JEE Physics Thermal Properties Of Matter MCQs Set 12

Answer: (c) 33

Explanation:
1. Power focused: \( 1400\times0.2=280\ \text{W} \).
2. Heat needed: \( 0.28\times3.3\times10^{5}=92{,}400\ \text{J} \).
3. \( t=\frac{92{,}400}{280}=330\ \text{s}=33\times10\ \text{s} \).

Teacher's Note:
1. Energy needed divided by power received gives the time.
2. Express the final time as a multiple of 10 s.

Answer: (d) 53 g

Explanation:
1. Without air resistance the bullet returns with the same speed. Its kinetic energy: \( \tfrac12(0.05)(840)^2=17{,}640\ \text{J}=4200\ \text{cal} \).
2. Heat released by the bullet cooling to 0 °C: \( 50\times0.02\times30=30\ \text{cal} \).
3. \( m=\frac{4230}{80}\approx53\ \text{g} \).

Teacher's Note:
1. The bullet loses kinetic energy and also heat as it cools to 0 °C; both melt ice.
2. Convert joules to calories with J = 4.2.

Answer: (b) 42 °C

Explanation:
1. Energy supplied: \( 90\times600=54{,}000\ \text{J}=12{,}857\ \text{cal} \).
2. Effective mass of water: \( 360+40=400\ \text{g} \).
3. \( \Delta T=\frac{12{,}857}{400}=32.1\ ^\circ\text{C}\Rightarrow T=42\ ^\circ\text{C} \).

Teacher's Note:
1. Include the water equivalent of the calorimeter.
2. Convert joules to calories before dividing.

Answer: (c) 8 cm

Explanation:
1. Each half is a straight rod of length \( \ell=2(1+25\times10^{-6}\times32)=2.0016\ \text{m} \) resting on the supports 2 m apart horizontally.
2. Pythagoras: \( x=\sqrt{2.0016^2-2^2}\approx0.080\ \text{m} \) = 8 cm.

Teacher's Note:
1. Each half forms the slanted side of a right triangle with base ℓ₀/2.
2. A tiny fractional expansion gives a large central rise because of the square root.

Answer: (a) 66

Explanation:
1. Length increases to \( L'=L(1+\alpha\Delta T)=1.1L \).
2. Correct period needs \( g_{eff}=1.1g=11\ \text{m/s}^2 \), so the lift accelerates upward with a = 1 m/s².
3. Apparent weight \( =60\times\frac{11}{10}=66\ \text{kgf} \).

Teacher's Note:
1. T = 2π√(L/g_eff): a longer string needs a larger effective g to keep the same period.
2. Upward acceleration increases the apparent weight.

Answer: (c) 15

Explanation:
1. Air volume constant means the mercury expands exactly as much as the flask: \( V_{Hg}\gamma_{Hg}=V_{flask}\gamma_{glass} \).
2. \( V_{Hg}=\frac{1\times3\times9\times10^{-6}}{1.8\times10^{-4}}=0.15\ \text{L} \), so X = 15.

Teacher's Note:
1. The empty space stays constant when the container and the liquid expand by equal volumes.
2. γ of glass = 3α.

Answer: (d) (2), (3) and (4)

Explanation:
1. Energy conservation never allows the final temperature to be outside the two initial temperatures, so (1) is impossible.
2. It can equal one of them when the other has negligible heat capacity or a phase change absorbs the heat (2), lie between them (3), and equal the average when the heat capacities are equal (4).

Teacher's Note:
1. Test each statement with extreme cases: negligible heat capacity, equal heat capacities, a phase change.
2. Heat flows only from hot to cold, so the final temperature is bracketed.

Answer: (a) (1), (2) and (3) only

Explanation:
1. The water can give at most \( m\times10=10m\ \text{cal} \) while melting all the ice needs \( 80m \) cal.
2. So the ice does not melt completely and the temperature is 0 °C: only (4) is true.
3. Statements (1), (2) and (3) are false.

