Measurements and Units

112 soru

Soru 101Soru

The coefficient of dynamic viscosity η\eta of a fluid can be defined using Newton's formula for viscous flow, η=FAΔvΔx\eta = \frac{F}{A \frac{\Delta v}{\Delta x}}, where FF is the viscous drag force, AA is the contact surface area, and ΔvΔx\frac{\Delta v}{\Delta x} is the velocity gradient perpendicular to the flow. Which of the following combinations of fundamental SI base units is equivalent to the coefficient of dynamic viscosity?

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Cevap: kgm1s1\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}

Cevap

The coefficient of dynamic viscosity expressed in fundamental SI base units is kgm1s1\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}.
The unit of dynamic viscosity is derived from η=FA(Δv/Δx)\eta = \frac{F}{A (\Delta v / \Delta x)}. Substituting base units yields kgms2m2s1=kgm1s1\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2 \cdot \text{s}^{-1}} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}, which correctly represents the unit in base SI quantities.

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1
Identify the fundamental SI base units for each quantity in the formula
Force FF has units kgms2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}; Area AA has units m2\text{m}^2; Velocity gradient ΔvΔx\frac{\Delta v}{\Delta x} has units ms1m=s1\frac{\text{m}\cdot\text{s}^{-1}}{\text{m}} = \text{s}^{-1}.
Decomposing derived physical quantities into SI base units is essential for dimensional reduction.
2
Substitute these fundamental units into the given formula for η\eta
Unit of η=kgms2m2s1\text{Unit of } \eta = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2 \cdot \text{s}^{-1}}.
Direct algebraic substitution yields the compound unit expression.
3
Simplify the powers of mass, length, and time
kg1m12s2(1)=kgm1s1\text{kg}^1 \cdot \text{m}^{1 - 2} \cdot \text{s}^{-2 - (-1)} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}.
Applying exponent rules simplifies the expression to base units.

Anahtar Kavram

Expressing derived units in terms of fundamental SI base units (kilogram, meter, second)
Soru 102Soru

The dynamic pressure PP exerted by a moving fluid depends on its density ρ\rho and flow velocity vv according to the relationship P=kρavbP = k \rho^a v^b, where kk is a dimensionless constant. What is the value of the exponent bb?

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Cevap: 2

Cevap

The value of the exponent bb is 2.
Equating the exponent of time (T) on both sides of the dimensional equation M L1T2=MaL3a+bTb\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b} yields 2=b-2 = -b, giving b=2b = 2.

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1
Determine the base dimensions of all physical quantities in the equation.
Pressure [P]=M L1T2[P] = \text{M L}^{-1} \text{T}^{-2}, Density [ρ]=M L3[\rho] = \text{M L}^{-3}, and Velocity [v]=L T1[v] = \text{L T}^{-1}.
Dimensional analysis requires converting derived physical quantities into fundamental dimensions of Mass (M), Length (L), and Time (T).
2
Apply dimensional homogeneity by substituting the dimensions into P=kρavbP = k \rho^a v^b.
\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b}
The principle of dimensional homogeneity states that exponents of M, L, and T must match on both sides of a physically correct equation.
3
Equate the corresponding exponents for time (T) and solve for bb.
-2 = -b \implies b = 2
Matching the powers of T directly yields the numerical value of exponent bb.

Anahtar Kavram

Principle of Dimensional Homogeneity and Dimensional Analysis
Soru 103Soru

Newton's law of universal gravitation states that the gravitational force FF between two point masses m1m_1 and m2m_2 separated by a distance rr is given by F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}, where GG is the universal gravitational constant. What is the dimensional formula of GG?

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Cevap: M1L3T2M^{-1} L^3 T^{-2}

Cevap

The dimensional formula of the universal gravitational constant GG is M1L3T2M^{-1} L^3 T^{-2}.
Rearranging F=Gm1m2r2F = \frac{G m_1 m_2}{r^2} gives G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. Substituting the dimensions of Force (MLT2M L T^{-2}), distance squared (L2L^2), and mass squared (M2M^2) gives (MLT2)(L2)M2=M1L3T2\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}.

