Measurements and Units

112 soru

Soru 81Soru

A student records the mass of a chemical sample using a digital balance as 0.02050 kg0.02050\text{ kg}. How many significant figures are contained in this measurement?

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Cevap: 4

Cevap

4 significant figures
In the measurement 0.02050 kg0.02050\text{ kg}, the leading zeros (0.00.0) are placeholders and do not count as significant. The non-zero digits (22 and 55), the captive zero between them, and the trailing zero following the decimal point are all significant. This gives a total of four significant figures.

Adım Adım Çözüm

1
Identify and evaluate leading zeros in 0.02050 kg0.02050\text{ kg}.
The zeros before the digit '2' (0.00.0) are leading zeros.
Leading zeros serve only as decimal placeholders and are never significant.
2
Identify non-zero digits and zeros between non-zero digits.
The digits '2' and '5' are non-zero, and the zero between them is a captive zero.
All non-zero digits and zeros trapped between non-zero digits are always significant.
3
Evaluate the trailing zero.
The zero after '5' is a trailing zero in a decimal number.
Trailing zeros to the right of a decimal point indicate measurement precision and are significant.
4
Sum the total number of significant figures.
Digits '2', '0', '5', and '0' give a total of 4 significant figures.
Combining two non-zero digits, one captive zero, and one trailing decimal zero gives four significant figures.

Anahtar Kavram

Rules for counting significant figures in physical measurements
Tahmini Süre:45s
Soru 82Soru

A simple pendulum on Earth (g=10 m/s2g = 10\text{ m/s}^2) completes 5050 full oscillations in 40 s40\text{ s}. The pendulum is then transferred to a lunar station where the acceleration due to gravity is 1.6 m/s21.6\text{ m/s}^2, and its length is reduced by 64%64\%. What is the time taken, in seconds, for this modified pendulum to complete 3030 oscillations on the lunar station?

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Cevap: 36

Cevap

The time taken for the modified pendulum to complete 30 oscillations on the lunar station is 36 s.
The initial period on Earth is T1=4050=0.8 sT_1 = \frac{40}{50} = 0.8\text{ s}. The formula for the period of a simple pendulum is T=2πLgT = 2\pi\sqrt{\frac{L}{g}}. When length decreases by 64%64\%, the remaining length ratio is L2L1=0.36\frac{L_2}{L_1} = 0.36. The ratio of gravity is g1g2=101.6=6.25\frac{g_1}{g_2} = \frac{10}{1.6} = 6.25. Taking the ratio gives T2T1=0.36×6.25=2.25=1.5\frac{T_2}{T_1} = \sqrt{0.36 \times 6.25} = \sqrt{2.25} = 1.5. Thus, the new period is T2=1.5×0.8 s=1.2 sT_2 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}. For 3030 oscillations, the total time is t=30×1.2 s=36 st = 30 \times 1.2\text{ s} = 36\text{ s}.

Adım Adım Çözüm

1
Calculate the initial period of oscillation on Earth
T1=0.8 sT_1 = 0.8\text{ s}
Period T1T_1 is total time divided by the number of oscillations: T1=40 s50=0.8 sT_1 = \frac{40\text{ s}}{50} = 0.8\text{ s}.
2
Set up the ratio for period under altered length and gravitational field
T2T1=1.5\frac{T_2}{T_1} = 1.5
Using T=2πLgT = 2\pi \sqrt{\frac{L}{g}}, we have T2T1=L2L1g1g2=(10.64)101.6=0.366.25=2.25=1.5\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1} \cdot \frac{g_1}{g_2}} = \sqrt{(1 - 0.64) \cdot \frac{10}{1.6}} = \sqrt{0.36 \cdot 6.25} = \sqrt{2.25} = 1.5.
3
Determine the new period of oscillation
T2=1.2 sT_2 = 1.2\text{ s}
T2=1.5×T1=1.5×0.8 s=1.2 sT_2 = 1.5 \times T_1 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}.
4
Calculate the total time required for 30 oscillations
t2=36 st_2 = 36\text{ s}
Total time t2=N2×T2=30×1.2 s=36 st_2 = N_2 \times T_2 = 30 \times 1.2\text{ s} = 36\text{ s}.

Anahtar Kavram

Period of a simple pendulum and its dependence on length and gravitational acceleration
Soru 83Soru

A student records the time taken for a simple pendulum to complete 2020 complete oscillations as 30 s30\text{ s}. What is the period of oscillation of the pendulum, in seconds?

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Cevap: 1.5

Cevap

The period of oscillation of the pendulum is 1.5 s1.5\text{ s}.
The period of oscillation (TT) is the time required for one complete cycle. Dividing the total time (30 s30\text{ s}) by the number of oscillations (2020) gives 1.5 s1.5\text{ s}.

