JEE Physics Thermal Properties Of Matter MCQs Set 04

Answer: (b) 5 : 3

Explanation:
1. Thermal resistance \( R = \dfrac{L}{KA} \).
2. Equal R and equal A give \( L \propto K \).
3. \( L_1 : L_2 = 5 : 3 \).

Teacher's Note:
1. Thermal resistance grows with length and falls with conductivity.
2. Equal resistance means length is proportional to K.

Answer: (b) \( Q_2 = 2Q_1 \)

Explanation:
1. Rate \( = \dfrac{KA\,\Delta T}{L} \).
2. Doubling all dimensions makes \( A \) four times and \( L \) twice.
3. \( Q_2 = \dfrac{4}{2}Q_1 = 2Q_1 \).

Teacher's Note:
1. Area scales as the square of the dimension, length as the first power.
2. So the heat current scales with the linear dimension.

Answer: (c) 100°C

Explanation:
1. \( \dfrac{Q}{t} = \dfrac{KA\,\Delta T}{L} \).
2. \( \Delta T = \dfrac{4000\times0.1}{400\times0.01} = 100 \)°C.

Teacher's Note:
1. Convert cm and cm² to m and m² first.
2. 100 cm² = 0.01 m².

Answer: (c) 75°C

Explanation:
1. In series the heat current is the same: \( \dfrac{9K A(100-T)}{18} = \dfrac{KA(T-0)}{6} \).
2. \( \dfrac{100-T}{2} = \dfrac{T}{6} \), so \( 300 - 3T = T \).
3. \( T = 75 \)°C.

Teacher's Note:
1. Most of the temperature drop occurs across the poorer conductor.
2. Equal heat current through both sections.

Answer: (a) \( 15K/16 \)

Explanation:
1. For equal lengths in series, \( \dfrac{3}{K_{eff}} = \dfrac{1}{K/2} + \dfrac{1}{5K} + \dfrac{1}{K} = \dfrac{2}{K} + \dfrac{0.2}{K} + \dfrac{1}{K} = \dfrac{3.2}{K} \).
2. \( K_{eff} = \dfrac{3K}{3.2} = \dfrac{15K}{16} \).

Teacher's Note:
1. Series combination: resistances add.
2. With equal lengths, the number of rods appears in the numerator.

Answer: (c) 5°C

Explanation:
1. The heat current is the same in each layer, so \( k\,\Delta T \) is constant.
2. First layer: \( k(20-10) = 10k \).
3. Second layer: \( 2k(10-\theta) = 10k \), so \( \theta = 5 \)°C.
4. Check with third layer: \( k(5+5) = 10k \).

Teacher's Note:
1. Layers with twice the conductivity have half the temperature drop.
2. Use the first layer to find the common heat current.

Answer: (b) 60°C

Explanation:
1. Heat flowing in from the two hot rods equals heat flowing out through the cold rod.
2. \( 2K A\dfrac{(90-T)}{\ell} = KA\dfrac{(T-0)}{\ell} \).
3. \( 180 - 2T = T \), so \( T = 60 \)°C.

Teacher's Note:
1. The two parallel rods act like a rod of double conductance.
2. Balance the heat currents at the junction.

Answer: (d) length 1 m, radius 2 cm

Explanation:
1. Heat current \( \propto \dfrac{r^2}{L} \).
2. Values: 1, 0.5, 2 and 4 in the given order.
3. The 1 m rod of radius 2 cm gives the largest value.

Teacher's Note:
1. Thicker and shorter rods conduct more.
2. Radius enters squared through the area.

Answer: (c) 8 : 9

Explanation:
1. Rate \( \propto \dfrac{r^2}{L} \).
2. First: \( \dfrac{4}{1} = 4 \); second: \( \dfrac{9}{2} = 4.5 \).
3. Ratio \( 4 : 4.5 = 8 : 9 \).

Teacher's Note:
1. Ratios of radii are squared, ratios of lengths are used directly.
2. Both rods have the same K and ΔT.

Answer: (b) 60°C

Explanation:
1. In steady state the temperature varies linearly along a uniform rod.
2. At the centre \( T = \dfrac{100+20}{2} = 60 \)°C.

