Questions & Answers: "units and measurement"

From Newton's law of gravitation, F = G m1 m2 / r². Thus, G = F r² / (m1 m2). Substituting dimensions: [G] = [MLT⁻²] [L²] / [M²] = [M⁻¹ L³ T⁻²]. The other options are incorrect as they do not balance the mass or length dimensions correctly according to the force equation.

In 0.007, trailing zeros after a decimal point but before a non-zero digit are not significant, so only '7' is significant (1 figure). In scientific notation (2.64 × 10²⁴), only the coefficients are counted, so '2', '6', and '4' are significant (3 figures). 4 and 3 is wrong because leading zeros are not significant. 3 and 3 is wrong because the zeros in 0.007 are not significant. 1 and 24 is wrong because the power of 10 does not affect significant figures.

Magnification = Observed size / Real size. Therefore, Real size = Observed size / Magnification = 3.5 mm / 100 = 0.035 mm. 3.5 mm is the observed size, not the actual size. 0.35 mm and 0.0035 mm result from incorrect division by 10 or 1000 respectively.

By the principle of homogeneity, dimensions of each term on the RHS must equal dimensions of the LHS. So, [bt²] = [x]. [b] [T²] = [L]. Thus, [b] = [L T⁻²]. The unit is meters per second squared (m/s²). m/s is the unit for 'a', m s and m are dimensionally inconsistent with the term bt² equaling a length.

Impulse = Force × Time = [MLT⁻²][T] = [MLT⁻¹]. Momentum = Mass × Velocity = [M][LT⁻¹] = [MLT⁻¹]. Work [ML²T⁻²] and Power [ML²T⁻³] are different. Stress [ML⁻¹T⁻²] and Strain [dimensionless] are different. Force [MLT⁻²] and Pressure [ML⁻¹T⁻²] are different.

Volume of a sphere V = (4/3)πr³. The relative error is ΔV/V = 3(Δr/r). Given Δr/r = 2%, the percentage error in volume is 3 × 2% = 6%. 2% is the error in radius, 4% would be the error in surface area (r²), and 8% is a calculation error.

Least Count (LC) = Pitch / No. of divisions = 1 mm / 100 = 0.01 mm. Diameter = Main Scale Reading + (Circular Scale Reading × LC) = 1 mm + (47 × 0.01 mm) = 1 + 0.47 = 1.47 mm. 1.047 mm and 1.0047 mm are errors in LC calculation. 1.47 cm uses incorrect units.

In multiplication and division, the result should have the same number of significant figures as the term with the least significant figures. 25.2 has 3, 1374 has 4, and 33.3 has 3. The result must have 3 significant figures. Calculation: 1039.78... Rounding to 3 sig figs gives 1040 (The zero is not significant unless specified). 1039.78 and 1039.8 have too many sig figs. 1000 has only 1 sig fig.

1 parsec is defined as the distance at which an arc of 1 AU length subtends an angle of 1 arc second. θ = l / r => 1" = 1 AU / 1 parsec. Since 1" = (1/3600) * (π/180) radians, 1 parsec = 1 AU / radians value ≈ 206265 AU. 3.08 × 10¹⁶ is the value in meters, not AU. 1.496 × 10¹¹ is the value of 1 AU in meters.

Solar constant = Energy / (Area × Time) = [ML²T⁻²] / ([L²][T]) = [ML⁰T⁻³]. [ML²T⁻²] is energy, [ML¹T⁻²] is force, and [ML⁰T⁻²] is surface tension/spring constant. These do not represent energy per area per time.

Let F = [P^a V^b T^c]. [MLT⁻²] = [ML⁻¹T⁻²]^a [LT⁻¹]^b [T]^c. Equating powers of M: a = 1. Equating powers of L: -a + b = 1 => -1 + b = 1 => b = 2. Equating powers of T: -2a - b + c = -2 => -2(1) - 2 + c = -2 => c = 2. So F = [P V² T²]. Other options fail the dimensional consistency check for mass, length, or time.

A light year is the distance travelled by light in vacuum in one year; it is a unit of distance ([L]), not time. Leap year, Lunar month, and Solar day are all measures of time intervals. This is a common trap in NCERT physics.

Least Count = 1 MSD - 1 VSD. Given 10 VSD = 9 MSD, so 1 VSD = 0.9 MSD. LC = 1 MSD - 0.9 MSD = 0.1 MSD. Since 1 MSD = 1 mm, LC = 0.1 mm. 0.01 mm is typical for screw gauges. 1 mm is the MSD value. 0.9 mm is the value of one VSD, not the least count.

