JEE Physics Thermal Properties Of Matter MCQs Set 03

Answer: (d) −40°

Explanation:
1. Put \( C = F \) in \( \dfrac{C}{5} = \dfrac{F-32}{9} \).
2. \( 9C = 5C - 160 \), so \( C = -40 \).

Teacher's Note:
1. The two scales coincide only at −40.
2. Use the conversion formula and set both readings equal.

Answer: (c) 45°F

Explanation:
1. One Celsius degree equals \( \dfrac{9}{5} \) Fahrenheit degrees.
2. \( \Delta F = \dfrac{9}{5} \times 25 = 45 \).

Teacher's Note:
1. For a temperature difference the 32 offset cancels.
2. Only the factor 9/5 is used.

Answer: (d) −297°F

Explanation:
1. \( F = \dfrac{9}{5}C + 32 \).
2. \( F = \dfrac{9}{5}(-183) + 32 = -329.4 + 32 = -297.4 \).

Teacher's Note:
1. Do not forget to add 32 after multiplying by 9/5.
2. The result is about −297°F.

Answer: (c) 100°

Explanation:
1. Fraction of the fundamental interval on X: \( \dfrac{60-15}{75-15} = \dfrac{45}{60} = 0.75 \).
2. On Y: \( 25 + 0.75\times(125-25) = 100 \).

Teacher's Note:
1. All linear thermometers agree on the fractional position between the fixed points.
2. Equate \((R-\text{ice})/(\text{steam}-\text{ice})\) for both scales.

Answer: (b) less than 25 cm

Explanation:
1. At 0°C the steel tape has contracted, so each marked centimetre is slightly shorter than 1 cm.
2. The wood therefore spans 25 of these shorter divisions, so its true length is slightly less than 25 cm.

Teacher's Note:
1. A contracted scale over-reads the length.
2. Real length = reading × (1 − α ΔT) for the tape.

Answer: (b) \( 2.0\times10^{-5}\,/^\circ\)C

Explanation:
1. \( \dfrac{\Delta V}{V} = 3\alpha\,\Delta T \).
2. \( \alpha = \dfrac{0.0015}{3\times24} = 2.08\times10^{-5} \), about \( 2.0\times10^{-5} \) per °C.

Teacher's Note:
1. Volume expansion coefficient is three times the linear one for an isotropic solid.
2. Convert the percentage to a fraction first.

Answer: (c) \( \dfrac{5}{3}\alpha \)

Explanation:
1. Total expansion \( = L\alpha\,\Delta T + 2L(2\alpha)\Delta T = 5L\alpha\,\Delta T \).
2. \( \alpha_{avg} = \dfrac{5L\alpha\,\Delta T}{3L\,\Delta T} = \dfrac{5}{3}\alpha \).

Teacher's Note:
1. Average coefficient is total expansion divided by total length and temperature rise.
2. Longer rod and larger alpha both contribute.

Answer: (d) A4

Explanation:
1. \( \alpha = \dfrac{\Delta\ell}{\ell_0\,\Delta T} \).
2. A1 = 0.01, A2 = 0.02, A3 = \( \dfrac{3}{75} = 0.04 \), A4 = \( \dfrac{4}{50} = 0.08 \).
3. A4 is the largest.

Teacher's Note:
1. Compare the ratio of change in length to the product of initial length and temperature change.
2. Large Δℓ alone does not decide it.

Answer: (b) \( 10^4 \) N

Explanation:
1. Thermal stress \( = Y\alpha\,\Delta T \), so force \( F = YA\alpha\,\Delta T \).
2. \( F = 10^{11}\times10^{-3}\times10^{-6}\times100 = 10^4 \) N.

Teacher's Note:
1. A rod that is prevented from expanding develops stress equal to Y times the thermal strain.
2. The length of the rod does not matter.

Answer: (d) 350 cal

Explanation:
1. Energy \( = mgh = 5\times9.8\times30 = 1470 \) J.
2. Heat in calories \( = \dfrac{1470}{4.2} = 350 \) cal.

Teacher's Note:
1. Convert joules to calories by dividing by 4.2.
2. The mass cancels if you only want heat per unit mass.

