All practice questions

Question 12741Question →

Oil spillage is a major environmental hazard associated with petroleum exploration in the Niger Delta. Which specialized agency was established by the Nigerian federal government with the primary mandate to detect, respond to, and clean up oil spills?

National Oil Spill Detection and Response Agency (NOSDRA)

National Environmental Standards and Regulations Enforcement Agency (NESREA)

Nigerian Conservation Foundation (NCF)

Federal Ministry of Water Resources (FMWR)

Show answer & explanation

Answer: National Oil Spill Detection and Response Agency (NOSDRA)

Answer

The National Oil Spill Detection and Response Agency (NOSDRA) is the specialized Nigerian federal agency mandated to detect, respond to, and coordinate the cleanup of oil spills.

The National Oil Spill Detection and Response Agency (NOSDRA) was created specifically by the Federal Government of Nigeria to coordinate emergency response, detection, and remediation for oil spill incidents across the country.

Step-by-Step Solution

1

Identify the specific environmental hazard mentioned in the stem.

The hazard is oil spillage caused by petroleum exploration and transport.

Targeted environmental management requires knowing the specific hazard type.

2

Distinguish between the statutory mandates of Nigerian environmental management bodies.

NOSDRA was enacted under Act 15 of 2006 specifically for oil spill detection and response, while NESREA handles non-petroleum environmental regulation.

Recalling exact regulatory jurisdiction ensures correct identification of the managing agency.

Key Concept

Environmental Hazard Management and Statutory Agency Mandates in Nigeria

Question 12742Question →

An electron in a hydrogen atom makes a transition from the third energy level ( $n = 3$ ) to the second energy level ( $n = 2$ ). If the energy of the electron at $n = 3$ is $-1.51\text{ eV}$ and at $n = 2$ is $-3.40\text{ eV}$ , what is the frequency of the emitted photon? ( $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

4.58 \times 10^{14}\text{ Hz}

2.86 \times 10^{33}\text{ Hz}

1.19 \times 10^{15}\text{ Hz}

8.24 \times 10^{14}\text{ Hz}

Show answer & explanation

Answer: $4.58 \times 10^{14}\text{ Hz}$

Answer

The frequency of the emitted photon is

4.58 \times 10^{14}\text{ Hz}

.

When an electron transitions from a higher energy state to a lower energy state, it emits a photon whose energy equals the difference between the two levels:

\Delta E = E_3 - E_2 = 1.89\text{ eV}

. Converting

1.89\text{ eV}

to Joules gives

3.024 \times 10^{-19}\text{ J}

. Dividing by Planck's constant (

6.6 \times 10^{-34}\text{ J}\cdot\text{s}

) gives a frequency of

4.58 \times 10^{14}\text{ Hz}

.

Step-by-Step Solution

1

Calculate the energy difference between the initial and final energy levels.

\Delta E = E_3 - E_2 = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

The energy of the emitted photon equals the difference in energy between the two states.

2

Convert the energy change from electron-volts (eV) to Joules (J).

\Delta E = 1.89 \times 1.6 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}

SI units (Joules) are required to calculate frequency using Planck's constant in

\text{J}\cdot\text{s}

.

3

Apply the photon energy formula

E = hf

to find the frequency

f

.

f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J}\cdot\text{s}} = 4.5818 \times 10^{14}\text{ Hz} \approx 4.58 \times 10^{14}\text{ Hz}

Dividing energy by Planck's constant yields the photon frequency.

Key Concept

Photon Emission and Energy Level Transition

Question 12743Question →

Match each major Nigerian soil group on the left with its primary geographical zone of distribution and pedological characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ferrallitic Soils match with the Southern Rainforest belt; Ferruginous Tropical Soils match with the Guinea Savanna on Basement Complex rocks; Hydromorphic Soils match with river basins and low-lying floodplains; Regosols and Semi-Arid Brown Soils match with the northern boundary zones.

Each Nigerian soil group reflects specific climatic, topographic, and lithological controls: Ferrallitic soils belong to the humid southern rainforest, Ferruginous tropical soils cover the central crystalline savanna, Hydromorphic soils occupy floodplains and valleys, and Regosols occur in the dry northern margins.

Step-by-Step Solution

1

Identify climate and vegetation zonal controls on Nigerian soils.

Heavy rainfall in the south produces deeply leached Ferrallitic soils, whereas moderate savanna rainfall produces Ferruginous soils.

Zonal soil formation in Nigeria directly follows the latitudinal rainfall and vegetation gradient.