Volume = length × breadth × thickness = 4.234 m × 1.005 m × 0.0201 m = 0.0855289... m³. Since the thickness (2.01 cm = 0.0201 m) has 3 significant figures, the result must be rounded to 3 significant figures, giving 0.0855 m³. 0.08553 and 0.085529 have too many significant figures. 0.086 has too few.

The seven SI base units are metre (length), kilogram (mass), second (time), ampere (electric current), Kelvin (thermodynamic temperature), mole (amount of substance), and candela (luminous intensity). Newton is the SI derived unit for force, which is defined in terms of base units as $\text{kg} \cdot \text{m} \cdot \text{s}^{-2}$.

Energy can be expressed as work done, which is defined as force multiplied by distance. The dimensional formula for force is $[MLT^{-2}]$ and for distance is $[L]$. Therefore, the dimensional formula for energy is $[MLT^{-2}] \times [L] = [ML^2T^{-2}]$.

According to Newton's law of universal gravitation, the force $F$ between two masses $M_1$ and $M_2$ separated by a distance $r$ is given by $F = \frac{GM_1M_2}{r^2}$. Rearranging this equation for $G$, we get $G = \frac{Fr^2}{M_1M_2}$. The dimensional formula for force $F$ is $[MLT^{-2}]$. The dimension for distance $r$ is $[L]$, so $r^2$ has dimensions $[L^2]$. The dimension for mass $M$ is $[M]$, so $M_1M_2$ has dimensions $[M^2]$. Substituting these into the formula for $G$: $[G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{1-2}L^{1+2}T^{-2}] = [M^{-1}L^3T^{-2}]$.

The principle of dimensional homogeneity states that an equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same. In the given equation:

1. The dimension of $x$ (displacement) is $[L]$.
2. The dimension of $vt$ (velocity $[LT^{-1}]$ multiplied by time $[T]$) is $[LT^{-1}][T] = [L]$.
3. The dimension of $\frac{1}{2}at^2$ (a dimensionless constant $\frac{1}{2}$ multiplied by acceleration $[LT^{-2}]$ and time squared $[T^2]$) is $[LT^{-2}][T^2] = [L]$. For the equation to be dimensionally correct, all terms must have the same dimensions, which in this case is $[L]$.

Let's assume $T = k \cdot l^a m^b g^c$, where $k$ is a dimensionless constant. Writing the dimensions for each quantity: $[T] = [T^1]$ (Period) $[l] = [L^1]$ (Length) $[m] = [M^1]$ (Mass) $[g] = [LT^{-2}]$ (Acceleration due to gravity) Substituting these into the assumed relation: $[M^0L^0T^1] = [L]^a [M]^b [LT^{-2}]^c$ $[M^0L^0T^1] = [L^a M^b L^c T^{-2c}]$ $[M^0L^0T^1] = [M^b L^{a+c} T^{-2c}]$ Comparing the powers of M, L, and T on both sides: For M: $b = 0$ For L: $a+c = 0 \implies a = -c$ For T: $1 = -2c \implies c = -1/2$ Substitute $c = -1/2$ into $a = -c$: $a = -(-1/2) = 1/2$. So, $T \propto l^{1/2} m^0 g^{-1/2} = l^{1/2} g^{-1/2} = \sqrt{\frac{l}{g}}$.

According to the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions. In the first parenthesis, $P$ is added to $\frac{a}{V^2}$. Therefore, the dimension of $\frac{a}{V^2}$ must be the same as the dimension of pressure ($P$). Dimensions of pressure $P = \frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$. Dimensions of volume $V = [L^3]$, so $V^2$ has dimensions $[L^6]$. Equating the dimensions: $[\frac{a}{V^2}] = [P]$ $[a] = [P][V^2]$ $[a] = [ML^{-1}T^{-2}][L^6] = [ML^{(-1+6)}T^{-2}] = [ML^5T^{-2}]$.

Given length $L = 12.4 \text{ cm}$. This measurement has 3 significant figures. Given width $W = 4.12 \text{ cm}$. This measurement also has 3 significant figures. The area is calculated by multiplying length and width: $A = L \times W = 12.4 \text{ cm} \times 4.12 \text{ cm} = 51.088 \text{ cm}^2$. When multiplying or dividing measurements, the final result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. In this case, both measurements have 3 significant figures. Therefore, the area should be rounded to 3 significant figures. $51.088$ rounded to 3 significant figures is $51.1$ (since the first dropped digit, 8, is greater than or equal to 5, the preceding digit is rounded up).

