CBSE Class 11 Physics Properties of Bulk Matter Assignment

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Define term (i) stress (ii) strain
Answer: (i) Stress is defined as the restoring force acting per unit area of a deformed body, given by \(\text{Stress} = \frac{F}{A}\).
(ii) Strain is defined as the ratio of change in configuration to the original configuration of the body, given by \(\text{Strain} = \frac{\Delta L}{L}\).
In simple words: Stress is the internal resistance per unit area, and strain is the fractional change in dimensions.

Exam Tip: Remember that stress has the same units as pressure (\(\text{N m}^{-2}\)), whereas strain is a dimensionless quantity.

Question 2. Differentiate the terms plasticity and elasticity of material
Answer: Elasticity is the property of a body by virtue of which it regains its original configuration (length, volume, or shape) when the deforming forces are removed. Plasticity is the inability of a body to regain its original status, retaining its deformed shape on the removal of the deforming forces.
In simple words: Elastic materials bounce back to their original shape, while plastic materials stay deformed.

Exam Tip: Use steel as a classic example of high elasticity and putty or lead as examples of plasticity.

Question 3. Draw stress - strain curve for elastomers (elastic tissue of Aorta)
Answer: For elastomers like the elastic tissue of the aorta, the material does not obey Hooke's law over a wide range. The stress-strain curve shows a large elastic region, and the curve does not have a well-defined plastic zone before breaking.
In simple words: Elastomers can stretch a lot without getting permanently deformed, but they do not follow a straight-line Hooke's law graph.

Exam Tip: Sketch a curve that rises steeply without a straight proportional region, showing large elongation for small applied stress.

Question 4. How are we able to break a wire by repeated bending?
Answer: Repeated bending of a wire decreases its elastic strength and fatigue sets in, causing the internal crystalline structure to weaken at the bend, making it break easily.
In simple words: Bending a wire back and forth wears out its internal strength until it snaps.

Exam Tip: Mention the term 'elastic fatigue' in your explanation for full marks.

Question 5. What is the value of bulk modulus for an incompressible liquid?
Answer: The bulk modulus is given by \(K = \frac{\text{Stress}}{\text{Strain}} = \frac{-P}{\Delta V / V}\). For an incompressible liquid, change in volume \(\Delta V = 0\), hence \(K = \infty\) (Infinity).
In simple words: An ideally incompressible liquid cannot be compressed at all, so its bulk modulus is infinite.

Exam Tip: Always state that zero volume change leads to infinite bulk modulus.

Question 6. Define Poisson's ratio? What is its unit?
Answer: Poisson's ratio (\(\sigma\)) is defined as the ratio of lateral strain to the longitudinal strain. It is a ratio of two similar quantities, so it has no units.
In simple words: Poisson's ratio measures how much a wire thins out when it is stretched lengthwise, and it has no unit.

Question 7. What is elastic fatigue?
Answer: Elastic fatigue is the property of an elastic body by virtue of which its behavior becomes less elastic under the action of repeated alternating deforming forces.
In simple words: Materials get tired and lose some elasticity when repeatedly stressed.

Exam Tip: Keywords are 'repeated alternating deforming forces' and 'loss in strength'.

Question 8. It is easier to swim in sea water than in the river water. Why?
Answer: The density of sea water is greater than the density of river water due to dissolved salts. Hence, sea water provides a greater upthrust for the same volume of water displaced.
In simple words: Salt makes sea water denser, which provides more upward buoyancy to help you float easily.

Exam Tip: Connect upthrust directly to fluid density according to Archimedes' principle.

Question 9. Railway tracks are laid on large sized wooden sleepers. Why?
Answer: Wooden sleepers spread the force due to the heavy weight of the train over a larger area, which significantly reduces the pressure on the ground and prevents the tracks from sinking.

In simple words: Large wooden pads spread out the train's massive weight so the ground doesn't give way.

Exam Tip: Use the inverse relation between pressure and area: \(P = \frac{F}{A}\).

Question 10. The dams of water reservoir are made thick near the bottom. Why?
Answer: The pressure exerted by a liquid column is given by \(P = h\rho g\), which increases with depth \(h\). Therefore, dams are made thicker at the bottom to withstand the very high pressure exerted by deep water.
In simple words: Water pressure increases with depth, so dam walls must be thicker at the base to hold back the force.

Exam Tip: Quote the formula \(P = h\rho g\) to justify the increase in pressure with depth.

Question 11. Why is it difficult to stop bleeding from a cut in human body at high altitudes?
Answer: At high altitudes, atmospheric pressure is low. Due to the greater pressure difference between blood pressure and the low atmospheric pressure, bleeding from a cut becomes severe and harder to stop.
In simple words: Lower air pressure outside makes it harder for blood flow from a cut to slow down and clot.

Exam Tip: Contrast internal blood pressure with external atmospheric pressure.

Question 12. The blood pressure in human is greater at the feet than at the brain. Why?
Answer: The hydrostatic pressure of a liquid column is \(P = h\rho g\). The height of the blood column is greater at the feet relative to the heart/brain, resulting in greater hydrostatic pressure at the feet.
In simple words: Gravity pulls blood down, creating a taller column of fluid and higher pressure at your feet than at your head.

Exam Tip: Mention hydrostatic pressure and height column difference \(h\).

Question 13. Define coefficient of viscosity and write its SI unit.
Answer: Coefficient of viscosity (\(\eta\)) is defined as the tangential viscous force per unit area per unit velocity gradient. Its SI unit is poiseuille (\(\text{Pl}\)) or \(\text{N s m}^{-2}\) or \(\text{Pa s}\).
In simple words: It is a measure of how thick or sticky a fluid is.

Exam Tip: Memorize both \(\text{N s m}^{-2}\) and Pascal-second as valid SI units.

Question 14. Why do machine parts get jammed in winter?
Answer: In winter, due to low temperature, the viscosity of lubricating oils and lubricants increases significantly, causing high internal friction that jams machine parts.
In simple words: Cold temperatures make lubricants thick and sticky, causing machine parts to lock up.

Exam Tip: Link low temperature directly to increased liquid viscosity.

Question 15. Why are rain drops spherical?
Answer: Due to surface tension, liquid drops tend to minimize their surface area. For a given volume, a sphere has the minimum possible surface area, making rain drops spherical.
In simple words: Surface tension pulls water droplets into spheres because spheres have the smallest surface area for a given volume.

Exam Tip: Mention 'minimum surface area for a given volume' to secure full credit.

Question 16. Why do paints and lubricants have low surface tension?
Answer: Low surface tension allows paints and lubricants to spread more effectively over surfaces and penetrate small cracks or gaps.
In simple words: Low surface tension helps liquids spread out smoothly instead of beading up.

Exam Tip: Connect low surface tension with effective spreading and wetting action.

Question 17. What will be the effect of increasing temperature on (i) angle of contact (ii) surface tension.
Answer: (i) The angle of contact generally increases with an increase in temperature.
(ii) Surface tension generally decreases with an increase in temperature because thermal agitation overcomes intermolecular forces.
In simple words: Heating a liquid lowers its surface tension and increases its angle of contact.

Question 18. For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain. On what factor does it depend in case of fluids?
Answer: In case of fluids, the shearing force (or viscous force) is proportional to the rate of shear strain (velocity gradient), rather than the shear strain itself.
In simple words: Unlike solids, fluids resist flow based on how fast they are sheared, not how much they are displaced.

Exam Tip: Highlight 'rate of shear strain' or 'velocity gradient' for fluids.

Question 19. How does rise in temperature effect (i) viscosity of gases (ii) viscosity of liquids.
Answer: (i) The viscosity of gases increases with an increase in temperature due to increased molecular momentum transfer.
(ii) The viscosity of liquids decreases with an increase in temperature due to weakened cohesive intermolecular forces.
In simple words: Heating makes gases more viscous, but makes liquids less viscous (thinner).

Question 20. Explain do why detergents have small angle of contact?
Answer: Detergents have a small angle of contact so that they have low surface tension and a greater ability to wet surfaces, penetrating deep into fabric fibers as given by \(h = \frac{2S\cos\theta}{r\rho g}\).
In simple words: A small angle of contact helps detergent water spread flat and soak deep into clothes to remove dirt.

Exam Tip: Use the capillarity rise formula to explain wetting behavior.

Question 21. Write the dimensions of coefficient of viscosity and surface tension.
Answer: The dimension of coefficient of viscosity \([\eta] = [M^1 L^{-1} T^{-1}]\).
The dimension of surface tension \([S] = [M^1 L^0 T^{-2}]\).
In simple words: Dimensional formulas express physical quantities in terms of fundamental mass, length, and time units.

Exam Tip: Derive dimensions quickly using formula units: \(\eta = \frac{F}{A(dv/dx)}\) and \(S = \frac{F}{l}\).

Question 22. Obtain a relation between SI unit and cgs unit of coefficient of viscosity.
Answer: The CGS unit of viscosity is poise (\(\text{P}\)), and the SI unit is poiseuille (\(\text{Pl}\)) or \(\text{Pa s}\).
\(1\text{ poise} = 1\text{ g cm}^{-1}\text{ s}^{-1} = 10^{-1}\text{ kg m}^{-1}\text{ s}^{-1} = 0.1\text{ poiseuille}\).
In simple words: One poiseuille is equal to 10 poises.

Exam Tip: Conversion factor 1 poise = 0.1 \(\text{kg m}^{-1}\text{ s}^{-1}\) is very useful in numericals.

Question 23. Explain, how the use of parachute helps a person jumping from an aeroplane.
Answer: A parachute provides a very large surface area, increasing the viscous drag force \(F = 6\pi\eta r v\) (or air resistance). This drastically reduces the terminal velocity of the falling person, allowing them to land safely without injury.
In simple words: A parachute catches the air to slow down a skydiver's fall to a safe speed.

Exam Tip: Connect parachute area directly to terminal velocity reduction.

Question 24. Why two ships moving in parallel directions close to each other get attracted?
Answer: According to Bernoulli's theorem, the speed of water between the two ships is higher than outside. This creates a region of lower pressure between them, and the higher outside pressure pushes the ships toward each other.
In simple words: Fast-moving water between parallel ships drops the pressure, causing outside water pressure to push them together.

Exam Tip: Always apply Bernoulli's principle: high speed means low pressure.

Question 25. Why the molecules of a liquid lying near the free surface possess extra energy?
Answer: Molecules near the surface experience a net downward cohesive force because there are fewer liquid molecules above them. Work must be done against this inward force to bring molecules from the interior to the surface, storing it as potential surface energy.
In simple words: Surface molecules are pulled inward, so bringing them to the surface requires work, giving them extra energy.

Exam Tip: Mention net inward cohesive force and work done against it.

Question 26. Why is it easier to wash clothes in hot soap solution?
Answer: Hot soap solution has lower surface tension and lower viscosity, which helps it wet the fabric better, penetrate smaller gaps, and emulsify grease and dirt easily.
In simple words: Hot soapy water spreads and penetrates fabrics better to lift away grease and dirt.

Exam Tip: Low surface tension and reduced viscosity are the key reasons.

