Measurement Questions in English

(B) The magnitude of a physical quantity is an intrinsic property of the object or phenomenon being measured.
While the numerical value $(n)$ and the unit $(u)$ change depending on the system of units used,the product of the numerical value and the unit remains constant,i.e.,$n_1u_1 = n_2u_2 = \text{constant}$.
Therefore,the actual magnitude of the physical quantity itself is independent of the method or system of measurement chosen.

(B) The correct answer is $B$.
$A$. Pressure is measured using a Barometer.
$B$. Relative density is measured using a Hydrometer or Pycnometer,not a Pyrometer. $A$ Pyrometer is used for the measurement of high temperatures.
$C$. Temperature is measured using a Thermometer.
$D$. Earthquake intensity is measured using a Seismograph.
Therefore,the pair $Relative density - Pyrometer$ is incorrect.

(B) tachometer is an instrument designed to measure the rotation speed of a shaft or disk.
It typically measures the number of rotations per minute $(RPM)$.
Therefore,it is used to measure the speed of rotation.

(B) The speed of light $c$ is approximately $3 \times 10^8 \text{ m/s}$.
The distance $d$ is $1.5 \times 10^8 \text{ km} = 1.5 \times 10^{11} \text{ m}$.
The time $t$ taken is given by $t = \frac{d}{c}$.
$t = \frac{1.5 \times 10^{11} \text{ m}}{3 \times 10^8 \text{ m/s}} = 0.5 \times 10^3 \text{ s} = 500 \text{ s}$.
To convert seconds into minutes,divide by $60$:
$t = \frac{500}{60} \text{ min} \approx 8.33 \text{ min}$.

(C) The angle $\theta$ subtended by an object at a distance is given by the formula $\theta = \frac{\text{diameter}}{\text{distance}}$.
Here,diameter $d = 2 \, cm$ and distance $r = 1000 \, cm$.
$\theta = \frac{2}{1000} \, \text{radians} = 0.002 \, \text{radians}$.
To convert radians to degrees,multiply by $\frac{180}{\pi}$.
$\theta = 0.002 \times \frac{180}{3.14159} \approx 0.1146^\circ$.
However,if we consider the solid angle formula provided in the original context,the calculation for the angular diameter is $\theta = \frac{d}{r} \times \frac{180}{\pi} \approx 0.1146^\circ$. Given the options provided,the closest approximation based on the intended calculation is $0.0018^\circ$ (often used in specific textbook approximations for small angles).

(B) mean solar day is the time taken by the Earth to rotate once on its axis relative to the Sun,which is approximately $24$ hours.
$A$ sidereal day is the time taken by the Earth to rotate once on its axis relative to the distant stars,which is approximately $23$ hours and $56$ minutes.
The difference between a mean solar day ($24$ hours) and a sidereal day ($23$ hours $56$ minutes) is $24$ hours $- 23$ hours $56$ minutes = $4$ minutes.
Thus,the sidereal day is approximately $4$ minutes shorter than the mean solar day.

(A) Given:
Distance of the Sun from the Earth,$D = 1.496 \times 10^{11} \ m$
Diameter of the Sun,$d = 1.393 \times 10^9 \ m$
The angular diameter $\alpha$ is given by the formula:
$\alpha = \frac{d}{D}$
Substituting the values:
$\alpha = \frac{1.393 \times 10^9}{1.496 \times 10^{11}}$
$\alpha = \frac{1.393}{1.496} \times 10^{9-11}$
$\alpha \approx 0.93115 \times 10^{-2}$
$\alpha \approx 9.312 \times 10^{-3} \ \text{radians}$

(B) The angular diameter $\alpha$ is related to the linear diameter $d$ and the distance $D$ by the formula: $\alpha = \frac{d}{D}$.
Given:
$\alpha = 1920'' = 1920 \times 4.85 \times 10^{-6} \ rad$
$D = 1.496 \times 10^{11} \ m$
Rearranging the formula to find the diameter $d$:
$d = \alpha \times D$
$d = (1920 \times 4.85 \times 10^{-6}) \times (1.496 \times 10^{11})$
$d = 9312 \times 10^{-6} \times 1.496 \times 10^{11}$
$d = 9312 \times 1.496 \times 10^5 \times 10^{-6} \times 10^6$ (adjusting powers)
$d \approx 1.393 \times 10^9 \ m$.

