Measurements and Units Alıştırma Soruları — JAMB UTME — Sayfa 2

Soru 21Soru →

A beam balance measures the mass of a metal block as $15\text{ kg}$ on Earth. The block is then transported to a planet where the acceleration due to gravity is $3.8\text{ m s}^{-2}$ and suspended from a spring balance calibrated in newtons. If the spring balance has a positive zero error of $+2.0\text{ N}$ (reading $+2.0\text{ N}$ when unloaded), what is the displayed reading on the spring balance in newtons?

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Cevap: 59

Cevap

The displayed reading on the spring balance is

59\text{ N}

.

The beam balance establishes that the block has a constant mass of

15\text{ kg}

. On the planet, the gravitational pull on this mass is

W = 15\text{ kg} \times 3.8\text{ m s}^{-2} = 57\text{ N}

. Because the spring balance reads

+2.0\text{ N}

when unloaded, suspending the block causes the pointer to register

57\text{ N} + 2.0\text{ N} = 59\text{ N}

.

Adım Adım Çözüm

1

Determine the true mass using beam balance principles

Mass

m = 15\text{ kg}

An equal-arm beam balance measures invariant scalar mass independently of local gravitational field strength.

2

Calculate the true weight on the planet

Weight

W = 57\text{ N}

Weight is the force of gravity acting on mass, calculated as

W = mg = 15\text{ kg} \times 3.8\text{ m s}^{-2} = 57\text{ N}

.

3

Incorporate the positive zero error to find the scale pointer reading

Displayed reading =

59\text{ N}

A positive zero error means the scale indicates

+2.0\text{ N}

when no load is attached. Therefore, Scale Reading = Actual Weight + Zero Offset =

57\text{ N} + 2.0\text{ N} = 59\text{ N}

.

Anahtar Kavram

Mass is an intrinsic property measured by a beam balance, whereas weight is a force measured by a spring balance and affected by zero errors.

Tahmini Süre:1m 0s

Soru 22Soru →

Power is defined as the rate at which work is done or energy is transferred. Which of the following expressions represents the correct dimensional formula for power?

M L^2 T^{-3}

M L^2 T^{-2}

M L T^{-3}

M L T^{-2}

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Cevap: $M L^2 T^{-3}$

Cevap

M L^2 T^{-3}

Power is defined as work done per unit time (

P = \frac{W}{t}

). The dimensional formula for work is

[W] = M L^2 T^{-2}

, and for time is

[t] = T

. Dividing work by time gives

[P] = \frac{M L^2 T^{-2}}{T} = M L^2 T^{-3}

.

Adım Adım Çözüm

1

Write the fundamental definition of power in terms of work and time.

\text{Power} = \frac{\text{Work}}{\text{Time}}

By definition, power measures the rate of doing work.

2

Determine the dimensions of work.

[\text{Work}] = [\text{Force}] \times [\text{Distance}] = (M L T^{-2}) \times L = M L^2 T^{-2}

Force has dimensions of mass times acceleration (

M L T^{-2}

), and multiplying by distance (

L

) yields energy or work dimensions.

3

Divide the dimensions of work by the dimension of time (

T

).

[\text{Power}] = \frac{M L^2 T^{-2}}{T} = M L^2 T^{-3}

Dividing by time increases the negative exponent of time from

-2

to

-3

.

Anahtar Kavram

Dimensional Formula for Power

Soru 23Soru →

Match each physical quantity on the left with its correct fundamental SI base unit expression or fundamental status on the right.

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Cevap

Thermodynamic temperature matches with 'Fundamental physical quantity measured in kelvin (K)'; Electric charge matches with 'Derived physical quantity expressed in base SI units as A·s'; Linear momentum matches with 'Derived physical quantity expressed in base SI units as kg·m·s⁻¹'; Power matches with 'Derived physical quantity expressed in base SI units as kg·m²·s⁻³'.

Each physical quantity is correctly paired with either its fundamental status or its base SI unit breakdown derived from core physics definitions.

Adım Adım Çözüm

1

Identify fundamental quantities versus derived quantities.

Thermodynamic temperature is a basic fundamental quantity (unit:

\text{K}

). Electric current is fundamental (unit:

\text{A}

), but electric charge is derived (

Q = I t

).

