IMAT Physics Mastery

Step 1: Understand Buoyant Force ($F_B$).
The apparent loss of weight in water is the buoyant force.
$F_B = W_{\text{air}} - W_{\text{water}} = 9.8 \text{ N} - 8.82 \text{ N} = 0.98 \text{ N}$.

Step 2: Relate $F_B$ to Volume.
According to Archimedes, $F_B = \rho_{\text{water}} \cdot V_{\text{crown}} \cdot g$.
$0.98 = (1000) \cdot V_{\text{crown}} \cdot (9.8)$.
$V_{\text{crown}} = \frac{0.98}{9800} = 1 \times 10^{-4} \text{ m}^3$.

Step 3: Calculate the Mass.
$W_{\text{air}} = m \cdot g \implies m = \frac{9.8}{9.8} = 1 \text{ kg}$.

Step 4: Calculate the Density.
$\rho_{\text{crown}} = \frac{m}{V} = \frac{1 \text{ kg}}{1 \times 10^{-4} \text{ m}^3} = 10,000 \text{ kg/m}^3$.

Conclusion: 10,000 is much less than 19,300. The crown is a fake!

Step 1: Set up Bernoulli's Equation.
Let Point 1 be the water surface at the top, and Point 2 be the hole.
$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

Step 2: Apply Boundary Conditions.
Both the top and the hole are open to the atmosphere, so $P_1 = P_2 = P_{atm}$. They cancel out.
The tank is large, meaning the water level drops very slowly compared to the exit velocity. Thus, $v_1 \approx 0$.
Let the hole be the reference height ($h_2 = 0$). Then $h_1 = 5 \text{ m}$.

Step 3: Simplify and Solve (Torricelli's Law).
$\rho g h_1 = \frac{1}{2}\rho v_2^2$
$g h_1 = \frac{1}{2}v_2^2$
$v_2 = \sqrt{2gh_1} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10 \text{ m/s}$.

Answer: The water exits at 10 m/s (equivalent to an object falling freely from a 5m height).

Step 1: Identify the variables.
$N_0 = 64 \text{ MBq}$
$N_t = 4 \text{ MBq}$
$T_{1/2} = 6 \text{ hours}$

Step 2: Determine the number of half-lives ($n$).
Each half-life halves the amount:
$64 \to 32$ (1 half-life)
$32 \to 16$ (2 half-lives)
$16 \to 8$ (3 half-lives)
$8 \to 4$ (4 half-lives)
Alternatively: $N_t = N_0 \times (1/2)^n \implies 4 = 64 \times (1/2)^n \implies 1/16 = (1/2)^n \implies n=4$.

Step 3: Calculate total time.
Total time $t = n \times T_{1/2} = 4 \times 6 \text{ hours} = 24 \text{ hours}$.

Answer: 24 hours.

Step 1: Apply the Principle of Dimensional Homogeneity.
All additive terms in an equation must have the same dimensions.
Therefore: $[F_D] = [A \cdot v] = [B \cdot v^2]$

Step 2: Write down known dimensions.
Force $[F] = [M][L][T]^{-2} \implies \text{kg} \cdot \text{m} \cdot \text{s}^{-2}$
Velocity $[v] = [L][T]^{-1} \implies \text{m} \cdot \text{s}^{-1}$

Step 3: Solve for [A].
$[A] \cdot [v] = [F] \implies [A] = \frac{[F]}{[v]} = \frac{[M][L][T]^{-2}}{[L][T]^{-1}} = [M][T]^{-1}$
SI Unit of A: $\text{kg/s}$ (or $\text{kg} \cdot \text{s}^{-1}$).

Step 4: Solve for [B].
$[B] \cdot [v^2] = [F] \implies [B] = \frac{[F]}{[v^2]} = \frac{[M][L][T]^{-2}}{([L][T]^{-1})^2} = \frac{[M][L][T]^{-2}}{[L]^2[T]^{-2}} = [M][L]^{-1}$
SI Unit of B: $\text{kg/m}$ (or $\text{kg} \cdot \text{m}^{-1}$).

Step 1: Write Poiseuille's Law.
The flow rate of a viscous fluid through a cylindrical pipe is given by:
$$Q = \frac{\pi r^4 \Delta P}{8 \eta L}$$

Step 2: Relate Flow Rate to Radius.
Under constant pressure difference ($\Delta P$), length ($L$), and viscosity ($\eta$), the flow rate is directly proportional to the fourth power of the radius:
$$Q \propto r^4$$

Step 3: Calculate the change.
If the new radius is $r' = 0.5r$, then:
$$Q' \propto (0.5r)^4 = 0.0625 r^4 = \frac{1}{16} Q$$

Answer: The blood flow rate decreases by a factor of 16 (it becomes only 6.25% of the original flow rate).

