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Soru 1241Soru →

Match each physical quantity to its correct physical quantity classification and corresponding SI base unit expression.

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Öğeler

Luminous intensity
Electric potential
Specific heat capacity
Thermodynamic temperature

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Cevap

Luminous intensity matches with fundamental quantity in cd\text{cd}; Electric potential matches with derived quantity in kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}; Specific heat capacity matches with derived quantity in m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}; Thermodynamic temperature matches with fundamental quantity in K\text{K}.
Luminous intensity and thermodynamic temperature are fundamental SI quantities with base units cd and K. Electric potential and specific heat capacity are derived quantities whose definitions reduce to kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1} and m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1} respectively.

Adım Adım Çözüm

1
Identify fundamental physical quantities
Luminous intensity and thermodynamic temperature are base SI quantities measured in candela (cd) and kelvin (K) respectively.
Base quantities cannot be defined in terms of other physical quantities.
2
Decompose derived quantities into base SI units
Electric potential V=WqV = \frac{W}{q} yields kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}. Specific heat capacity c=QmΔTc = \frac{Q}{m\Delta T} yields m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Derived quantities originate from mathematical combinations of fundamental quantities.

Anahtar Kavram

Classification of physical quantities into fundamental (base) and derived categories and resolution into SI base units
Soru 1242Soru →

Match each composite physical quantity or ratio on the left with its correct fundamental (SI base) unit decomposition on the right.

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Öğeler

Electric potential gradient
Coefficient of dynamic viscosity
Ratio of Planck's constant to moment of inertia
Specific latent heat divided by spatial temperature gradient

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Cevap

Electric potential gradient matches kg⋅m⋅s−3⋅A−1\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{A}^{-1}; Coefficient of dynamic viscosity matches kg⋅m−1⋅s−1\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}; Ratio of Planck's constant to moment of inertia matches s−1\text{s}^{-1}; Specific latent heat divided by spatial temperature gradient matches m3⋅s−2⋅K−1\text{m}^3\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Each physical quantity or ratio is systematically reduced to its SI base quantities (mass in kg, length in m, time in s, electric current in A, thermodynamic temperature in K) by substituting fundamental definitions of derived units.

Adım Adım Çözüm

1
Decompose electric potential gradient into fundamental SI base units
Electric potential gradient=Electric PotentialDistance=WorkCharge×Distance=kg⋅m2⋅s−2A⋅s×m=kg⋅m⋅s−3⋅A−1\text{Electric potential gradient} = \frac{\text{Electric Potential}}{\text{Distance}} = \frac{\text{Work}}{\text{Charge} \times \text{Distance}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s} \times \text{m}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{A}^{-1}.
Electric potential is defined as energy per unit charge, and potential gradient is its spatial rate of change.
2
Decompose coefficient of dynamic viscosity into fundamental SI base units
η=Force×DistanceArea×Velocity=(kg⋅m⋅s−2)×mm2×(m⋅s−1)=kg⋅m−1⋅s−1\eta = \frac{\text{Force} \times \text{Distance}}{\text{Area} \times \text{Velocity}} = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m}}{\text{m}^2 \times (\text{m}\cdot\text{s}^{-1})} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}.
Newton's law of viscosity relates shear force to surface area and velocity gradient.
3
Determine the base unit ratio of Planck's constant to moment of inertia
\frac{h}{I} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}}{\text{kg}\cdot\text{m}^2} = \text{s}^{-1}.
Planck's constant carries dimensions of angular momentum, while moment of inertia is mass multiplied by distance squared.
4
Decompose specific latent heat divided by spatial temperature gradient
\frac{L}{\frac{\Delta T}{\Delta x}} = \frac{\text{J}\cdot\text{kg}^{-1}}{\text{K}\cdot\text{m}^{-1}} = \frac{\text{m}^2\cdot\text{s}^{-2}}{\text{K}\cdot\text{m}^{-1}} = \text{m}^3\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Specific heat quantities represent thermal energy per unit mass, whereas temperature gradient represents thermal variation per unit displacement.

Anahtar Kavram

Fundamental SI base unit decomposition of derived physical quantities
Soru 1243Soru →

Match each temperature scale reference state on the left with its correct thermodynamic definition on the right.

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Öğeler

Absolute zero (0 K0\text{ K})
Ice point (0∘C0^\circ\text{C})
Steam point (100∘C100^\circ\text{C})
Triple point of water (273.16 K273.16\text{ K})

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Cevap

Absolute zero matches with the state of minimum molecular kinetic energy. The ice point matches with the lower fixed point of pure melting ice at standard pressure. The steam point matches with the upper fixed point of pure boiling water steam at standard pressure. The triple point of water matches with the thermodynamic equilibrium state of ice, liquid water, and water vapour.
Each temperature scale reference point is correctly paired with its defining physical state: absolute zero represents minimum molecular kinetic energy, the ice point represents melting ice at standard pressure, the steam point represents steam from boiling water at standard pressure, and the triple point represents the three-phase equilibrium of water.