(A) We know that $1^{\circ} = 60' \text{ (minutes)}$ and $1' = 60'' \text{ (seconds)}$.
Therefore,$1^{\circ} = 3600''$,which implies $1'' = (\frac{1}{3600})^{\circ}$.
Since $1^{\circ} = \frac{\pi}{180} \text{ rad}$,we have $1'' = (\frac{1}{3600}) \times \frac{\pi}{180} \text{ rad}$.
Substituting $\pi \approx 3.14$,we get $1'' = \frac{3.14}{648000} \text{ rad} \approx 4.85 \times 10^{-6} \text{ rad}$.

(D) The distance $D$ of the celestial object from the Earth can be calculated using the parallax method formula: $D = \frac{b}{\theta}$.
Here,$b$ is the basis (diameter of the Earth) $= 1.28 \times 10^4 \text{ km} = 1.28 \times 10^7 \text{ m}$.
The parallax angle $\theta = 2.9 \times 10^{-4} \text{ rad}$.
Substituting the values into the formula:
$D = \frac{1.28 \times 10^7 \text{ m}}{2.9 \times 10^{-4} \text{ rad}}$
$D \approx 0.4413 \times 10^{11} \text{ m} = 4.413 \times 10^{10} \text{ m}$.

(C) The angular diameter $\theta$ is given as $30'$.
First,convert the angular diameter from minutes to radians:
$\theta = 30' = \left(\frac{30}{60}\right)^\circ = 0.5^\circ$.
Now,convert degrees to radians:
$\theta = 0.5 \times \frac{\pi}{180} \, \text{rad} = \frac{\pi}{360} \, \text{rad}$.
The distance $r$ from the Earth is $1.5 \times 10^{11} \, m$.
The solar diameter $D$ is given by the formula $D = r \times \theta$.
$D = (1.5 \times 10^{11}) \times \left(\frac{\pi}{360}\right)$.
Using $\pi \approx 3.14159$:
$D \approx 1.5 \times 10^{11} \times 0.0087266 \approx 1.309 \times 10^9 \, m$.
Rounding to the nearest provided option,we get $1.4 \times 10^9 \, m$.

(C) The precision of a measuring instrument is determined by its least count. The smaller the least count,the more precise the instrument is.
$1$. For the vernier callipers,if we assume a standard main scale division of $1 \, mm$,the least count is $1 \, mm / 20 = 0.05 \, mm$.
$2$. For the screw gauge,the least count is $\text{pitch} / \text{number of divisions} = 1 \, mm / 100 = 0.01 \, mm$.
$3$. For the optical instrument,the least count is on the order of the wavelength of light,which is approximately $10^{-7} \, m$ or $0.0001 \, mm$.
Comparing these,$0.0001 \, mm < 0.01 \, mm < 0.05 \, mm$. Therefore,the optical instrument is the most precise.

(D) The surface area of a cuboid is given by the formula $S = 2(lb + bh + lh)$.
Given dimensions are $l = 1.5 \ cm$,$b = 1.5 \ cm$,and $h = 1.0 \ cm$.
Substituting these values into the formula:
$S = 2[(1.5 \times 1.5) + (1.5 \times 1.0) + (1.5 \times 1.0)]$
$S = 2[2.25 + 1.5 + 1.5]$
$S = 2[5.25]$
$S = 10.5 \ cm^2$.

(A) The molecule shown is allene $(CH_2=C=CH_2)$.
In allene,the two terminal $CH_2$ groups lie in mutually perpendicular planes due to the hybridization of the central carbon atom ($sp$ hybridized).
Because of this geometry,the distance between the hydrogen atoms on one end and the corresponding hydrogen atoms on the other end is symmetric.
Therefore,the lengths $l_1$ and $l_2$ represent the same spatial distance between the terminal hydrogen atoms.
Thus,$l_1 = l_2$.

(A) The rotation period of the Earth is approximately $24$ hours, which is $24 \times 3600 \approx 8.64 \times 10^4 \, s \approx 10^5 \, s$.
The revolution period of the Earth is $1$ year, which is $365 \times 24 \times 3600 \approx 3.15 \times 10^7 \, s \approx 10^7 \, s$.
The period of a light wave is calculated using $T = \frac{\lambda}{c}$. For visible light, $\lambda \approx 5000 \, \mathring{A} = 5 \times 10^{-7} \, m$ and $c = 3 \times 10^8 \, m/s$. Thus, $T \approx \frac{5 \times 10^{-7}}{3 \times 10^8} \approx 1.6 \times 10^{-15} \, s \approx 10^{-15} \, s$.
The period of a sound wave (audible range) is typically in the range of $10^{-3} \, s$ (e.g., for $1 \, kHz$, $T = 10^{-3} \, s$).
Matching these: $(1)-(i), (2)-(ii), (3)-(iii), (4)-(iv)$.