Step 1: Calculate Gauge Pressure ($P_{gauge}$).
Gauge pressure is the pressure relative to the atmosphere, caused purely by the water column weight:
$$P_{gauge} = \rho g h = 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 30 \text{ m} = 3.0 \times 10^5 \text{ Pa} = 3 \text{ bar}$$

Step 2: Calculate Absolute Pressure ($P_{abs}$).
Absolute pressure includes atmospheric pressure:
$$P_{abs} = P_{atm} + P_{gauge} = 1.0 \times 10^5 \text{ Pa} + 3.0 \times 10^5 \text{ Pa} = 4.0 \times 10^5 \text{ Pa} = 4 \text{ atm}$$

Answer: Gauge Pressure = 3.0 × 10⁵ Pa (3 atm); Absolute Pressure = 4.0 × 10⁵ Pa (4 atm).

Step 1: Calculate energy transition ($\Delta E$).
$$\Delta E = E_{high} - E_{low} = -1.5 \text{ eV} - (-13.6 \text{ eV}) = 12.1 \text{ eV}$$

Step 2: Calculate photon frequency ($f$).
Using Einstein's relation $\Delta E = hf$:
$$f = \frac{\Delta E}{h} = \frac{12.1 \text{ eV}}{4.1 \times 10^{-15} \text{ eV}\cdot\text{s}} \approx 2.95 \times 10^{15} \text{ Hz}$$

Answer: Photon energy = 12.1 eV; Frequency ≈ 2.95 × 10¹⁵ Hz (UV range).

Step 1: Find the number of half-lives ($n$).
$$n = \frac{t}{T_{1/2}} = \frac{32 \text{ days}}{8 \text{ days}} = 4 \text{ half-lives}$$

Step 2: Apply the fraction formula.
$$\text{Fraction remaining} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^4 = \frac{1}{16} = 0.0625$$

Answer: 1/16 (or 6.25%) of the radioactive Iodine-131 remains.

Step 1: Rearrange for $G$.
$$G = \frac{F \cdot r^2}{m_1 \cdot m_2}$$

Step 2: Express each variable in SI base units.
- Force $F$: $\text{N} = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}$
- Distance $r$: $\text{m} \implies r^2: \text{m}^2$
- Mass $m$: $\text{kg} \implies m_1 \cdot m_2: \text{kg}^2

Step 3: Substitute and simplify.
$$[G] = \frac{(\text{kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot \text{m}^2}{\text{kg}^2} = \text{kg}^{-1} \cdot \text{m}^3 \cdot \text{s}^{-2}$$

Answer: kg⁻¹ · m³ · s⁻².

Step 1: Apply hydrostatic equilibrium at the interface level.
The pressure at the interface level on both sides must be equal:
$$P_{interface, left} = P_{interface, right}$$ $$P_{atm} + \rho_{oil} g h_{oil} = P_{atm} + \rho_{water} g h_{water}$$

Step 2: Simplify the equation.
Cancel out $P_{atm}$ and $g$:
$$\rho_{oil} \cdot h_{oil} = \rho_{water} \cdot h_{water}$$

Step 3: Solve for $\rho_{oil}$.
$$\rho_{oil} = \rho_{water} \frac{h_{water}}{h_{oil}} = 1000 \text{ kg/m}^3 \times \frac{10 \text{ cm}}{12 \text{ cm}} \approx 833.3 \text{ kg/m}^3$$

Answer: The density of the oil is approximately 833 kg/m³.

Step 1: Understand Alpha Decay.
An alpha particle is a helium nucleus: $^{4}_{2}\text{He}$.
The emission of an alpha particle reduces the mass number by 4 and the atomic number by 2.

Step 2: Set up conservation equations.
- Conservation of nucleon number: $238 = A + 4 \implies A = 234$.
- Conservation of charge: $92 = Z + 2 \implies Z = 90$ (which corresponds to Thorium, Th).

Step 3: Write the reaction.
$$^{238}_{92}\text{U} \to ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$$

Answer: The daughter nucleus Thorium-234 has A = 234 and Z = 90.

Step 1: Identify constituent particles.
Helium-4 has $Z = 2$ protons and $N = 4 - 2 = 2$ neutrons.