Adım Adım Çözüm

1
Identify the given values for total time and total number of oscillations.
Total time t=30 st = 30\text{ s} and number of oscillations N=20N = 20.
The period is defined as the time taken for a single complete oscillation.
2
Divide the total time by the number of oscillations to calculate the period.
T=30 s20=1.5 sT = \frac{30\text{ s}}{20} = 1.5\text{ s}.
Applying the period formula T=tNT = \frac{t}{N} yields the duration of one period.

Anahtar Kavram

Period of Oscillation
Tahmini Süre:45s
Soru 84Soru

The electrical resistance RR of a uniform wire of length LL and diameter dd is given by R=4ρLπd2R = \frac{4\rho L}{\pi d^2}, where ρ\rho is the constant resistivity of the material. If the length LL is measured with a percentage error of 2.0%2.0\% and the diameter dd is measured with a percentage error of 1.5%1.5\%, what is the maximum percentage error in the calculated value of RR?

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Cevap: 5.0%5.0\%

Cevap

The maximum percentage error in the calculated resistance RR is 5.0%5.0\%.
For a physical quantity defined by R=4ρLπd2R = \frac{4\rho L}{\pi d^2}, the maximum percentage error is determined by adding the percentage error of LL to twice the percentage error of dd. Calculating 2.0%+2(1.5%)=5.0%2.0\% + 2(1.5\%) = 5.0\% yields the correct maximum percentage error.

Adım Adım Çözüm

1
Identify the relation for maximum percentage error propagation in a physical formula.
For R=4ρLπd2R = \frac{4\rho L}{\pi d^2}, the fractional error equation is ΔRR=ΔLL+2(Δdd)\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\left(\frac{\Delta d}{d}\right).
When quantities are multiplied or divided, their relative errors add, and any exponent becomes a multiplier for that quantity's relative error.
2
Express the relation in terms of percentage errors.
\% \text{ Error in } R = (\% \text{ Error in } L) + 2 \times (\% \text{ Error in } d)
Multiplying the fractional error expression by 100%100\% converts all relative errors into percentage errors.
3
Substitute the given values into the formula.
\% \text{ Error in } R = 2.0\% + 2 \times (1.5\%) = 2.0\% + 3.0\% = 5.0\%
Substituting 2.0%2.0\% for length error and 1.5%1.5\% for diameter error yields the combined maximum percentage error.

Anahtar Kavram

Propagation of errors in physical quantities involving exponents
Soru 85Soru

A ticker-tape timer connected to a power supply operates at a frequency of 50 Hz50\text{ Hz}. A continuous strip of paper tape pulled through the timer records a sequence of dots. Calculate the total time interval, in seconds, between the 1st1\text{st} dot and the 21st21\text{st} dot on the tape.

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Cevap: 0.4

Cevap

0.4 s
On a ticker tape, the time interval between consecutive dots represents one period T=1f=150 Hz=0.02 sT = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}. The total elapsed time for NN dots corresponds to (N1)(N - 1) spaces. For 21 dots, there are 20 spaces, giving a total time of 20×0.02 s=0.40 s20 \times 0.02\text{ s} = 0.40\text{ s}.

Adım Adım Çözüm

1
Find the number of spaces between dots
20 spaces
Between NN dots on a ticker-tape, there are (N1)(N - 1) time intervals.
2
Calculate the period per space
0.02 s
The period TT is the reciprocal of the operating frequency f=50 Hzf = 50\text{ Hz}.
3
Multiply the number of spaces by the period per space
0.4 s
Total time duration is the product of the total number of intervals and the time for one interval.

Anahtar Kavram

Calculation of time interval using ticker-tape timer frequency and dot count
Soru 86Soru

In an experiment to determine the density of a solid sphere, the mass of the sphere is measured as (50.0±0.5) g(50.0 \pm 0.5)\text{ g} and its radius is measured as (1.00±0.02) cm(1.00 \pm 0.02)\text{ cm}. What is the maximum percentage error in the calculated density of the sphere?

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Cevap: 7

Cevap

The maximum percentage error in the calculated density of the sphere is 7.0%7.0\%.
Density is related to mass and radius by ρ=m43πr3\rho = \frac{m}{\frac{4}{3}\pi r^3}. In error analysis, the maximum fractional error of a calculated quantity is the sum of the fractional errors of its components multiplied by their respective powers. The mass has a percentage error of 0.550.0×100%=1.0%\frac{0.5}{50.0} \times 100\% = 1.0\%, and the radius has a percentage error of 0.021.00×100%=2.0%\frac{0.02}{1.00} \times 100\% = 2.0\%. Multiplying the radius percentage error by 3 gives 6.0%6.0\%, and adding the mass percentage error of 1.0%1.0\% yields a maximum percentage error of 7.0%7.0\%.