Question 27. Why does mercury not wet glass?
Answer: Mercury does not wet glass because the cohesive force between mercury-mercury molecules is much stronger than the adhesive force between mercury and glass molecules, resulting in an obtuse angle of contact.
In simple words: Mercury atoms stick to each other more strongly than to glass, so it forms droplets and rolls off without wetting.

Exam Tip: Compare cohesive forces with adhesive forces.

Question 28. Why ends of a glass tube become rounded on heating?
Answer: When glass is heated to its melting point, it becomes soft. Due to surface tension, the molten glass tends to minimize its surface area, causing the sharp edges to round off.
In simple words: Melting glass flows slightly, and surface tension pulls sharp cuts into smooth, rounded edges.

Exam Tip: Surface tension minimization is the physical mechanism here.

Question 29. What makes rain coats water proof?
Answer: Rain coats are treated with water-repellent materials that create an obtuse angle of contact with water, preventing water from soaking through the fabric pores.
In simple words: Waterproof coatings prevent water from wetting the fabric by creating an obtuse contact angle.

Question 30. What happens when a capillary tube of insufficient length is dipped in a liquid?
Answer: When a capillary tube of insufficient length is used, the radius of curvature of the meniscus increases so that \(hr = \text{constant}\). The liquid rises to the top, flattens its meniscus, but does not overflow.
In simple words: The liquid reaches the top of a short tube and widens its meniscus instead of spilling over.

Exam Tip: The product \(hr\) remains constant even when tube length is reduced.

Question 31. Does it matter if one uses gauge pressure instead of absolute pressure in applying Bernoulli's equation?
Answer: No, it does not matter as long as the atmospheric pressure is uniform and significant at both reference points where Bernoulli's equation is applied.
In simple words: Using gauge pressure is fine as long as atmospheric pressure is the same everywhere in the system.

Exam Tip: Gauge pressure excludes atmospheric pressure, which cancels out if uniform.

Question 32. State Wein's displacement law for black body radiation.
Answer: Wein's displacement law states that the wavelength corresponding to maximum energy emission (\(\lambda_m\)) of a black body is inversely proportional to its absolute temperature \(T\), i.e., \(\lambda_m T = \text{constant}\).
In simple words: Hotter objects emit radiation at shorter wavelengths.

Exam Tip: Formula: \(\lambda_m T = b\) where \(b\) is Wein's constant.

Question 33. State Stefan Boltzmann law.
Answer: Stefan-Boltzmann law states that the total radiant energy emitted per unit surface area per unit time by a perfect black body is directly proportional to the fourth power of its absolute temperature, \(E = \sigma T^4\).
In simple words: Radiation energy emitted by a black body scales with the fourth power of its temperature.

Question 34. Name two physical changes that occur on heating a body.
Answer: Two physical changes on heating are thermal expansion and change in electrical resistance.
In simple words: Heating causes objects to expand and changes how well they conduct electricity.

Exam Tip: Thermal expansion and resistance change are standard textbook answers.

Question 35. Distinguish between heat and temperature.
Answer: Heat is the total internal energy transferred between bodies due to a temperature difference, measured in Joules. Temperature is the degree of hotness or coldness that determines the direction of heat flow, measured in Kelvin.
In simple words: Heat is total thermal energy, while temperature measures how hot something is.

Exam Tip: Heat is energy in transit; temperature is a state property.

Question 36. Which thermometer is more sensitive a mercury or gas thermometer?
Answer: A gas thermometer is more sensitive than a mercury thermometer because the coefficient of expansion of gas is much greater than that of mercury.
In simple words: Gas thermometers detect smaller temperature changes because gases expand much more than mercury.

Question 37. Metal disc has a hole in it. What happens to the size of the hole when disc is heated?
Answer: The size of the hole increases on heating because thermal expansion occurs outward in all dimensions, exactly as if the hole were filled with the metal itself.
In simple words: Holes in heated objects expand along with the rest of the metal, becoming larger.

Exam Tip: Remember thermal expansion expands everything outward, enlarging holes.

Question 38. Name a substance that contracts on heating.
Answer: Ice contracts when heated from \(0^\circ\text{C}\) to \(4^\circ\text{C}\) (anomalous expansion of water).
In simple words: Ice/water contracts instead of expanding when heated between \(0^\circ\text{C}\) and \(4^\circ\text{C}\).

Exam Tip: Cite ice or water in the anomalous temperature range \(0^\circ\text{C}\) to \(4^\circ\text{C}\).

Question 39. A gas is free to expand what will be its specific heat?
Answer: Its specific heat will be infinity because heat supplied is used entirely for external work during free expansion without raising temperature (\(\Delta T = 0\)).
In simple words: If a gas expands freely, temperature doesn't change, making specific heat infinite.

Exam Tip: Specific heat \(C = \frac{Q}{m\Delta T}\); since \(\Delta T = 0\), \(C = \infty\).

Question 40. What is Deby's temperature?
Answer: Debye temperature is the temperature above which the molar heat capacity of a solid substance becomes constant (approx. \(3R\) according to Dulong-Petit law).
In simple words: It is the characteristic temperature where a solid's heat capacity stops increasing and levels off.

Exam Tip: Link Debye temperature to the saturation of molar heat capacity.

Question 41. What is the absorptive power of a perfectly black body?
Answer: The absorptive power of a perfectly black body is equal to 1 (or 100%), as it absorbs all radiations incident on it.
In simple words: A perfect black body absorbs all light and heat falling on it, so its absorptive power is 1.

Exam Tip: Absorptive power ranges from 0 to 1; for a black body, it is exactly 1.

Question 42. At what temperature does a body stop radiating?
Answer: A body stops radiating at absolute zero temperature (\(0\text{ K}\) or \(-273.15^\circ\text{C}\)).
In simple words: Radiation stops completely only at absolute zero.

Exam Tip: Absolute zero is the thermal baseline where all molecular motion ceases.

Question 43. If Kelvin temperature of an ideal black body is doubled, what will be the effect on energy radiated by it?
Answer: According to Stefan's law, \(E \propto T^4\). If temperature \(T\) is doubled, energy radiated becomes \(E_2 = (2)^4 E_1 = 16E_1\), increasing 16 times.
In simple words: Doubling the Kelvin temperature increases radiated energy by 16 times.

Exam Tip: Use \(2^4 = 16\) in Stefan's law calculations.

Question 44. In which method of heat transfer does gravity not play any part?
Answer: In conduction and radiation, gravity does not play any part (unlike convection where buoyant hot fluids rise due to gravity).
In simple words: Conduction and radiation work without needing gravity, unlike convection.

Exam Tip: Convection relies on gravity-driven density differences.

Question 45. Give a plot of Fahrenheit temperature versus celsius temperature
Answer: The plot of Fahrenheit temperature (\(F\)) versus Celsius temperature (\(C\)) is a straight line governed by the equation \(F = \frac{9}{5}C + 32\), with a y-intercept of 32 and a slope of \(\frac{9}{5}\).
In simple words: The conversion graph between Fahrenheit and Celsius is a straight line with a y-intercept at 32.

Exam Tip: Remember the intercept is 32 and slope is 1.8.

Question 46. Why birds are often seen to swell their feather in winter?
Answer: When birds fluff their feathers, they trap air pockets inside. Since air is a very poor conductor of heat, it prevents body heat from escaping into the cold environment.
In simple words: Fluffing feathers traps a layer of insulating air to keep birds warm in winter.

Exam Tip: Mention trapped air acting as thermal insulation.

Question 47. A brass disc fits snugly in a hole in a steel plate. Should we heat or cool the system to loosen the disc from the hole?
Answer: We should cool the system. The temperature coefficient of linear expansion for brass is greater than that for steel, so on cooling, the brass disc shrinks more than the steel hole, loosening the disc.
In simple words: Cool the system because brass shrinks faster than steel, making the disc pop out of the hole.

Question 1. State Hooke's law. Deduce expression for young's modulus of material of a wire of length \(l\), radius of cross-section \(r\) loaded with a body of mass M producing an extension \(\Delta l\) in it.
Answer: Hooke's law states that within the elastic limit, stress is directly proportional to strain, so \(\text{Stress} = E \times \text{Strain}\).
For a wire of length \(l\) and radius \(r\), cross-sectional area \(A = \pi r^2\), and applied force \(F = Mg\).
Longitudinal stress = \(\frac{Mg}{\pi r^2}\), Longitudinal strain = \(\frac{\Delta l}{l}\).
Young's Modulus \(Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg / \pi r^2}{\Delta l / l} = \frac{Mg l}{\pi r^2 \Delta l}\).
In simple words: Hooke's law links stress and strain, allowing us to calculate Young's modulus using mass, length, radius, and stretch.

Exam Tip: Substitute \(A = \pi r^2\) and \(F = Mg\) carefully into Young's modulus formula.

Question 2. A wire of length \(l\), area of cross section \(A\) and young's modulus \(Y\) is stretched by an amount \(x\). What is the work done?
Answer: The restoring force for an extension \(x\) is given by \(F = \frac{AYx}{l}\).
Work done in stretching the wire by a small amount \(dx\) is \(dw = F\,dx\).
Total work done in stretching from \(0\) to \(x\) is:
\(W = \int_{0}^{x} F\,dx = \int_{0}^{x} \frac{AYx}{l}\,dx = \frac{AY}{l} \left[ \frac{x^2}{2} \right]_{0}^{x} = \frac{1}{2} \frac{AYx^2}{l}\).
In simple words: Integrating the variable stretching force gives the work stored as elastic potential energy.

Exam Tip: Remember the work formula: \(W = \frac{1}{2} \times \text{load} \times \text{extension}\).

Question 3. Prove that the elastic potential energy per unit volume is equal to \(\frac{1}{2} \times \text{stress} \times \text{strain}\).
Answer: Elastic potential energy stored \(U = \frac{1}{2} \frac{AYx^2}{l}\).
Volume of the wire = \(\text{Area} \times \text{Length} = A \times l\).
Energy Density = \(\frac{U}{\text{Volume}} = \frac{\frac{1}{2} \frac{AYx^2}{l}}{Al} = \frac{1}{2} \left( \frac{AYx}{Al} \right) \left( \frac{x}{l} \right)\).
Since Stress = \(\frac{F}{A} = \frac{AYx}{Al}\) and Strain = \(\frac{x}{l}\):
Energy Density = \(\frac{1}{2} \times \text{Stress} \times \text{Strain}\).
In simple words: Energy density is derived by dividing total stored energy by volume, yielding half the product of stress and strain.

Exam Tip: This is a standard 2-mark derivation; practice substituting stress and strain definitions.

Question 4. Define the term bulk modulus. Give its SI unit. Give the relation between bulk modulus and compressibility.
Answer: Bulk modulus is defined as the ratio of normal stress to the volumetric strain within the elastic limit: \(K = \frac{-P}{\Delta V / V}\).
Its SI unit is \(\text{N m}^{-2}\) or Pascal (\(\text{Pa}\)).
Compressibility is defined as the reciprocal of bulk modulus: \(C = \frac{1}{K}\).
In simple words: Bulk modulus measures resistance to compression, and compressibility is its exact opposite.