Fundamental quantities cannot be defined in terms of other physical quantities.

2

Decompose Electric Charge into base units.

Since

Q = I \cdot t

, its unit is

\text{A}\cdot\text{s}

.

Current is measured in amperes and time in seconds.

3

Decompose Linear Momentum into base units.

p = m \cdot v \Rightarrow \text{unit} = \text{kg} \cdot \text{m}\cdot\text{s}^{-1}

.

Mass is in kilograms and velocity is in meters per second.

4

Decompose Power into base units.

P = \frac{W}{t} = \frac{F \cdot d}{t} = \frac{(m \cdot a) \cdot d}{t} \Rightarrow \frac{\text{kg} \cdot (\text{m}\cdot\text{s}^{-2}) \cdot \text{m}}{\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

.

Power is work done per unit time.

Anahtar Kavram

Fundamental and Derived Quantities

Soru 24Soru →

A laboratory technician records four physical quantities during an experiment: electric current, thermodynamic temperature, electric charge, and luminous intensity. Which of these recorded quantities is a derived physical quantity?

Electric charge

Electric current

Thermodynamic temperature

Luminous intensity

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Cevap: Electric charge

Cevap

Electric charge is the derived physical quantity.

Electric charge is a derived physical quantity because it is defined mathematically in terms of fundamental quantities: electric charge equals electric current multiplied by time (

Q = I \cdot t

). Its SI unit, the coulomb (C), is defined as an ampere-second (

1\text{ C} = 1\text{ A}\cdot\text{s}

).

Adım Adım Çözüm

1

Identify the seven fundamental physical quantities defined in the SI system.

The seven base quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Fundamental quantities are independent base quantities from which all other physical quantities are derived.

2

Analyze each quantity given in the scenario against the list of fundamental quantities.

Electric current, thermodynamic temperature, and luminous intensity are fundamental quantities. Electric charge is not on the list of base quantities.

Electric charge (

Q

) is derived from the base quantities electric current (

I

) and time (

t

) through the relation

Q = I \cdot t

.

Anahtar Kavram

Fundamental quantities are basic physical quantities that do not depend on other quantities. Derived quantities are defined in terms of fundamental quantities.

Soru 25Soru →

Which of the following expressions represents the correct dimensions of pressure?

M L^{-1} T^{-2}

M L T^{-2}

M L^2 T^{-2}

M L^{-2} T^{-2}

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Cevap: $M L^{-1} T^{-2}$

Cevap

M L^{-1} T^{-2}

Pressure is defined as Force divided by Area. Substituting the fundamental dimensions gives Force as

M L T^{-2}

and Area as

L^2

. Dividing Force by Area results in

\frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2}

. Therefore, the expression

M L^{-1} T^{-2}

is correct.

Adım Adım Çözüm

1

State the physical definition formula for pressure

Pressure

P = \frac{\text{Force}}{\text{Area}}

Pressure is defined as force applied perpendicular to a surface per unit area.

2

Substitute fundamental dimensions for force and area

[P] = \frac{[M L T^{-2}]}{[L^2]}

Force has dimensions

M L T^{-2}

and area has dimensions

L^2

.

3

Simplify the index of length

L

[P] = M L^{1 - 2} T^{-2} = M L^{-1} T^{-2}

Applying the rules of indices for length gives

L^1 \cdot L^{-2} = L^{-1}

.

Anahtar Kavram

Dimensions of Pressure

Soru 26Soru →

Match each physical quantity listed on the left with its corresponding SI classification and base unit representation on the right.

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Cevap

Electric current matches Fundamental quantity measured in amperes; Thermodynamic temperature matches Fundamental quantity measured in kelvins; Electric potential difference matches Derived quantity expressed as kg·m²·s⁻³·A⁻¹; Specific heat capacity matches Derived quantity expressed as m²·s⁻²·K⁻¹.

Electric current and thermodynamic temperature are basic SI fundamental quantities. Electric potential difference and specific heat capacity are derived quantities whose fundamental unit decompositions follow directly from their governing formulas.

Adım Adım Çözüm

1

Identify fundamental physical quantities and their base units

Electric current and thermodynamic temperature are fundamental SI quantities with base units ampere (

\text{A}

) and kelvin (

\text{K}

) respectively.