2

Identify azonal/intrazonal soil types tied to localized drainage and parent material.

River valleys (Fadama) create waterlogged Hydromorphic soils, and dry northern margins create poorly developed Regosols.

Topography and moisture availability create localized hydromorphic soils regardless of regional climate zone.

3

Pair each soil group to its precise regional distribution profile.

All four soil types are matched accurately to their corresponding geographical locations.

Confirms complete alignment of pedological characteristics with Nigerian regional geography.

Key Concept

Zonal and Azonal Distribution of Soil Types in Nigeria

Question 12744Question →

A boutique owner in Enugu borrowed $\text{₦}80,000$ to expand her business at an annual interest rate of $15\%$ , compounded annually. If she repays the loan in full after $2$ years, what is the total interest she paid on the loan?

\text{₦}24,000

\text{₦}25,800

\text{₦}27,600

\text{₦}105,800

Show answer & explanation

Answer: $\text{₦}25,800$

Answer

The total interest paid on the loan after 2 years is

\text{₦}25,800

.

The compound interest is calculated year by year. In the first year, the interest paid is

15\%

of

\text{₦}80,000 = \text{₦}12,000

, raising the balance to

\text{₦}92,000

. In the second year, the interest paid is

15\%

of

\text{₦}92,000 = \text{₦}13,800

. Adding the two yearly interest payments gives

\text{₦}12,000 + \text{₦}13,800 = \text{₦}25,800

.

Step-by-Step Solution

1

Calculate the interest for the first year.

Interest for Year 1 =

15\%

of

\text{₦}80,000 = \frac{15}{100} \times 80,000 = \text{₦}12,000

.

In compound interest, interest for the first period is calculated on the initial principal.

2

Determine the amount at the end of the first year, which becomes the principal for the second year.

Principal for Year 2 =

\text{₦}80,000 + \text{₦}12,000 = \text{₦}92,000

.

Compound interest adds earned interest to the principal for subsequent period calculations.

3

Calculate the interest for the second year.

Interest for Year 2 =

15\%

of

\text{₦}92,000 = \frac{15}{100} \times 92,000 = \text{₦}13,800

.

Interest in year 2 is computed on the updated principal balance of

\text{₦}92,000

.

4

Sum the interest amounts from both years to find total interest paid.

Total Interest =

\text{₦}12,000 + \text{₦}13,800 = \text{₦}25,800

.

The total interest is the sum of interest accumulated across each compounding period.

Key Concept

Compound Interest Calculation

Estimated Time:1m 30s

Question 12745Question →

Match each chemical formula of the inorganic redox species on the left with its correct IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match

KMnO_4

to Potassium tetraoxomanganate(VII),

K_2Cr_2O_7

to Potassium heptaoxodichromate(VI),

NaClO_3

to Sodium trioxochlorate(V), and

Fe_2O_3

to Iron(III) oxide.

Each formula is correctly paired by evaluating the algebraic sum of oxidation numbers to find the oxidation state of the central transition metal or non-metal, then matching the corresponding oxo prefix and Roman numeral according to standard IUPAC conventions.

Step-by-Step Solution

1

Determine the oxidation state of Mn in

KMnO_4

Let oxidation state of Mn be

x

:

(+1) + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7

.

The sum of oxidation numbers in a neutral compound is equal to zero. This identifies the Roman numeral for manganese as (VII).

2

Determine the oxidation state of Cr in

K_2Cr_2O_7

Let oxidation state of Cr be

x

:

2(+1) + 2x + 7(-2) = 0 \implies 2x - 12 = 0 \implies x = +6

.

The compound contains seven oxygen atoms (heptaoxo) and two chromium atoms (dichromate), giving chromium an oxidation state of (VI).

3

Determine the oxidation state of Cl in

NaClO_3

Let oxidation state of Cl be

x

:

(+1) + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5

.

Three oxygen atoms give the prefix 'trioxo', and chlorine has an oxidation state of (V).

4

Determine the oxidation state of Fe in

Fe_2O_3

Let oxidation state of Fe be

x

:

2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies x = +3

.

Iron has an oxidation state of

+3

, naming the simple binary metal oxide as Iron(III) oxide.

Key Concept

Calculating central atom oxidation numbers and applying IUPAC nomenclature rules for oxoanions, oxoacids, and binary oxides.