Adım Adım Çözüm

1
Identify the definition of absolute zero.
Absolute zero (0 K0\text{ K}) corresponds to the state of minimum internal molecular kinetic energy.
At 0 K0\text{ K}, thermal motion of particles theoretically ceases.
2
Identify the definition of the ice point.
The ice point (0∘C0^\circ\text{C}) corresponds to pure melting ice at standard atmospheric pressure.
It serves as the standard lower fixed point on the Celsius temperature scale.
3
Identify the definition of the steam point.
The steam point (100∘C100^\circ\text{C}) corresponds to steam from pure boiling water at standard atmospheric pressure.
It serves as the standard upper fixed point on the Celsius temperature scale.
4
Identify the definition of the triple point of water.
The triple point (273.16 K273.16\text{ K}) is the unique thermodynamic state where ice, liquid water, and steam coexist in equilibrium.
It is used as a single fundamental reference point on the Kelvin thermodynamic scale.

Anahtar Kavram

Temperature Scale Fixed Points and Reference States
Tahmini Süre:1m 30s
Soru 1244Soru →

Historical developments in atomic physics led to several distinct models of atomic structure. Match each atomic model on the left with its defining structural feature or experimental basis on the right.

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Öğeler

Thomson's Model
Rutherford's Model
Bohr's Model
Quantum Mechanical Model

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Cevap

Thomson's Model matches with embedded electrons in a positive sphere; Rutherford's Model matches with a dense positive nucleus from alpha scattering; Bohr's Model matches with quantized non-radiating orbits; Quantum Mechanical Model matches with electron probability orbitals.
Each historical atomic model is correctly matched to its fundamental feature: Thomson proposed electrons suspended in positive mass; Rutherford used alpha scattering to discover the compact nucleus; Bohr quantized electron orbits to explain spectral lines; and the Quantum Mechanical Model represents electrons via three-dimensional probability orbitals.

Adım Adım Çözüm

1
Identify Thomson's contribution
Thomson proposed the 'plum pudding' model where negative electrons are embedded inside a uniform positive charge sphere.
This preceded the discovery of the atomic nucleus.
2
Identify Rutherford's contribution
Rutherford discovered the central positive nucleus through the alpha particle deflection experiment.
Large deflections meant most atomic mass and positive charge concentrated at a tiny core.
3
Identify Bohr's contribution
Bohr added quantum conditions to planetary orbits so electrons remain stable without continuously radiating energy.
Quantized angular momentum explains discrete emission line spectra.
4
Identify the Quantum Mechanical Model contribution
Modern quantum mechanics replaces fixed circular orbits with wave functions and 3D probability clouds (orbitals).
Heisenberg's uncertainty principle rules out precise circular orbits.

Anahtar Kavram

Evolution of Atomic Models
Soru 1245Soru →

Match each physical quantity listed on the left with its corresponding expression in fundamental SI base units on the right.

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Öğeler

Pressure
Surface Tension
Specific Heat Capacity
Electric Capacitance

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Cevap

Pressure matches with kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}, Surface Tension matches with kg⋅s−2\text{kg}\cdot\text{s}^{-2}, Specific Heat Capacity matches with m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}, and Electric Capacitance matches with kg−1⋅m−2⋅s4⋅A2\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^4\cdot\text{A}^2.
Each quantity on the left is matched correctly to its fundamental SI base unit equivalent derived directly from its defining physical formula.

Adım Adım Çözüm

1
Derive base SI units for Pressure
P=FA=kg⋅m⋅s−2m2=kg⋅m−1⋅s−2P = \frac{F}{A} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
Pressure is force per unit area.
2
Derive base SI units for Surface Tension
γ=FL=kg⋅m⋅s−2m=kg⋅s−2\gamma = \frac{F}{L} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}} = \text{kg}\cdot\text{s}^{-2}
Surface tension is force per unit length.
3
Derive base SI units for Specific Heat Capacity
c=QmΔT=kg⋅m2⋅s−2kg⋅K=m2⋅s−2⋅K−1c = \frac{Q}{m\Delta T} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}
Specific heat capacity is thermal energy per unit mass per kelvin.
4
Derive base SI units for Electric Capacitance
C=QV=A⋅skg⋅m2⋅s−3⋅A−1=kg−1⋅m−2⋅s4⋅A2C = \frac{Q}{V} = \frac{\text{A}\cdot\text{s}}{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}} = \text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^4\cdot\text{A}^2
Capacitance is electric charge divided by electric potential difference.

Anahtar Kavram

Expressing derived SI units in terms of fundamental base SI units
Soru 1246Soru →

Match each physical scenario involving scalar and vector quantities on the left with its corresponding resultant value or component magnitude on the right.

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Öğeler

A particle undergoes successive horizontal displacements of 10 m10\text{ m} East, 12 m12\text{ m} North, and 5 m5\text{ m} West. The magnitude of its net displacement.
Two equal coplanar forces, each of magnitude FF, act at an angle of 60∘60^\circ to each other. The magnitude of their resultant force.
A force vector of magnitude 40 N40\text{ N} is inclined at an angle of 60∘60^\circ to the vertical axis. The magnitude of its vertical component.
Two concurrent forces of magnitudes 8 N8\text{ N} and 15 N15\text{ N} act at an angle of 90∘90^\circ to one another. The magnitude of their resultant force.

