Measurement Questions in English

(N/A) Consider the diagram provided.
Let $D_{me}$ be the distance of the moon from the earth.
Let $D_{se}$ be the distance of the sun from the earth.
Let $R_m$ be the radius of the moon and $R_s$ be the radius of the sun.
Since the moon covers the sun,the angular diameter $\theta$ subtended by both at the earth is the same.
Using the formula for angular diameter,$\theta = \frac{\text{diameter}}{\text{distance}}$,we have:
$\theta = \frac{2R_s}{D_{se}} = \frac{2R_m}{D_{me}}$
Dividing both sides by $2$,we get:
$\frac{R_s}{D_{se}} = \frac{R_m}{D_{me}}$
Therefore,the relation is:
$\frac{R_s}{R_m} = \frac{D_{se}}{D_{me}}$
This shows that the ratio of the sizes of the sun and moon is equal to the ratio of their respective distances from the earth.

(A) The solid angle $\Omega$ subtended by an area $A$ at a point at distance $r$ is given by the formula $\Omega = \frac{A}{r^2}$.
Given,the area $A = 1 \, cm^2$ and the distance $r = 5 \, cm$.
Substituting these values into the formula:
$\Omega = \frac{1 \, cm^2}{(5 \, cm)^2} = \frac{1}{25} \, sr$.
$\Omega = 0.04 \, sr$.
Thus,the solid angle subtended is $0.04 \, sr$.

(C) Oleic acid is dissolved in alcohol because it does not dissolve in water,allowing it to form a distinct film on the water surface.
$(b)$ Lycopodium powder is sprinkled on the water surface to make the boundary of the oleic acid film visible. As the oleic acid spreads,it pushes the powder aside,creating a clear circular area that can be measured.
$(c)$ The concentration of the solution is: First dilution: $1\, mL$ in $20\, mL$ total. Second dilution: $1\, mL$ of the first solution in $20\, mL$ total. Thus,the volume of oleic acid per $mL$ of the final solution is $\frac{1}{20} \times \frac{1}{20} = \frac{1}{400}\, mL$.
$(d)$ The volume of $n$ drops can be calculated by using a burette or a measuring cylinder to find the total volume of a known number of drops and then dividing by $n$.
$(e)$ If $n$ drops make $1\, mL$,then the volume of one drop is $\frac{1}{n}\, mL$. Since the concentration is $\frac{1}{400}\, mL$ of oleic acid per $mL$ of solution,the volume of oleic acid in one drop is $\frac{1}{400n}\, mL$.

(N/A) By definition,$1$ parsec is the distance at which an arc of length $1 AU$ subtends an angle of $1^{\prime \prime}$.
Since $1^{\circ} = 3600^{\prime \prime}$ and $1^{\circ} = \frac{\pi}{180} \text{ rad}$,we have $1^{\prime \prime} = \frac{\pi}{180 \times 3600} \text{ rad}$.
Thus,$1 \text{ parsec} = \frac{1 AU}{1^{\prime \prime}} = \frac{180 \times 3600}{\pi} AU \approx 2.06 \times 10^{5} AU$.
$(b)$ The angular diameter of the Sun is $0.5^{\circ} = 30^{\prime}$ at $1 AU$. At a distance of $2$ parsecs $(2 \times 2.06 \times 10^{5} AU)$,the angular diameter $\alpha$ of the star is $\alpha = \frac{0.5^{\circ}}{2 \times 2.06 \times 10^{5}} \approx 1.21 \times 10^{-6} \text{ degrees} \approx 0.0044^{\prime \prime}$.
With $100$ magnification,the apparent size is $100 \times 0.0044^{\prime \prime} = 0.44^{\prime \prime}$. Since $0.44^{\prime \prime} < 1^{\prime}$,the star will not be resolved and will appear as a point source due to atmospheric turbulence.
$(c)$ Mars diameter is $0.5 \times D_{Earth}$. Angular size $\theta = \frac{\text{Diameter}}{\text{Distance}}$. At $0.5 AU$,$\theta_{Mars} = \frac{0.5 \times D_{Earth}}{0.5 AU} = \frac{D_{Earth}}{1 AU} \approx 0.5^{\circ} = 30^{\prime}$.
With $100$ magnification,the apparent size is $100 \times 30^{\prime} = 3000^{\prime} = 50^{\circ}$.

(B) We know that $1^{\circ} = 60^{\prime}$ (minutes of arc).
Therefore,$1^{\prime} = (1/60)^{\circ}$.
To convert degrees to radians,we multiply by $\frac{\pi}{180}$.
So,$1^{\prime} = \left(\frac{1}{60}\right) \times \left(\frac{\pi}{180}\right) \text{ rad}$.
Using $\pi \approx 3.14159$,we get:
$1^{\prime} = \frac{3.14159}{10800} \text{ rad} \approx 2.9088 \times 10^{-4} \text{ rad}$.
Rounding to three significant figures,we get $2.91 \times 10^{-4} \text{ rad}$.