Question 12746Question →

In a nuclear fusion process, two deuterium nuclei (

\text{^{2}_{1}H}

) fuse to form a helium-3 nucleus (

\text{^{3}_{2}He}

) and a neutron (

\text{^{1}_{0}n}

) according to the reaction equation:

\text{^{2}_{1}H} + \text{^{2}_{1}H} \rightarrow \text{^{3}_{2}He} + \text{^{1}_{0}n} + Q

Given the mass values:
- Mass of $\text{^{2}_{1}H} = 2.0141\text{ u}$
- Mass of $\text{^{3}_{2}He} = 3.0160\text{ u}$
- Mass of $\text{^{1}_{0}n} = 1.0087\text{ u}$

Using the conversion factor $1\text{ u} = 931.5\text{ MeV}$ , what is the energy released ( $Q$ ) in this fusion reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 3.26

Answer

The total energy released (

Q

) in the reaction is

3.26\text{ MeV}

.

The energy released in a nuclear fusion reaction is proportional to the decrease in total rest mass (mass defect). Summing the mass of two deuterium nuclei gives

4.0282\text{ u}

, while the sum of the product masses (

\text{^{3}_{2}He}

and a neutron) is

4.0247\text{ u}

. Subtracting these yields a mass defect of

0.0035\text{ u}

. Multiplying

0.0035\text{ u}

by

931.5\text{ MeV/u}

gives

3.26\text{ MeV}

of released energy.

Step-by-Step Solution

1

Calculate the total initial mass of the reacting deuterium nuclei.

m_{\text{reactants}} = 2 \times 2.0141\text{ u} = 4.0282\text{ u}

Two deuterium nuclei participate on the reactant side of the equation.

2

Calculate the total final mass of the products.

m_{\text{products}} = 3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

The reaction produces one helium-3 nucleus and one neutron.

3

Determine the mass defect (difference between reactant and product masses).

\Delta m = 4.0282\text{ u} - 4.0247\text{ u} = 0.0035\text{ u}

The mass lost during fusion is converted into nuclear kinetic energy and radiation.

4

Convert the mass defect into energy in MeV using the conversion factor.

Q = 0.0035\text{ u} \times 931.5\text{ MeV/u} = 3.26025\text{ MeV} \approx 3.26\text{ MeV}

Each atomic mass unit lost corresponds to

931.5\text{ MeV}

of energy.

Key Concept

Mass defect and energy release in nuclear fusion reactions (

E = \Delta m c^2

)

Question 12747Question →

In a nuclear fusion reaction, two deuterium nuclei ( ${^{2}_{1}\text{H}}$ ) combine to form a helium-3 nucleus ( ${^{3}_{2}\text{He}}$ ) and a neutron ( ${^{1}_{0}\text{n}}$ ). The total mass of the two reactant deuterium nuclei is $4.0282\text{ u}$ , while the total mass of the resulting helium-3 and neutron products is $4.0247\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , calculate the total energy released in this reaction in $\text{MeV}$ .

Show answer & explanation

Answer: 3.26

Answer

The total energy released in the nuclear fusion reaction is approximately 3.26 MeV.

The energy released in a nuclear fusion reaction is determined by the mass defect, which is the difference between the total mass of reactants and the total mass of products. Subtracting

4.0247\text{ u}

from

4.0282\text{ u}

yields a mass defect of

0.0035\text{ u}

. Multiplying this mass defect by the mass-energy conversion factor of

931.5\text{ MeV/u}

gives an energy release of approximately

3.26\text{ MeV}

.

Step-by-Step Solution

1

Calculate the mass defect (Δm)

Δm = 4.0282 u - 4.0247 u = 0.0035 u

Mass defect is the difference between the initial mass of reactants and the final mass of products in a nuclear reaction.

2

Calculate the energy released (E) in MeV

E = 0.0035 u × 931.5 MeV/u = 3.26025 MeV

According to mass-energy equivalence, 1 unified atomic mass unit (u) liberates 931.5 MeV of energy.

Key Concept

Mass defect and energy release in nuclear fusion

Question 12748Question →

Match each visual defect or optical condition listed in Column A with its corresponding cause and corrective lens in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Myopia pairs with rays focusing in front of the retina (diverging lens correction); Hypermetropia pairs with rays focusing behind the retina (converging lens correction); Astigmatism pairs with uneven corneal curvature (cylindrical lens correction); Presbyopia pairs with age-related loss of accommodation (bifocal lens correction).