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Cevap

The correct pairings are: (1) The particle's net displacement corresponds to 13 m; (2) The resultant of two equal forces of magnitude F at 60 degrees corresponds to F√3; (3) The vertical component of a 40 N force inclined at 60 degrees to the vertical corresponds to 20 N; (4) The resultant of perpendicular forces of 8 N and 15 N corresponds to 17 N.
Each scenario correctly applies vector algebra: 2D displacement resolution yields a 5-12-13 right triangle; the parallelogram rule for equal forces at 60 degrees produces F√3; resolving a force adjacent to the vertical axis uses cos(60°) to give 20 N; and perpendicular 8 N and 15 N forces synthesize to a 17 N resultant using the Pythagorean theorem.

Adım Adım Çözüm

1
Calculate net displacement for Item 1
Net x-component: 10 m−5 m=5 m10\text{ m} - 5\text{ m} = 5\text{ m} East. Net y-component: 12 m12\text{ m} North. Magnitude R=52+122=13 mR = \sqrt{5^2 + 12^2} = 13\text{ m}.
Displacements along parallel lines subtract scalar-wise, and perpendicular components combine via the Pythagorean theorem.
2
Determine the resultant of two equal forces at 60 degrees for Item 2
R=F2+F2+2(F)(F)cos⁡(60∘)=2F2+2F2(0.5)=3F2=F3R = \sqrt{F^2 + F^2 + 2(F)(F)\cos(60^\circ)} = \sqrt{2F^2 + 2F^2(0.5)} = \sqrt{3F^2} = F\sqrt{3}.
Applying the parallelogram law of vector addition.
3
Resolve the force vector along the vertical direction for Item 3
Fvertical=Fcos⁡(θvertical)=40cos⁡(60∘)=40×0.5=20 NF_{\text{vertical}} = F \cos(\theta_{\text{vertical}}) = 40 \cos(60^\circ) = 40 \times 0.5 = 20\text{ N}.
The component adjacent to the reference angle uses the cosine function.
4
Compute resultant magnitude of orthogonal forces for Item 4
R=82+152=64+225=289=17 NR = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ N}.
Vectors at right angles sum directly using Pythagorean synthesis.

Anahtar Kavram

Vector resolution, component synthesis, and parallelogram law of vector addition
Soru 1247Soru →

Match each type of wave listed on the left with its correct classification and propagation characteristic on the right.

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Öğeler

Sound wave in air
Radio wave in vacuum
Water ripple on a lake surface

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Cevap

Sound wave in air matches Mechanical longitudinal wave requiring a material medium; Radio wave in vacuum matches Electromagnetic transverse wave capable of traveling without a medium; Water ripple on a lake surface matches Mechanical transverse wave propagating along a liquid surface.
Each wave is correctly paired based on whether it needs a physical medium to propagate (mechanical waves require a medium, electromagnetic waves do not) and whether the displacement is parallel (longitudinal) or perpendicular (transverse) to the direction of energy propagation.

Adım Adım Çözüm

1
Identify the medium requirement and vibration direction for a sound wave in air.
Sound waves require a material medium (air) and vibrate parallel to the direction of wave movement, making them mechanical longitudinal waves.
Classification depends on whether a physical medium is needed and how particles oscillate relative to energy transport.
2
Identify the medium requirement and vibration direction for a radio wave in a vacuum.
Radio waves can travel through empty space without a material medium and consist of field oscillations perpendicular to propagation, making them electromagnetic transverse waves.
Electromagnetic waves propagate via mutually perpendicular electric and magnetic field oscillations and require no medium.
3
Identify the medium requirement and vibration direction for a water ripple.
Ripples require a material medium (water) and displace surface water up and down perpendicular to wave travel, making them mechanical transverse waves.
Surface water waves exhibit transverse displacement characteristics in a physical liquid medium.

Anahtar Kavram

Classification of waves based on medium requirement (mechanical vs. electromagnetic) and particle vibration direction relative to propagation (transverse vs. longitudinal).
Soru 1248Soru →

Match each physical quantity on the left with its corresponding SI unit expressed in terms of fundamental (base) units on the right.

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Öğeler

Frequency
Electric charge
Mass density
Acceleration

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Cevap

Frequency matches s−1\text{s}^{-1}, Electric charge matches A⋅s\text{A}\cdot\text{s}, Mass density matches kg⋅m−3\text{kg}\cdot\text{m}^{-3}, and Acceleration matches m⋅s−2\text{m}\cdot\text{s}^{-2}.
Frequency (f=1/Tf = 1/T) is expressed in reciprocal seconds (s−1\text{s}^{-1}). Electric charge (Q=I⋅tQ = I \cdot t) is current times time, giving A⋅s\text{A}\cdot\text{s}. Mass density (ρ=m/V\rho = m/V) is mass per volume, giving kg⋅m−3\text{kg}\cdot\text{m}^{-3}. Acceleration (a=Δv/Δta = \Delta v / \Delta t) is rate of velocity change, giving m⋅s−2\text{m}\cdot\text{s}^{-2}.