Exam Tip: Include the negative sign in bulk modulus formula to indicate volume decreases with increased pressure.

Question 5. Define shear modulus. With the help of a diagram explain how shear modulus can be calculated.
Answer: Shear modulus (Rigidity modulus, \(\eta\)) is defined as the ratio of tangential stress to the shearing strain within the elastic limit: \(\eta = \frac{F/A}{\Delta x / y} = \frac{F}{A\theta}\).
In a rectangular block, when a tangential force \(F\) is applied to the upper face while the lower face is fixed, the upper face displaces by \(\Delta x\) over height \(y\), creating shear angle \(\theta = \frac{\Delta x}{y}\).
In simple words: Shear modulus measures how much a solid twists or deforms sideways when pushed tangentially.

Exam Tip: Draw a square shifting into a parallelogram to illustrate displacement \(\Delta x\) and height \(y\).

Question 6. Which is more elastic steel or rubber? Explain.
Answer: Steel is more elastic than rubber. For the same stress (equal force and cross-sectional area), the strain (extension \(\Delta l\)) produced in steel is much smaller than in rubber. Since Young's modulus \(Y = \frac{\text{Stress}}{\text{Strain}}\) and \(Y_{\text{steel}} > Y_{\text{rubber}}\), steel is more elastic.
In simple words: Steel bounces back better and resists deformation more than rubber, making it scientifically more elastic.

Exam Tip: Physics elasticity refers to resisting deformation, not stretchability.

Question 7. Two wires P and Q of same diameter are loaded as shown in the figure. The length of wire P is L m and its young's modulus is \(Y\text{ N/m}^2\) while length of wire Q is twice that of P and its material has young's modulus half that of P. Compute the ratio of their elongation.
Answer: For wire P, tension force \(F_P = 3mg\), length \(l_P = L\), area \(A_P = A\), Young's modulus \(Y_P = Y\).
Elongation \(\Delta l_P = \frac{(3mg)L}{A Y}\).
For wire Q, tension force \(F_Q = 2mg\), length \(l_Q = 2L\), area \(A_Q = A\), Young's modulus \(Y_Q = \frac{Y}{2}\).
Elongation \(\Delta l_Q = \frac{(2mg)(2L)}{A (Y/2)} = \frac{8mg L}{A Y}\).
Ratio \(\frac{\Delta l_P}{\Delta l_Q} = \frac{3}{8}\).
In simple words: Calculate the tension and dimensions for each wire separately, then take their ratio to find \(3:8\).

Exam Tip: Be careful with hanging loads; wire P supports both masses (\(3m\)), while wire Q supports only the lower mass (\(2m\)).

Question 8. What is Reynold's number? Write its significance. On what factors does it depend?
Answer: Reynold's number (\(N_R\)) is a dimensionless number defined as \(N_R = \frac{\rho v D}{\eta}\), representing the ratio of inertial force to viscous force in fluid flow.
Significance: It determines whether fluid flow is streamline (\(N_R < 2000\)) or turbulent (\(N_R > 3000\)).
Factors: It depends on fluid density (\(\rho\)), flow velocity (\(v\)), tube diameter (\(D\)), and fluid viscosity (\(\eta\)).
In simple words: Reynold's number predicts if water flows smoothly or swirls turbulently.

Exam Tip: Memorize the critical thresholds: below 2000 is laminar; above 3000 is turbulent.

Question 9. Define surface tension and surface energy. Obtain a relation between them.
Answer: Surface tension (\(S\)) is the force acting per unit length on an imaginary line drawn tangentially on the liquid surface: \(S = \frac{F}{l}\). Surface energy is the potential energy stored per unit surface area of a liquid.
Relation: Consider a rectangular film on a wire frame where one side of length \(l\) is pulled by distance \(\Delta x\).
Work done \(W = F \times \Delta x = (S \times 2l) \times \Delta x\) (since film has two free surfaces).
Increase in area \(\Delta A = 2 \times (l \times \Delta x)\).
Surface energy per unit area \(E = \frac{W}{\Delta A} = \frac{S \times 2l \times \Delta x}{2l \times \Delta x} = S\). Thus, surface tension equals surface energy per unit area.
In simple words: Surface tension numerically equals the surface energy per unit area.

Exam Tip: Always account for two surfaces for liquid films in wire frames.

Question 10. State and prove Torricelli's theorem for velocity of efflux.
Answer: Torricelli's theorem states that the velocity of efflux (velocity with which liquid flows out of an orifice) is equal to the velocity that a freely falling body would acquire in falling through a vertical distance equal to the depth of the orifice below the free liquid surface.
Proof: By applying Bernoulli's theorem between the free surface (level 1) and the orifice (level 2):
\(P_0 + \rho g h + \frac{1}{2}\rho v_1^2 = P_0 + 0 + \frac{1}{2}\rho v_2^2\).
Since the tank area is large, \(v_1 \approx 0\).
\(\rho g h = \frac{1}{2}\rho v_2^2 \implies v_2 = \sqrt{2gh}\).
In simple words: Liquid shoots out of a hole at speed \(\sqrt{2gh}\), exactly like an object falling freely from the liquid surface.

Exam Tip: State Bernoulli's equation clearly before setting surface velocity to zero.

Question 11. Using dimensional method obtain Stoke's law expression for viscous force \(F = 6\pi\eta r v\).
Answer: Assuming viscous force \(F\) depends on coefficient of viscosity \(\eta\), sphere radius \(r\), and velocity \(v\):
\(F = k \eta^a r^b v^c\)
Substitute dimensions: \([MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c\)
Equating powers of M: \(a = 1\)
Equating powers of T: \(-a - c = -2 \implies c = 1\)
Equating powers of L: \(-a + b + c = 1 \implies -1 + b + 1 = 1 \implies b = 1\)
Therefore, \(F = k\eta r v\) (where dimensionless constant \(k = 6\pi\)).
In simple words: Dimensional analysis helps derive physical formulas by balancing mass, length, and time dimensions.

Exam Tip: Note that dimensional analysis cannot determine the dimensionless constant \(6\pi\), which comes from fluid dynamics integration.

Question 12. The fig (a) & (b) refer to the steady flow of a non viscous liquid. Which one of the two figures is incorrect? Why?
Answer: Fig. (b) is incorrect. In fig. (a), at the narrow constriction, area is small, velocity is large, so according to Bernoulli's principle, pressure is small, causing the liquid level in the manometric tube to be lower. In fig. (b), the opposite and wrong level is shown.
In simple words: Fig. (a) is correct because narrower sections have faster flow and lower pressure, resulting in lower liquid levels in the manometer.

Exam Tip: Constriction = high speed = low pressure = lower liquid column.

Question 13. The fig a below shows a thin liquid supporting a small weight \(4.5 \times 10^{-2}\text{ N}\). What is the weight supported by a film of same liquid at the same temperature in fig (b) & (c) Explain your answer.
Answer: The weight supported in figures (b) and (c) is the same as in (a), equal to \(4.5 \times 10^{-2}\text{ N}\).
Reason: The upward force due to surface tension is \(F = 2\sigma l\), where \(\sigma\) is surface tension and \(l\) is the total contact length of the slider. Since the length \(l\) and liquid are identical in all three cases, the supported weight remains the same.
In simple words: The supported weight stays identical because the contact length and surface tension are the same across all three figures.

Exam Tip: Total contact length \(l\) determines surface tension force, independent of container shape.

Question 14. Two soap bubbles of different diameter are in contact with a certain portion common to both the bubbles. What will be the shape of the common boundary as seen from inside the smaller bubble? Support your answer with a neat diagram and justify your answer.
Answer: When seen from inside the smaller bubble, the common boundary will appear concave.
Justification: Excess pressure inside a bubble is given by \(P = \frac{4T}{R}\). Since radius \(R\) is smaller for the smaller bubble, the excess pressure inside it is greater, forcing the common interface to bulge inward (concave) toward the larger bubble.
In simple words: The smaller bubble has higher internal pressure, making the shared wall bulge inward like a bowl toward the larger bubble.

Exam Tip: Remember excess pressure is inversely proportional to radius (\(P \propto \frac{1}{R}\)).

Question 15. During blood transfusion the needle is inserted in a vein where gauge pressure is \(p_g\) and atmospheric pressure is \(p\). At what height must the blood container be placed so that blood may just enter the vein. Given density of blood is \(\rho\).
Answer: The hydrostatic pressure exerted by the blood column of height \(h\) is \(P = h\rho g\).
For blood to just enter the vein, this hydrostatic pressure must balance the vein's gauge pressure: \(h\rho g = p_g \implies h = \frac{p_g}{\rho g}\).
In simple words: The container must be elevated high enough so its hydrostatic pressure matches the blood pressure in the vein.

Question 16. Why we cannot remove a filter paper from a funnel by blowing air into narrow end.
Answer: When air is blown into the narrow end, its velocity in the narrow passage between the filter paper and funnel increases. According to Bernoulli's principle, high velocity creates a region of low pressure, which holds the filter paper more firmly against the funnel wall.
In simple words: Blowing air speeds up airflow around the paper, dropping pressure and pinning the filter paper tighter against the funnel.

Exam Tip: High air velocity = low pressure = suction effect holding the paper in place.

Question 17. On a hot day, a car is left in sunlight with all windows closed. Explain why it is considerably warmer than outside, after some time?
Answer: Glass transmits about \(50\%\) of short-wavelength heat radiation from the hot sun, which heats the car's interior. The interior objects reradiate this as long-wavelength infrared radiation, which cannot pass back out through glass (greenhouse effect), trapping heat inside.
In simple words: Sunlight enters through glass and heats the interior, but trapped heat radiation cannot escape back out through the glass.

Question 18. Name the suitable thermometers to measure the following temperatures (a) \(-100^\circ\text{C}\) (b) \(80^\circ\text{C}\) (c) \(780^\circ\text{C}\) (d) \(2000^\circ\text{C}\)
Answer: (a) \(-100^\circ\text{C}\): Gas thermometer or Alcohol thermometer.
(b) \(80^\circ\text{C}\): Mercury thermometer.
(c) \(780^\circ\text{C}\): Platinum resistance thermometer.
(d) \(2000^\circ\text{C}\): Radiation pyrometer.
In simple words: Different temperature ranges require specific thermometric substances based on their working limits.

Exam Tip: Memorize temperature ranges: mercury freezes at \(-39^\circ\text{C}\) and boils at \(357^\circ\text{C}\).

Question 19. If a drop of water falls on a very hot iron, it does not evaporate for a long time. Why?
Answer: Due to the high temperature of the iron, a vapor film is formed beneath the water drop (Leidenfrost effect). Since vapor is a poor conductor of heat, it insulates the water drop, delaying its evaporation.
In simple words: An insulating layer of steam forms under the water droplet, keeping it floating and slowing down evaporation.

Question 20. The earth without its atmosphere would be in hospitably cold. Why?
Answer: Without an atmosphere, greenhouse gases would not trap outgoing terrestrial thermal radiation at night. Heat would radiate freely into space, causing surface temperatures to drop to extremely cold levels.
In simple words: The atmosphere acts like a thermal blanket; without it, all daytime heat would escape into space at night.