Each defect of vision is accurately matched to its optical cause and standard corrective device: Myopia uses diverging lenses to push the image focal plane onto the retina, Hypermetropia uses converging lenses to pull the image forward onto the retina, Astigmatism uses cylindrical lenses for non-spherical corneal curves, and Presbyopia uses bifocal/converging lenses to correct age-related accommodation loss.

Step-by-Step Solution

1

Analyze Myopia

Myopia causes distant rays to focus before reaching the retina because the eye lens is overly converging or the eye focal length is too short; a diverging (concave) lens spreads rays to push the focal point back onto the retina.

Identify optical cause and lens remedy for short-sightedness.

2

Analyze Hypermetropia

Hypermetropia causes near rays to focus behind the retina; a converging (convex) lens bends incoming light rays inwards to bring the focal point onto the retina.

Identify optical cause and lens remedy for long-sightedness.

3

Analyze Astigmatism

Astigmatism arises from non-uniform curvature of the refracting surfaces, requiring a cylindrical lens with differential curvature along different planes.

Identify refractive error causing multiple focal planes.

4

Analyze Presbyopia

Presbyopia is due to age-induced stiffening of the eye lens and loss of ciliary accommodation power, which is managed using bifocal or converging lenses.

Distinguish physiological aging effects on focal accommodation.

Key Concept

Defects of Vision and Corrective Lenses

Estimated Time:1m 30s

Question 12749Question →

Match each defect of vision on the left with its corresponding corrective optical lens on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Myopia matches with Concave (diverging) lens, Hypermetropia matches with Convex (converging) lens, Presbyopia matches with Bifocal lens, and Astigmatism matches with Cylindrical lens.

Each eye condition is paired with its specific optical correction: Myopia requires a concave lens to diverge light, Hypermetropia requires a convex lens to converge light, Presbyopia uses a bifocal lens to assist diminished accommodation, and Astigmatism relies on a cylindrical lens to correct asymmetrical curvature.

Step-by-Step Solution

1

Analyze Myopia (Short-sightedness)

Parallel rays focus in front of the retina due to an elongated eyeball or over-refractive lens.

Diverging (concave) lenses spread out incoming rays slightly before entering the eye so the focal point shifts back onto the retina.

2

Analyze Hypermetropia (Long-sightedness)

Light rays focus behind the retina due to a shortened eyeball or insufficient focal power.

Converging (convex) lenses provide additional converging power to focus rays directly on the retina.

3

Analyze Presbyopia

The eye lens loses elasticity with age, reducing its power to accommodate both near and far objects.

Bifocal lenses have two distinct focal lengths in a single glass unit to assist with both near and distant vision.

4

Analyze Astigmatism

Cornea or crystalline lens curvature is uneven along different axes, producing distorted vision.

Cylindrical lenses correct uneven refractive power by bending light along one axis without affecting the orthogonal axis.

Key Concept

Defects of Vision and Corrective Lenses

Estimated Time:45s

Question 12750Question →

If a sample of unsaturated air with a relative humidity of $40\%$ at $30^\circ\text{C}$ is compressed isothermally to one-third of its original volume, the final relative humidity of the air will be $100\%$ .

True

False

Show answer & explanation

Answer: True

Answer

True. The relative humidity cannot exceed 100% because water vapour condenses into liquid once its partial pressure reaches the saturated vapour pressure at that temperature.

The statement is true because relative humidity is the ratio of actual partial vapour pressure to saturated vapour pressure at a specific temperature. Although isothermal compression to one-third volume triples the partial pressure of water vapour (giving a theoretical 120%), partial vapour pressure is bounded by the saturated vapour pressure. When the relative humidity reaches 100%, condensation occurs, keeping the relative humidity at exactly 100%.

Step-by-Step Solution

1

Apply Boyle's law to calculate the theoretical partial vapour pressure after isothermal compression.

Since volume is reduced to

\frac{1}{3}V_1

at constant temperature, the partial pressure of water vapour would increase by a factor of 3 (

p_2 = 3p_1

).

For an unsaturated vapour at constant temperature, partial pressure is inversely proportional to volume.

2

Determine the theoretical relative humidity from the pressure ratio.

\text{Theoretical R.H.} = 3 \times 40\% = 120\%.

Relative humidity is defined as

\text{R.H.} = \frac{p}{p_s} \times 100\%

, where

p

is the partial pressure and

p_s

is the saturated vapour pressure.

3

Apply the physical limit imposed by saturated vapour pressure.

The actual relative humidity cannot exceed

100\%

; excess vapour condenses.