Adım Adım Çözüm

1
Identify the defining formula for each physical quantity
Frequency f=1Tf = \frac{1}{T}, Charge Q=I⋅tQ = I \cdot t, Density ρ=mV\rho = \frac{m}{V}, Acceleration a=ΔvΔta = \frac{\Delta v}{\Delta t}.
Relating derived quantities to their defining equations allows reduction into fundamental quantities.
2
Substitute the SI base units for mass (kg\text{kg}), length (m\text{m}), time (s\text{s}), and current (A\text{A})
Frequency: s−1\text{s}^{-1}; Charge: A⋅s\text{A}\cdot\text{s}; Density: kgm3=kg⋅m−3\frac{\text{kg}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-3}; Acceleration: m/ss=m⋅s−2\frac{\text{m/s}}{\text{s}} = \text{m}\cdot\text{s}^{-2}.
This expresses each derived unit strictly in terms of fundamental SI units.
3
Match each physical quantity to its calculated base unit representation
Frequency →s−1\rightarrow \text{s}^{-1}, Electric charge →A⋅s\rightarrow \text{A}\cdot\text{s}, Mass density →kg⋅m−3\rightarrow \text{kg}\cdot\text{m}^{-3}, Acceleration →m⋅s−2\rightarrow \text{m}\cdot\text{s}^{-2}.
Completes the pairing verification.

Anahtar Kavram

Expressing derived physical quantities in terms of SI fundamental (base) units
Tahmini Süre:45s
Soru 1249Soru →

Match each statistical chart term or parameter in Column A with its corresponding definition or formula in Column B.

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Öğeler

Class boundary
Frequency density
Cumulative frequency
Pie chart sector angle

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Cevap

Class boundary matches the exact continuous limit of a class interval; Frequency density matches the quotient of class frequency and class width; Cumulative frequency matches the running total of frequencies plotted on an ogive; Pie chart sector angle matches the central angle formula involving multiplication by 360 degrees.
Each chart parameter in Column A directly corresponds to its standard mathematical definition, formula, or geometric representation in Column B.

Adım Adım Çözüm

1
Identify the statistical definition for continuous class intervals in histograms.
Class boundaries remove gaps between non-overlapping class limits.
Histograms require continuous real class boundaries on the horizontal axis.
2
Recall the formula for histogram bar height when class intervals vary.
Frequency density = FrequencyClass Width\frac{\text{Frequency}}{\text{Class Width}}.
Bar area must remain proportional to frequency.
3
Determine the parameter used to plot an ogive curve.
Cumulative frequency tracks accumulated totals across upper class boundaries.
An ogive represents cumulative distribution.
4
Identify the angular calculation for circular charts.
Sector angle = Class FrequencyTotal Frequency×360∘\frac{\text{Class Frequency}}{\text{Total Frequency}} \times 360^\circ.
A complete pie chart represents 360 degrees.

Anahtar Kavram

Data Representation and Chart Properties
Soru 1250Soru →

Match each physical quantity listed on the left with its corresponding SI unit expressed strictly in terms of fundamental (base) units on the right.

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Öğeler

Electric permittivity of free space (ε0\varepsilon_0)
Magnetic permeability of free space (μ0\mu_0)
Thermal conductivity (kk)
Specific heat capacity (cc)

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Cevap

Electric permittivity of free space maps to kg−1⋅m−3⋅s4⋅A2\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2, Magnetic permeability of free space maps to kg⋅m⋅s−2⋅A−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}, Thermal conductivity maps to kg⋅m⋅s−3⋅K−1\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}, and Specific heat capacity maps to m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Each physical quantity is matched to its exact fundamental unit expression obtained by substituting basic formulas into SI base units (kg\text{kg}, m\text{m}, s\text{s}, A\text{A}, K\text{K}). Electric permittivity resolves to kg−1⋅m−3⋅s4⋅A2\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2, magnetic permeability resolves to kg⋅m⋅s−2⋅A−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}, thermal conductivity resolves to kg⋅m⋅s−3⋅K−1\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}, and specific heat capacity resolves to m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.

Adım Adım Çözüm

1
Derive the base SI units for Electric permittivity of free space (ε0\varepsilon_0).
From Coulomb's Law, F=q1q24πε0r2  ⟹  ε0=q1q24πFr2F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2} \implies \varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}. Expressing terms in SI fundamental units: charge q=A⋅sq = \text{A}\cdot\text{s}, force F=kg⋅m⋅s−2F = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}, distance r=mr = \text{m}. Thus, [ε0]=(A⋅s)2(kg⋅m⋅s−2)⋅m2=kg−1⋅m−3⋅s4⋅A2[\varepsilon_0] = \frac{(\text{A}\cdot\text{s})^2}{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2} = \text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2.
Relating derived electromagnetic quantities to fundamental SI units via governing physical equations.
2
Derive the base SI units for Magnetic permeability of free space (μ0\mu_0).
From the force per unit length between parallel current-carrying conductors, FL=μ0I1I22πr  ⟹  μ0=2πFrI1I2L\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \implies \mu_0 = \frac{2\pi F r}{I_1 I_2 L}. Units: [μ0]=(kg⋅m⋅s−2)⋅mA2⋅m=kg⋅m⋅s−2⋅A−2[\mu_0] = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}}{\text{A}^2\cdot\text{m}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}.
Using Ampère's force law to solve for magnetic permeability in base units.
3
Derive the base SI units for Thermal conductivity (kk).
From Fourier's Law of Heat Conduction, Qt=kAΔTΔx  ⟹  k=Q⋅ΔxtAΔT\frac{Q}{t} = k A \frac{\Delta T}{\Delta x} \implies k = \frac{Q \cdot \Delta x}{t A \Delta T}. Heat energy Q=kg⋅m2⋅s−2Q = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}, time t=st = \text{s}, area A=m2A = \text{m}^2, thickness Δx=m\Delta x = \text{m}, temperature difference ΔT=K\Delta T = \text{K}. Thus, [k]=(kg⋅m2⋅s−2)⋅ms⋅m2⋅K=kg⋅m⋅s−3⋅K−1[k] = \frac{(\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2})\cdot\text{m}}{\text{s}\cdot\text{m}^2\cdot\text{K}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}.
Connecting thermal conduction equations to fundamental mechanics and thermodynamic units.
4
Derive the base SI units for Specific heat capacity (cc).
From Q=mcΔT  ⟹  c=QmΔTQ = m c \Delta T \implies c = \frac{Q}{m \Delta T}. Units: [c]=kg⋅m2⋅s−2kg⋅K=m2⋅s−2⋅K−1[c] = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Applying the defining equation of heat capacity to reduce the unit to fundamental base units.

