A.

Total internal reflection occurs only when light attempts to pass from a medium of higher refractive index (denser) to a medium of lower refractive index (rarer).

3

Calculate the sine of the critical angle

\sin C = \frac{v_B}{v_A} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}

.

By Snell's law at the critical angle,

\sin C = \frac{n_A}{n_B} = \frac{v_B}{v_A}

.

Key Concept

Conditions for Total Internal Reflection and Critical Angle calculation from wave speeds

Estimated Time:1m 30s

Question 12953Question →

Match each ecological sampling instrument or technique with its most appropriate sampling application in field biology studies.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pooter pairs with collecting tiny insects via suction; Pitfall trap pairs with capturing crawling arthropods on the ground; Quadrat frame pairs with estimating population density of stationary plants; Line transect pairs with determining species distribution along an environmental gradient.

Each instrument maps directly to its specific field biological function: the pooter uses suction for fragile insects, the pitfall trap captures soil-crawling fauna in a sunken container, the quadrat isolates standard areas for non-motile population counts, and the line transect continuously tracks species shifts across environmental gradients.

Step-by-Step Solution

1

Analyze the functional mechanism of each instrument

Pooter uses suction for delicate leaf insects; Pitfall traps catch ground-level crawling species.

Apparatus design dictates which organism group can be effectively sampled.

2

Differentiate spatial area sampling from gradient transition sampling

Quadrat frames define standard area bounds for non-mobile species, whereas line transects map linear changes across gradients.

Quadrats yield area-based density measurements, while transects capture ecological zonation.

Key Concept

Ecological Sampling Instruments and Techniques

Question 12954Question →

Which of the following natural radioactive emissions has the least penetrating power and can be stopped by a thin sheet of paper?

Alpha particles

Beta particles

Gamma rays

Neutrons

Show answer & explanation

Answer: Alpha particles

Answer

Alpha particles have the lowest penetrating power of the three primary natural radioactive emissions and are stopped by a thin sheet of paper.

Alpha particles consist of heavy, doubly charged helium nuclei (

^4_2\text{He}

). Their relatively large size and strong positive charge lead to frequent ionizing collisions with surrounding atoms, causing them to deposit their energy over a very short distance and making them unable to pass through a simple sheet of paper.

Step-by-Step Solution

1

Identify the nature and physical properties of natural radioactive emissions (alpha particles, beta particles, and gamma rays).

Alpha particles consist of 2 protons and 2 neutrons (helium nucleus,

^4_2\text{He}

), carrying a heavy mass (

4\text{ u}

) and a

+2e

charge.

Large mass and high charge increase the likelihood of collisions with atoms in matter.

2

Compare ionizing ability and penetrating power across emission types.

Because alpha particles cause high ionization over short distances, they lose kinetic energy rapidly.

High rate of energy loss corresponds directly to very low penetrating depth.

3

Determine the stopping material required for alpha particles.

Alpha particles are absorbed by a thin layer of matter, such as a single sheet of paper or a few centimeters of air.

This confirms alpha radiation has the least penetrating power among natural emissions.

Key Concept

Relative Penetrating and Ionizing Capacities of Natural Radioactive Emissions

Estimated Time:45s

Question 12955Question →

A commercial fleet management enterprise in Port Harcourt that transports refined petroleum products from coastal refineries to inland filling stations is classified under secondary production because it handles processed industrial goods.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Transporting refined petroleum products is a logistics and distribution service creating place utility, which falls under tertiary production rather than secondary production.

Transportation services generate place utility by making goods available where they are needed. In economic classification, all distributive, logistics, and commercial service activities belong to tertiary production, regardless of whether the items transported are raw materials or manufactured products.

Step-by-Step Solution

1

Identify the primary economic activity described in the scenario.

The enterprise provides transportation and movement of already-refined petroleum products from refineries to distribution points.

Classifying an economic activity requires determining its specific functional role in the production chain.

2

Distinguish between the definitions of secondary and tertiary production.

Secondary production involves manufacturing, construction, and physical transformation of raw materials, whereas tertiary production covers commercial services, transport, and distribution activities.

Transporting finished or refined items does not physically alter the commodity; it alters its location.

3

Evaluate the truth value based on utility creation and sector boundaries.

Because logistics generates place utility through service delivery, it belongs to tertiary production.

The stage of the good being transported (refined petroleum) does not change the transport service itself into a secondary manufacturing activity.

Key Concept

Types of Production (Secondary vs. Tertiary Production)

Question 12956Question →

An $\alpha$ -particle and a $\beta^-$ -particle emitted from a radioactive source enter a uniform magnetic field perpendicularly with equal linear momenta. What is the ratio of the radius of curvature of the trajectory of the $\alpha$ -particle to that of the $\beta^-$ -particle in the magnetic field?