Answer: (d) 3028 J

Explanation:
1. Heating ice: \( 1\times0.5\times10 = 5 \) cal. Melting: 80 cal. Heating water: 100 cal. Vaporising: 536 cal.
2. Total \( = 721 \) cal.
3. In joules \( 721\times4.2 = 3028 \) J.

Teacher's Note:
1. Add the four stages separately: warm ice, melt, warm water, boil.
2. Check the unit asked for, J not cal.

Answer: (d) 6 kg

Explanation:
1. Heat to bring ice to 0°C: \( 2000\times0.5\times20 = 20000 \) cal.
2. Heat available when water cools to 0°C: \( 5000\times20 = 100000 \) cal.
3. Left for melting \( = 80000 \) cal, which melts \( 80000/80 = 1000 \) g.
4. Water \( = 5 + 1 = 6 \) kg.

Teacher's Note:
1. Check whether all ice melts before assuming a final temperature.
2. Here the mixture ends at 0°C with some ice left (1 kg).

Answer: (b) 1.5 kg

Explanation:
1. Heat given out \( = 2500\times0.1\times500 = 125000 \) cal.
2. Mass melted \( = \dfrac{125000}{80} \approx 1562 \) g, about 1.5 kg.

Teacher's Note:
1. The block cools only to 0°C on a large ice block.
2. Maximum melting corresponds to all the heat going into fusion.

Answer: (d) 8.7°C

Explanation:
1. Heat to warm ice to 0°C and melt it: \( 1\times2100\times10 + 3.36\times10^5 = 3.57\times10^5 \) J.
2. Heat from water cooling to 0°C: \( 4.4\times4200\times30 = 5.544\times10^5 \) J.
3. Excess \( 1.974\times10^5 \) J heats 5.4 kg: \( \Delta T = \dfrac{1.974\times10^5}{5.4\times4200} = 8.7 \)°C.

Teacher's Note:
1. Since the water has enough heat to melt all the ice, the final temperature is above 0°C.
2. Treat ice-melt water as part of the final mass.

Answer: (b) 716 cal

Explanation:
1. Melting: 80 cal; heating water: 100 cal; vaporising: 536 cal.
2. Total \( = 80 + 100 + 536 = 716 \) cal.

Teacher's Note:
1. No heating of the ice stage is needed because it starts at 0°C.
2. Latent heats change state without changing temperature.

Answer: (c) 0°C

Explanation:
1. Water can give at most \( 10\times1\times20 = 200 \) cal on cooling to 0°C.
2. Melting the ice needs \( 10\times80 = 800 \) cal.
3. Not all the ice melts, so the mixture stays at 0°C.

Teacher's Note:
1. Compare heat available with heat needed for melting before computing a mean temperature.
2. A mixture of ice and water coexists at 0°C.

Answer: (c) 10°C

Explanation:
1. \( \Delta T = \dfrac{Q}{ms} = \dfrac{420}{10\times4.2} = 10 \)°C.

Teacher's Note:
1. Use Q = m s ΔT with consistent units.
2. 4.2 J/g°C is the specific heat of water.

Answer: (c) 8 min 20 s

Explanation:
1. Heat needed \( = 2000\times4.2\times50 = 4.2\times10^5 \) J.
2. Net power \( = 1000 - 160 = 840 \) W.
3. Time \( = \dfrac{4.2\times10^5}{840} = 500 \) s = 8 min 20 s.

Teacher's Note:
1. Only the net power heats the water.
2. Two litres of water have a mass of 2000 g.

Answer: (b) 8 g

Explanation:
1. Total heat \( 420 \) J = heat to warm the ice to 0°C + heat to melt 1 g.
2. Warming: \( m\times10^{-3}\ \text{kg}\times2100\times5 = 10.5\,m \) J, with m in grams.
3. Melting 1 g needs 336 J.
4. \( 10.5m + 336 = 420 \), so \( m = 8 \) g.

Teacher's Note:
1. Split the heat into the warming part and the melting part.
2. The heat capacity here is per kilogram, so convert grams carefully.

Answer: (c) 35°C

Explanation:
1. Power \( = \dot m\,s\,\Delta T \), so \( 2100 = 20\times4.2\times\Delta T \).
2. \( \Delta T = 25 \)°C.
3. Outlet \( = 10 + 25 = 35 \)°C.

Teacher's Note:
1. In continuous flow, use the rate of mass flow.
2. Add the rise to the inlet temperature.