Units and Measurement

(a) Volume of a cube (side 1 cm):
Side $a = 1 \text{ cm} = 10^{-2} \text{ m}$.
Volume $V = a^3 = (10^{-2})^3 = \mathbf{10^{-6} \text{ m}^3}$.

(b) Surface area of solid cylinder:
$r = 2.0 \text{ cm} = 20 \text{ mm}, h = 10.0 \text{ cm} = 100 \text{ mm}$.
$S = 2\pi r(r+h) = 2 \times 3.14 \times 20(20+100)$
$S = 15072 \text{ mm}^2$. Rounding to 2 significant figures (due to 2.0):
$\mathbf{1.5 \times 10^4 \text{ mm}^2}$.

(c) Speed:
$18 \text{ km h}^{-1} = 18 \times \frac{5}{18} \text{ m s}^{-1} = 5 \text{ m s}^{-1}$.
Distance in 1s = $\mathbf{5 \text{ m}}$.

(d) Relative Density:
Relative density = 11.3. Density of water = $1 \text{ g cm}^{-3}$.
$\rho = 11.3 \times 1 = \mathbf{11.3 \text{ g cm}^{-3}}$.
In SI units: $\rho = 11.3 \times 10^3 = \mathbf{1.13 \times 10^4 \text{ kg m}^{-3}}$.

(a) $1 \text{ kg m}^2 \text{ s}^{-2} \to \text{g cm}^2 \text{ s}^{-2}$:
$1 \text{ kg} = 10^3 \text{ g}, 1 \text{ m}^2 = 10^4 \text{ cm}^2$.
Factor: $10^3 \times 10^4 = \mathbf{10^7}$.

(b) $1 \text{ m} \to \text{ly}$:
$1 \text{ ly} = 9.46 \times 10^{15} \text{ m}$.
$1 \text{ m} = (9.46 \times 10^{15})^{-1} \approx \mathbf{1.057 \times 10^{-16} \text{ ly}}$.

(c) $3.0 \text{ m s}^{-2} \to \text{km h}^{-2}$:
$3.0 \times \frac{10^{-3} \text{ km}}{(1/3600 \text{ h})^2} = 3.0 \times 10^{-3} \times (3600)^2 = \mathbf{3.9 \times 10^4}$.

(d) G in CGS units:
$G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.
Conversion factor = $10^5 \text{ (N to dyne)} \times 10^4 \text{ (m to cm)} \times 10^{-6} \text{ (kg to g)} = 10^3$.
$G = 6.67 \times 10^{-11} \times 10^3 = \mathbf{6.67 \times 10^{-8} \text{ cm}^3 \text{ s}^{-2} \text{ g}^{-1}}$.

Given: $1 \text{ cal} = 4.2 \text{ J} = 4.2 \text{ kg m}^2 \text{ s}^{-2}$.
Dimensional Formula for Energy: $[ML^2T^{-2}]$.

New System:
$M’ = \alpha \text{ kg} \Rightarrow 1 \text{ kg} = \alpha^{-1} M’$.
$L’ = \beta \text{ m} \Rightarrow 1 \text{ m} = \beta^{-1} L’$.
$T’ = \gamma \text{ s} \Rightarrow 1 \text{ s} = \gamma^{-1} T’$.

Substituting into the value:
$4.2 (\alpha^{-1} M’) (\beta^{-1} L’)^2 (\gamma^{-1} T’)^{-2}$
Result: $\mathbf{4.2 \alpha^{-1} \beta^{-2} \gamma^{2}}$ in new units.

Explanation: A physical quantity is large or small only when compared to a standard.

• (a) Atoms are very small objects compared to a tennis ball.
• (b) A jet plane moves with great speed compared to a train.
• (c) The mass of Jupiter is very large compared to the mass of Earth.
• (d) The air inside this room contains a large number of molecules compared to Avogadro’s number.
• (e) Already correct (proton vs electron).
• (f) Already correct (sound vs light).

