JEE Physics Thermal Properties Of Matter MCQs Set 11

Answer: (a) 590 kcal

Explanation:
1. Cooling the steam from 200 °C to 100 °C: \( 1000\times0.5\times100=50{,}000\ \text{cal} \).
2. Condensing at 100 °C: \( 1000\times540=540{,}000\ \text{cal} \).
3. Total \( H=590{,}000\ \text{cal}=590\ \text{kcal} \).

Teacher's Note:
1. Add the sensible heat of the steam and the latent heat of condensation.
2. The condensation term dominates, so the answer is slightly above 540 kcal.

Answer: (b) 136 km

Explanation:
1. Heat absorbed: \( \tfrac14 mgh = mL \).
2. \( h=\frac{4L}{g}=\frac{4\times3.4\times10^{5}}{10}=1.36\times10^{5}\ \text{m} \) = 136 km.

Teacher's Note:
1. Mass cancels from both sides, so the height depends only on L and g.
2. The large value shows how huge the latent heat is compared with ordinary mechanical energies.

Answer: (d) 1/40 °C

Explanation:
1. The block descends slowly, so all its loss of potential energy goes into stirring the liquid: \( W=mgh=10\times10\times1=100\ \text{J} \).
2. The container is adiabatic, so no heat enters; the work raises the internal energy: \( \Delta T=\frac{100}{4000}=\frac1{40}\ ^\circ\text{C} \).

Teacher's Note:
1. Work done on a thermally isolated liquid raises its temperature just like heat would (Joule's experiment).
2. Slow descent means the block gains no kinetic energy.

Answer: (a) 0 °C

Explanation:
1. Heat the water can give up while cooling to 0 °C: \( 300\times1\times25=7500\ \text{cal} \).
2. Heat needed to melt all the ice: \( 100\times80=8000\ \text{cal} \).
3. The water cannot melt all the ice, so ice and water remain together at 0 °C.

Teacher's Note:
1. Always compare the heat available with the heat needed to melt all the ice before assuming a final temperature above 0 °C.
2. If ice survives, the mixture must be at 0 °C.

Answer: (c) cooled

Explanation:
1. On heating, the hole expands exactly like a steel disc of the same size would, while the brass disc expands more, so the fit becomes tighter.
2. On cooling, brass shrinks more than the hole, so the disc becomes loose.

Teacher's Note:
1. A hole expands as if it were filled with the same material as the plate.
2. The material with the larger coefficient changes size more in either direction.

Answer: (d) \( A(1+2\alpha\Delta\theta) \)

Explanation:
1. Every linear dimension, including both radii, increases by the factor \( (1+\alpha\Delta\theta) \).
2. \( A'=\pi(R_2'^2-R_1'^2)=\pi(R_2^2-R_1^2)(1+\alpha\Delta\theta)^2\approx A(1+2\alpha\Delta\theta) \).

Teacher's Note:
1. The hole grows with the plate, so the annulus area scales as (length)².
2. Binomial approximation: (1 + x)² ≈ 1 + 2x for small x.

Answer: (a) the two changes are equal

Explanation:
1. A cavity expands exactly as if it were filled with the same material, so the outer volume of the hollow sphere changes as that of a solid sphere of the same radius.
2. \( \Delta V=\gamma V\Delta T \) is the same for both.

Teacher's Note:
1. Think of the hollow sphere as a solid one with a cavity that also expands.
2. Only the outer radius, material and temperature rise matter.

Answer: (c) 1.1 × 10⁻² % (reads short)

Explanation:
1. At 30 °C the tape's divisions are longer than marked, so it reads a smaller number than the true length.
2. \( \frac{\Delta\ell}{\ell}=\alpha\Delta T=11\times10^{-6}\times10=1.1\times10^{-4} \), which is \( 1.1\times10^{-2}\% \).

Teacher's Note:
1. A hot scale under-reads; a cold scale over-reads.
2. Percentage = fractional change × 100.

Answer: (d) 50 °X, 50 °Y, 50 °W

Explanation:
1. Each scale has 90 divisions between freezing and boiling, so every degree has the same size.
2. Distance above freezing: \( X:50-(-20)=70,\ W:50-30=20,\ Y:50-0=50 \).
3. Highest first: X (70), Y (50), W (20).

Teacher's Note:
1. Convert each reading to a common measure: divisions above the freezing point.
2. All three scales have the same degree size here, so the comparison reduces to offsets.

Answer: (b) \( \frac{mc\theta}{L} \)

Explanation:
1. The ice is large, so the mixture stays at 0 °C and the water just cools to 0 °C.
2. Heat released \( mc\theta \) melts mass x: \( xL=mc\theta \).
3. \( x=\frac{mc\theta}{L} \).

Teacher's Note:
1. Because M is large, ice survives and the final temperature is 0 °C.
2. Dimension check: mcθ/L has the dimension of mass.

Answer: (d) \( \frac{85}{32}\ \text{g} \)

Explanation:
1. Heat needed for the ice to become water at 0 °C: \( 20(0.5)(10)+20(80)=100+1600=1700\ \text{cal} \).
2. The minimum m condenses to water and cools to 0 °C: \( m(540+100)=1700 \).
3. \( m=\frac{1700}{640}=\frac{85}{32}\ \text{g} \).