1 : 2

2 : 1

4 : 1

1 : 4

Show answer & explanation

Answer: $1 : 2$

Answer

The ratio of the radius of curvature of the path of the

\alpha

-particle to that of the

\beta^-

-particle is

1 : 2

.

In a uniform magnetic field, the radius of curvature

r

of a moving charged particle is given by

r = \frac{p}{qB}

, where

p

is momentum,

q

is charge magnitude, and

B

is magnetic field flux density. Since both particles have equal momentum in the same field,

r \propto \frac{1}{q}

. The

\alpha

-particle carries charge magnitude

2e

while the

\beta^-

-particle carries charge magnitude

e

. Consequently, the ratio of their radii is

r_\alpha : r_\beta = \frac{1}{2} : 1 = 1 : 2

.

Step-by-Step Solution

1

Relate radius of trajectory to linear momentum and magnetic field

The magnetic force supplies centripetal force:

qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} = \frac{p}{qB}

, where

p

is linear momentum.

Expressing radius in terms of momentum directly utilizes the given condition that

p

is equal for both particles.

2

Identify the magnitude of charge for each emission

For the

\alpha

-particle (

^4_2\text{He}^{2+}

),

q_\alpha = 2e

. For the

\beta^-

-particle (

^0_{-1}\text{e}

),

q_\beta = e

.

Radius of curvature depends on the magnitude of charge carried by each radiation type.

3

Calculate the ratio of radii

r_\alpha / r_\beta

\frac{r_\alpha}{r_\beta} = \frac{\frac{p}{2eB}}{\frac{p}{eB}} = \frac{e}{2e} = \frac{1}{2}.

Since momentum

p

and field strength

B

are identical, the ratio simplifies directly to the inverse ratio of their charge magnitudes.

Key Concept

Deflection of radioactive emissions in magnetic fields and radius of curvature

Question 12957Question →

Industrial chemical manufacturing processes are broadly categorized into heavy chemical production and fine chemical production based on output volume, purity requirements, and process dynamics. Which of the following pairs correctly pairs a heavy chemical synthesized via continuous industrial operations with a fine chemical produced in batch operations?

Tetraoxosulfate(VI) acid (

H_2SO_4

) and chloramphenicol

Analytical reagent grade hydrochloric acid (

HCl

) and sodium hydroxide (

NaOH

)

Methyl orange indicator and synthetic ammonia (

NH_3

)

Azo dye pigment and analytical grade tetraoxosulfate(VI) acid (

H_2SO_4

)

Show answer & explanation

Answer: Tetraoxosulfate(VI) acid ( $H_2SO_4$ ) and chloramphenicol

Answer

Tetraoxosulfate(VI) acid (

H_2SO_4

) as the heavy chemical and chloramphenicol as the fine chemical.

The correct selection pairs tetraoxosulfate(VI) acid (

H_2SO_4

), an industrial bulk acid synthesized continuously in millions of tonnes, with chloramphenicol, a specialized pharmaceutical drug produced in high-purity batch operations.

Step-by-Step Solution

1

Analyze the operational features of heavy chemicals.

Heavy chemicals are manufactured in extremely large metric ton quantities using continuous production processes, sold relatively cheaply, and have standard industrial purity levels.

This establishes tetraoxosulfate(VI) acid (

H_2SO_4

) prepared via the Contact process as a primary example of a heavy chemical.

2

Analyze the operational features of fine chemicals.

Fine chemicals are produced in small batch quantities, possess high chemical purity, command high unit costs, and serve specific applications such as pharmaceuticals, dyes, and analytical reagents.

This establishes chloramphenicol (an antibiotic drug) as a fine chemical.

3

Evaluate the option choices against the required heavy-then-fine ordering.

The pair containing tetraoxosulfate(VI) acid and chloramphenicol places the continuous-process heavy chemical first and the batch-process fine chemical second.

This satisfies both the classification and order constraints specified in the stem.

Key Concept

Distinction between Heavy Chemicals and Fine Chemicals based on scale, purity, and manufacturing process

Question 12958Question →

In the classification of Spermatophytes, gymnosperms and angiosperms display distinct anatomical and reproductive features. Match each diagnostic feature listed on the left with its corresponding plant group on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Uncovered ovules match Gymnospermae; Double fertilization yielding triploid endosperm matches Angiospermae; Single cotyledon with parallel venation matches Monocotyledoneae; Two cotyledons with reticulate venation match Dicotyledoneae.

The correct pairings accurately reflect the defining evolutionary and structural traits of Spermatophytes: uncovered ovules on megasporophyll scales are characteristic of Gymnospermae; double fertilization producing triploid (

3n

) endosperm defines Angiospermae; single cotyledons paired with parallel leaf venation characterize Monocotyledoneae; and two cotyledons paired with reticulate leaf venation characterize Dicotyledoneae.