When adding or subtracting measurements, the result should be reported to the same number of decimal places as the measurement with the fewest decimal places. The given measurements are: $2.14 \text{ m}$ (2 decimal places) $0.2 \text{ m}$ (1 decimal place) $1.025 \text{ m}$ (3 decimal places) The measurement with the fewest decimal places is $0.2 \text{ m}$, which has 1 decimal place. The sum is $2.14 + 0.2 + 1.025 = 3.365 \text{ m}$. Rounding $3.365$ to 1 decimal place, we look at the second decimal place (6). Since 6 is greater than or equal to 5, we round up the first decimal place. So, $3.365 \text{ m}$ becomes $3.4 \text{ m}$.

The formula for kinetic energy is $KE = \frac{1}{2}mv^2$. The constant $\frac{1}{2}$ is exact and does not contribute to the error. The fractional error in a quantity $X = A^p B^q C^r$ is given by $\frac{\Delta X}{X} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C}$. For $KE = mv^2$, the fractional error is $\frac{\Delta KE}{KE} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$. Given: Mass $m = 5.0 \text{ kg}$ with absolute error $\Delta m = 0.1 \text{ kg}$. Velocity $v = 10.0 \text{ m/s}$ with absolute error $\Delta v = 0.2 \text{ m/s}$. Fractional error in mass: $\frac{\Delta m}{m} = \frac{0.1}{5.0} = 0.02$. Fractional error in velocity: $\frac{\Delta v}{v} = \frac{0.2}{10.0} = 0.02$. Substituting these values into the error formula: $\frac{\Delta KE}{KE} = 0.02 + 2(0.02) = 0.02 + 0.04 = 0.06$. To find the percentage error, multiply by $100\%$: Percentage error in $KE = 0.06 \times 100\% = 6.0\%$.

First, calculate the total number of seconds in a day: Seconds in a minute = 60 Minutes in an hour = 60 Hours in a day = 24 Total seconds in a day = $24 \times 60 \times 60 = 86400$ seconds. To find the order of magnitude, express the number in scientific notation: $86400 = 8.64 \times 10^4$. For a number written as $N \times 10^p$, where $1 \le N < 10$, the order of magnitude is $10^p$ if $N \sqrt{10} \approx 3.16$, we round up the power of 10. Therefore, the order of magnitude is $10^{4+1} = 10^5$.

The measured value is $L = 2.5 \text{ m}$. The absolute error is $\Delta L = 0.05 \text{ m}$. Fractional error is given by $\frac{\Delta L}{L} = \frac{0.05}{2.5}$. $\frac{0.05}{2.5} = \frac{5}{250} = \frac{1}{50} = 0.02$. Percentage error is the fractional error multiplied by $100\%$: Percentage error $= 0.02 \times 100\% = 2.0\%$.

When quantities are added or subtracted, their absolute errors add up. For resistors in series, the equivalent resistance is $R_{eq} = R_1 + R_2$. Mean value of $R_{eq} = 100 \Omega + 200 \Omega = 300 \Omega$. The maximum possible absolute error in the sum is the sum of the individual absolute errors: $\Delta R_{eq} = \Delta R_1 + \Delta R_2 = 3 \Omega + 4 \Omega = 7 \Omega$. Therefore, the equivalent resistance is $(300 \pm 7) \Omega$.

The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$. The constants $\frac{4}{3}$ and $\pi$ are exact and do not contribute to the error. For a quantity $X = r^p$, the fractional error is $\frac{\Delta X}{X} = p\frac{\Delta r}{r}$. In this case, for $V = r^3$, the fractional error in volume is $\frac{\Delta V}{V} = 3\frac{\Delta r}{r}$. Given radius $r = 2.0 \text{ cm}$ and absolute error $\Delta r = 0.1 \text{ cm}$. Fractional error in radius $\frac{\Delta r}{r} = \frac{0.1}{2.0} = 0.05$. Now, calculate the fractional error in volume: $\frac{\Delta V}{V} = 3 \times 0.05 = 0.15$. To convert this to percentage error, multiply by $100\%$: Percentage error in volume $= 0.15 \times 100\% = 15\%$.

For a quantity $X = \frac{A^p B^q}{C^r D^s}$, the maximum percentage error in $X$ is given by: $\%E_X = p(\%E_A) + q(\%E_B) + r(\%E_C) + s(\%E_D)$

In the given equation, $X = A^2 B^3 C^{-1} D^{-1/2}$. Note that the powers are always taken as positive when summing errors. $\%E_X = 2(\%E_A) + 3(\%E_B) + 1(\%E_C) + \frac{1}{2}(\%E_D)$

Given percentage errors:

• $\%E_A = 1\%$
• $\%E_B = 2\%$
• $\%E_C = 3\%$
• $\%E_D = 4\%$

Substitute these values into the formula: $\%E_X = 2(1\%) + 3(2\%) + 1(3\%) + \frac{1}{2}(4\%)$ $\%E_X = 2\% + 6\% + 3\% + 2\%$ $\%E_X = 13\%$

Dimensional analysis is a powerful tool with several applications, including checking dimensional consistency of equations, deriving relationships between physical quantities (up to a dimensionless constant), and converting units between different systems. However, a key limitation is that it cannot determine the value of dimensionless constants. For example, in the formula for the period of a simple pendulum, $T = 2\pi \sqrt{l/g}$, dimensional analysis can only determine that $T \propto \sqrt{l/g}$, but it cannot determine the value of $2\pi$.