Saturated vapour pressure

p_s

represents the maximum possible partial pressure of water vapour in air at a given temperature.

Key Concept

Saturated Vapour Pressure Limit and Relative Humidity under Isothermal Compression

Question 12751Question →

The standard enthalpy of formation of gaseous water ( $\text{H}_2\text{O}(g)$ ) at $298\text{ K}$ is equal to the sum of the standard enthalpy of combustion of hydrogen gas ( $\text{H}_2(g)$ ) and the standard enthalpy of vaporization of liquid water ( $\text{H}_2\text{O}(l)$ ).

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because adding the thermochemical equation for the combustion of hydrogen (which forms liquid water) to the equation for the vaporization of liquid water yields the formation equation for gaseous water from its constituent elements.

The statement accurately reflects Hess's law by combining the reaction for hydrogen combustion to form liquid water with the phase transformation of liquid water into water vapor.

Step-by-Step Solution

1

Write the thermochemical equation for the standard enthalpy of combustion of hydrogen gas.

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H_1^\circ = \Delta H_c^\circ[\text{H}_2(g)]

By definition, standard combustion of hydrogen produces water in its standard state at

298\text{ K}

, which is liquid water.

2

Write the thermochemical equation for the standard enthalpy of vaporization of water.

\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g) \quad \Delta H_2^\circ = \Delta H_{\text{vap}}^\circ[\text{H}_2\text{O}(l)]

Vaporization represents the phase change of one mole of liquid water into gaseous water.

3

Apply Hess's law to combine Step 1 and Step 2.

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(l) + \text{H}_2\text{O}(g)

, which simplifies to

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g)

with

\Delta H^\circ = \Delta H_1^\circ + \Delta H_2^\circ

.

Summing the two equations cancels out liquid water, leaving the formation reaction of gaseous water from its elemental gases.

Key Concept

Hess's Law and Thermochemical Cycles involving Standard States

Question 12752Question →

During a laboratory experiment, a student measures the thickness of a glass slide as $2.5\text{ mm}$ . If the actual thickness of the slide is $2.0\text{ mm}$ , what is the percentage error in the measurement?

20.0\%

25.0\%

0.25\%

5.0\%

Show answer & explanation

Answer: $25.0\%$

Answer

The percentage error in the measurement is

25.0\%

.

Percentage error is determined by dividing the absolute error (

|2.5\text{ mm} - 2.0\text{ mm}| = 0.5\text{ mm}

) by the actual value (

2.0\text{ mm}

) and multiplying by

100\%

. This gives

\frac{0.5}{2.0} \times 100\% = 25.0\%

.

Step-by-Step Solution

1

Calculate the absolute error in the measurement

\text{Absolute error} = |\text{Measured value} - \text{Actual value}| = |2.5\text{ mm} - 2.0\text{ mm}| = 0.5\text{ mm}

Absolute error measures the magnitude of discrepancy between measured and true values.

2

Calculate the percentage error relative to the actual value

\text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100\% = \frac{0.5\text{ mm}}{2.0\text{ mm}} \times 100\% = 25.0\%

Percentage error is defined as the absolute error divided by the accepted actual value, expressed as a percentage.

Key Concept

Percentage Error Calculation

Estimated Time:45s

Question 12753Question →

An object is placed at a distance of $15\text{ cm}$ in front of a concave mirror with a focal length of $10\text{ cm}$ . What is the distance of the image formed from the mirror?

30\text{ cm}

6\text{ cm}

25\text{ cm}

-30\text{ cm}

Show answer & explanation

Answer: $30\text{ cm}$

Answer

The distance of the image formed from the mirror is

30\text{ cm}

.

Applying the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with

f = 10\text{ cm}

and

u = 15\text{ cm}

gives

\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}

, yielding an image distance of

30\text{ cm}

.

Step-by-Step Solution

1

Identify given quantities and signs

Focal length

f = 10\text{ cm}

and object distance

u = 15\text{ cm}

.

For a concave mirror, the real focus and real object distances are both positive.

2

Set up the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{10} = \frac{1}{15} + \frac{1}{v}

The mirror formula relates object distance, image distance, and focal length.

3

Solve for the image distance

v

\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} \implies v = 30\text{ cm}

Subtracting the reciprocals and taking the inverse gives the image position.

Key Concept

Concave Mirror Formula

Question 12754Question →

In a laboratory experiment, the radius of a circular wire is measured as $(2.00 \pm 0.05)\text{ mm}$ . What is the percentage error in the calculated cross-sectional area of the wire?