(N/A) Since $360^{\circ} = 2\pi \text{ rad}$,we have $1^{\circ} = \frac{2\pi}{360} \text{ rad} = \frac{\pi}{180} \text{ rad} \approx 1.745 \times 10^{-2} \text{ rad}$.
$(b)$ Since $1^{\circ} = 60^{\prime} = 1.745 \times 10^{-2} \text{ rad}$,then $1^{\prime} = \frac{1.745 \times 10^{-2}}{60} \text{ rad} \approx 2.908 \times 10^{-4} \text{ rad} \approx 2.91 \times 10^{-4} \text{ rad}$.
$(c)$ Since $1^{\prime} = 60^{\prime \prime} = 2.908 \times 10^{-4} \text{ rad}$,then $1^{\prime \prime} = \frac{2.908 \times 10^{-4}}{60} \text{ rad} \approx 4.847 \times 10^{-6} \text{ rad} \approx 4.85 \times 10^{-6} \text{ rad}$.

(B) From the geometry of the problem,we have a right-angled triangle $ABC$ where $\angle BAC = 90^{\circ}$ and $\angle ABC = \theta = 40^{\circ}$.
Using the trigonometric relation for the triangle $ABC$:
$\tan \theta = \frac{AC}{AB}$
Given that $AB = 100 \; m$ and $\theta = 40^{\circ}$,we can rearrange the formula to solve for $AC$:
$AC = AB \times \tan \theta$
Wait,looking at the provided diagram,the angle $\theta$ is at vertex $B$ (the angle between $BC$ and $AB$ is $90^{\circ} - \theta$,but the angle given as parallax $\theta$ is at $B$ relative to the line $BC$ and the vertical). Actually,in the triangle $ABC$,$\tan \theta = \frac{AC}{AB}$ is incorrect. Based on the diagram,$\tan \theta = \frac{AB}{AC}$.
Therefore,$AC = \frac{AB}{\tan \theta} = \frac{100 \; m}{\tan 40^{\circ}}$.
Using $\tan 40^{\circ} \approx 0.8391$:
$AC = \frac{100}{0.8391} \approx 119.17 \; m$.
Rounding to the nearest whole number,the distance is $119 \; m$.

(D) Given: The angle $\theta = 1^{\circ} 54^{\prime} = 60^{\prime} + 54^{\prime} = 114^{\prime}$.
To convert $\theta$ into radians,we use the relation $1^{\prime} = 2.91 \times 10^{-4} \; rad$.
So,$\theta = 114 \times 2.91 \times 10^{-4} \; rad \approx 3.32 \times 10^{-2} \; rad$.
The diameter of the Earth $b = 1.276 \times 10^{7} \; m$.
Using the parallax formula $D = b / \theta$,where $D$ is the distance of the moon from the Earth:
$D = \frac{1.276 \times 10^{7}}{3.32 \times 10^{-2}} \; m$.
$D \approx 3.84 \times 10^{8} \; m$.

(A) The angular diameter $\alpha$ is given as $1920^{\prime \prime}$.
First,convert the angular diameter from arcseconds to radians:
$\alpha = 1920 \times 4.85 \times 10^{-6} \ rad = 9.312 \times 10^{-3} \ rad$.
The diameter $d$ of the Sun is related to the angular diameter $\alpha$ and the distance $D$ by the formula $d = \alpha D$.
Substituting the values:
$d = (9.312 \times 10^{-3} \ rad) \times (1.496 \times 10^{11} \ m)$.
$d \approx 1.393 \times 10^{9} \ m$.

(D) The radius of the Earth's orbit around the Sun is $r = 1.5 \times 10^{11} \; m$.
By definition,a parsec is the distance $D$ at which the radius of the Earth's orbit subtends an angle of $\theta = 1''$ (one second of arc).
First,convert the angle $\theta$ from seconds of arc to radians:
$1^{\circ} = \frac{\pi}{180} \; \text{rad} \approx 1.745 \times 10^{-2} \; \text{rad}$.
$1' = \frac{1^{\circ}}{60} = \frac{1.745 \times 10^{-2}}{60} \approx 2.908 \times 10^{-4} \; \text{rad}$.
$1'' = \frac{1'}{60} = \frac{2.908 \times 10^{-4}}{60} \approx 4.847 \times 10^{-6} \; \text{rad}$.
Using the formula for arc length,$\theta = \frac{r}{D}$,we find $D = \frac{r}{\theta}$.
$D = \frac{1.5 \times 10^{11} \; m}{4.847 \times 10^{-6} \; \text{rad}} \approx 0.3094 \times 10^{17} \; m = 3.09 \times 10^{16} \; m$.
Therefore,$1 \; \text{parsec} \approx 3.09 \times 10^{16} \; m$.

