Grade 10 Physics Second Semester Final Exam Summary Notes

True or False Assessment:
- Area as a Derivative Quantity: Area is classified as a derivative (secondary) physical quantity. (True)
- Amorphous Solids: Salts are classified as amorphous (non-crystalline) solids. (False - Salts are typically crystalline).
- Kinetic Energy and Velocity Relationship: If the kinetic energy of two objects is equal, their velocities must also be equal. (False - Kinetic energy also depends on mass $K = \frac{1}{2}mv^2$ ).
- Melting Point Definition: The constant temperature at which a solid substance turns into a liquid is called the melting point. (True)
Key Definitions and Fill-in-the-Blanks:
- Density: The ratio of mass to volume of a substance is called its density ( $\rho = \frac{m}{V}$ ).
- Gauge Pressure: The difference between absolute pressure ( $P$ ) and atmospheric pressure ( $P_0$ ) is known as gauge pressure ( $P_g = P - P_0$ ).
- Mechanical Energy: The sum of the potential and kinetic energies of an object is called mechanical energy ( $E = K + U$ ).
- Physical Quantities: A physical quantity described only by a numerical value and an appropriate unit is called a scalar quantity.
- Chain Conversion:
  - To convert 42 mg to nanograms (ng):
  - $42\,mg \times \frac{10^{-3}\,g}{1\,mg} \times \frac{10^9\,ng}{1\,g} = 4.2 \times 10^7\,ng$

Fluid Surface Phenomena:
- Surface Tension: This phenomenon is caused by the cohesion (attractive forces between like molecules) of liquid surface molecules.
Phase Changes and Pressure:
- Usually, an increase in pressure results in lowering the melting point (specifically for substances like ice that contract upon melting).
Conceptual Applications:
- Oil vs. Water Fires: Liquid water is not suitable for extinguishing burning gasoline. This is because the density of gasoline ( $\rho \approx 6.8 \times 10^2\,kg/m^3$ ) is lower than the density of water ( $\rho \approx 10^3\,kg/m^3$ ). Consequently, gasoline floats on top of the water and continues to burn while spreading.
- Solid Forms of Water: Snow and ice are both solid forms of water but appear different. This is due to their crystalline structure; snow forms via water vapor deposition into delicate, air-trapping hexagonal structures, whereas ice forms by the freezing of liquid water into a more compact solid.
- Evaporative Cooling: Clay pots (Koozeh) keep water cool through the process of evaporation. The porous walls of the pot allow some water to seep through to the outer surface, where it evaporates by absorbing heat from the remaining water inside.

Buoyancy Analysis:
- The state of an object in a fluid is determined by the net force (difference between the buoyant force $F_b$ and weight $W$ ):
  - Rising (Bala-raftan): If $F_b > W$ .
  - Submerged/Neutral Buoyancy (Ghotte-vari): If $F_b = W$ and the object is fully inside the fluid.
  - Floating (Shenavari): If $F_b = W$ and the object is partially above the surface.
  - Sinking (Foroo-raftan): If $F_b < W$ .
Gas Pressure in Manometers:
- To calculate gas pressure in a reservoir (manometer):
- Fluid Density ( $\rho$ ) = $6\,g/cm^3 = 6000\,kg/m^3$
- Atmospheric pressure ( $P_0$ ) = $10^5\,Pa$
- Gravitational acceleration ( $g$ ) = $10\,N/kg$
- Height difference ( $h$ ) = $35\,cm - 20\,cm = 15\,cm = 0.15\,m$
- $P_{gas} = P_0 + \rho gh = 10^5 + (6000)(10)(0.15) = 100,000 + 9,000 = 109,000\,Pa$
Fluid Dynamics (Continuity Equation):
- According to the equation of continuity: $A_1 v_1 = A_2 v_2$
- Given: $A_1 = 20\,cm^2$ , $A_2 = 4\,cm^2$ , and speed at section-2 $v_2 = 12\,m/s$
- Calculation: $(20)v_1 = (4)(12)$
- $20v_1 = 48 \rightarrow v_1 = \frac{48}{20} = 2.4\,m/s$