2.5\%

5.0\%

6.25\%

1.25\%

Show answer & explanation

Answer: $5.0\%$

Answer

The percentage error in the calculated area is

5.0\%

.

The cross-sectional area of a wire is given by

A = \pi r^2

. The percentage error in a calculated quantity

y = x^n

is given by

n \times (\text{percentage error in } x)

. Here, the percentage error in the radius

r

is

(0.05 / 2.00) \times 100\% = 2.5\%

. Multiplying by the power

n = 2

gives

2 \times 2.5\% = 5.0\%

.

Step-by-Step Solution

1

Calculate the percentage error in the measured radius

r

.

\text{Percentage error in } r = \left(\frac{0.05}{2.00}\right) \times 100\% = 2.5\%

Relative error is the absolute error divided by the measured value.

2

Apply the rule of error propagation for quantities raised to a power (

A = \pi r^2

).

\text{Percentage error in } A = 2 \times (\text{Percentage error in } r) = 2 \times 2.5\% = 5.0\%

When a physical quantity is raised to the power

n

, its fractional or percentage error is multiplied by

n

.

Key Concept

Error Propagation in Derived Quantities

Estimated Time:1m 30s

Question 12755Question →

An unknown object is evaluated inside a space probe descending towards the surface of a distant planet. A beam balance calibrated on Earth registers a reading of $15.0\text{ kg}$ for the object. The probe has a downward acceleration of $2.0\text{ m s}^{-2}$ in a region where the local gravitational acceleration of the planet is $6.0\text{ m s}^{-2}$ . What reading will a spring balance display when the same object is suspended from it inside the probe?

90.0\text{ N}

120.0\text{ N}

60.0\text{ N}

150.0\text{ N}

Show answer & explanation

Answer: $60.0\text{ N}$

Answer

60.0\text{ N}

A beam balance compares gravitational forces on equal balance arms, meaning local gravity and frame acceleration affect both sides equally. Therefore, the beam balance measures the true invariant mass of

15.0\text{ kg}

. A spring balance measures the tension or apparent weight

W_{\text{app}} = m(g - a)

. Substituting

m = 15.0\text{ kg}

, local gravity

g = 6.0\text{ m s}^{-2}

, and downward acceleration

a = 2.0\text{ m s}^{-2}

gives

W_{\text{app}} = 15.0 \times (6.0 - 2.0) = 60.0\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the object using the beam balance measurement

True mass

m = 15.0\text{ kg}

A beam balance compares unknown mass with standard masses under the same local gravitational field. Because local gravity acts equally on both pans, beam balance readings yield the true invariant mass regardless of local gravitational acceleration or uniform reference frame acceleration (provided net effective gravity is greater than zero).

2

Calculate the effective acceleration experienced inside the accelerating reference frame

g_{\text{eff}} = g_{\text{planet}} - a = 6.0\text{ m s}^{-2} - 2.0\text{ m s}^{-2} = 4.0\text{ m s}^{-2}

When a reference frame accelerates downward at rate

a

, the effective apparent gravitational acceleration felt by objects suspended inside is reduced by

a

.

3

Calculate the apparent weight registered by the spring balance

W_{\text{app}} = m \cdot g_{\text{eff}} = 15.0\text{ kg} \times 4.0\text{ m s}^{-2} = 60.0\text{ N}

A spring balance measures tension force (apparent weight) exerted on its spring, which equals

m(g - a)

in a downward accelerating system.

Key Concept

Mass measurement via beam balance vs. apparent weight measurement via spring balance in accelerating frames

Question 12756Question →

A physics laboratory utilizes four specialized instruments to detect different regions of the electromagnetic spectrum: an aerial antenna, a thermopile, a photographic plate sensitive to sun-tanning radiation, and a Geiger-Müller tube.

Arrange these detectors in order of INCREASING frequency of the electromagnetic radiation they are primarily designed to detect (from lowest frequency to highest frequency).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest frequency to highest frequency is: Aerial antenna (Radio waves) → Thermopile (Infrared) → Photographic plate sensitive to sun-tanning radiation (Ultraviolet) → Geiger-Müller tube (Gamma rays).

The correct sequence arranges the instruments according to the increasing frequency of the radiation they detect. Radio waves (detected by an aerial antenna) have the lowest frequency, followed by infrared radiation (detected by a thermopile), ultraviolet radiation (detected by photographic plates sensitive to sun-tanning rays), and gamma rays (detected by a Geiger-Müller tube) which have the highest frequency.