Teacher's Note:
1. Uniform rod and steady state mean a constant temperature gradient.
2. The midpoint temperature is the mean of the end temperatures.

Answer: (b) \( \dfrac{4}{3}K \)

Explanation:
1. The layers are in series, so \( \dfrac{2d}{K_{eq}} = \dfrac{d}{K} + \dfrac{d}{2K} = \dfrac{3d}{2K} \).
2. \( K_{eq} = \dfrac{4K}{3} \).

Teacher's Note:
1. Equivalent conductivity of a series stack is a harmonic-type mean.
2. It lies between K and 2K.

Answer: (b) \( 2\times10^4 \) J

Explanation:
1. \( \dfrac{Q}{t} = \dfrac{KA\,\Delta T}{d} = \dfrac{0.2\times10\times20}{2\times10^{-3}} \).
2. \( = 2\times10^4 \) J per second.

Teacher's Note:
1. Thickness 2 mm is 0.002 m.
2. MKS conductivity is in W/m K.

Answer: (c) 20°C/cm

Explanation:
1. Heat flux \( \dfrac{Q}{At} = K\dfrac{dT}{dx} \).
2. \( \dfrac{dT}{dx} = \dfrac{10}{0.5} = 20 \)°C/cm.

Teacher's Note:
1. Heat flux is heat per unit area per second.
2. Gradient = flux divided by conductivity.

Answer: (b) \( M^{-1}L^{-2}T^{3}\theta \)

Explanation:
1. Thermal resistance \( = \dfrac{\Delta T}{\text{heat current}} \).
2. Heat current has dimensions of power, \( ML^2T^{-3} \).
3. So \( R = \theta\,M^{-1}L^{-2}T^{3} \).

Teacher's Note:
1. Thermal resistance is temperature difference per watt.
2. Power has dimensions ML²T⁻³.

Answer: (d) 16 : 1

Explanation:
1. Rate \( \propto \dfrac{d^2}{L} \).
2. Ratio \( = \dfrac{4/1}{1/4} = 16 \).

Teacher's Note:
1. Diameter ratio is squared because area depends on d².
2. A shorter, thicker rod conducts far more.

Answer: (c) 60°C

Explanation:
1. Thermal resistances: \( R_1 \propto \dfrac{1}{K} \), \( R_2 \propto \dfrac{2}{3K} \).
2. Temperature drops are in the ratio \( 1 : \dfrac{2}{3} = 3 : 2 \).
3. The K bar takes \( \dfrac{3}{5}\times100 = 60 \)°C, so the interface is at 60°C.

Teacher's Note:
1. The temperature drop across each section is proportional to its resistance.
2. Measure the interface temperature from the 0°C end.

Answer: (c) 1 : 1

Explanation:
1. Power \( \propto r^2T^4 \).
2. First: \( 1\times4^4 = 256 \) (units of \( 10^{12} \)).
3. Second: \( 16\times2^4 = 256 \).
4. The ratio is 1 : 1.

Teacher's Note:
1. Radiated power depends on surface area and the fourth power of temperature.
2. A bigger but cooler body can radiate equally.

Answer: (c) \( \dfrac{3}{16}R \)

Explanation:
1. Cooling rate \( \propto T^4 - T_0^4 \).
2. At 600 K: \( 600^4 - 200^4 \propto 1296 - 16 = 1280 \).
3. At 400 K: \( 256 - 16 = 240 \).
4. Ratio \( = \dfrac{240}{1280} = \dfrac{3}{16} \).

Teacher's Note:
1. Use the fourth-power difference because the body also absorbs radiation from the surroundings.
2. Newton's law would give a different answer.

Answer: (d) 800 K

Explanation:
1. \( 400 \)°C \( = 673 \) K.
2. \( E \propto T^4 \), so \( T' = 673\times2^{1/4} = 673\times1.189 \approx 800 \) K.

Teacher's Note:
1. Temperatures in the Stefan law must be absolute.
2. The factor for doubling is the fourth root of 2.

Answer: (c) bad conductor of heat

Explanation:
1. Mud is a poor conductor of heat.
2. It slows the flow of heat from outside to inside in summer and from inside to outside in winter.

Teacher's Note:
1. Thick walls of poor conductors insulate the room.