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1
Rearrange the gravitational force equation to solve for GG.
G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}
Isolating the physical constant allows us to substitute the dimensions of each constituent quantity.
2
Substitute the fundamental dimensions for force, distance, and mass.
[F]=MLT2[F] = M L T^{-2}, [r]=L[r] = L, [m1]=[m2]=M[m_1] = [m_2] = M
Force is mass times acceleration (MLT2M \cdot L T^{-2}), distance is length (LL), and mass is [M][M].
3
Substitute these fundamental dimensions into the expression for GG and simplify the exponents.
[G]=(MLT2)(L2)M2=M12L1+2T2=M1L3T2[G] = \frac{(M L T^{-2}) (L^2)}{M^2} = M^{1-2} L^{1+2} T^{-2} = M^{-1} L^3 T^{-2}
Applying standard exponent rules yields the final dimensional formula.

Anahtar Kavram

Deriving Dimensions of Physical Constants
Soru 104Soru

A Vernier caliper with a least count of 0.01 cm0.01\text{ cm} is used to measure the internal diameter of a cylindrical pipe. When the jaws are fully closed without any object between them, the zero mark of the Vernier scale lies to the right of the zero mark on the main scale, and the 3rd3\text{rd} Vernier division coincides with a main scale division. During measurement, the main scale reads 2.40 cm2.40\text{ cm} and the 6th6\text{th} Vernier division coincides with a main scale mark. What is the corrected internal diameter of the pipe?

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Cevap: 2.43 cm2.43\text{ cm}

Cevap

The corrected internal diameter of the pipe is 2.43 cm2.43\text{ cm}.
The Vernier caliper has a positive zero error because the zero mark on the Vernier scale lies to the right of the zero mark on the main scale when closed. The magnitude of this error is 3×0.01 cm=+0.03 cm3 \times 0.01\text{ cm} = +0.03\text{ cm}. The observed reading is 2.40 cm+(6×0.01 cm)=2.46 cm2.40\text{ cm} + (6 \times 0.01\text{ cm}) = 2.46\text{ cm}. Subtracting the zero error from the observed reading yields Corrected reading=2.46 cm0.03 cm=2.43 cm\text{Corrected reading} = 2.46\text{ cm} - 0.03\text{ cm} = 2.43\text{ cm}.

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1
Determine the zero error of the Vernier caliper.
Zero error =+(3×0.01 cm)=+0.03 cm= + (3 \times 0.01\text{ cm}) = +0.03\text{ cm}.
Since the zero of the Vernier scale lies to the right of the main scale zero mark, the instrument has a positive zero error.
2
Calculate the uncorrected (observed) reading.
\text{Observed reading} = 2.40\text{ cm} + (6 \times 0.01\text{ cm}) = 2.46\text{ cm}$.
The observed reading is the sum of the main scale reading and the Vernier scale coincidence value.
3
Apply the zero error correction formula.
\text{Corrected reading} = 2.46\text{ cm} - (+0.03\text{ cm}) = 2.43\text{ cm}$.
Corrected reading is given by subtracting the zero error (including its sign) from the observed reading.

Anahtar Kavram

Zero Error Correction in Length Measurement Instruments
Soru 105Soru

The force constant (stiffness) kk of a helical spring measures its resistance to elastic deformation and is defined by the relationship k=Fek = \frac{F}{e}, where FF represents the restoring force and ee represents the extension. Which of the following represents the SI unit of kk expressed strictly in terms of fundamental SI base units?

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Cevap: kgs2\text{kg} \cdot \text{s}^{-2}

Cevap

kgs2\text{kg} \cdot \text{s}^{-2}
The force constant kk is defined as force per unit extension (k=Fek = \frac{F}{e}). Since force has base units of kgms2\text{kg} \cdot \text{m} \cdot \text{s}^{-2} and extension has base units of m\text{m}, dividing force by extension yields kgms2m=kgs2\frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{m}} = \text{kg} \cdot \text{s}^{-2}.

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1
Identify the defining formula and constituent quantities.
The force constant formula is k=Fek = \frac{F}{e}, where force FF is measured in newtons (N\text{N}) and extension ee is measured in metres (m\text{m}).
Determining SI base units requires substituting base dimensions into the governing physical equation.
2
Express force in fundamental SI base units using Newton's second law (F=maF = ma).
1 N=1 kgms21\text{ N} = 1\text{ kg} \cdot \text{m} \cdot \text{s}^{-2}.
Mass has fundamental unit kg\text{kg}, and acceleration has fundamental unit ms2\text{m} \cdot \text{s}^{-2}.
3
Substitute the base unit expression of force into the formula for kk and simplify.
\text{SI unit of } k = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{m}} = \text{kg} \cdot \text{s}^{-2}.
The unit of length (m\text{m}) in the numerator cancels out with the unit of length in the denominator.