Adım Adım Çözüm

1
Determine the percentage error in the measurement of mass (mm).
Percentage error in mass = 0.5 g50.0 g×100%=1.0%\frac{0.5\text{ g}}{50.0\text{ g}} \times 100\% = 1.0\%.
Relative error multiplied by 100 gives the percentage error of a measurement.
2
Determine the percentage error in the measurement of radius (rr).
Percentage error in radius = 0.02 cm1.00 cm×100%=2.0%\frac{0.02\text{ cm}}{1.00\text{ cm}} \times 100\% = 2.0\%.
Relative error in radius multiplied by 100 gives its percentage error.
3
Apply the error propagation formula for the density of a sphere.
Maximum percentage error in density = 1.0%+3(2.0%)=7.0%1.0\% + 3(2.0\%) = 7.0\%.
Density is given by ρ=m43πr3\rho = \frac{m}{\frac{4}{3}\pi r^3}. For a formula of the form X=AaBbX = A^a B^b, the fractional error propagates as ΔXX=aΔAA+bΔBB\frac{\Delta X}{X} = a\frac{\Delta A}{A} + b\frac{\Delta B}{B}. Here, the exponent of rr is 3, so its percentage error is multiplied by 3.

Anahtar Kavram

Error propagation in fractional powers and derived physical quantities
Soru 87Soru

A student measures the time taken for 4040 complete oscillations of a simple pendulum using a digital stopwatch that has a negative zero error of 0.40 s-0.40\text{ s}. If the stopwatch displays a reading of 47.60 s47.60\text{ s} for the oscillations, what will be the correct period of oscillation when the pendulum's length is reduced to one-fourth (1/41/4) of its initial length?

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Cevap: 0.60 s0.60\text{ s}

Cevap

The correct period of oscillation for the shortened pendulum is 0.60 s0.60\text{ s}.
The correct answer is obtained by first adjusting the measured time for negative zero error (47.60 s(0.40 s)=48.00 s47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}), calculating the initial period per oscillation (48.00/40=1.20 s48.00 / 40 = 1.20\text{ s}), and then taking into account that period varies with the square root of pendulum length. Reducing length to 1/41/4 reduces period by a factor of 4=2\sqrt{4} = 2, yielding 0.60 s0.60\text{ s}.

Adım Adım Çözüm

1
Correct the recorded total time for instrument zero error
True total time t=Uncorrected TimeZero Error=47.60 s(0.40 s)=48.00 st = \text{Uncorrected Time} - \text{Zero Error} = 47.60\text{ s} - (-0.40\text{ s}) = 48.00\text{ s}
Negative zero error means the timer reads less than the actual elapsed time, so the absolute value of the error must be added back.
2
Calculate the initial period of oscillation
Initial period T1=tN=48.00 s40=1.20 sT_1 = \frac{t}{N} = \frac{48.00\text{ s}}{40} = 1.20\text{ s}
Period is defined as the time per single oscillation.
3
Apply the pendulum scaling relationship for length and period
New period T2=T1×L2L1=1.20 s×14=1.20 s×0.5=0.60 sT_2 = T_1 \times \sqrt{\frac{L_2}{L_1}} = 1.20\text{ s} \times \sqrt{\frac{1}{4}} = 1.20\text{ s} \times 0.5 = 0.60\text{ s}
The period of a simple pendulum is proportional to L\sqrt{L}, so quartering the length reduces the period to half of its original value.

Anahtar Kavram

Measurement of time using a stopwatch with zero error correction combined with simple pendulum period dependence on length.
Tahmini Süre:2m 0s
Soru 88Soru

In a physics experiment to determine the acceleration due to gravity gg using a free-fall apparatus, the distance of fall is measured as h=(2.00±0.04) mh = (2.00 \pm 0.04)\text{ m} and the duration of fall is measured as t=(0.50±0.01) st = (0.50 \pm 0.01)\text{ s}. Given that g=2ht2g = \frac{2h}{t^2}, what is the percentage error in the calculated value of gg?

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Cevap: 6

Cevap

The percentage error in the calculated value of gg is 6%.
For a physical quantity defined by g=2ht2g = \frac{2h}{t^2}, the maximum percentage error is determined by adding the percentage error in hh to twice the percentage error in tt. The percentage error in hh is 0.042.00×100%=2%\frac{0.04}{2.00} \times 100\% = 2\% and in tt is 0.010.50×100%=2%\frac{0.01}{0.50} \times 100\% = 2\%. Therefore, the total percentage error in gg is 2%+2(2%)=6%2\% + 2(2\%) = 6\%.