Teacher's Note:
1. For the minimum steam, the final state is water at exactly 0 °C.
2. Steam gives up latent heat plus the heat released while its condensate cools from 100 °C to 0 °C.

Answer: (a) 6 kg

Explanation:
1. Heat to warm the ice to 0 °C: \( 2000(0.5)(20)=20{,}000\ \text{cal} \). Heat released by the water cooling to 0 °C: \( 5000(1)(20)=100{,}000\ \text{cal} \).
2. Remaining 80,000 cal melts \( 80{,}000/80=1000\ \text{g} \) of ice, which is less than 2 kg, so the final temperature is 0 °C.
3. Water = 5 + 1 = 6 kg.

Teacher's Note:
1. Check each stage in turn: warm ice to 0 °C, then see how much latent heat is available.
2. If less than the full ice mass melts, the mixture ends at 0 °C.

Answer: (c) decreases the density of the material

Explanation:
1. Matter expands on heating, so the volume increases while the mass stays the same.
2. \( \rho=m/V \) therefore decreases.
3. Liquids and gases also expand, and different substances expand at different rates.

Teacher's Note:
1. Density is mass divided by volume, and mass does not change on heating.
2. The only common exception is the anomalous behaviour of water between 0 °C and 4 °C.

Answer: (d) bend towards the metal with the lower coefficient of expansion

Explanation:
1. The metal with the larger α lengthens more.
2. The strip is bound together, so it bends with the more expanding metal on the outside (convex side).
3. So it bends towards the metal with the lower coefficient.

Teacher's Note:
1. The longer side must lie on the outer arc of the bend.
2. Thermostats use this principle to make or break a contact.

Answer: (a) \( \gamma>3\alpha \)

Explanation:
1. The container's volume grows with coefficient \( 3\alpha \) (its cavity expands like the solid).
2. Overflow occurs when the liquid's volume grows faster than the container's, so \( \gamma>3\alpha \).

Teacher's Note:
1. Compare the volume expansion of the liquid with that of the container's inner volume.
2. The difference γ − 3α is the apparent expansion of the liquid.

Answer: (c) 45 °F

Explanation:
1. Differences scale by the size ratio of the degrees: \( \Delta F=\tfrac95\Delta C \).
2. \( \tfrac95\times25=45\ ^\circ\text{F} \). (The offset of 32 cancels in a difference.)

Teacher's Note:
1. Do not add 32 to a temperature difference.
2. One Celsius degree equals 1.8 Fahrenheit degrees.

Answer: (d) I-p, II-r, III-s, IV-q

Explanation:
1. Floating: displaced liquid weighs the same as the cylinder, so the immersed volume is \( V_{in}=M/d_T \), and it stays constant only if the liquid density does not change: \( \gamma=0 \) (I-p).
2. Outside volume constant needs the total volume growth \( 3\alpha/\rho \) to match the immersed growth \( \gamma/d \): \( \gamma=3\alpha d/\rho \) (II-r).
3. Immersed height = V_in / area changes by \( (\gamma-2\alpha) \); the outer height stays constant when \( \alpha=\tfrac{\rho}{d}(\gamma-2\alpha) \), i.e. \( \gamma=2\alpha+\alpha d/\rho \) (III-s).
4. Immersed height constant needs \( \gamma=2\alpha \) (IV-q).

Teacher's Note:
1. Write each quantity as (initial value) × (1 + coefficient × ΔT) and set its net change to zero.
2. Areas expand with 2α, volumes with 3α, and the density of a liquid falls with γ.

Answer: (b) 80 °C

Explanation:
1. Let the ball's volume be V cm³. Its mass is 4.5V g. The ice melted occupies the sunk volume V/2, so its mass is \( 0.9\times V/2=0.45V \) g.
2. Heat balance: \( 4.5V(0.1)\theta=0.45V\times80 \).
3. \( 0.45\theta=36\Rightarrow\theta=80\ ^\circ\text{C} \).

Teacher's Note:
1. The cavity melted in the ice has the same volume as the part of the ball that sank.
2. V cancels, so only the ratios of densities matter.

Answer: (d) 0.130 kg

Explanation:
1. Heat gained: \( (1100+20)(1)(80-15)=72{,}800\ \text{cal} \).
2. Heat lost per gram of steam: \( 540+(100-80)=560\ \text{cal} \).
3. \( m=\frac{72{,}800}{560}=130\ \text{g}=0.130\ \text{kg} \).

Teacher's Note:
1. Condensed steam gives up latent heat and then cools as water to 80 °C.
2. Add the water equivalent of the calorimeter to the water mass.

Answer: (a) \( 2\alpha I\Delta T \)

Explanation:
1. Moment of inertia scales as mass × (length)², and mass is constant.
2. So \( I'=I(1+\alpha\Delta T)^2\approx I(1+2\alpha\Delta T) \) and \( \Delta I=2\alpha I\Delta T \).