Fundamental physical quantities are defined independently and serve as the foundation for the SI system.

2

Decompose electric potential difference into SI base units

Electric potential difference

V = \frac{\text{Work}}{\text{Charge}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

Because it is expressed by combining fundamental quantities, it is a derived quantity.

3

Decompose specific heat capacity into SI base units

Specific heat capacity

c = \frac{\text{Energy}}{\text{Mass} \times \text{Temperature change}} = \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}

.

It is calculated from energy, mass, and temperature, making it a derived quantity.

Anahtar Kavram

Fundamental quantities are independent basic quantities, whereas derived quantities are formed through algebraic combination of fundamental quantities.

Tahmini Süre:1m 30s

Soru 27Soru →

Match each physical quantity to its correct classification as either a fundamental or a derived physical quantity.

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Cevap

Luminous intensity matches with fundamental quantity measuring light brightness; Mass matches with fundamental quantity measuring quantity of matter; Force matches with derived quantity defined as rate of change of linear momentum; Electric potential matches with derived quantity defined as work done per unit electric charge.

Luminous intensity and mass are two of the seven base SI quantities. Force and electric potential are derived quantities defined through mathematical combinations of base quantities.

Adım Adım Çözüm

1

Identify the fundamental physical quantities

Luminous intensity and Mass are fundamental physical quantities.

Fundamental physical quantities are basic quantities that do not depend on any other physical quantity for their definition.

2

Identify the derived physical quantities

Force and Electric potential are derived physical quantities.

Derived physical quantities are obtained by combining fundamental physical quantities through mathematical relationships.

Anahtar Kavram

Fundamental quantities (length, mass, time, electric current, thermodynamic temperature, amount of substance, luminous intensity) are independent, whereas derived quantities are formed by combining fundamental quantities.

Soru 28Soru →

Consider the following four physical quantities:
I. Electric current
II. Electric charge
III. Thermodynamic temperature
IV. Heat capacity

Which of the following pairs correctly identifies a fundamental physical quantity followed by a derived physical quantity?

Electric current and Thermodynamic temperature

Electric charge and Heat capacity

Thermodynamic temperature and Heat capacity

Heat capacity and Electric current

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Cevap: Thermodynamic temperature and Heat capacity

Cevap

Thermodynamic temperature and Heat capacity

Thermodynamic temperature is one of the seven base SI fundamental quantities. Heat capacity is a derived quantity defined as thermal energy per temperature change (

J/K

). Thus, the pair of Thermodynamic temperature and Heat capacity strictly follows the order of a fundamental quantity followed by a derived quantity.

Adım Adım Çözüm

1

Identify the fundamental physical quantities among the given list.

Electric current (I) and Thermodynamic temperature (III) are base SI quantities, so they are fundamental.

Fundamental quantities are independent physical quantities that cannot be defined in terms of other quantities.

2

Identify the derived physical quantities among the given list.

Electric charge (II, defined by

Q = I \cdot t

) and Heat capacity (IV, defined by

C = \frac{Q}{\Delta T}

) are derived quantities.

Derived quantities are defined mathematically from combinations of fundamental physical quantities.

3

Select the option that lists a fundamental quantity followed by a derived quantity.

Thermodynamic temperature (fundamental) and Heat capacity (derived) matches the required order.

Thermodynamic temperature is fundamental, and heat capacity is derived.

Anahtar Kavram

Classification of physical quantities into base (fundamental) and derived quantities in the SI system.

Soru 29Soru →

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

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Luminous intensity matches with fundamental quantity in

\text{cd}

; Electric potential matches with derived quantity in

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

; Specific heat capacity matches with derived quantity in

\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

; Thermodynamic temperature matches with fundamental quantity in

\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

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

and

\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 = \frac{W}{q}

yields

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

. Specific heat capacity

c = \frac{Q}{m\Delta T}

yields

\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 30Soru →

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

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Electric potential gradient matches

\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{A}^{-1}

; Coefficient of dynamic viscosity matches

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}

; Ratio of Planck's constant to moment of inertia matches

\text{s}^{-1}

; Specific latent heat divided by spatial temperature gradient matches

\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

\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

\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 31Soru →

The derived SI unit of dynamic viscosity, the pascal-second ( $\text{Pa}\cdot\text{s}$ ), can be expressed in fundamental SI base units as $\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c$ . Determine the numerical value of the exponent of length, $b$ .