Adım Adım Çözüm

1
Calculate the percentage error in the distance measurement hh
2%
The fractional error in height is Δhh=0.042.00=0.02\frac{\Delta h}{h} = \frac{0.04}{2.00} = 0.02, which corresponds to 2%2\%.
2
Calculate the percentage error in the time measurement tt
2%
The fractional error in time is Δtt=0.010.50=0.02\frac{\Delta t}{t} = \frac{0.01}{0.50} = 0.02, which corresponds to 2%2\%.
3
Apply the power-law error propagation rule to find the maximum percentage error in gg
6%
For g=2ht2g = \frac{2h}{t^2}, the total relative error is Δgg=Δhh+2Δtt=2%+2(2%)=6%\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t} = 2\% + 2(2\%) = 6\%, since the exponent of tt is 2.

Anahtar Kavram

Error propagation in physical formulas involving powers and quotients
Tahmini Süre:2m 0s
Soru 89Soru

In an experiment to determine the Young's modulus YY of a metal wire using the formula Y=4FLπd2eY = \frac{4FL}{\pi d^2 e}, the force FF, length LL, diameter dd, and extension ee are measured with maximum percentage errors of 1.2%1.2\%, 0.8%0.8\%, 1.5%1.5\%, and 2.0%2.0\% respectively. What is the maximum percentage error in the calculated value of YY?

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Cevap: 7

Cevap

The maximum percentage error in the calculated value of Young's modulus YY is 7.0%7.0\%.
The maximum percentage error is determined by adding the percentage error of each measured quantity, scaled by the absolute value of its power. For Y=4FLπd2eY = \frac{4FL}{\pi d^2 e}, the calculation is 1.2%+0.8%+2(1.5%)+2.0%=7.0%1.2\% + 0.8\% + 2(1.5\%) + 2.0\% = 7.0\%.

Adım Adım Çözüm

1
Set up the relative error propagation formula for the derived quantity YY.
ΔYY=ΔFF+ΔLL+2(Δdd)+Δee\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta e}{e}
Mathematical constants (44 and π\pi) carry no measurement error, and exponents multiply the fractional error of their respective variables.
2
Convert fractional errors into percentage errors by multiplying each term by 100%100\%.
Percentage Error(Y)=Percentage Error(F)+Percentage Error(L)+2×Percentage Error(d)+Percentage Error(e)\text{Percentage Error}(Y) = \text{Percentage Error}(F) + \text{Percentage Error}(L) + 2 \times \text{Percentage Error}(d) + \text{Percentage Error}(e)
Percentage error is directly proportional to fractional error.
3
Substitute the given percentage errors into the formula and sum them up.
Percentage Error(Y)=1.2%+0.8%+2(1.5%)+2.0%=7.0%\text{Percentage Error}(Y) = 1.2\% + 0.8\% + 2(1.5\%) + 2.0\% = 7.0\%
To find the worst-case (maximum) error, all individual percentage contributions are added regardless of whether the variable appears in the numerator or denominator.

Anahtar Kavram

Error Propagation in Derived Physical Quantities
Soru 90Soru

An experimenter uses a digital stopwatch to record the duration of 4040 complete oscillations of a simple pendulum. The stopwatch displays a reading of 50.8 s50.8\text{ s}, but it has a known positive zero error of +0.8 s+0.8\text{ s}. What is the correct period of oscillation of the pendulum?

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Cevap: 1.25 s1.25\text{ s}

Cevap

The correct period of oscillation is 1.25 s1.25\text{ s}.
To find the true duration of the oscillations, the positive zero error of +0.8 s+0.8\text{ s} must be subtracted from the stopwatch display of 50.8 s50.8\text{ s}, giving an actual elapsed time of 50.0 s50.0\text{ s}. Dividing this corrected time by the 4040 complete oscillations gives T=50.0 s/40=1.25 sT = 50.0\text{ s} / 40 = 1.25\text{ s}.

Adım Adım Çözüm

1
Calculate the true time interval by correcting for the instrument zero error.
tactual=50.8 s0.8 s=50.0 st_{\text{actual}} = 50.8\text{ s} - 0.8\text{ s} = 50.0\text{ s}
A positive zero error means the timer reads higher than the true time, so the error value must be subtracted.
2
Calculate the period of a single oscillation by dividing the total corrected time by the number of oscillations.
T=tactualN=50.0 s40=1.25 sT = \frac{t_{\text{actual}}}{N} = \frac{50.0\text{ s}}{40} = 1.25\text{ s}
The period TT is defined as the time required to complete one full oscillation cycle.

Anahtar Kavram

Measurement of Time with Zero Error Correction
Tahmini Süre:1m 30s
Soru 91Soru

The frequency of a wave is measured as (200±4) Hz(200 \pm 4)\text{ Hz} and its wavelength is measured as (0.50±0.01) m(0.50 \pm 0.01)\text{ m}. What is the maximum percentage error in the calculated speed of the wave?