Question 21. The coolant used in a nuclear plant should have high specific heat. Why?
Answer: A coolant with high specific heat can absorb a massive amount of heat energy with only a comparatively small rise in its own temperature, making it highly efficient at cooling reactor cores.
In simple words: High specific heat lets coolants absorb huge amounts of heat without getting dangerously hot themselves.

Exam Tip: Use formula \(Q = mc\Delta T\); high \(c\) means low \(\Delta T\) for large \(Q\).

Question 22. A sphere, a cube and a disc made of same material and of equal masses heated to same temperature of \(200^\circ\text{C}\). These bodies are then kept at same lower temperature in the surrounding, which of these will cool (i) fastest, (ii) slowest. Explain.
Answer: (i) The disc will cool fastest because for a given mass and material, the disc has the largest surface area, giving it the highest rate of heat emission.
(ii) The sphere will cool slowest because it has the minimum surface area for the same mass.
In simple words: Objects with larger surface areas radiate heat faster, so the flat disc cools fastest and the sphere cools slowest.

Exam Tip: Rate of cooling is directly proportional to surface area.

Question 23. (a) Why pendulum clocks generally go faster in winter and slow in summer?
(b) Why the brake drums of a car are heated when it moves down a hill at constant speed?
Answer: (a) In winter, cold temperatures shorten the pendulum length \(l\). Since time period \(T = 2\pi\sqrt{l/g}\), a shorter pendulum has a smaller time period, making the clock go faster. In summer, expansion lengthens the pendulum, making it run slow.
(b) When a car moves down a hill at constant speed, its gravitational potential energy is converted into work against friction between the brake shoes and drum, appearing as heat.
In simple words: Pendulums shrink in winter (ticking faster) and expand in summer (ticking slower), while brakes convert gravitational energy into frictional heat.

Exam Tip: Connect thermal expansion/contraction directly to pendulum length and time period formula.

Question 24. The plots of intensity versus wavelength for three blackbodies at temperature \(T_1, T_2\) and \(T_3\) respectively are shown. Arrange the temperature in decreasing order. Justify your answer.
Answer: According to Wein's displacement law, \(\lambda_m T = \text{constant}\), implying \(\lambda_m \propto \frac{1}{T}\). From the given graph, the wavelengths corresponding to peak emission satisfy \(\lambda_{m1} < \lambda_{m3} < \lambda_{m2}\). Therefore, the temperatures in decreasing order are \(T_1 > T_3 > T_2\).
In simple words: Shorter peak wavelength means higher temperature, giving the order \(T_1 > T_3 > T_2\).

Exam Tip: Peak wavelength shifts left toward shorter wavelengths for hotter bodies.

Question 25. The triple point of water is a standard fixed point in modern thermometry. Why? Why melting point of ice or boiling point of water not used as standard fixed fixed points.
Answer: The triple point of water is a unique temperature (\(273.16\text{ K}\) and pressure \(0.46\text{ cm}\) of Hg) where ice, water, and water vapor coexist in equilibrium, and it is completely independent of external atmospheric pressure. In contrast, the melting point of ice and boiling point of water depend heavily on external atmospheric pressure variations.
In simple words: The triple point is a fixed, unchangeable natural reference point, whereas ice and boiling points change with air pressure.

Exam Tip: Emphasize pressure independence for the triple point of water.

SHORT ANSWER TYPE QUESTION (3 MARKS)

Question 1. How is the knowledge of elasticity useful in selecting metal ropes used in cranes for lifting heavy loads.
Answer: The ultimate stress of a material should not exceed its elastic limit. For steel, breaking stress is around \(30 \times 10^7\text{ N/m}^2\).
To lift a heavy load of mass \(M\) (e.g., \(10^4\text{ kg}\)), using \(\text{Stress} = \frac{Mg}{\pi r^2}\), we calculate the required radius \(r\) of the rope.
To impart flexibility and prevent permanent damage, cranes use ropes made of a large number of thin wires braided together rather than one thick solid rod.
In simple words: Calculating elastic limits helps engineers design thick, flexible steel wire ropes that can safely lift heavy loads without snapping.

Exam Tip: Braiding thin wires provides flexibility while maintaining high tensile strength.

Question 2. The torque required to produce unit twist in a solid shaft of radius r, length l and made of material of modulus of rigidity \(\eta\) is given by \(\tau = \frac{\pi\eta r^4}{2l}\). Explain why hollow shafts are preferred to solid shafts for transmitting torque?
Answer: The torque required for unit twist in a hollow shaft of internal radius \(r_1\) and external radius \(r_2\) is \(\tau' = \frac{\pi\eta(r_2^4 - r_1^4)}{2l}\).
Taking the ratio of hollow to solid shaft torque for equal cross-sectional area (\(\pi r^2 = \pi(r_2^2 - r_1^2)\)):
\(\frac{\tau'}{\tau} = \frac{r_2^4 - r_1^4}{r^4} = \frac{(r_2^2 + r_1^2)(r_2^2 - r_1^2)}{r^4}\).
Since \((r_2^2 + r_1^2) > r^2\), we get \(\tau' > \tau\). Therefore, hollow shafts are much stronger and stiffer than solid shafts of the same weight and material.
In simple words: Hollow shafts can transmit more twisting force with less material, making them stronger and lighter than solid shafts.

Exam Tip: Use the difference of fourth powers to prove hollow shafts have higher torsional rigidity.

Question 3. Stress strain curve for two wires of material A and B are as shown in Fig. (a) which material is more ductile? (b) which material has greater value of young's modulus? (c) which of the two is stronger material? (d) which material is more brittle?
Answer: (a) Material A is more ductile because it has a larger plastic region before breaking.
(b) Material A has a greater Young's modulus because its stress-strain curve has a steeper initial slope (\(Y = \text{Stress}/\text{Strain}\)).
(c) Material A is stronger because its ultimate tensile strength point is higher on the stress axis.
(d) Material B is more brittle because it has a very small plastic region and breaks soon after the elastic limit.
In simple words: Material A stretches further and is stronger, while Material B snaps easily with little deformation.

Exam Tip: Steeper slope = higher Young's modulus; longer curve = higher ductility.

Question 4. State Pascal's law for fluids with the help of a neat labelled diagram. Explain the principle and working of hydraulic brakes.
Answer: Pascal's law states that the increase in pressure at one point of an enclosed liquid in equilibrium at rest is transmitted undiminished in all directions.
Working of Hydraulic Brakes: When the brake pedal is pushed, a small piston exerts force on brake oil, creating pressure. According to Pascal's law, this pressure is transmitted equally to larger wheel cylinders, pushing larger pistons outward with magnified force to clamp the brake shoes against the rotating brake drum and stop the vehicle.
In simple words: Pascal's law allows a small push on the brake pedal to transmit equal pressure and multiply force at the wheels to stop a car.

Exam Tip: Key terms: pressure transmission undiminished, force multiplication.

Question 5. A manometer reads the pressure of a gas in an enclosure as shown in the fig (a) when some of the gas is removed by a pump, the manometer reads as in fig (b) The liquid used in manometer is mercury and the atmospheric pressure is \(76\text{ cm}\) of mercury. (i) Give absolute and gauge pressure of the gas in the enclosure for cases (a) and (b).
Answer: (a) In case (a), Pressure head \(h = +20\text{ cm}\) of Hg.
Gauge pressure = \(h = 20\text{ cm}\) of Hg.
Absolute pressure = \(P_{\text{atm}} + h = 76 + 20 = 96\text{ cm}\) of Hg.
(b) In case (b), Pressure head \(h = -18\text{ cm}\) of Hg.
Gauge pressure = \(h = -18\text{ cm}\) of Hg.
Absolute pressure = \(76 - 18 = 58\text{ cm}\) of Hg.
In simple words: Add atmospheric pressure to the mercury height difference for absolute pressure, and use the height alone for gauge pressure.

Exam Tip: Absolute pressure equals atmospheric pressure plus/minus gauge pressure.

Question 6. How would the levels change in (b) if \(13.6\text{ cm}\) of \(\text{H}_2\text{O}\) (immiscible with mercury) are poured into the right limb of the manometer in the above numerical.
Answer: Using hydrostatic balance condition \(h_1 \rho_1 g = h_2 \rho_2 g\):
\(h_{\text{water}} \times 1 = h_{\text{Hg}} \times 13.6 \implies 13.6 \times 1 = h_{\text{Hg}} \times 13.6 \implies h_{\text{Hg}} = 1\text{ cm}\).
Therefore, pouring \(13.6\text{ cm}\) of water in the right limb will displace the mercury level by \(1\text{ cm}\) in the left limb, making the level difference \(19\text{ cm}\).
In simple words: Due to mercury's high density (\(13.6\) times water), a 13.6 cm column of water depresses the mercury level by only 1 cm.

Exam Tip: Always equate pressure heads using relative density: \(h_w \rho_w = h_m \rho_m\).

Question 7. Define Capillarity and angle of contact. Derive an expression for the ascent of a liquid in a capillary tube.
Answer: Capillarity is the phenomenon of rise or fall of liquid in a capillary tube. The angle of contact is the angle between tangents drawn at the point of contact to the liquid surface and solid surface inside the liquid.
Derivation for ascent: Upward surface tension force = \(2\pi r S \cos\theta\).
Downward weight of liquid column = Volume \(\times\) density \(\times\) gravity \(= (\pi r^2 h + \text{meniscus volume}) \rho g \approx \pi r^2 h \rho g\).
Equating upward and downward forces: \(2\pi r S \cos\theta = \pi r^2 h \rho g \implies h = \frac{2S\cos\theta}{r\rho g}\).
In simple words: Equating surface tension pulling upward to the weight of the water column gives the capillary rise formula.

Exam Tip: Resolve the vertical component of surface tension as \(S\cos\theta\).

Question 8. The terminal velocity of a tiny droplet is v. N number of such identical droplets combine together forming a bigger drop. Find the terminal velocity of the bigger drop.
Answer: Terminal velocity is given by \(v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta} \implies v \propto r^2\).
If \(N\) small droplets of radius \(r\) combine to form a big drop of radius \(R\):
Volume of big drop = Volume of \(N\) small droplets: \(\frac{4}{3}\pi R^3 = N \left(\frac{4}{3}\pi r^3\right) \implies R = N^{1/3}r\).
Terminal velocity of bigger drop: \(V = \left(\frac{R}{r}\right)^2 v = (N^{1/3})^2 v = N^{2/3}v\).
In simple words: A combined drop falls faster because terminal velocity increases with the square of the radius.

Exam Tip: Establish volume conservation first to find the radius relation \(R = N^{1/3}r\).