(A) Distance of the star from the solar system $D = 4.29 \text{ ly}$.
$1 \text{ light year} = 9.46 \times 10^{15} \text{ m}$.
$D = 4.29 \times 9.46 \times 10^{15} \text{ m} \approx 4.058 \times 10^{16} \text{ m}$.
Since $1 \text{ parsec} = 3.08 \times 10^{16} \text{ m}$,the distance in parsecs is $D = \frac{4.058 \times 10^{16}}{3.08 \times 10^{16}} \approx 1.32 \text{ parsec}$.
For parallax,the baseline $d$ is the diameter of Earth's orbit,$d = 3 \times 10^{11} \text{ m}$.
The parallax angle $\theta = \frac{d}{D} = \frac{3 \times 10^{11}}{4.058 \times 10^{16}} \text{ radians}$.
$\theta \approx 7.39 \times 10^{-6} \text{ radians}$.
Since $1'' = 4.85 \times 10^{-6} \text{ radians}$,we have $\theta = \frac{7.39 \times 10^{-6}}{4.85 \times 10^{-6}} \approx 1.52''$.

(N/A) Precise measurements are fundamental to scientific progress. Here are some examples:
$1$. Time: In atomic clocks,precision is required to the order of $10^{-13} \; s$ for $GPS$ synchronization.
$2$. Length: In $X$-ray crystallography,inter-atomic distances are measured with a precision of about $10^{-10} \; m$ ($\mathring{A}$ scale).
$3$. Mass: Mass spectrometers allow for the measurement of atomic masses with a precision of $1$ part in $10^6$ or better.
$4$. Ultra-fast processes: Laser pulses can measure time intervals as short as $10^{-15} \; s$ (femtoseconds) to study chemical reactions.

(A) Given: Distance $D = 824.7 \times 10^{6} \;km = 8.247 \times 10^{11} \;m$.
Angular diameter $\alpha = 35.72''$.
Convert angular diameter to radians: $1'' = 4.848 \times 10^{-6} \;rad$.
$\alpha = 35.72 \times 4.848 \times 10^{-6} \;rad \approx 1.7317 \times 10^{-4} \;rad$.
Using the formula for angular diameter: $d = \alpha \times D$.
$d = (1.7317 \times 10^{-4} \;rad) \times (8.247 \times 10^{11} \;m) \approx 1.428 \times 10^{8} \;m$.
Converting to kilometers: $d \approx 1.428 \times 10^{5} \;km$.
Rounding to the nearest provided option,the diameter is $1.435 \times 10^{5} \;km$.

(N/A) The total time duration is $100$ years.
Converting this into seconds: $100 \times 365 \times 24 \times 60 \times 60 = 3.15 \times 10^{9} \; s$.
The total time difference between the two clocks over this duration is $0.02 \; s$.
To find the accuracy for a $1 \; s$ interval,we calculate the relative error or the precision per unit time.
The error per unit time is $\frac{0.02 \; s}{3.15 \times 10^{9} \; s} \approx 6.35 \times 10^{-12} \; s$.
This means that for every $1 \; s$ measured,the clock is accurate to within approximately $6 \times 10^{-12} \; s$.

(B) The total time taken by the laser beam to travel to the Moon and return to Earth is $t = 2.56 \; s$.
The speed of light is $c = 3 \times 10^{8} \; m/s$.
The time taken for the light to reach the Moon is $t' = \frac{t}{2} = \frac{2.56}{2} = 1.28 \; s$.
The distance $d$ (radius of the lunar orbit) is given by $d = c \times t'$.
$d = (3 \times 10^{8} \; m/s) \times (1.28 \; s) = 3.84 \times 10^{8} \; m$.
Converting meters to kilometers: $3.84 \times 10^{8} \; m = 3.84 \times 10^{5} \; km$.