Given: Speed of light $c = 1$ unit.
Time $t = 8 \text{ min} 20 \text{ s} = (8 \times 60) + 20 = 500 \text{ s}$.

Distance: $d = c \times t = 1 \times 500 = \mathbf{500 \text{ new units}}$.

Correct Answer: (c) An optical instrument.

Reasoning: Precision depends on the limit of resolution (Least Count).

• Vernier Callipers: $\approx 10^{-4} \text{ m}$.
• Screw Gauge: $\approx 10^{-5} \text{ m}$.
• Optical Instrument: Uses wavelength of light $\approx 10^{-7} \text{ m}$.

Since the wavelength of light is smaller than the mechanical least counts, the optical instrument is more precise.

Magnification $m = 100$.
Observed width $y = 3.5 \text{ mm}$.

$x = \frac{3.5}{100} = \mathbf{0.035 \text{ mm}}$.

Object Area (Slide) $A_o = 1.75 \text{ cm}^2$.
Image Area (Screen) $A_i = 1.55 \text{ m}^2 = 1.55 \times 10^4 \text{ cm}^2$.

Areal Magnification $m_A = \frac{A_i}{A_o} = \frac{1.55 \times 10^4}{1.75} \approx 8857$.

Linear Magnification $m = \sqrt{m_A} = \sqrt{8857} \approx \mathbf{94.1}$.

• (a) $0.007 \text{ m}^2$: 1 (leading zeros insignificant).
• (b) $2.64 \times 10^{24} \text{ kg}$: 3 (power of 10 irrelevant).
• (c) $0.2370 \text{ g cm}^{-3}$: 4 (trailing zero after decimal is significant).
• (d) $6.320 \text{ J}$: 4.
• (e) $6.032 \text{ N m}^{-2}$: 4.
• (f) $0.0006032 \text{ m}^2$: 4.

$l = 4.234 \text{ m}, b = 1.005 \text{ m}, h = 0.0201 \text{ m}$.
Least significant figures = 3 (from thickness $h$).

Area: $A = 2(lb + bh + hl)$.
Calc: $A \approx 8.7209 \text{ m}^2$. Rounding to 3 sig figs: $8.72 \text{ m}^2$.

Volume: $V = l \times b \times h$.
Calc: $V \approx 0.0855289 \text{ m}^3$. Rounding to 3 sig figs: $0.0855 \text{ m}^3$.

Box $M = 2.30 \text{ kg}$. Pieces: $0.02015 \text{ kg}$ and $0.02017 \text{ kg}$.

(a) Total Mass: Add all. Rule: Keep least decimal places (2).
Total = $2.34032 \text{ kg}$. Rounded: $2.34 \text{ kg}$.

(b) Difference: $20.17 – 20.15 = 0.02 \text{ g}$.
Rounded to least decimal places (2): $0.02 \text{ g}$.

Formula given: $m = m_0(1 – v^2)^{1/2}$.

The term inside the bracket $(1 – v^2)$ must be dimensionless. Since $v$ has dimensions $[LT^{-1}]$, $v^2$ must be divided by $c^2$ (speed of light squared) to cancel dimensions.

Correct Formula: $m = \frac{m_0}{(1 – v^2/c^2)^{1/2}}$.

This phenomenon is due to Parallax or the rate of change of the line of sight.

• Nearby Objects: The line of sight changes rapidly as the train moves, creating a large angular displacement in the opposite direction. They appear to rush backwards.
• Distant Objects (Stars/Moon): The angular displacement is negligible because the distance is huge. They appear stationary relative to you.

Mass $M = 2.0 \times 10^{30} \text{ kg}$. Radius $R = 7.0 \times 10^8 \text{ m}$.

$\rho = \frac{2.0 \times 10^{30}}{1.437 \times 10^{27}} \approx \mathbf{1.39 \times 10^3 \text{ kg m}^{-3}}$.

Conclusion: The density is in the range of solids/liquids, not gases. This high density is due to the intense gravitational compression of the plasma in the core.