Anahtar Kavram

Deriving SI units of physical quantities from fundamental base units using defining equations.
Soru 106Soru

A chemical balance is used to determine the mass of a metallic sphere, giving a reading of 6.0 kg6.0\text{ kg}. The sphere is then suspended from a spring balance at a laboratory where the local acceleration due to gravity is 10 m s210\text{ m s}^{-2}. If the spring balance reads 3.0 N-3.0\text{ N} before any load is attached, what is the indicated reading on the spring balance when the sphere is suspended?

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Cevap: 57.0 N57.0\text{ N}

Cevap

The indicated reading on the spring balance is 57.0 N57.0\text{ N}.
The true weight of the metallic sphere is calculated as W=mg=6.0 kg×10 m s2=60.0 NW = mg = 6.0\text{ kg} \times 10\text{ m s}^{-2} = 60.0\text{ N}. Because the spring balance has an initial zero reading of 3.0 N-3.0\text{ N} before any load is attached, the indicated reading when the sphere is hung is 60.0 N3.0 N=57.0 N60.0\text{ N} - 3.0\text{ N} = 57.0\text{ N}.

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1
Calculate the true weight of the metallic sphere
W=mg=6.0 kg×10 m s2=60.0 NW = mg = 6.0\text{ kg} \times 10\text{ m s}^{-2} = 60.0\text{ N}
Weight is the gravitational force acting on mass.
2
Set up the zero error relationship for instrument measurement
\text{Actual Weight} = \text{Indicated Reading} - \text{Zero Reading}
The actual physical value equals the observed pointer reading minus the zero reading of the unloaded instrument.
3
Substitute known values and solve for the indicated reading (RR)
60.0 N=R(3.0 N)    R=60.03.0=57.0 N60.0\text{ N} = R - (-3.0\text{ N}) \implies R = 60.0 - 3.0 = 57.0\text{ N}
Solving the linear relation yields an indicated pointer position of 57.0 N57.0\text{ N}.

Anahtar Kavram

Mass versus Weight Measurement and Zero Error Correction
Tahmini Süre:1m 0s
Soru 107Soru

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

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Cevap: 5.0 kg5.0\text{ kg}

Cevap

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

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

Anahtar Kavram

Mass vs. Weight and Instrument Principles
Tahmini Süre:1m 0s
Soru 108Soru

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

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Cevap: 3.545 cm3.545\text{ cm}

Cevap

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

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

Anahtar Kavram

Measurement of length using Vernier calipers and zero error correction
Tahmini Süre:1m 30s
Soru 109Soru

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

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Cevap: 32.4 kg32.4\text{ kg}

Cevap

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

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

Anahtar Kavram

Mass is an intrinsic property measured independently of gravity by a beam balance, whereas weight is a force measured by a spring balance, causing mass-calibrated spring balances to give different readings under varying local gravitational accelerations.
Tahmini Süre:1m 0s
Soru 110Soru

A spring balance suspended inside an elevator reads 48 N48\text{ N} when supporting an object while the elevator accelerates downwards at 2 m s22\text{ m s}^{-2}. Taking acceleration due to gravity g=10 m s2g = 10\text{ m s}^{-2}, what is the mass of the object as measured by an equal-arm beam balance?

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Cevap: 6.0 kg6.0\text{ kg}

Cevap

The mass of the object measured by an equal-arm beam balance is 6.0 kg6.0\text{ kg}.
In a downward accelerating elevator, the apparent weight indicated by a spring balance is Wapp=m(ga)W_{app} = m(g - a). Substituting 48 N=m(10 m s22 m s2)=8m48\text{ N} = m(10\text{ m s}^{-2} - 2\text{ m s}^{-2}) = 8m gives a mass of 6.0 kg6.0\text{ kg}. Because a beam balance compares masses directly and both sides experience the exact same effective acceleration, it measures true mass regardless of the frame's acceleration.