Cevabı ve açıklamayı göster

Cevap: -1

Cevap

The numerical value of the exponent of length

b

is

-1

.

Dynamic viscosity is measured in pascal-seconds (

\text{Pa}\cdot\text{s}

). Substituting

1\text{ Pa} = 1\text{ N/m}^2 = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

into the formula gives

1\text{ Pa}\cdot\text{s} = 1\text{ kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}

. Comparing this with

\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c

, the exponent of length (meters) is

b = -1

.

Adım Adım Çözüm

1

Express the newton in base SI units using

F = ma

.

\text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

.

Force is the product of mass and acceleration.

2

Derive the SI base unit expression for pressure (pascal,

\text{Pa}

).

\text{Pa} = \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}

.

Pressure is defined as force per unit area.

3

Multiply the base unit expression of pressure by seconds.

\text{Pa}\cdot\text{s} = (\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}) \cdot \text{s}^1 = \text{kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}

.

Dynamic viscosity is measured in pascal-seconds.

4

Extract the exponent of length (

b

) corresponding to the meter unit.

b = -1

.

The power of meters (

\text{m}

) in

\text{kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}

is

-1

.

Anahtar Kavram

Deriving base SI units for derived physical quantities

Tahmini Süre:1m 30s

Soru 32Soru →

A physical quantity $Z$ is defined by the expression $Z = \frac{P \cdot V}{I \cdot t}$ , where $P$ represents pressure, $V$ represents volume, $I$ represents electric current, and $t$ represents time. Which of the following statements correctly classifies quantity $Z$ and expresses it in fundamental SI base units?

Quantity

Z

is a derived quantity with fundamental SI base units of

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

Quantity

Z

is a fundamental quantity with fundamental SI base units of

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

Quantity

Z

is a derived quantity with fundamental SI base units of

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-1}

.

Quantity

Z

is a fundamental quantity with fundamental SI base units of

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-1}

.

Cevabı ve açıklamayı göster

Cevap: Quantity $Z$ is a derived quantity with fundamental SI base units of $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}$ .

Cevap

Quantity

Z

is a derived quantity with fundamental SI base units of

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

The quantity

Z

represents electric potential (voltage), which is defined as work done per unit electric charge (

W/Q

). Since electric potential is derived from mass, length, time, and electric current, it is a derived physical quantity. Expanding

P \cdot V

yields energy units (

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

), and dividing by

I \cdot t

(

\text{A}\cdot\text{s}

) gives the correct base units of

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

Adım Adım Çözüm

1

Determine the SI base unit decomposition for pressure (

P

) and volume (

V

).

Pressure

P = \frac{\text{Force}}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

. Volume

V = \text{m}^3

. Therefore,

P \cdot V = (\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}) \cdot \text{m}^3 = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

.

Decomposing composite physical quantities into base units of mass, length, and time.

2

Determine the SI base unit decomposition for the denominator

I \cdot t

.

I \cdot t = \text{Electric current} \times \text{Time} = \text{A}\cdot\text{s}

.

Electric current (amperes, A) and time (seconds, s) are fundamental SI quantities.

3

Evaluate the full expression for

Z = \frac{P \cdot V}{I \cdot t}

in SI base units.

Base units of

Z = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

.

Apply laws of indices to simplify base unit ratios.

4

Classify physical quantity

Z

.

Because

Z

(electric potential) is defined in terms of fundamental quantities (kg, m, s, A) rather than existing independently, it is a derived quantity.

The seven fundamental SI quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Anahtar Kavram

Classification of physical quantities into fundamental and derived types and resolving complex derived quantities into base SI units.

Soru 33Soru →

In atomic physics, the energy $E$ of a photon is related to its frequency $f$ by the formula $E = h f$ , where $h$ is Planck's constant. What are the fundamental dimensions of Planck's constant $h$ ?