Question 9. Two spherical soap bubble coalesce. If v be the change in volume of the contained air, A is the change in total surface area then show that \(3PV + 4AT = 0\) where T is the surface tension and P is atmospheric pressure.
Answer: Pressures inside bubbles of radii \(r_1\) and \(r_2\) are \(P_1 = P + \frac{4T}{r_1}\) and \(P_2 = P + \frac{4T}{r_2}\).
When they coalesce into a bubble of radius \(r\), mass of air is conserved: \(P_1 V_1 + P_2 V_2 = PV\).
Substituting pressure and volume relations, and using change in volume \(V\) and change in area \(A = 4\pi(r^3 - r_1^3 - r_2^3)\), algebraic simplification yields \(3PV + 4AT = 0\).
In simple words: Applying Boyle's law and surface energy conservation during bubble coalescence leads to the relation \(3PV + 4AT = 0\).

Exam Tip: This is an advanced conservation problem combining ideal gas pressure with surface energy.

Question 10. Give the principle of working of venturimeter. Obtain an expression for volume of liquid flowing through the tube per second.
Answer: Venturimeter works on Bernoulli's principle. When fluid flows from a wide tube into a narrow constriction, velocity increases and pressure decreases.
Let \(A_1, A_2\) be cross-sectional areas and \(v_1, v_2\) be velocities. By equation of continuity: \(A_1 v_1 = A_2 v_2\).
Applying Bernoulli's theorem: \(P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)\).
Volume flow rate per second \(V = A_1 A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho(A_1^2 - A_2^2)}}\).
In simple words: A venturimeter measures fluid flow speed using pressure differences created by a narrow constriction.

Exam Tip: Combine Bernoulli's equation with the equation of continuity.

Question 11. A big size balloon of mass M is held stationary in air with the help of a small block of mass \(M/2\) tied to it by a light string such that both float in mid air. Describe the motion of the balloon and the block when the string is cut. Support your answer with calculations.
Answer: When the system is stationary, Upthrust \(U = \text{Weight of balloon} + \text{Tension} \implies U = Mg + T\).
For the small block: \(T = \frac{M}{2}g\).
Therefore, Total Upthrust \(U = Mg + \frac{M}{2}g = \frac{3}{2}Mg\).
When the string is cut (\(T = 0\)), the block falls freely under gravity (\(a_{\text{block}} = g\) downward).
For the balloon, net upward force = \(U - Mg = \frac{3}{2}Mg - Mg = \frac{1}{2}Mg = Ma_{\text{balloon}}\).
Acceleration of the balloon \(a_{\text{balloon}} = \frac{g}{2}\) upward.
In simple words: Cutting the string causes the block to fall freely downward while the balloon accelerates upward at \(g/2\).

Exam Tip: Calculate total upthrust first using static equilibrium conditions.

Question 12. Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same? Why do the vessels filled to same height give different reading on weighing scale.
Answer: (i) Yes, the force exerted by water on the base is the same because force = pressure \(\times\) base area = \((h\rho g) \times A\), which depends only on height \(h\) and base area \(A\).
(ii) Weighing scales give different readings because vessel 1 contains a larger total volume of water than vessel 2.
In simple words: Base pressure and force depend only on water height, but total weight on the scale depends on total water volume.

Question 13. A liquid drop of diameter D breaks up into 27 tiny drops. Find the resulting change in energy. Take surface tension of liquid as \(\sigma\).
Answer: Radius of large drop \(R = D/2\). Radius of each small drop \(r\).
Volume conservation: \(27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies 3r = R \implies r = R/3 = D/6\).
Initial surface area of large drop = \(4\pi R^2 = \pi D^2\).
Final surface area of 27 small drops = \(27 \times (4\pi r^2) = 27 \times 4\pi \left(\frac{D}{6}\right)^2 = 3\pi D^2\).
Increase in surface area = \(3\pi D^2 - \pi D^2 = 2\pi D^2\).
Change in energy = Increase in area \(\times \sigma = 2\pi D^2 \sigma\).
In simple words: Breaking a large drop into smaller ones increases total surface area, requiring energy equal to \(2\pi D^2\sigma\).

Question 14. Define the coefficients of linear expansion. Deduce relation between it and coefficient of superficial expansion and volume expansion.
Answer: Coefficient of linear expansion (\(\alpha\)) is fractional increase in length per degree rise in temperature: \(\Delta l = l \alpha \Delta T\).
Superficial expansion \(\beta = 2\alpha\) (area expansion coefficient).
Volume expansion \(\gamma = 3\alpha\) (volume expansion coefficient).
Ratio relation: \(\alpha : \beta : \gamma = 1 : 2 : 3\).
In simple words: Linear, superficial, and volume expansion coefficients relate as \(1:2:3\).

Exam Tip: Remember standard ratio \(\alpha : \beta : \gamma = 1 : 2 : 3\).

Question 15. Describe the different types of thermometers commonly used. Give the relation between temperature on different scales. Give four reasons for using mercury in a thermometer.
Answer: Types: Liquid-in-glass (mercury/alcohol), Gas thermometers, Platinum resistance thermometers, and Radiation pyrometers.
Relation: \(\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273}{5}\).
Reasons for mercury: (1) It is opaque and shines brightly, (2) It does not wet glass, (3) It has a uniform coefficient of expansion, (4) It remains liquid over a wide temperature range.
In simple words: Mercury is ideal for thermometers because it's opaque, doesn't stick to glass, and stays liquid across wide temperatures.

Question 16. Two rods of different metals of coefficient of linear expansion \(\alpha_1\) and \(\alpha_2\) and initial length \(l_1\) and \(l_2\) respectively are heated to the same temperature. Find relation in \(\alpha_1, \alpha_2, l_1\) and \(l_2\) such that that difference between their lengths remain constant.
Answer: Lengths after heating: \(l_1' = l_1(1 + \alpha_1 \Delta T)\) and \(l_2' = l_2(1 + \alpha_2 \Delta T)\).
Given difference in length remains constant: \(l_2' - l_1' = l_2 - l_1\).
\(l_2(1 + \alpha_2 \Delta T) - l_1(1 + \alpha_1 \Delta T) = l_2 - l_1\)
\(l_2 \alpha_2 \Delta T = l_1 \alpha_1 \Delta T \implies l_2 \alpha_2 = l_1 \alpha_1\).
In simple words: For the difference in rod lengths to stay constant, the product of initial length and expansion coefficient must be equal for both rods.

Question 17. Explain the principle of platinum resistance thermometer.
Answer: A platinum resistance thermometer works on the principle that the electrical resistance of metals (like platinum) varies directly with temperature according to \(R_t = R_0(1 + \alpha t)\). By measuring the electrical resistance, temperature can be accurately determined.
In simple words: Electrical resistance of platinum increases predictably with temperature, allowing precise temperature measurement.

Exam Tip: Mention resistance-temperature linearity: \(R_t = R_0(1 + \alpha t)\).

Question 18. Draw a graph to show the anomalous behaviour of water. Explain its importance for sustaining life under water.
Answer: Water contracts when heated from \(0^\circ\text{C}\) to \(4^\circ\text{C}\) and expands above \(4^\circ\text{C}\), reaching maximum density at \(4^\circ\text{C}\).
Importance: In winter, as pond surface water cools to \(4^\circ\text{C}\), it sinks to the bottom. Ice forms on top at \(0^\circ\text{C}\). Since ice is a poor conductor of heat and water at \(4^\circ\text{C}\) stays at the bottom, aquatic life survives safely underneath.
In simple words: Water's maximum density at \(4^\circ\text{C}\) keeps deep lakes unfrozen, allowing fish to survive winter.

Exam Tip: Sketch volume vs temperature curve showing a minimum at \(4^\circ\text{C}\) (density maximum).

Question 19. A steel rail of length \(5\text{ m}\) and area of cross section \(40\text{ cm}^2\) is prevented from expanding while the temperature rises by \(10^\circ\text{C}\). Given coefficient of linear expansion of steel is \(1.2 \times 10^{-5}\text{ K}^{-1}\). Explain why space needs to be given for thermal expansion.
Answer: Compressive strain = \(\alpha \Delta T = 1.2 \times 10^{-5} \times 10 = 1.2 \times 10^{-4}\).
Thermal stress = \(Y \times \text{strain} = 2 \times 10^{11} \times 1.2 \times 10^{-4} = 2.4 \times 10^7\text{ N m}^{-2}\).
Thermal force \(\Delta F = \text{Stress} \times \text{Area} = 2.4 \times 10^7 \times (40 \times 10^{-4}) = 9.6 \times 10^4\text{ N}\).
Explanation: Such massive forces can bend or buckle steel rails, so expansion gaps are essential.
In simple words: Preventing thermal expansion creates huge mechanical forces that can warp steel tracks, making expansion gaps necessary.

Exam Tip: Combine thermal strain with Young's modulus to calculate thermal stress \(F = YA\alpha\Delta T\).

Question 20. Define (i) Specific heat capacity (ii) Heat capacity (iii) Molar specific heat capacity at constant pressure and at constant volume and write their units.
Answer: (i) Specific heat capacity: Heat required to raise the temperature of unit mass by \(1^\circ\text{C}\). Unit: \(\text{J kg}^{-1}\text{ K}^{-1}\).
(ii) Heat capacity: Heat required to raise the temperature of the entire body by \(1^\circ\text{C}\). Unit: \(\text{J K}^{-1}\).
(iii) Molar specific heat capacity: Heat required to raise the temperature of 1 mole of gas by \(1^\circ\text{C}\) (at constant pressure \(C_p\) or constant volume \(C_v\)). Unit: \(\text{J mol}^{-1}\text{ K}^{-1}\).
In simple words: These terms measure how much heat substances absorb per unit mass, per entire body, or per mole.

Question 21. Plot a graph of temperature versus time showing the change in the state of ice on heating and hence explain the process (with reference to latent heat)
Answer: The graph shows temperature rising from negative values to \(0^\circ\text{C}\), pausing flat during melting (latent heat of fusion), rising again to \(100^\circ\text{C}\), and pausing flat during vaporization (latent heat of vaporization).
Explanation: During phase changes, heat supplied is entirely used to break intermolecular bonds without increasing kinetic energy or temperature.
In simple words: Flat plateaus on a heating graph show temperature pausing while latent heat changes ice to water and water to steam.

Question 22. What is the effect of pressure on melting point of a substance? What is regelation. Give a practical application of it.
Answer: For substances that contract on melting (like ice), melting point decreases with an increase in pressure. For substances that expand on melting, melting point increases with pressure.
Regelation: The phenomenon in which ice melts under increased pressure and refreezes when pressure is removed.
Application: Skating on ice is possible because pressure from ice skates melts a thin water film underneath, lubricating the skate blades.
In simple words: Pressure melts ice into water at sub-zero temperatures, which refreezes once pressure is removed (regelation).

Question 23. What is the effect of pressure on the boiling point of a liquid. Describe a simple experiment to demonstrate the boiling of \(\text{H}_2\text{O}\) at a temperature much lower than \(100^\circ\text{C}\). Give a practical application of this phenomenon.
Answer: The boiling point of a liquid increases with an increase in pressure and decreases with a decrease in pressure.
Experiment: Half-boil water in a round-bottom flask, remove from heat, cork it tightly, and invert it. Pouring cold water over the flask condenses internal steam, lowering pressure and causing the warm water inside to boil vigorously again.
Application: Cooking food in a pressure cooker takes less time because higher pressure raises the boiling point of water.
In simple words: Lowering air pressure makes water boil at lower temperatures, which is how pressure cookers work in reverse.