Step-by-Step Solution

1

Identify the primary seed enclosure characteristic

Unenclosed (naked) ovules on megasporophylls define Gymnospermae.

Gymnosperms lack an ovary wall surrounding their ovules.

2

Identify the reproductive fertilization hallmark of flowering plants

Double fertilization yielding a triploid (

3n

) nutritive tissue defines Angiospermae.

Angiosperms undergo a secondary fertilization event where one sperm fuses with two polar nuclei.

3

Distinguish between the two main classes of angiosperms

Single cotyledon with parallel leaf venation corresponds to Monocotyledoneae, while two cotyledons with reticulate venation correspond to Dicotyledoneae.

Morphological traits such as cotyledon count, leaf venation, and root system architecture divide angiosperms into monocots and dicots.

Key Concept

Diagnostic reproductive and structural distinctions between Gymnosperms, Angiosperms, Monocotyledons, and Dicotyledons

Estimated Time:1m 30s

Question 12959Question →

Match each seral stage of hydrarch succession (hydrosere) in a freshwater habitat on the left with its corresponding characteristic vegetation or ecological role on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Phytoplankton stage pairs with microscopic algae; Submerged macrophyte stage pairs with rooted aquatic plants like Vallisneria; Floating macrophyte stage pairs with broad surface-leaved plants like Nymphaea; Reed-swamp stage pairs with emergent amphibious plants like Typha.

In hydrarch primary succession, microscopic phytoplankton act as pioneers in deep water. As organic debris accumulates on the bed, submerged plants establish, followed by floating-leaved species that block sunlight, and finally emergent reed-swamp plants that convert shallow waters into wet soil.

Step-by-Step Solution

1

Identify the pioneer community of a hydrosere.

The pioneer stage consists of unattached microscopic producers (phytoplankton) that start organic sedimentation.

Deep open water bodies lack soil anchorages for rooted plants initially.

2

Trace the sequence of rooted vegetation growth as water depth decreases.

Fully submerged rooted species grow first, followed by floating-leaved species as mud layers thicken, and finally emergent amphibious reeds near the water edges.

Each community alters light penetration and substrate depth, facilitating the establishment of the next seral stage.

Key Concept

Hydrarch succession stages (hydrosere)

Question 12960Question →

An organic compound has the following condensed structural formula:

\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{C}_2\text{H}_5)\text{CH}_2\text{C}\equiv\text{CH}

What is the correct IUPAC name for this compound?

4-ethyl-5-methylhex-1-yne

4-isopropylhex-1-yne

3-ethyl-2-methylhex-5-yne

3-isopropylhex-5-yne

Show answer & explanation

Answer: 4-ethyl-5-methylhex-1-yne

Answer

4-ethyl-5-methylhex-1-yne

The compound is numbered starting from the terminal alkyne carbon to give the triple bond the lowest locant (

C1

). At

C4

, there are two possible 6-carbon chains. Following IUPAC tie-breaking guidelines, the continuous chain that yields the greater number of substituents (ethyl at

C4

and methyl at

C5

) is chosen as the parent chain, resulting in 4-ethyl-5-methylhex-1-yne.

Step-by-Step Solution

1

Identify the principal functional group and assign lowest locant numbering

The alkyne triple bond (

-C\equiv C-

) is at position 1, so numbering starts from the rightmost carbon: C1 is

HC\equiv

, C2 is

-C-

, C3 is

-CH_2-

, C4 is

-CH(C_2H_5)-

.

IUPAC rules state that principal functional groups receive the lowest possible locants.

2

Determine the longest continuous carbon chain containing the triple bond

From C4, continuing straight through

-CH(CH_3)CH_3

gives 6 carbons (hex-1-yne). Going down the ethyl group

-CH_2CH_3

also gives 6 carbons (hex-1-yne).

Both potential paths yield a parent chain length of 6 carbons.

3

Apply the tie-breaking rule for chains of equal length

The chain straight through

-CH(CH_3)CH_3

has 2 substituents (ethyl at C4, methyl at C5). The path through the ethyl group has only 1 substituent (isopropyl at C4). Thus, the chain with 2 substituents is selected.

When two chains of equal length compete for parent status, IUPAC rules dictate selecting the chain with the maximum number of substituents.

4

Assemble the IUPAC name in alphabetical order

Alphabetize 'ethyl' before 'methyl' to give 4-ethyl-5-methylhex-1-yne.

Substituents are listed alphabetically regardless of their locant numbers.

Key Concept

IUPAC Nomenclature of Alkynes with Branched Chains and Tie-Breaking Rules

Estimated Time:2m 0s

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