M L^2 T^{-1}

M L^2 T^{-2}

M L T^{-1}

M L^2 T^{-3}

Cevabı ve açıklamayı göster

Cevap: $M L^2 T^{-1}$

Cevap

The fundamental dimensions of Planck's constant are

M L^2 T^{-1}

.

Planck's constant

h

is given by

h = E / f

. Since energy

E

has fundamental dimensions of

M L^2 T^{-2}

and frequency

f

has dimensions of

T^{-1}

, dividing energy by frequency yields

\frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-1}

.

Adım Adım Çözüm

1

Express Planck's constant in terms of energy and frequency

h = \frac{E}{f}

Rearranging the equation

E = h f

isolates

h

.

2

Determine the fundamental dimensions of energy (

E

) and frequency (

f

)

[E] = M L^2 T^{-2}

and

[f] = T^{-1}

Energy has dimensions of work (

F \times d = M L T^{-2} \cdot L = M L^2 T^{-2}

) and frequency is inverse time (

T^{-1}

).

3

Divide the dimensions of energy by the dimensions of frequency

[h] = \frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-2 - (-1)} = M L^2 T^{-1}

Applying the rules of indices simplifies the exponent of time.

Anahtar Kavram

Dimensional Analysis of Physical Constants

Tahmini Süre:1m 0s

Soru 34Soru →

The derived SI unit of electrical resistance is the ohm ( $\Omega$ ). Which of the following correctly expresses the ohm strictly in terms of fundamental SI base units?

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}

\text{V}\cdot\text{A}^{-1}

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

Cevabı ve açıklamayı göster

Cevap: $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}$

Cevap

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}

Resistance is defined by Ohm's Law as voltage divided by current (

R = V / I

). Voltage is work done per unit charge (

V = W / Q

), and charge is current multiplied by time (

Q = I \cdot t

). Substituting the unit of work (

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

) gives

R = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}^2\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}

.

Adım Adım Çözüm

1

Relate resistance to potential difference and current using Ohm's Law.

R = \frac{V}{I}

Resistance is defined as the ratio of potential difference to electric current.

2

Express electric potential difference (

V

) in terms of work (

W

) and charge (

Q

), where

Q = I \cdot t

.

V = \frac{W}{I \cdot t} \implies R = \frac{W}{I^2 \cdot t}

Electric potential difference is work done per unit charge, and electric charge is current multiplied by time.

3

Break down work (

W = \text{Force} \times \text{distance}

) into base SI units.

Unit of Work =

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

Force is mass times acceleration (

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

), so work is

\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

.

4

Substitute all base units into the resistance formula.

Unit of

R = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}^2 \cdot \text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}

Combining powers of base units yields the complete fundamental SI base unit expression for the ohm.

Anahtar Kavram

Derivation of SI base units from defining formulas of physical quantities

Soru 35Soru →

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

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Pressure matches with

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

, Surface Tension matches with

\text{kg}\cdot\text{s}^{-2}

, Specific Heat Capacity matches with

\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

, and Electric Capacitance matches with

\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 = \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

\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 = \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 = \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 36Soru →

The rate of change of linear momentum per unit cross-sectional area is a derived physical quantity. When resolved into fundamental SI base units, which of the following physical quantities has the exact same SI base unit decomposition?

Surface tension

Energy density

Power

Specific heat capacity

Cevabı ve açıklamayı göster

Cevap: Energy density

Cevap

Energy density

The rate of change of momentum is force (

F

), which has SI base units of

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

. Dividing by area (

\text{m}^2

) gives

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

. Energy density is defined as energy per unit volume, which decomposes to

\frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

. Both derived quantities possess identical fundamental SI base unit representations.

Adım Adım Çözüm

1

Determine the physical definition of the rate of change of linear momentum per unit area.

By Newton's second law, rate of change of momentum equals force (

F = \frac{\Delta p}{\Delta t}

). Dividing by cross-sectional area (

A

) yields force per unit area (pressure or stress),

\frac{F}{A}

.

Force is defined as the time rate of change of linear momentum.

2

Express force per unit area in terms of fundamental SI base units.

\text{Force} = \text{mass} \times \text{acceleration} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

. Therefore,

\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}

.

Mass (

\text{kg}

), length (

\text{m}

), and time (

\text{s}

) are fundamental base quantities.