Teacher's Note:
1. Compare the heat available with the heat needed to melt all the ice first.
2. Only a small fraction (1/8) of the ice melts.

Answer: (c) (2) and (4)

Explanation:
1. The differential extension is \( |\alpha_B-\alpha_C|\,\Delta T\,\ell \), and this is spread over the thickness d: \( R=\frac{d}{|\alpha_B-\alpha_C|\,\Delta T} \).
2. So R varies inversely with both ΔT and |α_B − α_C|.

Teacher's Note:
1. A larger temperature rise or a larger mismatch of coefficients produces a tighter bend.
2. Tighter bend means smaller R.

Answer: (d) (1), (2), (3) and (4)

Explanation:
1. Every linear dimension of a body, including the size of a hole and the distance between two points on the plate, increases by the factor (1 + αΔT).
2. So r, a, b and R all increase, hence the area πr², the area ab, and R and b increase.

Teacher's Note:
1. Think of the plate as a uniform photographic enlargement.
2. Holes behave like the surrounding material would.

Answer: (b) 42.0 W

Explanation:
1. In 40 min the water gives \( 0.60\times4200\times40=100{,}800\ \text{J} \).
2. \( P=\frac{100{,}800}{2400}=42\ \text{W} \).

Teacher's Note:
1. Use the part of the graph where only the water's temperature changes.
2. Rate = energy / time, with time in seconds.

Answer: (d) 0.15 kg

Explanation:
1. From 40 min to 60 min the water stays at 0 °C while freezing, losing heat at 42 W: \( 42\times1200=50{,}400\ \text{J} \).
2. \( m=\frac{50{,}400}{3.3\times10^{5}}=0.15\ \text{kg} \).

Teacher's Note:
1. The flat part of the water curve is the phase change.
2. Mass frozen = heat lost / latent heat.

Answer: (a) 100 °C

Explanation:
1. Heat needed to melt the ice and bring all the water to 100 °C: \( 16{,}000+200(100)+200(45)=45{,}000\ \text{cal} \).
2. Condensing all the steam would release \( 100\times540=54{,}000\ \text{cal} \), which is more than 45,000 cal.
3. So the steam is only partly condensed and the final temperature is 100 °C.

Teacher's Note:
1. Test whether complete condensation of the steam would overheat the mixture.
2. Mixtures containing steam and water at 1 atm settle at 100 °C.

Answer: (b) 483.3 g

Explanation:
1. The steam that condenses supplies 45,000 cal: \( x=\frac{45{,}000}{540}=83.3\ \text{g} \).
2. Total water \( =200+200+83.3=483.3\ \text{g} \).

Teacher's Note:
1. Only the condensed steam adds to the water mass.
2. All the ice has melted.

Answer: (d) \( \gamma_L=2\alpha_s \)

Explanation:
1. Floating: weight of the cube equals the weight of liquid displaced, so \( h_{in}=\frac{\rho_s}{\rho_L}H \), where H is the side of the cube.
2. On heating, \( \rho_s'=\frac{\rho_s}{1+3\alpha_s\Delta T} \), \( \rho_L'=\frac{\rho_L}{1+\gamma_L\Delta T} \) and \( H'=H(1+\alpha_s\Delta T) \).
3. Constant depth requires \( \frac{(1+\gamma_L\Delta T)(1+\alpha_s\Delta T)}{1+3\alpha_s\Delta T}=1 \), so \( \gamma_L+\alpha_s=3\alpha_s \) i.e. \( \gamma_L=2\alpha_s \).

Teacher's Note:
1. The depth is constant when the change in the relative density of the cube exactly offsets the change in its height.
2. Solid density changes as 3α, height as α, liquid density as γ.

Answer: (b) 273 K

Explanation:
1. Heat needed to bring the ice to 273 K and melt it all: \( 0.45(2100)(20)+0.45(336{,}000)=18{,}900+151{,}200=170{,}100\ \text{J} \).
2. Maximum heat the steam can give while condensing and cooling to 273 K: \( 0.05(2{,}268{,}000)+0.05(4200)(100)=113{,}400+21{,}000=134{,}400\ \text{J} \).
3. This is less than 170,100 J, so the ice is not fully melted and the mixture stays at 273 K.