Question 24. State and explains the three modes of transfer of heat. Explain how the loss of heat due to these three modes is minimised in a thermos flask.
Answer: Modes: Conduction (transfer via molecular contact), Convection (transfer via bulk fluid movement), and Radiation (transfer via electromagnetic waves).
Thermos Flask minimization: (1) Double-walled glass container with a vacuum between walls eliminates conduction and convection. (2) Silvered walls reflect thermal radiation back inward. (3) Insulating stopper prevents heat loss from the top.
In simple words: Thermos flasks use a vacuum to stop conduction and convection, and silvered walls to block radiation.

Exam Tip: Break down flask features by which heat transfer mode they block.

Question 25. Define coefficient of thermal conductivity. Two metal slabs of same area of cross-section, thickness \(d_1\) and \(d_2\) having thermal conductivities \(K_1\) and \(K_2\) respectively are kept in contact. Deduce expression for equivalent thermal conductivity.
Answer: Coefficient of thermal conductivity is rate of heat flow per unit area per unit temperature gradient. Rate of heat flow in steady state:
\(\frac{Q}{t} = \frac{K_1 A (T_1 - T)}{d_1} = \frac{K_2 A (T - T_2)}{d_2} = \frac{K A (T_1 - T_2)}{d_1 + d_2}\).
Solving for equivalent thermal conductivity \(K\):
\(K = \frac{K_1 K_2 (d_1 + d_2)}{K_2 d_1 + K_1 d_2}\).
In simple words: Combining series thermal slabs gives an equivalent thermal conductivity formula based on individual thicknesses and conductivities.

Exam Tip: Equate heat flow rates across series slabs to derive equivalent conductivity.

LONG ANSWER TYPE QUESTIONS (5 MARKS)

Question 1. Draw and discuss stress versus strain graph, explaining the terms elastic limit, permanent set, proportionality limit, elastic hysteresis, tensible strength.
Answer: The stress-strain graph illustrates how a material deforms under increasing load.
- Proportionality limit (Point A): Region where stress is directly proportional to strain (Hooke's law holds).
- Elastic limit (Point B): Maximum stress up to which a material fully recovers its original shape upon load removal.
- Permanent set: Permanent deformation remaining when load is removed beyond the elastic limit.
- Tensile strength (Ultimate tensile strength, Point D): Maximum stress the material can endure before necking and breaking.
- Elastic hysteresis: The delayed recovery of strain in cyclic loading, represented by the area of the loading-unloading stress-strain loop.
In simple words: The stress-strain graph maps out elastic limits, yield points, ultimate strength, and permanent deformation zones.

Exam Tip: Sketch a clear curve labeling O-A (proportional), B (elastic limit), C (yield), D (ultimate tensile strength), and fracture point.

Question 2. Show that there is always an excess pressure on the concave side of the meniscus of a liquid. Obtain an expression for the excess pressure inside (i) a liquid drop (ii) liquid bubble (iii) air bubble inside a liquid.
Answer: (i) Liquid drop: Has only one surface film. Excess pressure \(P = \frac{2S}{R}\).
(ii) Liquid (soap) bubble: Has two surface films (inner and outer). Excess pressure \(P = \frac{4S}{R}\).
(iii) Air bubble inside a liquid: Has only one surface film. Excess pressure \(P = \frac{2S}{R}\).
Derivation involves equating work done by excess pressure to increase in surface energy: \(P \times (\Delta V) = S \times (\Delta A)\).
In simple words: Concave surfaces always have higher pressure, with drops and air bubbles having \(2S/R\) and soap bubbles having \(4S/R\).

Exam Tip: Remember soap bubbles have two free surfaces, hence \(4S/R\).

Question 3. State and prove Bernoullis theorem. Give its limitation. Name any two applications of the principle.
Answer: Statement: For streamline flow of an ideal (incompressible and non-viscous) liquid, the total energy (sum of pressure energy, potential energy, and kinetic energy) per unit volume remains constant at every cross-section throughout the tube.
Equation: \(P + \rho g h + \frac{1}{2}\rho v^2 = \text{constant}\).
Proof: Derived using work-energy theorem applied to fluid flow across unequal cross-sectional areas.
Limitations: Assumes non-viscous, incompressible, and steady streamline flow.
Applications: Working of venturimeter and aerodynamic lift of aeroplane wings.
In simple words: Bernoulli's theorem states that total energy in steady fluid flow is conserved, linking pressure, speed, and height.

Exam Tip: State all three limitations: non-viscous, incompressible, and irrotational flow.

Question 4. Define terminal velocity. Obtain an expression for terminal velocity of a sphere falling through a viscous liquid. Use the formula to explain the observed rise of air bubbles in a liquid.
Answer: Terminal velocity is the maximum constant velocity acquired by a body while falling freely in a viscous medium.
Derivation: Upward forces on sphere = Viscous force (\(6\pi\eta r v\)) + Buoyant force = Downward gravitational force (\(\frac{4}{3}\pi r^3 \rho g\)).
Equating forces: \(6\pi\eta r v + \frac{4}{3}\pi r^3 \sigma g = \frac{4}{3}\pi r^3 \rho g \implies v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}\).
Air bubbles rise because the density of air (\(\sigma\)) is much less than liquid density (\(\rho\)), making \((\rho - \sigma)\) negative, resulting in upward terminal velocity.
In simple words: When drag equals net downward weight, a falling object reaches a constant terminal velocity.

Exam Tip: Balance viscous drag plus upthrust against true weight.

Question 5. State Newton's law of cooling. Deduce the relations: \(\log_e(T - T_0) = -kt + c\) and \(T - T_0 = C e^{-kt}\).
Answer: Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings, provided the difference is small.
Mathematical formulation: \(-\frac{dT}{dt} = k(T - T_0) \implies \frac{dT}{T - T_0} = -k\,dt\).
Integrating both sides: \(\log_e(T - T_0) = -kt + c\).
Taking exponential on both sides: \(T - T_0 = e^{-kt + c} = e^c e^{-kt} = C e^{-kt}\).
In simple words: Hot objects cool down at a rate proportional to the temperature gap between them and the room.

Exam Tip: Integrate differential form \(\int \frac{dT}{T-T_0} = \int -k\,dt\) to get logarithmic and exponential solutions.

Question 6. On what factors does the rate of heat conduction in a metallic rod in the steady state depend. Write the necessary expression and hence define the coefficient of thermal conductivity. Write its unit and dimensions.
Answer: Rate of heat conduction \(\frac{dQ}{dt}\) is directly proportional to cross-sectional area \(A\) and temperature difference \((T_1 - T_2)\), and inversely proportional to rod length \(d\):
\(\frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{d}\).
Coefficient of thermal conductivity \(K\) is defined as the rate of heat flow per unit area per unit temperature gradient.
SI Unit: \(\text{J s}^{-1}\text{ m}^{-1}\text{ K}^{-1}\) or \(\text{W m}^{-1}\text{ K}^{-1}\).
Dimensions: \([M^1 L^1 T^{-3} \theta^{-1}]\).
In simple words: Thermal conductivity measures how well a material conducts heat, depending on area, length, and temperature difference.

Question 7. Distinguish between conduction, convection and radiation.
Answer:
- Conduction: Transfer of heat through stationary matter without actual movement of particles; occurs mainly in solids.
- Convection: Transfer of heat through actual bulk movement of fluid particles; occurs in liquids and gases.
- Radiation: Transfer of heat via electromagnetic waves requiring no intervening material medium.
In simple words: Conduction passes heat through contact, convection moves fluid, and radiation travels through space as waves.

Exam Tip: Highlight medium requirement: conduction/convection need matter, radiation works in vacuum.

Question 8. What is meant by a block body. Explain how a black body may be achieved in practice. State and explain Stefan's law?
Answer: A black body is an idealized physical body that absorbs all incident electromagnetic radiation regardless of frequency or angle of incidence.
Practical realization: Ferry's black body consists of a double-walled hollow sphere with a small pinhole aperture and a conical projection inside to trap entering radiation through multiple internal reflections.
Stefan's Law: Total radiant energy emitted per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature: \(E = \sigma T^4\).
In simple words: A black body absorbs all radiation falling on it, and Ferry's hollow sphere mimics this in practice.

Exam Tip: Mention Ferry's cavity and multiple reflections for practical realization.

Question 9. Explain the construction and working of a constant volume gas thermometer. What are its main draw backs?
Answer: Construction: Consists of a gas bulb connected to a mercury manometer via a capillary tube, where gas volume is kept constant by adjusting mercury levels.
Working: Based on Charles's law, pressure of gas at constant volume is directly proportional to absolute temperature (\(P \propto T\)).
Drawbacks: Bulky size, slow thermal response time, and glass bulb softening at very high temperatures.
In simple words: Gas thermometers measure temperature by tracking pressure changes in a constant-volume gas bulb.

Question 10. Discuss energy distribution of a black body radiation spectrum. Explain Wein's displacement law of radiation and Stefan's law of heat radiation.
Answer: The black body radiation spectrum shows that at a given temperature, energy is emitted across a continuous range of wavelengths with a distinct peak wavelength (\(\lambda_m\)). As temperature increases, the peak shifts toward shorter wavelengths.
- Wein's Law: \(\lambda_m T = \text{constant}\), showing peak wavelength is inversely proportional to temperature.
- Stefan's Law: Total emissive power \(E = \sigma T^4\), showing total radiated energy scales with the fourth power of temperature.
In simple words: Black body spectra show how radiant energy is distributed across wavelengths at various temperatures.

Exam Tip: Summarize both Wein's peak shift and Stefan's \(T^4\) total energy law.

NUMERICALS

Question 1. An aluminium wire 1m in length and radius 1mm is loaded with a mass of 40 kg hanging vertically. Young's modulus of Al is \(7.0 \times 10^{10}\text{ N/m}^2\) Calculate (a) tensile stress (b) change in length (c) tensile strain and (d) the force constant of such a wire.
Answer:
(a) Stress = \(\frac{F}{A} = \frac{mg}{\pi r^2} = \frac{40 \times 9.8}{\pi (1 \times 10^{-3})^2} = 1.25 \times 10^8\text{ N m}^{-2}\).
(b) \(\Delta L = \frac{FL}{AY} = \frac{40 \times 9.8 \times 1}{\pi (1 \times 10^{-3})^2 \times 7.0 \times 10^{10}} = 1.78 \times 10^{-3}\text{ m}\).
(c) Strain = \(\frac{\Delta L}{L} = \frac{1.78 \times 10^{-3}}{1} = 1.78 \times 10^{-3}\).
(d) Force constant \(K = \frac{F}{\Delta L} = \frac{40 \times 9.8}{1.78 \times 10^{-3}} = 2.2 \times 10^5\text{ N m}^{-1}\).
In simple words: Standard elasticity formulas yield stress, elongation, strain, and force constant step-by-step.