Step 2: Calculate the expected mass of separated nucleons.
$$m_{nucleons} = 2 \times m_{proton} + 2 \times m_{neutron}$$ $$m_{nucleons} = 2(1.0073 \text{ u}) + 2(1.0087 \text{ u}) = 2.0146 \text{ u} + 2.0174 \text{ u} = 4.0320 \text{ u}$$

Step 3: Find the mass defect ($\Delta m$).
$$\Delta m = m_{nucleons} - m_{actual} = 4.0320 \text{ u} - 4.0015 \text{ u} = 0.0305 \text{ u}$$

Answer: The mass defect of the Helium nucleus is 0.0305 u (which equals its binding energy when multiplied by c²).

Step 1: Rearrange the equation for $k_B$.
Since the coefficient $\frac{3}{2}$ is dimensionless:
$$[k_B] = \frac{[E_k]}{[T]}$$

Step 2: Write down dimensions.
- Energy $E_k$: $\text{J} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$
- Temperature $T$: $\text{K}$ (Kelvin)

Step 3: Combine dimensions.
$$[k_B] = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \text{K}^{-1}$$

Answer: kg · m² · s⁻² · K⁻¹ (or Joules per Kelvin, J/K).

Step 1: Use the Equation of Continuity.
Since the fluid is incompressible, the volume flow rate is conserved:
$$A_A \cdot v_A = A_B \cdot v_B$$

Step 2: Solve for $v_B$.
$$v_B = v_A \frac{A_A}{A_B} = 2 \text{ m/s} \times \frac{10 \text{ cm}^2}{2 \text{ cm}^2} = 10 \text{ m/s}$$

Answer: The fluid velocity in the constricted section is 10 m/s.

Step 1: Rearrange the equation for $\varepsilon_0$.
Ignore the dimensionless constant $4\pi$:
$$\varepsilon_0 \propto \frac{q_1 \cdot q_2}{F \cdot r^2}$$

Step 2: Identify base SI units for all terms.
- Charge $q$: $\text{C} = \text{A} \cdot \text{s}$
- Force $F$: $\text{N} = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}$
- Distance $r$: $\text{m} \implies r^2: \text{m}^2$

Step 3: Combine and simplify.
$$[\varepsilon_0] = \frac{(\text{A} \cdot \text{s})^2}{(\text{kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot \text{m}^2} = \frac{\text{A}^2 \cdot \text{s}^2}{\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2}} = \text{kg}^{-1} \cdot \text{m}^{-3} \cdot \text{s}^4 \cdot \text{A}^2$$

Answer: kg⁻¹ · m⁻³ · s⁴ · A² (also known as Farad per meter, F/m).

Step 1: Write de Broglie's wavelength equation.
$$\lambda = \frac{h}{p}$$

Step 2: Insert values.
$$\lambda = \frac{6.6 \times 10^{-34} \text{ kg}\cdot\text{m}^2\cdot\text{s}^{-1}}{6.6 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 10^{-10} \text{ m} = 0.1 \text{ nm}$$

Answer: The wavelength is 10⁻¹⁰ m (or 0.1 nm, which is on the scale of atomic diameters).

Step 1: Write the capillary rise formula.
$$h = \frac{2\gamma \cos\theta}{\rho g r}$$

Step 2: Convert radius to meters.
$$r = 0.5 \text{ mm} = 5.0 \times 10^{-4} \text{ m}$$

Step 3: Calculate the height.
$$h = \frac{2 \times 0.07 \text{ N/m} \times 1}{1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 5.0 \times 10^{-4} \text{ m}}$$ $$h = \frac{0.14}{5} = 0.028 \text{ m} = 2.8 \text{ cm}$$

Answer: The water rises by 2.8 cm.

Step 1: Solve for $B$.
$$B = \frac{F}{q \cdot v}$$

Step 2: Write units.
- Force $F$: $\text{kg} \cdot \text{m} \cdot \text{s}^{-2}$
- Charge $q$: $\text{A} \cdot \text{s}$
- Velocity $v$: $\text{m} \cdot \text{s}^{-1}$

Step 3: Simplify the ratio.
$$[B] = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{(\text{A} \cdot \text{s}) \cdot (\text{m} \cdot \text{s}^{-1})} = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{A} \cdot \text{m}} = \text{kg} \cdot \text{s}^{-2} \cdot \text{A}^{-1}$$

Answer: kg · s⁻² · A⁻¹ (equivalent to N / (A·m) or Tesla, T).