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1
Formulate the equation for apparent weight in a downward accelerating reference frame.
Wapp=m(ga)W_{app} = m(g - a)
When an elevator accelerates downward, the effective acceleration experienced by an object inside is (ga)(g - a).
2
Substitute the given values (Wapp=48 NW_{app} = 48\text{ N}, g=10 m s2g = 10\text{ m s}^{-2}, a=2 m s2a = 2\text{ m s}^{-2}) into the formula to find mass mm.
48=m(102)    48=8m    m=6.0 kg48 = m(10 - 2) \implies 48 = 8m \implies m = 6.0\text{ kg}
This yields the true scalar mass of the object.
3
Determine the reading on an equal-arm beam balance.
The beam balance reads 6.0 kg6.0\text{ kg}.
A beam balance compares unknown mass against standard counter-weights; both experience identical effective acceleration, making its measurement independent of motion or local gravity.

Anahtar Kavram

Apparent weight in an accelerating frame vs invariant mass measurement using a beam balance
Tahmini Süre:1m 0s
Soru 111Soru

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

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Metre rule
Vernier caliper
Micrometer screw gauge
Measuring tape

Eşleşmeler

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Cevap

Metre rule pairs with measuring laboratory desk height (precision 1 mm1\text{ mm}); Vernier caliper pairs with measuring internal tube diameter (precision 0.1 mm0.1\text{ mm}); Micrometer screw gauge pairs with measuring glass cover slip thickness (precision 0.01 mm0.01\text{ mm}); Measuring tape pairs with measuring distances over several metres.
Each instrument is correctly matched to its standard precision and intended application: the metre rule measures moderate lengths to within 1 mm1\text{ mm}; the Vernier caliper measures internal and external dimensions to within 0.1 mm0.1\text{ mm}; the micrometer screw gauge measures small thicknesses to within 0.01 mm0.01\text{ mm}; and the measuring tape measures long or curved distances.

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1
Determine the least count (precision) and functional design of each measuring instrument.
Metre rule has a least count of 1 mm1\text{ mm}; Vernier caliper has a least count of 0.1 mm0.1\text{ mm} (0.01 cm0.01\text{ cm}) and possesses internal jaws; Micrometer screw gauge has a least count of 0.01 mm0.01\text{ mm}; Measuring tape is flexible and suitable for long distances.
Each instrument is engineered for a specific range of dimensions and required degree of precision.
2
Match each instrument to the scenario requiring its specific precision and physical capability.
Metre rule matches desk height; Vernier caliper matches internal tube diameter; Micrometer screw gauge matches thin glass sheet thickness; Measuring tape matches large room dimensions.
Selecting the proper instrument minimizes measurement uncertainty and matches physical constraints.

Anahtar Kavram

Instrument selection based on least count, precision requirements, and physical geometry
Soru 112Soru

Match each physical quantity or measuring instrument related to mass and weight in Column A with its corresponding physical property or operational principle in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Beam balance
Spring balance
Mass
Weight

Eşleşmeler

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Cevap

Beam balance matches with comparing gravitational forces using the principle of moments; Spring balance matches with measuring gravitational pull directly based on Hooke's law; Mass matches with an intrinsic scalar quantity measured in kilograms; Weight matches with a downward vector quantity measured in newtons.
Beam balances measure mass via the principle of moments independently of gravitational variation. Spring balances measure weight force using spring extension under Hooke's law. Mass is a constant scalar quantity measured in kilograms, while weight is a variable vector force measured in newtons.

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1
Identify the working principles of mass and weight measuring instruments.
The beam balance uses equal arms to balance moments (m1gL=m2gLm_1 g L = m_2 g L), canceling gg, so it measures invariant mass. The spring balance measures the force pulling a spring (F=kx=mgF = kx = mg), depending directly on local gg.
Instrument operation dictates whether mass or weight is measured.
2
Distinguish between the physical properties of mass and weight.
Mass is a scalar measure of quantity of matter (SI unit: kg\text{kg}) and is constant. Weight is the gravitational force acting on mass (SI unit: N\text{N}) and is a vector quantity.
Fundamental physical definitions set the units and vector/scalar characteristics of mass versus weight.

Anahtar Kavram

Operating principles of balances and fundamental properties distinguishing mass from weight.
Tahmini Süre:1m 0s
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Measurements and Units Alıştırma Soruları — JAMB UTME — Sayfa 6 | Examkin