Teacher's Note:
1. I ∝ L², so the fractional change is twice that of a length.
2. Neglect the term (αΔT)² for small ΔT.

Answer: (b) \( \frac{\ell_1\alpha_1+\ell_2\alpha_2}{\ell_1+\ell_2} \)

Explanation:
1. Total extension: \( \Delta\ell=(\ell_1\alpha_1+\ell_2\alpha_2)\Delta T \).
2. Define \( \alpha_{eq}(\ell_1+\ell_2)\Delta T=\Delta\ell \).
3. \( \alpha_{eq}=\frac{\ell_1\alpha_1+\ell_2\alpha_2}{\ell_1+\ell_2} \).

Teacher's Note:
1. The effective coefficient is a length-weighted average of the two coefficients.
2. Check the case ℓ₂ = 0: it must reduce to α₁.

Answer: (c) 1/T

Explanation:
1. At constant pressure \( V\propto T \).
2. \( \gamma=\frac1V\frac{dV}{dT}=\frac1T \).

Teacher's Note:
1. Charles's law gives a linear V–T relation through the origin.
2. The coefficient falls as the gas gets hotter.

Answer: (a) \( \frac{1+\gamma_2\Delta T}{1+\gamma_1\Delta T} \)

Explanation:
1. For floating, the submerged fraction is \( \rho_{metal}/\rho_{Hg} \).
2. Densities change as \( \rho'=\rho/(1+\gamma\Delta T) \).
3. \( \frac{f'}{f}=\frac{1/(1+\gamma_1\Delta T)}{1/(1+\gamma_2\Delta T)}=\frac{1+\gamma_2\Delta T}{1+\gamma_1\Delta T} \).

Teacher's Note:
1. Fraction submerged = density of the floating body / density of the liquid.
2. Each density decreases by the factor 1/(1 + γΔT).

Answer: (b) \( \frac{\Delta h}{h_0t} \)

Explanation:
1. The pressures at the bottom of the two columns are equal: \( h_0\rho_0=(h_0+\Delta h)\rho_t \).
2. With \( \rho_t=\rho_0/(1+\gamma t) \): \( 1+\gamma t=\frac{h_0+\Delta h}{h_0} \).
3. \( \gamma=\frac{\Delta h}{h_0t} \).

Teacher's Note:
1. Equal pressure at the bottom means heights are inversely proportional to densities.
2. This is the Dulong–Petit method of measuring the absolute expansion of a liquid.

Answer: (c) 20 days

Explanation:
1. For ice of thickness x, the heat conducted through it freezes more ice: \( \rho L\,\frac{dx}{dt}=\frac{K\Delta T}{x} \).
2. Integrating from 0 to x: \( t=\frac{\rho Lx^2}{2K\Delta T} \).
3. \( t=\frac{900\times3.4\times10^5\times0.25}{2\times2.2\times10}\approx1.74\times10^{6}\ \text{s}\approx20\ \text{days} \).

Teacher's Note:
1. The growing ice layer acts as an ever thicker insulator, so thickness grows as √t.
2. Convert seconds to days by dividing by 86,400.

Answer: (a) 200 cm

Explanation:
1. The two central planes are at radii R and R + d, so their lengths differ by the angle times d: \( \ell_{Al}-\ell_{Fe}=\theta d \).
2. The difference in extension is \( \ell(\alpha_{Al}-\alpha_{Fe})\Delta T \), and \( \ell=R\theta \).
3. \( R=\frac{d}{(\alpha_{Al}-\alpha_{Fe})\Delta T}=\frac{2\times10^{-3}}{20\times10^{-6}\times50}=2\ \text{m} \).

Teacher's Note:
1. Both strips subtend the same angle at the centre of curvature.
2. R is inversely proportional to the difference of the coefficients.

Answer: (c) 6 min

Explanation:
1. Heat lost by the water: \( 20\times4200\times5=4.2\times10^{5}\ \text{J} \).
2. Mass that must evaporate: \( \frac{4.2\times10^5}{2.27\times10^6}=0.185\ \text{kg}=185\ \text{g} \).
3. Time \( =\frac{185}{0.5}=370\ \text{s}\approx6\ \text{min} \).

Teacher's Note:
1. The latent heat of the evaporating water is drawn from the remaining water.
2. Convert seconds into minutes at the end.

Answer: (d) 3 min

Explanation:
1. Heat required: \( 150\times4200\times17=1.071\times10^{7}\ \text{J} \).
2. Time \( =\frac{1.071\times10^7}{59{,}000}=181.5\ \text{s}\approx3\ \text{min} \).

Teacher's Note:
1. Power = energy / time.
2. Convert the answer from seconds to minutes.

Answer: (b) 8 °C

Explanation:
1. Melting all the ice needs \( 50\times80=4000\ \text{cal} \).
2. The remaining \( 6000-4000=2000\ \text{cal} \) heats 250 g of water: \( \Delta T=\frac{2000}{250}=8\ ^\circ\text{C} \).

Teacher's Note:
1. The temperature stays at 0 °C until all the ice has melted.
2. Only the leftover heat raises the temperature of the total water.