Units and Measurements

Concept. A physical quantity does not change when we change units: $n_1u_1=n_2u_2$. Density has dimensions $[ML^{-3}]$, so the unit of density is (unit of mass)/(unit of length)$^3$. Converting means comparing the old and new units of mass and length.

Step 1. Write the conversion rule for density. $$n_2=n_1\left(\frac{M_1}{M_2}\right)\left(\frac{L_1}{L_2}\right)^{-3}$$ Here $n_1=128$, $M_1=1$ kg, $L_1=1$ m, $M_2=50$ g, $L_2=25$ cm.

Step 2. Mass ratio. Put both in the same unit: $$\frac{M_1}{M_2}=\frac{1000\ \text{g}}{50\ \text{g}}=20$$

Step 3. Length ratio. $$\frac{L_1}{L_2}=\frac{100\ \text{cm}}{25\ \text{cm}}=4,\qquad\left(\frac{L_1}{L_2}\right)^{-3}=\frac{1}{64}$$

Step 4. Combine. $$n_2=128\times20\times\frac{1}{64}=40$$

Answer. $40$ (Option C). Check the direction: the new mass unit is smaller (raises the number $20$ times) but the new volume unit is $\tfrac1{64}$ m$^3$ (lowers it $64$ times), so the number drops overall.

Concept. Astronomical distances are measured in three special units. To judge Statement II we need their sizes in metres:

• Astronomical unit (mean Earth–Sun distance): $1\ \text{AU}\approx1.5\times10^{11}$ m.
• Light year (distance light travels in one year): $1\ \text{ly}=3\times10^8\times3.15\times10^7\approx9.46\times10^{15}$ m.
• Parsec (distance at which $1$ AU subtends $1''$): $1\ \text{pc}\approx3.08\times10^{16}$ m $\approx3.26$ ly.

Step 1. Statement I. AU, ly and pc are all units of length used for astronomical distances. Statement I is correct.

Step 2. Statement II. Comparing the values above, $$1\ \text{AU}\ (10^{11}\ \text{m})<1\ \text{ly}\ (10^{16}\ \text{m})<1\ \text{pc}\ (3\times10^{16}\ \text{m})$$ The statement puts the parsec below the light year, which is wrong. Statement II is incorrect.

Answer. Option B. Remember $1$ pc $\approx3.26$ ly, so the parsec is the largest of the three.

Concept. For a distant object, its size $D$, its distance $d$ and the angle $\theta$ it subtends are related by $D=d\,\theta$, where $\theta$ must be in radians (an arc of a circle of radius $d$).

Step 1. Convert one second of arc to radians. $1^\circ=\dfrac{\pi}{180}$ rad and $1^\circ=3600''$, so $$1''=\frac{\pi}{180\times3600}\ \text{rad}\approx4.85\times10^{-6}\ \text{rad}$$

Step 2. The angular diameter in radians. $$\theta=2000\times4.85\times10^{-6}=9.70\times10^{-3}\ \text{rad}$$

Step 3. Diameter. $$D=d\,\theta=1.5\times10^{11}\times9.70\times10^{-3}\approx1.45\times10^{9}\ \text{m}$$

Answer. $1.45\times10^{9}$ m (Option C). The common slip is to use the angle in degrees or seconds directly; $D=d\theta$ works only in radians.