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Cevap: 4.0%4.0\%

Cevap

The maximum percentage error in the calculated speed of the wave is 4.0%4.0\%.
For a calculated quantity given by the product of two variables v=fλv = f\lambda, the fractional error is Δvv=Δff+Δλλ\frac{\Delta v}{v} = \frac{\Delta f}{f} + \frac{\Delta \lambda}{\lambda}. Converting each term to percentage gives 2.0%+2.0%=4.0%2.0\% + 2.0\% = 4.0\%.

Adım Adım Çözüm

1
Calculate the percentage error in the frequency measurement.
4 Hz200 Hz×100%=2.0%\frac{4\text{ Hz}}{200\text{ Hz}} \times 100\% = 2.0\%
Percentage error of a measurement is given by absolute errormeasured value×100%\frac{\text{absolute error}}{\text{measured value}} \times 100\%.
2
Calculate the percentage error in the wavelength measurement.
0.01 m0.50 m×100%=2.0%\frac{0.01\text{ m}}{0.50\text{ m}} \times 100\% = 2.0\%
The relative uncertainty of the wavelength must be expressed as a percentage.
3
Sum the individual percentage errors to find the total maximum percentage error for v=fλv = f\lambda.
2.0%+2.0%=4.0%2.0\% + 2.0\% = 4.0\%
When physical quantities are multiplied together, their maximum percentage errors add up.

Anahtar Kavram

Propagation of Errors in Products
Soru 92Soru

A ticker-tape timer operates at an alternating current frequency of 50 Hz50\text{ Hz}. During a mechanics experiment, a paper tape pulled through the device records a section containing 1111 consecutive dots. What is the total time elapsed for this section of the tape?

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Cevap: 0.20 s0.20\text{ s}

Cevap

The total time elapsed for this section of the tape is 0.20 s0.20\text{ s}.
The period of a 50 Hz50\text{ Hz} ticker-tape timer is T=150 Hz=0.02 sT = \frac{1}{50\text{ Hz}} = 0.02\text{ s}. A sequence of 1111 consecutive dots contains 1010 time intervals (111=1011 - 1 = 10). Multiplying 1010 intervals by 0.02 s0.02\text{ s} yields 0.20 s0.20\text{ s}.

Adım Adım Çözüm

1
Determine the time interval (period) between consecutive dots recorded by the timer.
T=1f=150 Hz=0.02 sT = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}
The ticker-tape timer makes one dot every period TT of the operating frequency.
2
Calculate the number of intervals (spaces) between the first and last dot in the sequence.
\text{Number of intervals } n = N - 1 = 11 - 1 = 10
Time elapses between dots, so NN dots define N1N-1 spaces.
3
Multiply the number of intervals by the time period of one interval to get total time elapsed.
t = 10 \times 0.02\text{ s} = 0.20\text{ s}
Total duration is the product of the number of spaces and the duration of a single space.

Anahtar Kavram

Ticker-Tape Timer Period and Time Interval Calculation
Soru 93Soru

The fundamental frequency ff of a stretched vibrating string depends on the tension force FF, the length of the string ll, and its linear mass density μ\mu (mass per unit length) according to the relation f=kFalbμcf = k F^a l^b \mu^c, where kk is a dimensionless constant. Which of the following sets of exponents (a,b,c)(a, b, c) correctly satisfies dimensional homogeneity?

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Cevap: (12,1,12)(\frac{1}{2}, -1, -\frac{1}{2})

Cevap

The correct set of exponents is (12,1,12)(\frac{1}{2}, -1, -\frac{1}{2}).
Equating the base dimensions on both sides gives T1=Ma+cLa+bcT2aT^{-1} = M^{a+c} L^{a+b-c} T^{-2a}. Solving for the powers gives a=12a = \frac{1}{2}, b=1b = -1, and c=12c = -\frac{1}{2}, matching the option specifying (12,1,12)(\frac{1}{2}, -1, -\frac{1}{2}).

Adım Adım Çözüm

1
Express the dimensions of all physical quantities involved in fundamental dimensions MM, LL, and TT.
[f]=T1[f] = T^{-1}, [F]=MLT2[F] = M L T^{-2}, [l]=L[l] = L, and [μ]=ML1[\mu] = M L^{-1}.
Linear mass density μ\mu is mass per unit length (mass/length\text{mass}/\text{length}).
2
Substitute dimensions into the given formula f=kFalbμcf = k F^a l^b \mu^c (ignoring the dimensionless constant kk).
T1=(MLT2)a(L)b(ML1)c=Ma+cLa+bcT2aT^{-1} = (M L T^{-2})^a (L)^b (M L^{-1})^c = M^{a+c} L^{a+b-c} T^{-2a}.
Dimensional homogeneity requires both sides of the equation to have identical dimensional powers.
3
Equate exponents of MM, LL, and TT from both sides to form algebraic equations.
For TT: 2a=1    a=12-2a = -1 \implies a = \frac{1}{2}. For MM: a+c=0    c=a=12a + c = 0 \implies c = -a = -\frac{1}{2}. For LL: a+bc=0    12+b(12)=0    b+1=0    b=1a + b - c = 0 \implies \frac{1}{2} + b - (-\frac{1}{2}) = 0 \implies b + 1 = 0 \implies b = -1.
Solving the system yields the unique set of exponents (a,b,c)=(12,1,12)(a, b, c) = (\frac{1}{2}, -1, -\frac{1}{2}).