Step-by-Step Solution

1

Identify the type of electromagnetic radiation detected by each device.

Aerial antenna detects radio waves; Thermopile detects infrared radiation; Photographic plate for tanning radiation detects ultraviolet radiation; Geiger-Müller tube detects gamma rays.

Each detector operates on specific physical properties characteristic of a particular band of the electromagnetic spectrum.

2

Recall the order of the electromagnetic spectrum in terms of frequency (

f

).

Radio waves (

< 10^9\text{ Hz}

) < Infrared (

10^{11} - 10^{14}\text{ Hz}

) < Ultraviolet (

10^{15} - 10^{17}\text{ Hz}

) < Gamma rays (

> 10^{19}\text{ Hz}

).

Frequency increases continuously across the spectrum from radio waves to gamma rays.

3

Sequence the detectors based on their associated radiation frequencies from lowest to highest.

Aerial antenna

\rightarrow

Thermopile

\rightarrow

Photographic plate sensitive to sun-tanning radiation

\rightarrow

Geiger-Müller tube.

This directly matches the increasing frequency order of radio waves, infrared, ultraviolet, and gamma rays.

Key Concept

Detection mechanisms and frequency distribution across the electromagnetic spectrum

Estimated Time:2m 0s

Question 12757Question →

If the kinetic energy of a non-relativistic electron is increased by a factor of $4$ , how does its associated de Broglie wavelength change?

It decreases to one-half of its original value

It decreases to one-fourth of its original value

It increases by a factor of

2

It increases by a factor of

4

Show answer & explanation

Answer: It decreases to one-half of its original value

Answer

The de Broglie wavelength decreases to one-half of its original value.

The de Broglie wavelength

\lambda

of a particle is given by

\lambda = \frac{h}{p}

, where momentum

p = \sqrt{2m E_k}

. Substituting momentum into the wavelength equation yields

\lambda = \frac{h}{\sqrt{2m E_k}}

. If kinetic energy

E_k

quadruples, the denominator increases by a factor of

\sqrt{4} = 2

, which reduces the wavelength to half of its initial value.

Step-by-Step Solution

1

Relate de Broglie wavelength to momentum and kinetic energy.

The de Broglie wavelength is given by

\lambda = \frac{h}{p}

. Since kinetic energy

E_k = \frac{p^2}{2m}

, momentum is

p = \sqrt{2m E_k}

. Thus,

\lambda = \frac{h}{\sqrt{2m E_k}}

.

Establishing the mathematical relationship between wavelength

\lambda

and kinetic energy

E_k

.

2

Apply the scaling factor of 4 to the kinetic energy.

When

E_k' = 4 E_k

, the new wavelength

\lambda'

is

\lambda' = \frac{h}{\sqrt{2m (4 E_k)}} = \frac{h}{2\sqrt{2m E_k}} = \frac{\lambda}{2}

.

Evaluating the square root factor

\sqrt{4} = 2

in the denominator.

3

Conclude the final ratio.

The new wavelength is half the original wavelength.

The de Broglie wavelength is inversely proportional to the square root of kinetic energy.

Key Concept

Relationship between de Broglie wavelength and kinetic energy

Question 12758Question →

Match each metallurgical process or extraction stage on the left with its corresponding chemical principle or operational method on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Concentration of low-grade sulfide ores matches Froth flotation using oil collectors; Extraction of highly electropositive metals matches Electrolysis of fused salts; Reduction of haematite matches Chemical reduction by carbon monoxide gas; Refining of crude blister copper matches Electrolytic dissolution of impure anode and deposition of pure metal on cathode.

The paired principles directly correspond to standard industrial metallurgy: froth flotation uses oil wettability to concentrate sulfide ores; fused salt electrolysis extracts top-series electropositive metals; carbon monoxide reduces iron oxide in the blast furnace; and electro-refining purifies crude metal using crude anode oxidation and pure cathode deposition.

Step-by-Step Solution

1

Determine the physical concentration technique suitable for sulfide minerals.

Sulfide ores like galena and chalcopyrite selectively attach to air bubbles created by oil collectors and froth up, separating from waste gangue.

Difference in surface wettability between sulfide ore and siliceous gangue.

2

Evaluate the extraction strategy based on the reactivity series of metals.

Metals positioned near the top of the reactivity series (

Al

,

Na

,

Ca

) require electrical energy for reduction because carbon cannot displace them from their oxides.