Teacher's Note:
1. Compare the maximum heat available with the heat needed to melt all the ice.
2. If ice survives, the final temperature is the melting point.

Answer: (d) 3 kg

Explanation:
1. Contraction on cooling is cancelled by the stretch under the load: \( \frac{mg}{AY}=\alpha\Delta T \).
2. \( m=\frac{YA\alpha\Delta T}{g}=\frac{10^{11}\times\pi\times10^{-6}\times10^{-5}\times10}{10}\approx3.14\ \text{kg} \).

Teacher's Note:
1. Thermal strain αΔT equals mechanical strain mg/(AY).
2. Area = πr² with r = 1 mm.

Answer: (b) 0.78 mm

Explanation:
1. In steady state the same heat current flows through both wires: \( 2k(T-10)=k(400-T) \) (equal lengths and areas).
2. \( T=140\ ^\circ\text{C} \) at the junction.
3. Mean temperature of PQ is 75 °C, a rise of 65 °C above 10 °C: \( \Delta\ell=1\times1.2\times10^{-5}\times65=0.78\ \text{mm} \).

Teacher's Note:
1. Steady conduction in series: temperature drops are inversely proportional to the conductivities.
2. The temperature varies linearly along PQ, so use the average rise.

Answer: (d) between 0.148 kJ and 0.028 kJ

Explanation:
1. Heat to be removed: \( Q=m\int_4^{20}32\left(\frac{T}{400}\right)^3dT=0.1\times\frac{32}{400^3}\cdot\frac{20^4-4^4}{4}\approx0.002\ \text{kJ} \).
2. A refrigerator removing dQ at temperature T needs work \( dW=dQ\left(\frac{300}{T}-1\right) \) at best.
3. Integrating gives \( W\approx0.0377\ \text{kJ} \), which lies between 0.028 kJ and 0.148 kJ (real refrigerators need at least this much).

Teacher's Note:
1. Work is always greater than the heat removed because the refrigerator must lift heat to 300 K.
2. Use the Carnot relation at each temperature, then integrate.

Answer: (b) 28.9 cc

Explanation:
1. \( V=\tfrac43\pi(10)^3=4188.8\ \text{cm}^3 \).
2. \( \Delta V=3\alpha V\Delta T=3\times23\times10^{-6}\times4188.8\times100\approx28.9\ \text{cm}^3 \).

Teacher's Note:
1. γ = 3α for an isotropic solid.
2. Radius is 10 cm, not 20 cm.

Answer: (d) 25 °C; 1.85 × 10⁻⁵ /°C

Explanation:
1. Fractional change in period: \( \frac{\Delta T_p}{T_p}=\tfrac12\alpha(\theta-\theta_0) \), where θ₀ is the temperature of correct time.
2. At 40 °C: \( \tfrac12\alpha(40-\theta_0)=\frac{12}{86400} \); at 20 °C: \( \tfrac12\alpha(\theta_0-20)=\frac{4}{86400} \).
3. Adding: \( 10\alpha=\frac{16}{86400}\Rightarrow\alpha=1.85\times10^{-5} \). Dividing: \( \frac{40-\theta_0}{\theta_0-20}=3\Rightarrow\theta_0=25\ ^\circ\text{C} \).

Teacher's Note:
1. A clock loses time when its period is too long (hot) and gains time when the period is too short (cold).
2. Two equations give two unknowns, α and θ₀.

Answer: (a) 885 °C

Explanation:
1. Heat lost by the ball: \( 100(0.1)(T-75) \).
2. Heat gained by the calorimeter and water: \( (100\times0.1+170)\times45=8100\ \text{cal} \).
3. \( T-75=810\Rightarrow T=885\ ^\circ\text{C} \).