Anahtar Kavram

Dimensional Analysis and Method of Dimensions
Soru 94Soru

Match each physical quantity listed on the left with its correct classification and SI unit definition on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Electric current
Electric potential difference
Luminous intensity
Specific heat capacity

Eşleşmeler

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Cevap

Electric current matches fundamental quantity measured in amperes (A); Electric potential difference matches derived quantity measured in volts (V) or kgm2s3A1kg\cdot m^2\cdot s^{-3}\cdot A^{-1}; Luminous intensity matches fundamental quantity measured in candelas (cd); Specific heat capacity matches derived quantity measured in joules per kilogram per kelvin (Jkg1K1J\cdot kg^{-1}\cdot K^{-1}).
Electric current and luminous intensity belong to the seven basic SI fundamental quantities with units ampere and candela respectively. Electric potential difference (V=W/QV = W/Q) and specific heat capacity (c=Q/(mΔT)c = Q / (m\Delta T)) are derived physical quantities expressed in terms of fundamental SI units.

Adım Adım Çözüm

1
Identify the fundamental physical quantities from the list.
Electric current and luminous intensity are base SI quantities.
Fundamental quantities are independent basic quantities defined by international standards.
2
Identify the derived physical quantities and determine their SI units.
Electric potential difference and specific heat capacity are derived quantities.
Derived quantities are formed from combinations of fundamental quantities through mathematical relationships.
3
Pair each physical quantity on the left with its exact classification and unit definition on the right.
Match left_1 to right_1, left_2 to right_4, left_3 to right_3, and left_4 to right_2.
Direct comparison with standard SI fundamental and derived quantity definitions confirms these pairings.

Anahtar Kavram

Fundamental and Derived Quantities
Soru 95Soru

Match each derived physical quantity in Column I with its equivalent SI unit expressed in terms of fundamental (base) units in Column II.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Gravitational Potential
Electrical Conductance
Self-Inductance
Impulse

Eşleşmeler

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Cevap

Gravitational Potential matches m2s2\text{m}^2\cdot\text{s}^{-2}; Electrical Conductance matches kg1m2s3A2\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^3\cdot\text{A}^2; Self-Inductance matches kgm2s2A2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}; Impulse matches kgms1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}.
Each derived unit is systematically expressed in terms of the fundamental SI base units (kilogram, meter, second, ampere) by substituting defining formulas. Gravitational potential is energy per unit mass (Jkg1=m2s2\text{J}\cdot\text{kg}^{-1} = \text{m}^2\cdot\text{s}^{-2}). Electrical conductance is reciprocal resistance (Siemens =AV1=kg1m2s3A2= \text{A}\cdot\text{V}^{-1} = \text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^3\cdot\text{A}^2). Self-Inductance is electromotive force per rate of change of current (Henry =VsA1=kgm2s2A2= \text{V}\cdot\text{s}\cdot\text{A}^{-1} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}). Impulse is force times time (Ns=kgms1\text{N}\cdot\text{s} = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}).

Adım Adım Çözüm

1
Derive base units for Gravitational Potential
m2s2\text{m}^2\cdot\text{s}^{-2}
Gravitational potential is potential energy per unit mass (V=Epm=kgm2s2kgV = \frac{E_p}{m} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}}).
2
Derive base units for Electrical Conductance
kg1m2s3A2\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^3\cdot\text{A}^2
Conductance is current divided by voltage (G=IV=Akgm2s3A1G = \frac{I}{V} = \frac{\text{A}}{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}}).
3
Derive base units for Self-Inductance
kgm2s2A2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}
Inductance is voltage multiplied by time divided by current (L=VtI=kgm2s3A1sAL = \frac{V \cdot t}{I} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1} \cdot \text{s}}{\text{A}}).
4
Derive base units for Impulse
kgms1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}
Impulse is force multiplied by time (I=Ft=(kgms2)sI = F \cdot t = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{s}).