Kinetic Energy Calculation:
- Mass ( $m$ ) = $200\,g = 0.2\,kg$
- Kinetic Energy ( $K$ ) = $250\,J$
- Formula: $K = \frac{1}{2}mv^2$
- $250 = \frac{1}{2}(0.2)v^2$
- $250 = 0.1v^2 \rightarrow v^2 = 2500 \rightarrow v = 50\,m/s$
Jet Engine Power:
- Force per engine ( $F$ ) = $2.4 \times 10^5\,N$
- Displacement ( $d$ ) = $15\,km = 15,000\,m$
- Time ( $t$ ) = $1\,minute = 60\,s$
- Work done ( $W$ ) = $F \times d = (2.4 \times 10^5)(1.5 \times 10^4) = 3.6 \times 10^9\,J$
- Average Power ( $P_{avg}$ ) = $\frac{W}{t} = \frac{3.6 \times 10^9}{60} = 6 \times 10^7\,Watt$
Work-Energy Theorem with Air Resistance:
- A package ( $m = 30\,kg$ ) is released from a balloon at height $h = 50\,m$ with initial speed $v_1 = 4\,m/s$ , hitting the ground at speed $v_2 = 25\,m/s$ .
- Conservation of Energy excluding conservative work: $E_1 + W_f = E_2$
- $(mgh_1 + \frac{1}{2}m v_1^2) + W_{air} = \frac{1}{2}m v_2^2$
- Initial Mechanical Energy ( $E_1$ ): $(30)(10)(50) + \frac{1}{2}(30)(4^2) = 15,000 + 240 = 15,240\,J$
- Final Mechanical Energy ( $E_2$ ): $\frac{1}{2}(30)(25^2) = 15 \times 625 = 9,375\,J$
- Work of air resistance ( $W_{air}$ ) = $E_2 - E_1 = 9,375 - 15,240 = -5,865\,J$

Temperature Measurement:
- Pyrometer (Taf-sanj): Operates based on thermal radiation.
- Electric Circuit Thermometers: Thermocouples and resistance thermometers are suitable for electrical sensors.
- Greenhouse Thermometer: The Maximum-Minimum thermometer (Six's thermometer) is typically used in botanical centers.
Temperature Conversion:
- Convert $122^\circ F$ to Ceslius ( $C$ ) and Kelvin ( $K$
- $C = \frac{5}{9}(F - 32) = \frac{5}{9}(122 - 32) = \frac{5}{9}(90) = 50^\circ C$
- $K = C + 273.15 = 50 + 273.15 = 323.15\,K$
Linear Expansion:
- Initial Length ( $L_0$ ) = $12\,m$
- Initial Temperature ( $T_1$ ) = $0^\circ C$ , Final Temperature ( $T_2$ ) = $50^\circ C$
- Coefficient of linear expansion ( $\alpha_{iron}$ ) = $1.2 \times 10^{-5}\, ^\circ C^{-1}$
- $\Delta L = L_0 \alpha \Delta T = (12)(1.2 \times 10^{-5})(50) = 0.0072\,m$
- Increase in length = $7.2\,mm$
Volume Expansion and Overfill:
- Glass beaker filled with olive oil at $20^\circ C$ . Beaker capacity $V_0 = 300\,cm^3$ . Temperature rises to $65^\circ C$
- $\Delta T = 65 - 20 = 45\,K$
- $\alpha_{glass} = 1 \times 10^{-5}\,, \beta_{oil} = 7 \times 10^{-4}\,K^{-1}$
- $\beta_{glass} = 3\alpha_{glass} = 3 \times 10^{-5}\,K^{-1}$
- $\Delta V_{oil} = V_0 \beta_{oil} \Delta T = (300)(7 \times 10^{-4})(45) = 9.45\,cm^3$
- $\Delta V_{glass} = V_0 \beta_{glass} \Delta T = (300)(3 \times 10^{-5})(45) = 0.405\,cm^3$
- Volume spilled = $\Delta V_{oil} - \Delta V_{glass} = 9.45 - 0.405 = 9.045\,cm^3$

Cooling and Freezing Water:
- Determine total heat ( $Q$ ) removed from $500\,g = 0.5\,kg$ of water at $20^\circ C$ to turn it into ice at $-5^\circ C$
- $c_{water} = 4200\,J/(kg\cdot ^\circ C)$ , $c_{ice} = 2100\,J/(kg\cdot ^\circ C)$ , $L_f = 334\,kJ/kg$
- Phase 1 (Cooling water 20 to 0): $Q_1 = mc\Delta T = (0.5)(4200)(0 - 20) = -42,000\,J$
- Phase 2 (Freezing at 0): $Q_2 = -m L_f = -(0.5)(334,000) = -167,000\,J$
- Phase 3 (Cooling ice 0 to -5): $Q_3 = mc\Delta T = (0.5)(2100)(-5 - 0) = -5,250\,J$
- Total Heat ( $Q_{total}$ ) = $-(42,000 + 167,000 + 5,250) = -214,250\,J$
- Heat to be removed: $214.25\,kJ$