Step 1: Understand Terminal Velocity.
At terminal velocity, the net downward force equals the upward drag force given by Stokes' Law:
$$F_{net} = F_d = 6\pi\eta r v$$

Step 2: Solve for $v$.
$$v = \frac{F_{net}}{6\pi\eta r}$$

Step 3: Calculate using $r = 10 \times 10^{-6} \text{ m}$.
$$v = \frac{4.0 \times 10^{-11}}{6 \times 3.14 \times (1.8 \times 10^{-5}) \times (1.0 \times 10^{-5})} \approx \frac{4.0 \times 10^{-11}}{3.39 \times 10^{-9}} \approx 1.18 \times 10^{-2} \text{ m/s} = 1.18 \text{ cm/s}$$

Answer: The settling terminal velocity is approximately 1.2 cm/s.

Step 1: Explain Beta-minus Decay.
In $\beta^-$ decay, a neutron in the nucleus turns into a proton, emitting an electron ($^0_{-1}e$ or $\beta^-$) and an electron antineutrino ($\bar{\nu}_e$):
$$n \to p^+ + e^- + \bar{\nu}_e$$

Step 2: Balance Mass ($A$) and Charge ($Z$).
- Mass: $14 = A + 0 \implies A = 14$.
- Charge: $6 = Z - 1 \implies Z = 7$ (Nitrogen, N).

Step 3: Write the equation.
$$^{14}_6\text{C} \to ^{14}_7\text{N} + ^0_{-1}e + \bar{\nu}_e$$

Answer: Carbon-14 decays into stable Nitrogen-14 (A=14, Z=7).

Step 1: Rearrange for $\eta$.
$$\eta \propto \frac{r^4 \cdot \Delta P}{Q \cdot L}$$

Step 2: Substitute base dimensions.
- Radius $r$: $[L]$
- Pressure $\Delta P$: $[M][L]^{-1}[T]^{-2}$
- Flow rate $Q$ (Volume/time): $[L]^3[T]^{-1}$
- Length $L$: $[L]$

Step 3: Simplify dimensions.
$$[\eta] = \frac{[L]^4 \cdot [M][L]^{-1}[T]^{-2}}{[L]^3[T]^{-1} \cdot [L]} = \frac{[M][L]^3[T]^{-2}}{[L]^4[T]^{-1}} = [M][L]^{-1}[T]^{-1}$$

Answer: kg · m⁻¹ · s⁻¹ (or Pascal-seconds, Pa·s).

Step 1: Convert mass defect to kg.
$$\Delta m = 0.0188 \times 1.66 \times 10^{-27} \text{ kg} \approx 3.12 \times 10^{-29} \text{ kg}$$

Step 2: Use mass-energy equivalence $E = \Delta m \cdot c^2$.
$$E = (3.12 \times 10^{-29} \text{ kg}) \times (3.0 \times 10^8 \text{ m/s})^2 = 2.81 \times 10^{-12} \text{ J}$$

Step 3: Convert Joules to electronvolts (MeV).
$$E = \frac{2.81 \times 10^{-12} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 1.76 \times 10^7 \text{ eV} = 17.6 \text{ MeV}$$

Answer: The reaction releases 17.6 MeV of energy.

Step 1: Write down Continuity Equation for branching pipes.
The total volume flow rate entering the capillary bed must equal that in the aorta:
$$A_{aorta} \cdot v_{aorta} = N_{capillaries} \cdot A_{capillary} \cdot v_{capillary}$$

Step 2: Substitute area terms ($\pi r^2$).
$$\pi r_{aorta}^2 \cdot v_{aorta} = N \cdot \pi r_{capillary}^2 \cdot v_{capillary}$$ $$r_{aorta}^2 \cdot v_{aorta} = N \cdot r_{capillary}^2 \cdot v_{capillary}$$

Step 3: Solve for $v_{capillary}$ using uniform units.
- $r_{aorta} = 10^{-2} \text{ m}$
- $r_{capillary} = 4 \times 10^{-6} \text{ m}$
- $v_{aorta} = 0.3 \text{ m/s}$
$$(10^{-2})^2 \times 0.3 = (3 \times 10^9) \times (4 \times 10^{-6})^2 \times v_{capillary}$$ $$10^{-4} \times 0.3 = 3 \times 10^9 \times (16 \times 10^{-12}) \times v_{capillary}$$ $$3.0 \times 10^{-5} = 0.048 \times v_{capillary}$$ $$v_{capillary} = \frac{3.0 \times 10^{-5}}{4.8 \times 10^{-2}} = 6.25 \times 10^{-4} \text{ m/s} = 0.625 \text{ mm/s}$$

Answer: The average velocity in a single capillary is 0.625 mm/s (very slow to allow gas exchange).