Anahtar Kavram

Break down derived physical quantities into fundamental SI quantities (mass in kg, length in m, time in s, electric current in A) using defining physical formulas.
Tahmini Süre:1m 30s
Soru 96Soru

A micrometer screw gauge has a pitch of 0.5 mm0.5\text{ mm} and 5050 divisions on its circular thimble scale. When its anvil and spindle are fully brought together without any object, the zero mark of the thimble scale lies 44 divisions past the main scale index line in the tightening direction (a positive zero error). When the wall thickness of a hollow metal pipe is measured with this micrometer, the main scale reading is 3.5 mm3.5\text{ mm} and the 32nd32\text{nd} thimble division aligns with the index line. If the outer diameter of the pipe is separately measured to be 25.40 mm25.40\text{ mm}, what is the corrected inner diameter of the metal pipe in millimeters?

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Cevap: 17.84

Cevap

17.84 mm
The least count of the micrometer screw gauge is 0.5 mm50=0.01 mm\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}. The observed wall thickness is 3.5 mm+(32×0.01 mm)=3.82 mm3.5\text{ mm} + (32 \times 0.01\text{ mm}) = 3.82\text{ mm}. Accounting for the positive zero error of +0.04 mm+0.04\text{ mm} gives a true wall thickness of 3.82 mm0.04 mm=3.78 mm3.82\text{ mm} - 0.04\text{ mm} = 3.78\text{ mm}. Subtracting twice the wall thickness from the outer diameter yields the inner diameter: 25.40 mm2(3.78 mm)=17.84 mm25.40\text{ mm} - 2(3.78\text{ mm}) = 17.84\text{ mm}.

Adım Adım Çözüm

1
Calculate the least count (precision) of the micrometer screw gauge
Least count = 0.01 mm
Least count is defined as pitch divided by total circular scale divisions: 0.5 mm / 50 = 0.01 mm.
2
Determine the observed wall thickness from the instrument readings
Observed thickness = 3.82 mm
Observed reading = Main scale reading + (Thimble division * Least count) = 3.5 mm + (32 * 0.01 mm) = 3.82 mm.
3
Correct the observed thickness for positive zero error
Corrected wall thickness = 3.78 mm
Positive zero error (+0.04 mm) must be subtracted from the observed reading: 3.82 mm - 0.04 mm = 3.78 mm.
4
Calculate the corrected inner diameter of the hollow pipe
Inner diameter = 17.84 mm
The inner diameter equals the outer diameter minus twice the wall thickness: 25.40 mm - 2(3.78 mm) = 25.40 mm - 7.56 mm = 17.84 mm.

Anahtar Kavram

Measurement of length, micrometer zero error correction, and geometry of hollow cylinders
Tahmini Süre:2m 30s
Soru 97Soru

Which of the following combinations of fundamental SI base units is equivalent to the volt (V\text{V}), the SI unit of electric potential difference?

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Cevap: kgm2s3A1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

Cevap

kgm2s3A1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}
Electric potential difference (VV) is defined as work done (WW) per unit charge (QQ), giving V=WQV = \frac{W}{Q}. In fundamental SI units, work is kgm2s2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2} and electric charge is As\text{A}\cdot\text{s}. Dividing work by charge yields kgm2s2As=kgm2s3A1\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}.

Adım Adım Çözüm

1
State the defining physical equation for electric potential difference.
V=WQV = \frac{W}{Q}, where VV is electric potential, WW is work done in joules, and QQ is electric charge in coulombs.
Electric potential difference is defined as work done per unit electric charge.
2
Express work (WW) in terms of fundamental SI base units.
W=Force×distance=(kgms2)×m=kgm2s2W = \text{Force} \times \text{distance} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}.
Work is the product of force and displacement, where force is mass times acceleration.
3
Express electric charge (QQ) in terms of fundamental SI base units.
Q=I×t=AsQ = I \times t = \text{A}\cdot\text{s}.
Electric charge is defined as electric current multiplied by time.
4
Divide the base units of work by the base units of electric charge.
V=kgm2s2As=kgm2s3A1V = \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}.
Simplifying the exponents gives the equivalent combination of fundamental SI base units.

Anahtar Kavram

Expressing derived SI units in terms of fundamental (base) SI units.
Soru 98Soru

A rocket carries a payload of mass 25 kg25\text{ kg} resting on a spring balance. During its vertical ascent, the spring balance indicates a reading of 350 N350\text{ N}. Taking the acceleration due to gravity as g=10 m s2g = 10\text{ m s}^{-2}, what is the magnitude of the upward acceleration of the rocket in m s2\text{m s}^{-2}?

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Cevap: 4

Cevap

The upward acceleration of the rocket is 4.0 m s24.0\text{ m s}^{-2}.
When an object of mass mm accelerates upward at rate aa, the scale must exert an upward force RR that overcomes the weight mgmg and provides net acceleration mama, such that R=m(g+a)R = m(g + a). Substituting R=350 NR = 350\text{ N}, m=25 kgm = 25\text{ kg}, and g=10 m s2g = 10\text{ m s}^{-2} yields 350=25(10+a)350 = 25(10 + a), which simplifies to a=4.0 m s2a = 4.0\text{ m s}^{-2}.