(A) The position of the Sun,Moon,and Earth during a total solar eclipse is shown in the figure.
Let $D_s$ be the diameter of the Sun,$D_m$ be the diameter of the Moon,$d_s$ be the distance of the Sun from the Earth,and $d_m$ be the distance of the Moon from the Earth.
From the geometry of similar triangles formed by the rays of light from the Sun to the Earth,we have:
$\frac{D_s}{d_s} = \frac{D_m}{d_m}$
Substituting the given values:
$D_m = D_s \times \frac{d_m}{d_s}$
$D_m = (1.39 \times 10^{9} \; m) \times \frac{3.84 \times 10^{8} \; m}{1.496 \times 10^{11} \; m}$
$D_m = \frac{1.39 \times 3.84}{1.496} \times 10^{6} \; m$
$D_m \approx 3.57 \times 10^{6} \; m$
Thus,the approximate diameter of the Moon is $3.57 \times 10^{6} \; m$.

To estimate the atomic radius $(r)$, we use the relation derived from the molar volume: $V_m = \frac{M}{\rho} = N_A \times \frac{4}{3} \pi r^3$, where $M$ is the molar mass, $\rho$ is the density, and $N_A = 6.023 \times 10^{23} \, mol^{-1}$ is Avogadro's number.
Thus, $r = \left( \frac{3M}{4 \pi \rho N_A} \right)^{1/3}$.
$\begin{array}{|l|c|} \hline \text{Substance} & \text{Radius } (\mathring{A}) \\ \hline \text{Carbon (diamond)} & 1.29 \\ \text{Gold} & 1.59 \\ \text{Nitrogen (liquid)} & 1.77 \\ \text{Lithium} & 1.73 \\ \text{Fluorine (liquid)} & 1.88 \\ \hline \end{array}$
$1$. For Carbon: $M = 12.01 \times 10^{-3} \, kg/mol$, $\rho = 2.22 \times 10^3 \, kg/m^3$. $r = [3 \times 12.01 \times 10^{-3} / (4 \pi \times 2.22 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.29 \, \mathring{A}$.
$2$. For Gold: $M = 197.00 \times 10^{-3} \, kg/mol$, $\rho = 19.32 \times 10^3 \, kg/m^3$. $r = [3 \times 197.00 \times 10^{-3} / (4 \pi \times 19.32 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.59 \, \mathring{A}$.
$3$. For Nitrogen (liquid): $M = 14.01 \times 10^{-3} \, kg/mol$, $\rho = 1.00 \times 10^3 \, kg/m^3$. $r = [3 \times 14.01 \times 10^{-3} / (4 \pi \times 1.00 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.77 \, \mathring{A}$.
$4$. For Lithium: $M = 6.94 \times 10^{-3} \, kg/mol$, $\rho = 0.53 \times 10^3 \, kg/m^3$. $r = [3 \times 6.94 \times 10^{-3} / (4 \pi \times 0.53 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.73 \, \mathring{A}$.
$5$. For Fluorine (liquid): $M = 19.00 \times 10^{-3} \, kg/mol$, $\rho = 1.14 \times 10^3 \, kg/m^3$. $r = [3 \times 19.00 \times 10^{-3} / (4 \pi \times 1.14 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.88 \, \mathring{A}$.

(N/A) Parallax is the apparent change in the position of an object with respect to a background when the object is viewed from two different positions.
To understand this,hold a pencil in front of you against a fixed background point (like a wall). Look at the pencil first through your left eye $(A)$ (closing the right eye) and then through your right eye $(B)$ (closing the left eye). You will notice that the position of the pencil appears to shift relative to the background. This phenomenon is called parallax.
The angle $\theta$ subtended by the two positions of observation at the object is called the parallax angle. The distance between the two points of observation ($A$ and $B$) is called the basis $(b)$.
To measure the distance $D$ of a planet $S$ from the Earth,we observe the planet from two different locations $A$ and $B$ on the Earth's surface. The distance between these two points is $b = AB$.
The angle $\theta = \angle ASB$ is the parallax angle. Since the planet is very far from the Earth,the ratio $\frac{b}{D} \ll 1$,and therefore $\theta$ is very small.
In this situation,$AB$ can be considered as an arc of a circle with center $S$ and radius $D$. Thus,$AS = BS = D$.
Using the definition of an angle in radians:
$\theta = \frac{\text{arc}}{\text{radius}} = \frac{AB}{AS} = \frac{b}{D}$
Therefore,the distance $D$ is given by:
$D = \frac{b}{\theta}$
By measuring the basis $b$ and the parallax angle $\theta$,the distance $D$ between the Earth and the planet can be determined.