Anahtar Kavram

Derivation of complex derived SI units from fundamental base units using fundamental physical laws.
Soru 1251Soru →

In the study of electrical discharge through gases and cathode ray behavior, specific physical setups and pressure conditions produce distinct observable phenomena. Match each experimental condition or observation on the left with its corresponding underlying physical mechanism or property on the right.

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Öğeler

Extension of the Crookes dark space to fill the entire discharge tube at approximately 0.01 mmHg0.01\text{ mmHg}
Casting of a sharp shadow when an opaque metal Maltese cross is placed in the path of the rays
Deflection of the beam into a circular arc when passing through a uniform magnetic field directed perpendicularly to its motion
Breakdown of gas column into luminous striations separated by dark spaces at intermediate pressures (~1 mmHg1\text{ mmHg})

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Cevap

1. Extension of the Crookes dark space matches with cessation of gas ionization causing direct glass fluorescence. 2. Maltese cross shadow matches with rectilinear propagation of cathode rays. 3. Magnetic field deflection into circular arc matches with centripetal magnetic Lorentz force (F=qvBF = qvB). 4. Luminous striations match with periodic excitation, ionization, and recombination of gas molecules.
Each matching pair directly connects an observable discharge tube phenomenon with its fundamental physical principle: extreme evacuation (0.01 mmHg0.01\text{ mmHg}) allows unimpeded electron stream travel to fluoresce glass; obstacle shadows confirm rectilinear propagation; transverse magnetic fields induce circular motion via evBevB; and periodic energy exchange of electrons with gas molecules yields striations.

Adım Adım Çözüm

1
Analyze the pressure condition at 0.01 mmHg0.01\text{ mmHg}
At very low pressure (0.01 mmHg0.01\text{ mmHg}), gas collisions drop significantly, allowing cathode rays to reach the tube walls directly, extending the Crookes dark space throughout the tube and exciting glass fluorescence.
Mean free path increases beyond tube dimensions when gas density drops.
2
Analyze ray propagation using obstacle shadow formation
The sharp shadow cast by a Maltese cross demonstrates that cathode rays propagate in straight lines normal to the cathode.
Diffraction is negligible and ray trajectories do not bend around macroscopic obstacles.
3
Evaluate magnetic field interaction with cathode rays
The Lorentz force F=q(v⃗×B⃗)F = q(\vec{v} \times \vec{B}) acts as a centripetal force (evB=mv2revB = \frac{mv^2}{r}), bending the negatively charged particle trajectory into a circle.
Moving electric charges experience magnetic forces perpendicular to velocity.
4
Identify the mechanism behind positive column striations
Striations represent repeating regions of inelastic electron collisions with gas atoms resulting in excitation and emission of light, followed by dark zones where electrons re-accelerate.
Quantized energy transfer during gas excitation creates spatial periodicity in luminescence.

Anahtar Kavram

Physical mechanisms of gaseous conduction across pressure stages and properties of cathode rays
Soru 1252Soru →

Match each physical quantity to its correct SI unit.

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Öğeler

Electric current
Force
Thermodynamic temperature
Energy

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Cevap

Electric current matches Ampere (A); Force matches Newton (N); Thermodynamic temperature matches Kelvin (K); Energy matches Joule (J).
Electric current and thermodynamic temperature are fundamental quantities measured in amperes and kelvins respectively. Force and energy are derived quantities whose units (newton and joule) are combinations of base SI units.

Adım Adım Çözüm

1
Identify fundamental physical quantities and their base SI units.
Electric current is measured in amperes (A), and thermodynamic temperature is measured in kelvins (K). Both are fundamental SI units.
Fundamental units are basic units that are defined independently of other quantities.
2
Identify derived physical quantities and their derived SI units.
Force is measured in newtons (N), and energy is measured in joules (J). Both are derived SI units.
Derived units are obtained by combining fundamental SI base units according to physical equations.

Anahtar Kavram

Classification of fundamental and derived SI units
Tahmini Süre:45s
Soru 1253Soru →

In discharge tube experiments, electrical conduction in gases transitions through distinct physical regimes as the internal gas pressure is progressively reduced. Match each discharge phenomenon with its corresponding physical cause or operational pressure condition.