Exam Tip: Convert all units to SI meters and kilograms before plugging into formulas.

Question 2. The average depth of ocean is \(2500\text{ m}\). Calculate the fractional compression \((\Delta V / V)\) of water at the bottom of ocean, given that the bulk modulus of water is \(2.3 \times 10^9\text{ N/m}^2\).
Answer: Pressure at depth \(h = 2500\text{ m}\) is \(P = h\rho g = 2500 \times 1000 \times 9.8 = 2.45 \times 10^7\text{ N m}^{-2}\).
Bulk modulus \(K = \frac{P}{\Delta V / V} \implies \frac{\Delta V}{V} = \frac{P}{K} = \frac{2.45 \times 10^7}{2.3 \times 10^9} = 1.06 \times 10^{-2}\) (or \(1.06\%\)).
In simple words: High water pressure at 2500 m depth compresses water by about 1 percent.

Exam Tip: Calculate hydrostatic pressure first using \(P = h\rho g\).

Question 3. A force of \(5 \times 10^3\text{ N}\) is applied tangentially to the upper face of a cubical block of steel of side \(30\text{ cm}\). Find the displacement of the upper face relative to the lower one, and the angle of shear. The shear modulus of steel is \(8.3 \times 10^{10}\text{ Pa}\).
Answer: Side \(y = 0.30\text{ m}\), Area \(A = (0.30)^2 = 0.09\text{ m}^2\).
Shear modulus \(\eta = \frac{F/A}{\Delta x / y} \implies \Delta x = \frac{F y}{\eta A} = \frac{5 \times 10^3 \times 0.30}{8.3 \times 10^{10} \times 0.09} = 2.0 \times 10^{-7}\text{ m}\).
Angle of shear \(\theta = \frac{\Delta x}{y} = \frac{2.0 \times 10^{-7}}{0.30} = 6.67 \times 10^{-7}\text{ radians}\).
In simple words: Tangential force causes a microscopic sideways displacement and a tiny shear angle in the steel block.

Question 4. How much should the pressure on one litre of water be changed to compress it by \(0.10\%\)?
Answer: Fractional volume change \(\frac{\Delta V}{V} = 0.10\% = 0.001\).
Bulk modulus of water \(K \approx 2.2 \times 10^9\text{ N m}^{-2}\).
Pressure change \(\Delta P = K \left(\frac{\Delta V}{V}\right) = 2.2 \times 10^9 \times 0.001 = 2.2 \times 10^6\text{ N m}^{-2}\).
In simple words: Compressing water by just 0.1 percent requires a huge pressure of \(2.2 \times 10^6\text{ Pa}\).

Question 5. Calculate the pressure at a depth of \(10\text{ m}\) in an Ocean. The density of sea water is \(1030\text{ kg/m}^3\). The atmospheric pressure is \(1.01 \times 10^5\text{ Pa}\).
Answer: Hydrostatic pressure \(P_h = h\rho g = 10 \times 1030 \times 9.8 = 1.009 \times 10^5\text{ N m}^{-2}\).
Total absolute pressure = \(P_h + P_{\text{atm}} = 1.009 \times 10^5 + 1.01 \times 10^5 = 2.02 \times 10^5\text{ Pa}\).
In simple words: Total pressure at 10 meters depth equals atmospheric pressure plus water column pressure.

Exam Tip: Do not forget to add atmospheric pressure to gauge pressure for absolute pressure.

Question 6. In a hydraulic lift air exerts a force F on a small piston of radius \(5\text{ cm}\). The pressure is transmitted to the second piston of radius \(15\text{ cm}\). If a car of mass \(1350\text{ kg}\) is to be lifted, calculate force F that is to be applied.
Answer: By Pascal's law, pressure is equal: \(\frac{F_1}{A_1} = \frac{F_2}{A_2}\).
\(F = F_2 \left(\frac{r_1}{r_2}\right)^2 = (1350 \times 9.8) \times \left(\frac{5}{15}\right)^2 = 13230 \times \frac{1}{9} = 1470\text{ N}\).
In simple words: A hydraulic lift uses piston area ratios to multiply force, letting 1470 N lift a heavy car.

Exam Tip: Radii ratio squared equals area ratio since \(A = \pi r^2\).

Question 7. How much pressure will a man of weight \(80\text{ kg f}\) exert on the ground when (1) he is lying and (2) he is standing on his feet. Given area of the body of the man is \(0.6\text{ m}^2\) and that of his feet is \(80\text{ cm}^2\).
Answer: Force \(F = mg = 80 \times 9.8 = 784\text{ N}\).
(1) Lying: Area \(A_1 = 0.6\text{ m}^2\).
\(P_1 = \frac{784}{0.6} = 1.31 \times 10^3\text{ N m}^{-2}\).
(2) Standing: Area \(A_2 = 80\text{ cm}^2 = 80 \times 10^{-4}\text{ m}^2 = 0.008\text{ m}^2\).
\(P_2 = \frac{784}{0.008} = 9.8 \times 10^4\text{ N m}^{-2}\).
In simple words: Standing on your feet concentrates your weight onto a small area, creating much higher pressure than lying down.

Question 8. The manual of a car instructs the owner to inflate the tyres to a pressure of \(200\text{ k Pa}\). (a) What is the recommended gauge pressure? (b) What is the recommended absolute pressure? (c) If, after the required inflation of the tyres, the car is driven to a mountain peak where the atmospheric pressure is \(10\%\) below that at sea level, what will the tyre gauge read?
Answer:
(a) Recommended gauge pressure = \(200\text{ kPa}\).
(b) Absolute pressure = Gauge pressure + Atmospheric pressure = \(200 + 101 = 301\text{ kPa}\).
(c) On mountain, new atmospheric pressure \(P_a' = 101 - 10.1 = 90.9\text{ kPa}\).
Assuming absolute air in tyre remains constant: New gauge pressure = \(301 - 90.9 \approx 211\text{ kPa}\).
In simple words: Tyre gauge pressure rises slightly on mountains because outside atmospheric pressure drops.

Exam Tip: Gauge pressure equals absolute pressure minus external atmospheric pressure.

Question 9. Calculate excess pressure in an air bubble of radius \(6\text{ mm}\). Surface tension of liquid is \(0.58\text{ N/m}\).
Answer: An air bubble inside liquid has only one surface film. Radius \(r = 6\text{ mm} = 6 \times 10^{-3}\text{ m}\).
Excess pressure \(P = \frac{2S}{r} = \frac{2 \times 0.58}{6 \times 10^{-3}} = 193.3\text{ N m}^{-2}\).
In simple words: Using the single-surface excess pressure formula gives \(193.3\text{ Pa}\).

Exam Tip: Use \(2S/r\) for air bubbles in liquid and liquid drops; use \(4S/r\) for soap bubbles.

Question 10. Terminal velocity of a copper ball of radius \(2\text{ mm}\) through a tank of oil at \(20^\circ\text{C}\) is \(6.0\text{ cm/s}\). Compare coefficient of viscosity of oil. Given \(\rho_{\text{copper}} = 8.9 \times 10^3\text{ kg/m}^3, \rho_{\text{oil}} = 1.5 \times 10^3\text{ kg/m}^3\).
Answer: Terminal velocity formula: \(v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}\).
Rearranging for \(\eta\):
\(\eta = \frac{2}{9} \frac{(2 \times 10^{-3})^2 (8.9 \times 10^3 - 1.5 \times 10^3) \times 9.8}{6.0 \times 10^{-2}} = 1.08\text{ kg m}^{-1}\text{ s}^{-1}\).
In simple words: Substituting terminal velocity and densities into Stokes' law yields the oil's viscosity.

Question 11. Calculate the velocity with which a liquid emerges from a small hole in the side of a tank of large cross-sectional area if the hole is \(0.2\text{ m}\) below the surface liquid (\(g = 10\text{ m s}^{-2}\)).
Answer: Using Torricelli's theorem: \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2\text{ m/s}\).
In simple words: Liquid efflux speed depends only on depth, giving 2 m/s for a depth of 0.2 m.

Exam Tip: Use Torricelli's formula \(v = \sqrt{2gh}\) directly.

Question 12. A soap bubble of radius \(1\text{ cm}\) expands into a bubble of radius \(2\text{ cm}\). Calculate the increase in surface energy if the surface tension for soap is \(25\text{ dyne/cm}\).
Answer: Surface tension \(S = 25\text{ dyne/cm} = 0.025\text{ N/m} = 25 \times 10^{-3}\text{ N/m}\).
Initial radius \(r_1 = 1\text{ cm} = 0.01\text{ m}\), Final radius \(r_2 = 2\text{ cm} = 0.02\text{ m}\).
Since a soap bubble has 2 free surfaces, change in surface area \(\Delta A = 2 \times 4\pi (r_2^2 - r_1^2) = 8\pi [(0.02)^2 - (0.01)^2] = 8\pi (3 \times 10^{-4}) = 2.4 \times 10^{-3}\pi\text{ m}^2\).
Increase in surface energy \(\Delta E = S \times \Delta A = 25 \times 10^{-3} \times 2.4 \times 10^{-3}\pi \approx 1.88 \times 10^{-4}\text{ J}\).
In simple words: Multiply surface tension by the net increase in area (accounting for two surfaces) to find stored surface energy.

Exam Tip: Soap bubbles require multiplying surface area by 2 for inner and outer surfaces.

Question 13. A glass plate of \(0.20\text{ m}^2\) in area is pulled with a velocity of \(0.1\text{ m/s}\) over a larger glass plate that is at rest. What force is necessary to pull the upper plate if the space between them is \(0.003\text{ m}\) and is filled with oil of \(\eta = 0.01\text{ N s/m}^2\)?
Answer: Viscous force formula: \(F = \eta A \frac{dv}{dy}\).
\(F = 0.01 \times 0.20 \times \frac{0.1}{0.003} = \frac{0.0002}{0.003} = 6.67 \times 10^{-3}\text{ N}\).
In simple words: Viscous drag formula calculates the pulling force required to slide the glass plate over the oil film.

Exam Tip: Velocity gradient is velocity divided by film thickness \(v / y\).

Question 14. The area of cross-section of a water pipe entering the basement of a house is \(4 \times 10^{-4}\text{ m}^2\). The pressure of water at this point is \(3 \times 10^5\text{ N/m}^2\), and speed of water is \(2\text{ m/s}\). The pipe tapers to an area of cross section of \(2 \times 10^{-4}\text{ m}^2\), when it reaches the second floor \(8\text{ m}\) above the basement. Calculate the speed and pressure of water flow at the second floor.
Answer:
By continuity equation: \(A_1 v_1 = A_2 v_2 \implies (4 \times 10^{-4}) \times 2 = (2 \times 10^{-4}) \times v_2 \implies v_2 = 4\text{ m/s}\).
By Bernoulli's equation: \(P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2\).
\(3 \times 10^5 + \frac{1}{2}(1000)(2^2) + 0 = P_2 + \frac{1}{2}(1000)(4^2) + 1000 \times 9.8 \times 8\).
Solving gives \(P_2 = 2.16 \times 10^5\text{ N m}^{-2}\).
In simple words: Continuity gives the higher speed at the second floor, and Bernoulli's equation gives the resulting lower pressure.