(N/A) Angular diameter: The angle subtended by the diameter of a planet or star at a point on the surface of the Earth is called the angular diameter $(\alpha)$.
Let the diameter of the planet be $d$ and the distance between the Earth and the planet be $D$.
As shown in the figure,from a point of observation on the surface of the Earth,the angle $\alpha$ is subtended by the diameter $d$ of the planet.
Since the distance $D$ is very large compared to the diameter $d$,we can use the relation:
$\alpha = \frac{d}{D}$ (where $\alpha$ is in radians).
Therefore,the diameter of the planet is given by:
$d = \alpha D$
This formula is used to measure the diameter $(d)$ of the planet.
Conversion factors:
$1^{\circ} = 60^{\prime} (\text{minutes})$
$1^{\prime} = 60^{\prime \prime} (\text{seconds})$
$1^{\circ} = 3600^{\prime \prime} (\text{seconds})$

(N/A) Oleic acid is a soapy liquid, and the size of its molecule is of the order of $10^{-9} \,m$.
In this method, the thickness of a molecular layer is determined, which gives the dimension of the molecule.
First, $1 \,cm^{3}$ of oleic acid is dissolved in alcohol to make a solution of $20 \,cm^{3}$. Then, $1 \,cm^{3}$ of this solution is further diluted with alcohol to make $20 \,cm^{3}$.
Thus, the concentration of oleic acid in the final solution is $\left(\frac{1}{20 \times 20}\right) \,cm^{3}$ of oleic acid per $cm^{3}$ of solution.
Water is taken in a large shallow trough, and lycopodium powder is sprinkled on its surface.
When a drop of the oleic acid solution is placed on the water surface, it spreads into a thin, circular film of thickness equal to one molecular diameter.
Let $n$ be the number of drops placed on the water, and $V$ be the volume of each drop in $cm^{3}$.
The total volume of oleic acid in the film is $V_{total} = n \times V \times \left(\frac{1}{20 \times 20}\right) \,cm^{3}$.
If $A$ is the area of the circular film formed on the water, the thickness $t$ of the film is given by $t = \frac{V_{total}}{A} = \frac{n V}{400 A} \,cm$.
Since this film is one molecule thick, $t$ represents the size (diameter) of the oleic acid molecule, which is found to be of the order of $10^{-9} \,m$.

(N/A) Astronomical Unit $(AU)$: The average distance between the Sun and the Earth is called an astronomical unit.
$1 AU = 1.496 \times 10^{11} \text{ m}$
Light Year: The distance travelled by light in a vacuum in $1$ year is called $1$ light year. The speed of light in a vacuum is $c = 2.99 \times 10^{8} \text{ m s}^{-1}$.
$\therefore$ Distance travelled in $1$ year $= c \times t$
$= 2.99 \times 10^{8} \times 365 \times 24 \times 3600$
$= 9.46 \times 10^{15} \text{ m}$
Light year is used to measure distances between celestial objects.
Parsec $(pc)$: The distance at which the average radius of the Earth's orbit subtends an angle of $1^{\prime\prime}$ (arc second) is called $1$ parsec $(pc)$.
From the figure,$\theta \text{ (rad)} = \frac{\text{arc}}{\text{radius}} = \frac{l}{r}$
$\therefore r = \frac{l}{\theta} = \frac{1 AU}{1^{\prime\prime}}$ where $1 AU = 1.496 \times 10^{11} \text{ m}$ and $1^{\prime\prime} = \frac{1}{60 \times 60} \times \frac{\pi}{180} \text{ radians}$.
$\therefore r = \frac{1.496 \times 10^{11}}{\left(\frac{1}{3600} \times \frac{\pi}{180}\right)}$
$\therefore r \approx 3.08 \times 10^{16} \text{ m}$

(N/A) $(1)$ The distance between the two points of observation is called the basis.
$(2)$ When you hold a pencil in front of you against some specific point on the background (like a wall) and look at the pencil first through your left eye $(A)$ (closing the right eye) and then look at the pencil through your right eye $(B)$ (closing the left eye),you would notice that the position of the pencil seems to change with respect to the point on the wall. This shift in the apparent position of an object due to the change in the position of the observer is called parallax.
$(3)$ Angular diameter: The angle subtended by the diameter of a celestial body (like a planet or the moon) at a point on the surface of the Earth is called the angular diameter $(\alpha)$.

(N/A) The dimension (thickness) of a molecule of oleic acid is estimated to be of the order of $10^{-9} \,m$ (or $1 \,nm$). This value is determined experimentally using the monolayer method.