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Öğeler

Formation of the luminous positive column
Appearance and expansion of the Crookes dark space
Emission of high-velocity cathode rays
Complete cessation of electric current flow

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Cevap

The correct pairings are: (1) Formation of the luminous positive column matches continuous de-excitation and radiative recombination of gas atoms at pressures around 1.0 mmHg1.0\text{ mmHg}; (2) Appearance and expansion of the Crookes dark space matches increase in the electron mean free path at pressures around 0.01 mmHg0.01\text{ mmHg}; (3) Emission of high-velocity cathode rays matches bombardment of the cathode by energetic positive ions releasing secondary electrons below 0.01 mmHg0.01\text{ mmHg}; and (4) Complete cessation of electric current flow matches extensive evacuation below 10−4 mmHg10^{-4}\text{ mmHg} leaving insufficient gas molecules for ionization.
Conduction through gases relies heavily on pressure. At moderate low pressure (1.0 mmHg1.0\text{ mmHg}), excited gas atoms emit light forming the positive column. Decreasing pressure to 0.01 mmHg0.01\text{ mmHg} increases the electron mean free path to produce the Crookes dark space and generates energetic cathode rays through ion bombardment. Extreme evacuation (<10−4 mmHg< 10^{-4}\text{ mmHg}) removes all gas charge carriers, halting electric conduction.

Adım Adım Çözüm

1
Analyze the pressure regime of 1.0 mmHg1.0\text{ mmHg} in a discharge tube.
Identified the positive column as the main luminous region filling most of the tube due to atom excitation and light emission.
At this pressure, gas density is sufficient to undergo repeated inelastic collisions that excite atoms and produce visible glow.
2
Examine the physical origin of the Crookes dark space at 0.01 mmHg0.01\text{ mmHg}.
Understood that lower gas density increases electron mean free path.
Electrons near the cathode travel a longer distance before hitting gas particles, creating a dark gap where collisions do not occur.
3
Determine how cathode rays are emitted at very low pressures.
Linked cathode ray emission to positive ion impact on the cathode surface.
High electric fields accelerate residual positive ions to strike the cathode, causing secondary electron emission.
4
Evaluate the extreme vacuum limit below 10−4 mmHg10^{-4}\text{ mmHg}.
Concluded that current stops when gas particles are virtually absent.
Gases conduct electricity via ion and electron production from collision ionization; eliminating gas molecules prevents charge transport.

Anahtar Kavram

Pressure-dependent regimes of gas conduction and cathode ray generation
Tahmini Süre:2m 0s
Soru 1254Soru →

Match each length measuring instrument listed on the left with its standard precision or suitable measurement application on the right.

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Öğeler

Micrometer screw gauge
Vernier caliper
Metre rule
Tape measure

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Cevap

Micrometer screw gauge matches with precision 0.01 mm0.01\text{ mm} for fine wire diameter; Vernier caliper matches with precision 0.01 cm0.01\text{ cm} for tube diameters; Metre rule matches with precision 0.1 cm0.1\text{ cm} for laboratory lengths; Tape measure matches with flexible long-distance measuring.
Each instrument is correctly matched to its standard least count and characteristic physical measurement task: Micrometer screw gauge (0.01 mm0.01\text{ mm}), Vernier caliper (0.01 cm0.01\text{ cm}), Metre rule (0.1 cm0.1\text{ cm}), and Tape measure for long flexible lengths.

Adım Adım Çözüm

1
Identify the least count and primary application for each measuring tool.
Micrometer screw gauge: 0.01 mm0.01\text{ mm} (wires); Vernier caliper: 0.01 cm0.01\text{ cm} (internal/external tube diameters); Metre rule: 0.1 cm0.1\text{ cm} (standard lab objects); Tape measure: large flexible distances.
Matching instruments requires recalling their resolution (least count) and physical design features.

Anahtar Kavram

Instrument least counts and appropriate selection based on object dimensions.
Soru 1255Soru →

Match each kinetic theory parameter of an ideal gas on the left with its corresponding microscopic physical description on the right.

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Öğeler

Temperature of a gas
Gas pressure on container walls
Root-mean-square (r.m.s.) speed of gas molecules

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Cevap

Temperature corresponds to the measure of average translational kinetic energy; Gas pressure corresponds to the rate of momentum transfer per unit area from wall collisions; Root-mean-square speed corresponds to the square root of the mean of squared speeds.
Temperature is directly linked to the average kinetic energy of molecules, gas pressure arises from wall collisions delivering impulse per unit area, and r.m.s. speed is the square root of mean square velocity.

Adım Adım Çözüm

1
Identify the microscopic origin of temperature.
Temperature represents the average translational kinetic energy of gas molecules.
From the kinetic theory equation 12mv2‾=32kBT\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T, absolute temperature directly measures molecular kinetic energy.
2
Identify the microscopic origin of pressure.
Pressure is caused by molecular collisions with container walls.
Each collision transfers momentum to the wall; force is the time rate of momentum change, and force per unit area defines pressure.
3
Identify the definition of root-mean-square speed.
r.m.s. speed is the square root of the average of squared molecular speeds.
It accounts for the statistical distribution of molecular velocities in a gas sample.