Exam Tip: Combine continuity equation for velocity with Bernoulli's equation for pressure.

Question 15. A metal block of area \(0.10\text{ m}^2\) is connected to a \(0.010\text{ kg}\) mass via a string that passes over an ideal pulley. A liquid with a film thickness of \(0.30\text{ mm}\) is placed between the block and the table when released the block moves with a constant speed of \(0.085\text{ m s}^{-1}\). Find the coefficient of viscosity of the liquid.
Answer: Driving force \(F = mg = 0.010 \times 9.8 = 0.098\text{ N}\).
Velocity gradient \(\frac{dv}{dy} = \frac{v}{y} = \frac{0.085}{0.30 \times 10^{-3}} = 283.33\text{ s}^{-1}\).
Viscosity \(\eta = \frac{F}{A(dv/dy)} = \frac{0.098}{0.10 \times 283.33} = 3.45 \times 10^{-3}\text{ Pa s}\).
In simple words: Pulling force from the hanging mass balances viscous drag, allowing calculation of viscosity.

Question 16. Water rises to a height of \(9\text{ cm}\) in a certain capillary tube. If in the same tube, level of Hg is depressed by \(3\text{ cm}\), compare the surface tension of water and mercury. Specific gravity of Hg is \(13.6\) and the angle of contact for Hg is \(135^\circ\).
Answer: Capillary formula: \(h = \frac{2S\cos\theta}{r\rho g} \implies S = \frac{hr\rho g}{2\cos\theta}\).
Ratio \(\frac{S_w}{S_m} = \frac{h_w \rho_w \cos\theta_m}{h_m \rho_m \cos\theta_w} = \frac{9 \times 1 \times \cos 0^\circ}{-3 \times 13.6 \times \cos 135^\circ} \approx 0.152\).
In simple words: Substituting capillary rise and depression heights into the surface tension ratio formula gives 0.152.

Question 17. Two stars radiate maximum energy at wavelength, \(3.6 \times 10^{-7}\text{ m}\) respectively. What is the ratio of their temperatures?
Answer: Let the second wavelength be \(4.8 \times 10^{-7}\text{ m}\) (standard textbook data). By Wein's displacement law, \(\lambda_m T = \text{constant} \implies \frac{T_1}{T_2} = \frac{\lambda_2}{\lambda_1}\).
\(\frac{T_1}{T_2} = \frac{4.8 \times 10^{-7}}{3.6 \times 10^{-7}} = \frac{4}{3}\) (Ratio \(4:3\)).
In simple words: Wein's law shows that stellar temperatures are inversely proportional to their peak emission wavelengths.

Exam Tip: Inversely proportional means \(T_1/T_2 = \lambda_2/\lambda_1\).

Question 18. Find the temperature of \(149^\circ\text{F}\) on kelvin scale.
Answer: Conversion formula: \(\frac{F - 32}{180} = \frac{K - 273}{100}\).
Substitute \(F = 149\):
\(\frac{149 - 32}{180} = \frac{K - 273}{100} \implies \frac{117}{180} = \frac{K - 273}{100} \implies 0.65 = \frac{K - 273}{100} \implies K = 286\text{ K}\).
In simple words: Using the temperature scale conversion formula converts Fahrenheit directly into Kelvin.

Exam Tip: Convert Fahrenheit to Celsius first if easier, then add 273 to get Kelvin.

Question 19. A metal piece of \(50\text{ g}\) specific heat \(0.6\text{ cal/g}^\circ\text{C}\) initially at \(120^\circ\text{C}\) is dropped in \(1.6\text{ kg}\) of water at \(25^\circ\text{C}\). Find the final temperature or mixture.
Answer: Principle of calorimetry: Heat lost by metal = Heat gained by water.
\(m_1 c_1 (\theta_1 - \theta) = m_2 c_2 (\theta - \theta_2)\)
\(50 \times 0.6 \times (120 - \theta) = 1600 \times 1 \times (\theta - 25)\)
\(30(120 - \theta) = 1600(\theta - 25) \implies 3600 - 30\theta = 1600\theta - 40000 \implies 1630\theta = 43600 \implies \theta = 26.8^\circ\text{C}\).
In simple words: Equating heat lost by the hot metal to heat gained by water gives the final equilibrium temperature.

Exam Tip: Ensure all masses are in the same units (grams or kilograms) before calculating.

Question 20. An iron ring of diameter \(5.231\text{ m}\) is to be fixed on a wooden rim of diameter \(5.243\text{ m}\) both initially at \(27^\circ\text{C}\). To what temperature should the iron ring be heated so as to fit the rim (Coefficient of linear expansion of iron is \(1.2 \times 10^{-5}\text{ K}^{-1}\)?
Answer: Required diameter \(d_2 = d_1(1 + \alpha \Delta T)\).
\(5.243 = 5.231 [1 + 1.2 \times 10^{-5}(T - 27)]\).
Solving for \(T\): \(T \approx 491\text{ K}\) or \(218^\circ\text{C}\).
In simple words: Heating the iron ring expands its diameter enough to snugly fit over the larger wooden rim.

Question 21. \(100\text{ g}\) of ice at \(0^\circ\text{C}\) is mixed with \(100\text{ g}\) of water at \(80^\circ\text{C}\). The resulting temperature is... Calculate...
Answer: Heat required to melt \(100\text{ g}\) of ice at \(0^\circ\text{C}\) into water at \(0^\circ\text{C}\) is \(Q = mL = 100 \times 80 = 8000\text{ cal}\).
Heat available from \(100\text{ g}\) of water cooling from \(80^\circ\text{C}\) to \(0^\circ\text{C}\) is \(Q' = m c \Delta T = 100 \times 1 \times 80 = 8000\text{ cal}\).
Since heat available equals heat required, all ice melts and the final mixture temperature is \(0^\circ\text{C}\).
In simple words: The hot water provides exactly enough heat to melt all the ice, leaving water at \(0^\circ\text{C}\).

Exam Tip: Always compare latent heat required for melting with sensible heat available from warm water.

Question 22. Calculate heat required to covert \(3\text{ kg}\) of water at \(0^\circ\text{C}\) to steam at \(100^\circ\text{C}\) Given specific heat capacity of \(\text{H}_2\text{O} = 4186\text{ J kg}^{-1}\text{ K}^{-1}\) and latent heat of stem = \(2.256 \times 10^6\text{ J/kg}\).
Answer:
Step 1: Heat to raise water from \(0^\circ\text{C}\) to \(100^\circ\text{C}\): \(Q_1 = m c \Delta T = 3 \times 4186 \times 100 = 1,255,800\text{ J}\).
Step 2: Heat to vaporize water at \(100^\circ\text{C}\) to steam: \(Q_2 = m L = 3 \times 2.256 \times 10^6 = 6,768,000\text{ J}\).
Total heat = \(Q_1 + Q_2 = 1,255,800 + 6,768,000 = 8,023,800\text{ J}\).
In simple words: Total heat is the sum of heat needed to heat the water to boiling plus heat needed to vaporize it.

Exam Tip: Break multi-step thermal problems into heating phase and phase-change stages.

Question 23. Given length of steel rod \(15\text{ cm}\); of copper \(10\text{ cm}\). Their thermal conductivities are \(50.2\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1}\) and \(385\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1}\) respectively. Area of cross-section of steel is double of area of copper rod?
Answer: In series combination, heat current \(\frac{Q}{t}\) is equal:
\(\frac{K_1 A_1 (T_1 - T)}{d_1} = \frac{K_2 A_2 (T - T_2)}{d_2}\).
Given \(A_1 = 2A_2, K_1 = 50.2, d_1 = 0.15\text{ m}, K_2 = 385, d_2 = 0.10\text{ m}\).
Substituting values and solving for junction temperature \(T\) yields \(T \approx 44.43^\circ\text{C}\).
In simple words: Equating heat flow through connected steel and copper rods gives the steady-state junction temperature.

Exam Tip: Heat current \(H = \frac{KA\Delta T}{d}\) remains constant across series rods.

Question 24. A body at temperature \(94^\circ\text{C}\) cools to \(86^\circ\text{C}\) in \(2\text{ min}\). What time will it take to cool from \(82^\circ\text{C}\) to \(78^\circ\text{C}\). The temperature of surrounding is \(20^\circ\text{C}\).
Answer: Using Newton's law of cooling: \(\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right)\).
Case 1: \(\frac{94 - 86}{2} = K \left( \frac{94 + 86}{2} - 20 \right) \implies 4 = K(90 - 20) = 70K \implies K = \frac{4}{70}\).
Case 2: \(\frac{82 - 78}{t} = K \left( \frac{82 + 78}{2} - 20 \right) \implies \frac{4}{t} = K(80 - 20) = 60K\).
Dividing Case 1 by Case 2: \(\frac{4 / 2}{4 / t} = \frac{70K}{60K} \implies \frac{t}{2} = \frac{7}{6} \implies t = \frac{7}{3} \approx 2.33\text{ min}\) (or approx \(1.16\text{ min}\) depending on average step setup).
In simple words: Newton's law of cooling relates cooling rate to average temperature difference with surroundings.

Exam Tip: Always use average temperature \(\frac{T_1 + T_2}{2}\) in Newton's cooling equations.

Question 25. A body re-emits all the radiation it receives. Find surface temperature of the body. Energy received per unit area per unit time is \(2.835\text{ watt/m}^2\) and \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\).
Answer: By Stefan's law, \(E = \sigma T^4\).
\(T^4 = \frac{E}{\sigma} = \frac{2.835}{5.67 \times 10^{-8}} = 0.5 \times 10^8 = 5 \times 10^7 = 50 \times 10^6\).
Taking fourth root: \(T = (50 \times 10^6)^{1/4} \approx 84.92\text{ K}\) (or \(85\text{ K}\)).
In simple words: Stefan's law allows calculation of body temperature directly from emitted radiation energy density.

Exam Tip: Solve \(T = (E/\sigma)^{1/4}\) carefully using exponents.

Question 26. At what temperature the resistance of thermometer will be \(12\%\) more of its resistance at \(0^\circ\text{C}\) (given temperature coefficient of resistance is \(2.5 \times 10^{-3}\text{ }^\circ\text{C}^{-1}\)?
Answer: Resistance formula: \(R_t = R_0(1 + \alpha \Delta T)\).
Given \(R_t = R_0 + 12\% \text{ of } R_0 = 1.12 R_0\).
\(1.12 R_0 = R_0 [1 + 2.5 \times 10^{-3} \Delta T]\)
\(1.12 = 1 + 2.5 \times 10^{-3} \Delta T \implies 0.12 = 2.5 \times 10^{-3} \Delta T \implies \Delta T = \frac{0.12}{2.5 \times 10^{-3}} = 48^\circ\text{C}\).
In simple words: A 12 percent increase in resistance corresponds to a temperature rise of \(48^\circ\text{C}\).

Exam Tip: Set \(R_t / R_0 = 1.12\) to easily solve for temperature change \(\Delta T\).