Teacher's Note:
1. Treat the calorimeter as extra water equivalent of 10 g.
2. Equate heat lost to heat gained.

Answer: (c) 1764 J

Explanation:
1. The vessel is insulated, so no heat is transferred, but shaking does work on the system.
2. Heat capacity of system: \( 0.1(420)+0.2(4200)=882\ \text{J/K} \).
3. \( \Delta U=882\times2=1764\ \text{J} \), equal to the work done (Q = 0).

Teacher's Note:
1. First law: ΔU = Q + W with Q = 0.
2. Both the water and the vessel warm up.

Answer: (d) 169 J

Explanation:
1. Work at constant pressure: \( W=P\Delta V \).
2. \( \Delta V=(1671-1)\times10^{-6}=1.67\times10^{-3}\ \text{m}^3 \).
3. \( W=1.013\times10^5\times1.67\times10^{-3}=169.2\ \text{J} \).

Teacher's Note:
1. Convert cm³ to m³ before multiplying.
2. Only about 8% of the 539 cal of latent heat goes into expansion work.

Answer: (b) 75.04 cm

Explanation:
1. At 27 °C the scale divisions have lengthened, so each marked cm is longer than a true cm.
2. True length \( =75(1+2\times10^{-5}\times27)=75.0405\ \text{cm} \).

Teacher's Note:
1. An expanded scale under-reads; the true value is larger than the reading.
2. The question assumes the mercury column length is read by the scale, as in the original problem.

Answer: (c) lose 10.4 s

Explanation:
1. \( \alpha=\gamma/3=1.2\times10^{-5}\ /^\circ\text{C} \).
2. Fractional increase in period: \( \tfrac12\alpha\Delta T=\tfrac12(1.2\times10^{-5})(20)=1.2\times10^{-4} \).
3. Time lost per day: \( 86{,}400\times1.2\times10^{-4}=10.37\ \text{s} \).

Teacher's Note:
1. A longer pendulum has a longer period, so the clock runs slow.
2. Linear coefficient is one third of the cubical coefficient.

Answer: (a) 1.105 × 10¹¹ N/m²

Explanation:
1. Copper's extension: \( 300\times1.7\times10^{-5}\times100=0.51\ \text{mm} \), so the second rod (70 cm) extends 1.40 mm: \( \alpha_2=\frac{1.40}{700\times100}=2\times10^{-5}\ /^\circ\text{C} \).
2. Neither part changes length, so each experiences the same force and ends up with the same stress: \( Y_1\alpha_1=Y_2\alpha_2 \).
3. \( Y_2=\frac{1.3\times10^{11}\times1.7}{2}=1.105\times10^{11}\ \text{N/m}^2 \).

Teacher's Note:
1. First find α₂ from the total extension.
2. Equal stress in the two parts when both are held at fixed length.

Answer: (b) 1/43200 per °C

Explanation:
1. Volume from the apparent loss of weight: at 27 °C \( V_1=\frac{46-30}{1.24}=12.903\ \text{cm}^3 \); at 42 °C \( V_2=\frac{46-30.5}{1.20}=12.917\ \text{cm}^3 \).
2. Volume expansion coefficient \( \gamma=\frac{V_2-V_1}{V_1\times15}=\frac{1}{14400}\ /^\circ\text{C} \).
3. Linear coefficient \( \alpha=\gamma/3=\frac{1}{43200}\ /^\circ\text{C} \).

Teacher's Note:
1. Buoyant force in grams-weight equals the volume in cm³ times the specific gravity.
2. Convert γ to α by dividing by three.

Answer: (d) 9.996 cm

Explanation:
1. Each division on the scale is now longer: \( 1\ \text{cm}\to1(1+2\times10^{-5}\times20)=1.0004\ \text{cm} \).
2. An object of true length 10 cm then covers fewer divisions: \( \frac{10}{1.0004}\approx9.996\ \text{divisions} \).

Teacher's Note:
1. Hot scales read low.
2. So the answer must be slightly less than 10 cm.