Anahtar Kavram

Microscopic interpretation of macroscopic gas properties via Kinetic Theory of Matter
Soru 1256Soru →

Match each length measuring instrument listed on the left with its standard precision and appropriate physical measurement application on the right.

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Öğeler

Metre rule
Vernier caliper
Micrometer screw gauge
Flexible measuring tape

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Cevabı ve açıklamayı göster

Cevap

Metre rule matches precision 0.1 cm0.1\text{ cm} for pendulum rod; Vernier caliper matches precision 0.01 cm0.01\text{ cm} for tube diameters; Micrometer screw gauge matches precision 0.01 mm0.01\text{ mm} for wire diameter; Flexible measuring tape matches precision 0.1 cm0.1\text{ cm} for flexible/large distances.
Each instrument is matched according to its fundamental physical least count and specialized geometry: Metre rule measures rigid lengths to 0.1 cm0.1\text{ cm}, Vernier caliper measures internal/external diameters to 0.01 cm0.01\text{ cm}, Micrometer screw gauge measures fine dimensions to 0.01 mm0.01\text{ mm}, and flexible measuring tape measures long or curved distances to 0.1 cm0.1\text{ cm}.

Adım Adım Çözüm

1
Identify the least count (precision) of each measuring instrument.
Metre rule: 0.1 cm0.1\text{ cm}, Vernier caliper: 0.01 cm0.01\text{ cm}, Micrometer screw gauge: 0.01 mm0.01\text{ mm}, Measuring tape: 0.1 cm0.1\text{ cm}.
Least count defines the minimum measurable dimension and precision limit for each tool.
2
Match each instrument to its specific physical design feature and suitable application.
Metre rule →\rightarrow rigid straight measurements (pendulum rod); Vernier caliper →\rightarrow internal/external diameters (jaws); Micrometer screw gauge →\rightarrow very thin objects (wire diameter); Tape measure →\rightarrow curved/large dimensions.
The mechanical structure of the instrument dictates its intended physical measurement domain.

Anahtar Kavram

Instrument least count and application domain in length measurement
Soru 1257Soru →

Match each vibrating acoustic system setup on the left with the correct mathematical expression for its resonant frequency (ff) on the right, where vv is the speed of sound in air, LL is the physical length of the pipe or string, ee is the end correction per open end, TT is tension, and μ\mu is linear mass density.

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Öğeler

Fundamental mode of a pipe closed at one end, taking into account end correction
Fundamental mode of a uniform stretched string fixed at both ends
Fundamental mode of a pipe open at both ends, taking into account end corrections at both open ends
First overtone of a pipe closed at one end, neglecting end correction

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Cevap

The fundamental mode of a pipe closed at one end with end correction matches f=v4(L+e)f = \frac{v}{4(L + e)}; the fundamental mode of a stretched string matches f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}; the fundamental mode of a pipe open at both ends with end correction at both ends matches f=v2(L+2e)f = \frac{v}{2(L + 2e)}; and the first overtone of a closed pipe without end correction matches f=3v4Lf = \frac{3v}{4L}.
Each setup corresponds directly to its derived wave equation: closed pipes produce fundamental frequency f=v4(L+e)f = \frac{v}{4(L+e)} for one open end, open pipes produce f=v2(L+2e)f = \frac{v}{2(L+2e)} for two open ends, stretched strings depend on tension and mass per unit length as f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, and the first overtone of a closed pipe is its third harmonic f=3v4Lf = \frac{3v}{4L}.

Adım Adım Çözüm

1
Analyze boundary conditions and effective acoustic length for closed and open pipes.
A closed pipe has one displacement antinode at the open end and one node at the closed end, adding an effective end correction ee to its physical length LL (Leff=L+eL_{\text{eff}} = L + e). An open pipe has two open ends, giving an effective length Leff=L+2eL_{\text{eff}} = L + 2e.
Air displacement antinodes occur slightly outside open pipe boundaries by a distance ee per open end.
2
Derive the frequency formula for the fundamental mode of a closed pipe with end correction.
For the fundamental mode, L+e=λ4  ⟹  λ=4(L+e)L + e = \frac{\lambda}{4} \implies \lambda = 4(L + e). Frequency f=vλ=v4(L+e)f = \frac{v}{\lambda} = \frac{v}{4(L + e)}.
The distance between a node and an adjacent antinode is one-quarter of a wavelength.
3
Derive the fundamental frequency for a stretched string fixed at both ends.
L=λ2  ⟹  λ=2LL = \frac{\lambda}{2} \implies \lambda = 2L. Using wave velocity v=Tμv = \sqrt{\frac{T}{\mu}}, f=v2L=12LTμf = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}.
Nodes exist at both fixed ends in a vibrating string, making the fundamental wavelength twice the length.
4
Derive the fundamental frequency of an open pipe considering both end corrections.
L+2e=λ2  ⟹  λ=2(L+2e)L + 2e = \frac{\lambda}{2} \implies \lambda = 2(L + 2e), so f=v2(L+2e)f = \frac{v}{2(L + 2e)}.
Antinodes occur at both open ends, placing half a wavelength within the effective acoustic length.
5
Determine the first overtone frequency for a closed pipe without end correction.
The first overtone is the third harmonic (n=3n = 3), so L=3λ4  ⟹  λ=4L3L = \frac{3\lambda}{4} \implies \lambda = \frac{4L}{3}, which gives f=3v4Lf = \frac{3v}{4L}.
Closed pipes support only odd integer multiples of the fundamental frequency.