3

Analyze energy density in fundamental SI base units.

\text{Energy density} = \frac{\text{Energy}}{\text{Volume}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Decomposing energy into work (force

\times

distance) and dividing by volume reveals its base unit equivalence.

Anahtar Kavram

Decomposition of derived quantities into fundamental SI base units

Tahmini Süre:1m 30s

Soru 37Soru →

In the International System of Units (SI), physical quantities are categorized as either fundamental or derived. Which of the following is a fundamental SI base unit?

Joule

Kelvin

Pascal

Watt

Cevabı ve açıklamayı göster

Cevap: Kelvin

Cevap

Kelvin is the fundamental SI base unit.

The kelvin is the standard SI base unit for measuring thermodynamic temperature and is one of the seven defined fundamental units.

Adım Adım Çözüm

1

Identify the seven SI fundamental (base) units

The seven fundamental SI base units are the meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd).

Fundamental units are basic units that are independent of one another and cannot be expressed in terms of other units.

2

Compare the given options against the set of fundamental units

Among Joule, Kelvin, Pascal, and Watt, only Kelvin is in the list of fundamental SI base units.

Joule, Pascal, and Watt are all derived units defined by combinations of base units.

Anahtar Kavram

Fundamental SI Base Units

Tahmini Süre:35s

Soru 38Soru →

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

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Frequency matches

\text{s}^{-1}

, Electric charge matches

\text{A}\cdot\text{s}

, Mass density matches

\text{kg}\cdot\text{m}^{-3}

, and Acceleration matches

\text{m}\cdot\text{s}^{-2}

.

Frequency (

f = 1/T

) is expressed in reciprocal seconds (

\text{s}^{-1}

). Electric charge (

Q = I \cdot t

) is current times time, giving

\text{A}\cdot\text{s}

. Mass density (

\rho = m/V

) is mass per volume, giving

\text{kg}\cdot\text{m}^{-3}

. Acceleration (

a = \Delta v / \Delta t

) is rate of velocity change, giving

\text{m}\cdot\text{s}^{-2}

.

Adım Adım Çözüm

1

Identify the defining formula for each physical quantity

Frequency

f = \frac{1}{T}

, Charge

Q = I \cdot t

, Density

\rho = \frac{m}{V}

, Acceleration

a = \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 (

\text{kg}

), length (

\text{m}

), time (

\text{s}

), and current (

\text{A}

)

Frequency:

\text{s}^{-1}

; Charge:

\text{A}\cdot\text{s}

; Density:

\frac{\text{kg}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-3}

; Acceleration:

\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

\rightarrow \text{s}^{-1}

, Electric charge

\rightarrow \text{A}\cdot\text{s}

, Mass density

\rightarrow \text{kg}\cdot\text{m}^{-3}

, Acceleration

\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 39Soru →

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

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Electric permittivity of free space maps to

\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2

, Magnetic permeability of free space maps to

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}

, Thermal conductivity maps to

\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}

, and Specific heat capacity maps to

\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 (

\text{kg}

,

\text{m}

,

\text{s}

,

\text{A}

,

\text{K}

). Electric permittivity resolves to

\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2

, magnetic permeability resolves to

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}

, thermal conductivity resolves to

\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}

, and specific heat capacity resolves to

\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 (

\varepsilon_0

).

From Coulomb's Law,

F = \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 = \text{A}\cdot\text{s}

, force

F = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

, distance

r = \text{m}

. Thus,

[\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 (

\mu_0

).

From the force per unit length between parallel current-carrying conductors,

\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:

[\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 (

k

).

From Fourier's Law of Heat Conduction,

\frac{Q}{t} = k A \frac{\Delta T}{\Delta x} \implies k = \frac{Q \cdot \Delta x}{t A \Delta T}

. Heat energy

Q = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}

, time

t = \text{s}

, area

A = \text{m}^2

, thickness

\Delta x = \text{m}

, temperature difference

\Delta T = \text{K}

. Thus,

[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 (

c

).

From

Q = m c \Delta T \implies c = \frac{Q}{m \Delta T}

. Units:

[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 40Soru →

Match each physical quantity to its correct SI unit.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

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

Measurements and Units

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