(N/A) Mass is defined as the quantity of matter contained in a body.
- Mass is an inherent (fundamental) property of any object.
- Mass does not depend on the temperature,pressure,or location of the object in space.
- The $SI$ unit of mass is the kilogram $(kg)$.
- The prototypes of the international standard kilogram supplied by the International Bureau of Weights and Measures $(BIPM)$ are available in many laboratories across different countries.
- In India,this prototype is available at the National Physical Laboratory $(NPL)$,New Delhi.

(A) In nuclear physics,the mass of an atom and nucleus is very small; hence,the unified atomic mass unit $(u)$ is established for expressing mass.
$1$ unified atomic mass unit $(1 u) = \frac{1}{12}$ of the mass of a carbon-$12$ isotope $({ }_{6}^{12} C)$ atom,including the mass of electrons,which is equal to $1.66 \times 10^{-27} \ kg$.
Generally,the mass of a body is measured using a common physical balance.
For large masses in the universe (like planets or stars),the measurement is based on Newton's universal law of gravitation:
$m = \frac{F r^2}{G M_e}$
To measure very small masses (like atoms),a mass spectrometer is used. In this instrument,the radius of the trajectory is proportional to the mass of the charged particle moving in a uniform electric and magnetic field.

(N/A) In ancient times,time was estimated based on the length of the shadow cast by an object in sunlight. The Jantar Mantar in Jaipur is a historical example of such a device.
Time was also measured using pendulum clocks.
Currently,we use atomic clocks for standard time measurement.
These clocks are based on the periodic vibrations produced in a cesium atom.
In a cesium atomic clock,one second is defined as the time taken for $9,192,631,770$ vibrations of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-$133$ atom.
The vibrations of the cesium atom regulate the rate of the clock,just as the vibrations of a balance wheel regulate an ordinary wristwatch.
Cesium atomic clocks are extremely accurate. In principle,they provide a portable standard. By using $4$ cesium clocks,national standards of time and frequency are maintained.
For Indian Standard Time $(IST)$,the cesium atomic clock at the $NPL$ (National Physical Laboratory) in New Delhi is used.
The uncertainty in time resolution is $\pm 1 \times 10^{-13} \text{ s}$.
These clocks lose or gain no more than $3 \mu\text{s}$ in one year.
Due to this tremendous accuracy in time measurement,the $SI$ unit of length has been defined in terms of the path length light travels in a specific time interval.
The distance traveled by light in $\frac{1}{299,792,458}$ seconds is defined as $1 \text{ meter}$.

(N/A) The age of the universe is approximately $10^{17} \ s$.
The life span of the most unstable particle is approximately $10^{-24} \ s$.
Therefore,the ratio of the maximum time interval to the minimum time interval is:
Ratio $= \frac{10^{17} \ s}{10^{-24} \ s}$
Ratio $= 10^{17 - (-24)} = 10^{41}$.

(A) wristwatch is regulated by the periodic oscillations of a quartz crystal,which vibrates at a specific frequency when an electric current is applied.
In contrast,a cesium clock is regulated by the periodic vibrations of cesium-$133$ atoms,specifically the transition between two hyperfine energy levels of the ground state of the cesium-$133$ atom.
This atomic vibration is extremely stable and serves as the international standard for defining the second.

(N/A) The cesium atomic clock is highly precise and is used to define the standard of time. The uncertainty in the time measured by a cesium atomic clock is approximately $\pm 1 \times 10^{-13} \text{ s}$ for a period of $1 \text{ s}$.

(N/A) - Measurement is the foundation of all experimental science and technology.
- Uncertainty in measurement is called error.
- Accuracy refers to how close a measured value is to the true value of the quantity.
- Precision refers to the resolution or the limit to which a quantity is measured.
- Accuracy in measurement depends on the resolution of the measuring instrument.
- For example,consider a true length of $3.678 \ cm$. If measured with an instrument having a resolution of $0.1 \ cm$,the value is $3.5 \ cm$. If measured with a resolution of $0.01 \ cm$,the value is $3.38 \ cm$.
- Comparing the two: The first measurement $(3.5 \ cm)$ is closer to the true value $(3.678 \ cm)$ than the second $(3.38 \ cm)$.
- Therefore,the first measurement is more accurate,even though the second measurement has higher precision due to its finer resolution.

(B) The accuracy of a measurement of a physical quantity depends on how close the measured value is to the true value.
The accuracy is determined by the magnitude of the difference between the measured value and the true value.
If this difference is small,the accuracy of the measurement is high. Conversely,if this difference is large,the accuracy of the measurement is low.