High electropositivity means these metals have higher affinity for oxygen than carbon has.

3

Identify the primary chemical reducing agent in the blast furnace.

Coke burns to form

CO_2

, which reacts further with coke to give

CO

.

CO

gas then reduces

Fe_2O_3

step-by-step to molten iron.

CO

is a effective gaseous reducing agent that penetrates porous ore charges.

4

Analyze the electrolytic purification mechanism for crude metals.

Impurities stay in solution or form anode sludge while metal ions migrate and plate onto the cathode as pure copper metal.

Anodic oxidation releases

Cu^{2+}

ions while cathode reduction ensures selective plating of pure copper.

Key Concept

General Principles of Metallurgy and Metal Extraction

Estimated Time:1m 30s

Question 12759Question →

A spring balance and an equal-arm beam balance are used to measure an object at a location where the local acceleration due to gravity is $8.0\text{ m s}^{-2}$ . If the spring balance registers a weight of $40\text{ N}$ , what mass will the equal-arm beam balance register at this location?

5.0\text{ kg}

4.0\text{ kg}

320\text{ kg}

50\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The equal-arm beam balance will register a mass of

5.0\text{ kg}

.

Weight is given by

W = mg

. Given

W = 40\text{ N}

and

g = 8.0\text{ m s}^{-2}

, the mass of the object is

m = \frac{40}{8.0} = 5.0\text{ kg}

. An equal-arm beam balance balances the unknown mass against standard masses under the exact same local gravitational field, so local gravity cancels out and it accurately measures the object's mass as

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the true mass of the object from the spring balance reading

m = \frac{W}{g} = \frac{40\text{ N}}{8.0\text{ m s}^{-2}} = 5.0\text{ kg}

Weight is the gravitational force acting on a mass (

W = mg

), so mass equals weight divided by local acceleration due to gravity.

2

Determine the reading on the equal-arm beam balance

Mass registered =

5.0\text{ kg}

An equal-arm beam balance compares the gravitational force on the unknown mass against standard masses. Because local gravity affects both sides equally, it measures true mass independent of local gravitational acceleration.

Key Concept

Measurement of Mass and Weight

Estimated Time:1m 0s

Question 12760Question →

A straight copper rod of length $0.50\text{ m}$ and mass $40\text{ g}$ carries a steady current and is suspended horizontally in a uniform magnetic field of $0.40\text{ T}$ . The field is directed horizontally at an angle of $30^\circ$ to the length of the rod. If the upward magnetic force acting on the rod exactly balances its weight, what is the magnitude of the current flowing through the rod? (Take $g = 10\text{ m/s}^2$ )

0.25\text{ A}

2.0\text{ A}

4.0\text{ A}

40\text{ A}

Show answer & explanation

Answer: $4.0\text{ A}$

Answer

The magnitude of the current required to balance the weight of the rod is

4.0\text{ A}

.

For the rod to remain suspended in equilibrium, the upward magnetic force

F_B = BIL \sin \theta

must balance the downward gravitational force

W = mg

. Substituting

m = 0.040\text{ kg}

,

g = 10\text{ m/s}^2

,

B = 0.40\text{ T}

,

L = 0.50\text{ m}

, and

\theta = 30^\circ

yields

0.40 = 0.40 \times I \times 0.50 \times 0.50

, which solves to

I = 4.0\text{ A}

.

Step-by-Step Solution

1

Convert the mass of the rod to kilograms and calculate its weight.

m = 40\text{ g} = 0.040\text{ kg}

, so

W = mg = 0.040\text{ kg} \times 10\text{ m/s}^2 = 0.40\text{ N}

.

Standard SI units must be used to ensure dimensional consistency.

2

Express the magnetic force acting on the conductor in terms of current

I

.

F_B = B I L \sin \theta = (0.40\text{ T}) \times I \times (0.50\text{ m}) \times \sin 30^\circ = 0.10 I\text{ N}

.

The magnetic force on a current-carrying conductor at an angle

\theta

to a magnetic field is given by

F = BIL \sin \theta

.

3

Equate the upward magnetic force to the downward weight to find

I

.

0.10 I = 0.40 \implies I = \frac{0.40}{0.10} = 4.0\text{ A}

.

For vertical equilibrium, the net vertical force must equal zero.

Key Concept

Equilibrium of a current-carrying conductor in a uniform magnetic field (

F = BIL \sin \theta = mg

)

Estimated Time:2m 0s

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