(B) The total volume of $100$ drops is $V_{total} = 100 \times \frac{4}{3} \pi r^3$.
Given $r = \left(\frac{3}{40 \pi}\right)^{1/3} \times 10^{-3} \, cm$,so $r^3 = \frac{3}{40 \pi} \times 10^{-9} \, cm^3$.
$V_{total} = 100 \times \frac{4}{3} \pi \times \frac{3}{40 \pi} \times 10^{-9} = 10^{-8} \, cm^3$.
The thickness of the film $t_T$ is given by $Area \times t_T = V_{total}$.
$4 \, cm^2 \times t_T = 10^{-8} \, cm^3 \implies t_T = 0.25 \times 10^{-8} = 25 \times 10^{-10} \, cm$.
Converting to meters: $t_T = 25 \times 10^{-12} \, m$.
The concentration of oleic acid is $0.01$,so the thickness of the oleic acid layer $t_0 = 0.01 \times t_T$.
$t_0 = 0.01 \times 25 \times 10^{-12} \, m = 25 \times 10^{-14} \, m$.
Comparing with $x \times 10^{-14} \, m$,we get $x = 25$.

(B) The radius $R$ of the circle is given by $R = \frac{\ell}{\theta}$,where $\ell$ is the arc length and $\theta$ is the angle in radians.
Given $\ell = 4.4 \; ly = 4.4 \times 9.46 \times 10^{15} \; m$.
The angle $\theta = 4'' = \frac{4}{3600} \times \frac{\pi}{180} \; rad$.
Substituting these,$R = \frac{4.4 \times 9.46 \times 10^{15}}{\frac{4}{3600} \times \frac{\pi}{180}} \; m$.
The circumference of the circle is $C = 2\pi R$.
For $4$ revolutions,the total distance $D = 4 \times 2\pi R = 8\pi R$.
The speed $v = 8 \; AU/s = 8 \times 1.5 \times 10^{11} \; m/s$.
The time taken $t = \frac{D}{v} = \frac{8\pi R}{v} = \frac{8\pi}{v} \times \frac{\ell}{\theta}$.
Substituting the values: $t = \frac{8 \times \pi \times 4.4 \times 9.46 \times 10^{15}}{8 \times 1.5 \times 10^{11} \times (\frac{4}{3600} \times \frac{\pi}{180})}$.
Simplifying the expression,we get $t \approx 4.5 \times 10^{10} \; s$.

(C) The angular diameter $\theta$ is given by the formula $\theta = \frac{d}{r}$,where $d$ is the diameter of the Sun and $r$ is the distance from the Earth.
First,convert the angular diameter from seconds $(s)$ to radians:
$\theta = 2000 \,s = \frac{2000}{60 \times 60} \text{ degrees} = \frac{2000}{3600} \times \frac{\pi}{180} \text{ radians}$.
Given $r = 1.5 \times 10^{11} \,m$.
Substituting the values into the formula $d = \theta \times r$:
$d = \left( \frac{2000}{3600} \times \frac{\pi}{180} \right) \times (1.5 \times 10^{11})$
$d = \left( \frac{20}{36} \times \frac{\pi}{180} \right) \times 1.5 \times 10^{11}$
$d \approx (0.555 \times 0.01745) \times 1.5 \times 10^{11}$
$d \approx 0.00969 \times 1.5 \times 10^{11} \approx 1.45 \times 10^{9} \,m$.

(C) The distance of the planet from the Earth is $D$ and the angle subtended by the diameter $d$ of the planet at the observatory is $\theta$.
Using the definition of plane angle,we have:
$\theta = \frac{\text{Arc length}}{\text{Radius}}$
Here,the arc length is the diameter $d$ of the planet and the radius is the distance $D$.
Therefore,$\theta = \frac{d}{D}$.
Rearranging the formula to solve for the diameter $d$,we get:
$d = D \theta$.

(B) Given,true length of wire $l_0 = 3.678 \,cm$.
Measurement with instrument $A$ is $l_A = 3.5 \,cm$.
Measurement with instrument $B$ is $l_B = 3.38 \,cm$.
Accuracy is determined by how close the measured value is to the true value. The absolute error for $A$ is $|3.678 - 3.5| = 0.178 \,cm$,and for $B$ is $|3.678 - 3.38| = 0.298 \,cm$. Since $0.178 < 0.298$,measurement $A$ is more accurate.
Precision is determined by the resolution or the number of decimal places of the instrument. Instrument $A$ measures up to one decimal place,whereas instrument $B$ measures up to two decimal places. Therefore,instrument $B$ is more precise.

(A) The least count of an instrument is the smallest value that can be measured by it.
Given the readings $1.24 \ mm, 1.25 \ mm, 1.23 \ mm$,and $1.21 \ mm$,all values are recorded up to two decimal places in millimeters.
This implies that the instrument can resolve changes in the hundredths place $(0.01 \ mm)$.
Therefore,the least count of the instrument is $0.01 \ mm$.