Step 1: Express the dimensions of each variable.
- Force $[F] = [M][L][T]^{-2}$
- Density $[\rho] = [M][L]^{-3}$
- Area $[A] = [L]^2$
- Velocity $[v] = [L][T]^{-1} \implies [v^2] = [L]^2[T]^{-2}$

Step 2: Isolate $C_d$ in terms of dimensions.
$$[C_d] = \frac{[F]}{[\rho][A][v^2]}$$

Step 3: Simplify.
$$[C_d] = \frac{[M][L][T]^{-2}}{([M][L]^{-3}) \cdot ([L]^2) \cdot ([L]^2[T]^{-2})} = \frac{[M][L][T]^{-2}}{[M][L]^1[T]^{-2}} = [1] \ (\text{Dimensionless})$$

Answer: Cd is dimensionless. The equation is homogeneous.

Step 1: Apply Einstein's mass-energy equivalence.
$$E = m \cdot c^2 = (1.6605 \times 10^{-27} \text{ kg}) \times (2.9979 \times 10^8 \text{ m/s})^2$$ $$E \approx 1.6605 \times 10^{-27} \times 8.9874 \times 10^{16} = 1.4924 \times 10^{-10} \text{ J}$$

Step 2: Convert to eV and MeV.
$$E_{\text{eV}} = \frac{1.4924 \times 10^{-10} \text{ J}}{1.6022 \times 10^{-19} \text{ J/eV}} \approx 9.3147 \times 10^8 \text{ eV} = 931.5 \text{ MeV}$$

Answer: 1 atomic mass unit is equivalent to approximately 931.5 MeV.

Step 1: Calculate hydrostatic gauge pressure at depth.
$$P_{gauge} = \rho g h = 1025 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 80 \text{ m} = 8.2 \times 10^5 \text{ Pa}$$

Step 2: Calculate Force using $F = P \cdot A$.
$$F = P_{gauge} \cdot A = 8.2 \times 10^5 \text{ Pa} \times 0.4 \text{ m}^2 = 3.28 \times 10^5 \text{ N} = 328 \text{ kN}$$

Answer: The net inward hydrostatic force on the window is 328 kN.

Step 1: Isolate Planck's constant.
$$h = \frac{E}{f}$$

Step 2: Substitue dimensions.
- Energy $E$: $\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$
- Frequency $f$ ($1/\text{time}$): $\text{s}^{-1}$

Step 3: Solve for $h$.
$$[h] = \frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{s}^{-1}} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}$$

Answer: kg · m² · s⁻¹ (or Joule-seconds, J·s).

Step 1: Calculate the energy of a single photon ($E_p$).
$$E_p = \frac{hc}{\lambda} = \frac{(6.6 \times 10^{-34} \text{ J}\cdot\text{s}) \cdot (3.0 \times 10^8 \text{ m/s})}{660 \times 10^{-9} \text{ m}}$$ $$E_p = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-7}} = 3.0 \times 10^{-19} \text{ J}$$

Step 2: Relate Power to emission rate.
Power is total energy emitted per second ($P = n \cdot E_p$, where $n$ is photons/sec):
$$n = \frac{P}{E_p} = \frac{3.0 \times 10^{-3} \text{ W}}{3.0 \times 10^{-19} \text{ J}} = 1.0 \times 10^{16} \text{ photons/s}$$

Answer: The laser pointer emits 1.0 × 10¹⁶ photons per second.

Step 1: Identify surface area change ($\Delta A$).
A spherical bubble has surface area $A = 4\pi r^2$.
Because a soap bubble has two boundaries (inner and outer), the total surface area is doubled:
$$\Delta A = 2 \times (4\pi r^2) = 8\pi r^2$$

Step 2: Calculate area in square meters.
$$r = 0.05 \text{ m}$$ $$\Delta A = 8 \times 3.14 \times (0.05)^2 = 8 \times 3.14 \times 0.0025 = 0.0628 \text{ m}^2$$

Step 3: Calculate Work ($W$).
$$W = \gamma \cdot \Delta A = 0.03 \text{ N/m} \times 0.0628 \text{ m}^2 = 1.884 \times 10^{-3} \text{ J}$$

Answer: The work done is approximately 1.88 mJ (milliJoules).

Step 1: Isolate $R$ in the equation.
$$R = \frac{PV}{nT}$$

Step 2: Substitute units for each term.
- Pressure $P$: $\text{Pa} = \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}$
- Volume $V$: $\text{m}^3$
- Amount $n$: $\text{mol}$
- Temperature $T$: $\text{K}$

Step 3: Simplify.
$$[R] = \frac{(\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}) \cdot \text{m}^3}{\text{mol} \cdot \text{K}} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$$

Answer: kg · m² · s⁻² · K⁻¹ · mol⁻¹ (equivalent to J / (mol·K)).