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1
Formulate the equation of motion for apparent weight during upward acceleration
Rmg=ma    R=m(g+a)R - mg = ma \implies R = m(g + a)
The spring balance supports the mass against gravity while simultaneously providing the force for upward acceleration.
2
Substitute the given numerical parameters into the relation
350=25(10+a)350 = 25(10 + a)
The measured reading R=350 NR = 350\text{ N}, mass m=25 kgm = 25\text{ kg}, and standard gravity g=10 m s2g = 10\text{ m s}^{-2} are supplied.
3
Solve the algebraic equation for the rocket's acceleration aa
a=4.0 m s2a = 4.0\text{ m s}^{-2}
Dividing 350 N350\text{ N} by 25 kg25\text{ kg} gives an effective total acceleration of 14 m s214\text{ m s}^{-2}; subtracting g=10 m s2g = 10\text{ m s}^{-2} isolates the upward acceleration aa.

Anahtar Kavram

Apparent Weight and Mass Measurement in Accelerating Frames
Soru 99Soru

The derived SI unit of electric capacitance, the farad (F\text{F}), can be expressed in terms of fundamental SI base units as kgambscAd\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d. What is the numerical value of the sum of the exponents a+b+c+da + b + c + d?

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Cevap: 3

Cevap

The numerical value of the sum of the exponents a+b+c+da + b + c + d is 3.
The farad (F\text{F}), when resolved into fundamental SI base units, is kg1m2s4A2\text{kg}^{-1} \cdot \text{m}^{-2} \cdot \text{s}^4 \cdot \text{A}^2. Adding the exponents yields a=1a = -1, b=2b = -2, c=4c = 4, and d=2d = 2, giving a total sum of (1)+(2)+4+2=3(-1) + (-2) + 4 + 2 = 3.

Adım Adım Çözüm

1
Relate electric capacitance to fundamental physical quantities
Capacitance is defined as C=QVC = \frac{Q}{V}, where QQ is electric charge and VV is electric potential difference.
This fundamental relation connects capacitance to charge and energy per unit charge.
2
Express charge and potential difference in terms of SI base units
Electric charge QQ has base units As\text{A} \cdot \text{s}. Potential difference V=WorkQV = \frac{\text{Work}}{Q} has base units kgm2s2As=kgm2s3A1\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}.
Work is force times distance (kgms2×m=kgm2s2\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \times \text{m} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}) and electric current is a fundamental base quantity.
3
Determine the base unit representation of the farad
The farad is expressed as Askgm2s3A1=kg1m2s4A2\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.
Dividing the unit of charge by the unit of potential difference yields the derived base unit expression.
4
Calculate the sum of the exponents
a+b+c+d=(1)+(2)+4+2=3a + b + c + d = (-1) + (-2) + 4 + 2 = 3.
Summing the powers corresponding to kilograms, meters, seconds, and amperes gives the final scalar value.

Anahtar Kavram

Expressing derived SI units in terms of fundamental SI base units
Soru 100Soru

A spring balance calibrated in newtons shows an initial pointer reading of +0.4 N+0.4\text{ N} before any load is suspended. When a stone is attached to the balance hook, the pointer indicates 18.4 N18.4\text{ N}. Taking acceleration due to gravity g=10 m s2g = 10\text{ m s}^{-2}, what is the actual mass of the stone?

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Cevap: 1.8 kg1.8\text{ kg}

Cevap

The actual mass of the stone is 1.8 kg1.8\text{ kg}.
The spring balance has a systematic positive zero error of +0.4 N+0.4\text{ N}. To find the true weight exerted by the stone, this offset must be subtracted from the observed scale reading: 18.4 N0.4 N=18.0 N18.4\text{ N} - 0.4\text{ N} = 18.0\text{ N}. Dividing this true weight by the acceleration due to gravity (10 m s210\text{ m s}^{-2}) yields the correct mass of 1.8 kg1.8\text{ kg}.

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1
Determine the true weight by applying zero error correction
Wtrue=18.4 N0.4 N=18.0 NW_{\text{true}} = 18.4\text{ N} - 0.4\text{ N} = 18.0\text{ N}
A positive zero error means the scale overstates the load and must be subtracted from the observed reading.
2
Calculate mass using the relation between weight, mass, and gravitational field strength
m=Wtrueg=18.0 N10 m s2=1.8 kgm = \frac{W_{\text{true}}}{g} = \frac{18.0\text{ N}}{10\text{ m s}^{-2}} = 1.8\text{ kg}
Weight is equal to mass multiplied by acceleration due to gravity (W=mgW = mg).

Anahtar Kavram

Measurement of Mass and Weight with Zero Error Correction
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