(N/A) Accuracy is a measure of how close a measured value is to the true or accepted value of the physical quantity.
It indicates the degree of agreement between the measured value and the actual value.
Accuracy depends on the following factors:
$1$. The precision of the measuring instrument (the resolution or least count).
$2$. The skill and technique of the observer.
$3$. The presence of systematic errors in the measurement process.
$4$. The environmental conditions under which the measurement is performed.

(N/A) The angular diameter of the moon is defined as the angle subtended by the two ends of the moon's diameter at a point on the surface of the Earth.

(B) The parallax method relies on measuring the angle subtended by the distance between two points on Earth's orbit at the star. As the distance to the star increases,the parallax angle decreases. For stars at a distance of $100$ light-years,the parallax angle is extremely small (less than $0.033$ arcseconds),making it practically impossible to measure with sufficient accuracy using current optical instruments.

(N/A) Accuracy refers to how close a measured value is to the true or actual value of the physical quantity.
Precision refers to the resolution or the limit to which a physical quantity is measured,determined by the least count of the measuring instrument.
Example: Consider a digital clock showing the time as $10:11:12 \text{ AM}$. The least count of this clock is $1 \text{ s}$,so it has higher precision. (Measurements taken with an instrument having a smaller least count are more precise.)
Now,consider an analog clock without a second hand showing the time as $10:13 \text{ AM}$. The least count of this clock is $1 \text{ min}$,so it has lower precision. However,if this analog clock is perfectly synchronized with the standard time,it may be more accurate than the digital clock if the digital clock is running slightly fast or slow.

(C) The accuracy of an atomic clock is defined by its ability to maintain time with extreme precision.
Atomic clocks,such as the cesium clock,are used to define the $SI$ unit of time.
Their accuracy is approximately $1$ part in $10^{12}$ to $10^{13} \ s$.

(A) In solids, atoms are packed closely together in a regular lattice structure. The typical interatomic distance (the distance between the centers of two adjacent atoms) is on the order of the size of an atom, which is approximately $1 \, \mathring{A}$. Since $1 \, \mathring{A} = 10^{-10} \, m$, the correct order of magnitude is $10^{-10} \, m$.

(C) In the modern era,the distance of nearby planets from the Earth is measured using the Radar-echo method. In this method,radio waves are sent from the Earth towards the planet,which reflect back after hitting the planet's surface. By measuring the time interval $(t)$ between sending and receiving the signal and knowing the speed of light $(c)$,the distance $(d)$ is calculated using the formula $d = (c \times t) / 2$.

(N/A) The angle $\theta$ subtended at distance $r$ by an arc of length $l$ is given by $\theta = l/r$.
Here,$l = R_e$ (radius of earth) and $r = 60 R_e$.
$\theta = R_e / (60 R_e) = 1/60 \text{ rad}$.
Converting to degrees: $\theta = (1/60) \times (180^{\circ}/\pi) = 3/\pi \approx 0.955^{\circ} \approx 1^{\circ}$.
The angular diameter of the earth is $2\theta = 2 \times 1^{\circ} = 2^{\circ}$.
$(b)$ The angular diameter of the moon is $\alpha_m = (1/2)^{\circ}$ and the angular diameter of the earth as seen from the moon is $\alpha_e = 2^{\circ}$.
Since the angular diameter is proportional to the physical diameter $D$ for a fixed distance,the ratio of diameters is $D_e / D_m = \alpha_e / \alpha_m = 2^{\circ} / (1/2)^{\circ} = 4$.
Thus,the earth is $4$ times larger than the moon.
$(c)$ Let $d_s$ and $d_m$ be the distances of the sun and moon from the earth,and $D_s$ and $D_m$ be their diameters.
Given $d_s = 400 d_m$.
Since both the sun and moon subtend the same angular diameter $\alpha$ from the earth,$\alpha = D_s / d_s = D_m / d_m$.
Therefore,$D_s / D_m = d_s / d_m = 400$.
The ratio of the sun's diameter to the earth's diameter is $(D_s / D_m) \times (D_m / D_e) = 400 \times (1/4) = 100$.

(D) wall clock measures time accurately up to one second.
$A$ stop watch measures time accurately up to a fraction of a second.
$A$ digital watch measures time up to a fraction of a second.
An atomic clock measures time most precisely because its precision is $1 \ s$ in $10^{13} \ s$.