Let's compare the approximate values of these units:

1. Kilometre (km): $1 \text{ km} = 10^3 \text{ m}$
2. Astronomical Unit (AU): The average distance between the Earth and the Sun, $1 \text{ AU} \approx 1.496 \times 10^{11} \text{ m}$
3. Light-year (ly): The distance light travels in one year in a vacuum, $1 \text{ ly} \approx 9.461 \times 10^{15} \text{ m}$
4. Parsec (pc): The distance at which one astronomical unit subtends an angle of one arcsecond, $1 \text{ pc} \approx 3.086 \times 10^{16} \text{ m}$ Comparing these values, $1 \text{ km} < 1 \text{ AU} < 1 \text{ ly} < 1 \text{ pc}$. Therefore, the Parsec is the largest unit of distance among the given options.

The International System of Units (SI) defines seven base units: metre (for length), kilogram (for mass), second (for time), ampere (for electric current), Kelvin (for thermodynamic temperature), mole (for amount of substance), and candela (for luminous intensity). Joule is a derived unit of energy or work, defined as $\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$.

In an LC circuit, the resonant angular frequency $\omega$ is given by $\omega = \frac{1}{\sqrt{LC}}$. The dimension of angular frequency $\omega$ is $[T^{-1}]$ (since $\omega = 2\pi/T$, where $T$ is period). From the formula, we have $\frac{1}{\sqrt{[LC]}} = [T^{-1}]$. This implies $\sqrt{[LC]} = [T]$. Squaring both sides gives $[LC] = [T^2]$. Alternatively, one can find the dimensions of L and C separately: Inductance $L$: From Faraday's law, $V = L \frac{dI}{dt}$, so $[L] = \frac{[V]}{[I][T^{-1}]}$. Since $V = W/Q = W/(IT)$, $[V] = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$. Therefore, $[L] = \frac{[ML^2T^{-3}A^{-1}]}{[A][T^{-1}]} = [ML^2T^{-2}A^{-2}]$. Capacitance $C$: From $Q = CV$, so $C = Q/V$. Since $Q = IT$, $[C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]} = [M^{-1}L^{-2}T^4A^2]$. Multiplying $[L]$ and $[C]$: $[LC] = [ML^2T^{-2}A^{-2}] \times [M^{-1}L^{-2}T^4A^2] = [M^{1-1}L^{2-2}T^{-2+4}A^{-2+2}] = [M^0L^0T^2A^0] = [T^2]$.

For any trigonometric function, its argument must be dimensionless. Therefore, the quantity $(\omega t - kx)$ must be dimensionless ($[M^0L^0T^0]$). This implies that both $\omega t$ and $kx$ must individually be dimensionless. Considering the term $kx$: $[kx] = [M^0L^0T^0]$ Since $x$ is distance, its dimension is $[L]$. So, $[k][L] = [M^0L^0T^0]$. Therefore, $[k] = [L^{-1}]$. (Here $k$ is the wave number).

The least count (LC) of a screw gauge is defined as the ratio of its pitch to the total number of divisions on its circular scale. The formula for least count is: $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$ Given: Pitch = $0.5 \text{ mm}$ Number of divisions on circular scale = 50 Substituting these values: $\text{LC} = \frac{0.5 \text{ mm}}{50} = \frac{5}{500} \text{ mm} = \frac{1}{100} \text{ mm} = 0.01 \text{ mm}$.

Let's determine the dimensions for each quantity:

1. Planck's constant ($h$): From $E = h\nu$, where $E$ is energy and $\nu$ is frequency. $[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$.
2. Energy ($E$): $[ML^2T^{-2}]$.
3. Momentum ($p$): $p = mv$, so $[p] = [M][LT^{-1}] = [MLT^{-1}]$.
4. Angular Momentum ($L$ or $J$): $L = I\omega = mr^2 \omega$, or $L = r \times p$. $[L] = [L][MLT^{-1}] = [ML^2T^{-1}]$.
5. Power ($P$): $P = W/t$, so $[P] = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$. Comparing the dimensions, Planck's constant $[ML^2T^{-1}]$ has the same dimensions as Angular Momentum $[ML^2T^{-1}]$.