Anahtar Kavram

Standing Waves and Resonance in Air Columns and Strings
Soru 1258Soru →

Match each of the physical quantities given on the left with its corresponding fundamental SI status or base unit resolution on the right.

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Öğeler

Luminous intensity
Linear momentum
Thermodynamic temperature
Specific heat capacity

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Cevap

Luminous intensity matches with Fundamental quantity measured in candela (cd); Linear momentum matches with Derived quantity expressed in kg·m·s⁻¹; Thermodynamic temperature matches with Fundamental quantity measured in kelvin (K); Specific heat capacity matches with Derived quantity expressed in m²·s⁻²·K⁻¹.
Luminous intensity and thermodynamic temperature are fundamental physical quantities with SI units candela (cd\text{cd}) and kelvin (K\text{K}) respectively. Linear momentum (p=mvp = mv) and specific heat capacity (c=QmΔTc = \frac{Q}{m \Delta T}) are derived physical quantities whose fundamental SI base unit resolutions are kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1} and m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1} respectively.

Adım Adım Çözüm

1
Identify fundamental physical quantities and their base SI units
Luminous intensity (measured in cd\text{cd}) and thermodynamic temperature (measured in K\text{K}) are base physical quantities that cannot be expressed in terms of other quantities.
The standard SI system establishes 7 base independent quantities.
2
Resolve derived quantities into fundamental SI base units
Linear momentum (p=mvp = mv) has units kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}. Specific heat capacity (c=QmΔTc = \frac{Q}{m\Delta T}) has units kg⋅m2⋅s−2kg⋅K=m2⋅s−2⋅K−1\frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Derived physical quantities are formed by combining fundamental quantities algebraically according to physical laws.
3
Match each physical quantity to its correct description
Luminous intensity →\rightarrow candela (cd\text{cd}); Linear momentum →\rightarrow kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}; Thermodynamic temperature →\rightarrow kelvin (K\text{K}); Specific heat capacity →\rightarrow m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Each pair directly aligns the physical quantity with its base classification and unit derivation.

Anahtar Kavram

Classification of fundamental and derived physical quantities and their resolution into SI base units
Soru 1259Soru →

Pair each physical quantity in List I with its equivalent SI unit expressed purely in terms of fundamental base units in List II.

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Öğeler

Electric Charge
Linear Momentum
Young's Modulus
Magnetic Flux Density

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Cevap

Electric Charge matches A⋅s\text{A}\cdot\text{s}; Linear Momentum matches kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}; Young's Modulus matches kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}; Magnetic Flux Density matches kg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.
Each physical quantity is resolved to fundamental SI base units using defining equations: Electric charge (Q=ItQ=It) yields A⋅s\text{A}\cdot\text{s}, linear momentum (p=mvp=mv) yields kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}, Young's modulus (stress/strain) yields kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}, and magnetic flux density (B=FILB=\frac{F}{IL}) yields kg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.

Adım Adım Çözüm

1
Derive base units for Electric Charge
A⋅s\text{A}\cdot\text{s}
From Q=ItQ = I t, electric current has the fundamental unit ampere (A) and time has the fundamental unit second (s).
2
Derive base units for Linear Momentum
kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}
From p=mvp = m v, mass is in kilograms (kg) and velocity is in meters per second (m/s).
3
Derive base units for Young's Modulus
kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
Tensile strain is dimensionless. Tensile stress is force per area: Nm2=kg⋅m⋅s−2m2=kg⋅m−1⋅s−2\frac{\text{N}}{\text{m}^2} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}.
4
Derive base units for Magnetic Flux Density
kg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}
From magnetic force F=ILBF = I L B, solving for B=FILB = \frac{F}{I L} gives kg⋅m⋅s−2A⋅m=kg⋅s−2⋅A−1\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{A}\cdot\text{m}} = \text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.

Anahtar Kavram

Expressing derived physical quantities in terms of fundamental SI base units
Soru 1260Soru →

Match each physical quantity listed on the left with its corresponding classification and physical description on the right.

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Öğeler

Electric potential difference
Thermodynamic temperature
Impulse
Luminous intensity

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Cevap

Electric potential difference matches derived quantity defined as work done per unit electric charge; Thermodynamic temperature matches fundamental quantity representing thermal state, measured in kelvins; Impulse matches derived quantity defined as force times time interval; Luminous intensity matches fundamental quantity measuring perceived light power per unit solid angle.
Thermodynamic temperature and luminous intensity are two of the seven fundamental SI quantities. Electric potential difference and impulse are derived quantities because they are expressed through equations involving base physical quantities.

Adım Adım Çözüm

1
Identify fundamental physical quantities
Thermodynamic temperature and luminous intensity are basic independent physical quantities defined by SI standards.
Fundamental quantities cannot be defined in terms of other physical quantities.
2
Identify derived physical quantities and their defining expressions
Electric potential difference (V=WQV = \frac{W}{Q}) and impulse (I=FΔtI = F \Delta t) are derived from basic quantities.
Derived quantities are defined by mathematical combinations of fundamental quantities.

Anahtar Kavram

Fundamental and Derived Quantities
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