JEE Physics Thermal Properties Of Matter MCQs Set 09

Answer: (c) Pyrometer

Explanation:
1. The sun is extremely hot and very far away, so any thermometer that must touch the body is useless.
2. A pyrometer measures temperature from the thermal radiation received from the body, without contact.
3. Hence a pyrometer is used.

Teacher's Note:
1. Pyrometers work on the laws of radiation and are used for very high temperatures.
2. Contact thermometers have a limited range because the sensing material melts or changes its behaviour.

Answer: (b) if \( \alpha_{rod}>\alpha_{frame} \), then the rod may touch the floor

Explanation:
1. The gap equals the height of the frame minus the length of the rod.
2. On heating, the rod lengthens by about \( L\alpha_{rod}\Delta T \) and the frame height increases by about \( L\alpha_{frame}\Delta T \).
3. The gap closes only if the rod expands more than the frame, that is if \( \alpha_{rod}>\alpha_{frame} \).

Teacher's Note:
1. Compare the two expansions: the gap shrinks only when the rod expands more than the frame.
2. If the rod expanded less, the gap would simply increase.

Answer: (a) always increases

Explanation:
1. On heating, every linear dimension of the plate expands in the same ratio, as if it were enlarged photographically.
2. The hole behaves as if it were filled with copper of the same material, so its diameter also increases.

Teacher's Note:
1. A hole expands exactly like the material that was removed from it.
2. The same logic applies to the gap in a ring or the inner radius of a washer.

Answer: (a) \( \frac{\alpha_s}{\alpha_a+\alpha_s} \)

Explanation:
1. Equal expansions: \( \ell_1\alpha_a\Delta T=\ell_2\alpha_s\Delta T \Rightarrow\frac{\ell_1}{\ell_2}=\frac{\alpha_s}{\alpha_a} \).
2. \( \frac{\ell_1}{\ell_1+\ell_2}=\frac{1}{1+\ell_2/\ell_1}=\frac{1}{1+\alpha_a/\alpha_s}=\frac{\alpha_s}{\alpha_a+\alpha_s} \).

Teacher's Note:
1. The rod with the smaller coefficient must be longer to expand by the same amount.
2. Express everything through the ratio l1 : l2 first.

Answer: (a) 10,000 N

Explanation:
1. Thermal stress \( =Y\alpha\Delta T=10^{11}\times10^{-6}\times100=10^7\ \text{N/m}^2 \).
2. Force \( =\text{stress}\times A=10^7\times10^{-3}=10^4 \) N.

Teacher's Note:
1. When expansion is prevented, the force is independent of the length of the bar.
2. Force = Y A alpha (change in temperature).

Answer: (c) \( \frac5T \)

Explanation:
1. For an ideal gas \( P=\frac{nRT}{V} \), so \( PT^4=\frac{nRT^5}{V}=\text{constant} \), that is \( V\propto T^5 \).
2. Coefficient of volume expansion \( \gamma=\frac1V\frac{dV}{dT}=\frac5T \).

Teacher's Note:
1. Eliminate P using the ideal gas equation to get V as a power of T.
2. For a power law V proportional to T to the n, the coefficient is n/T.

Answer: (a) \( \frac{v^2}{4S} \)

Explanation:
1. Heat produced \( =\frac12\left(\frac12mv^2\right)=\frac14mv^2 \).
2. \( mS\Delta T=\frac14mv^2\Rightarrow\Delta T=\frac{v^2}{4S} \).

Teacher's Note:
1. The mass of the bullet cancels out.
2. Take care that only half of the kinetic energy becomes heat.

Answer: (a) \( \frac{L}{5g} \)

Explanation:
1. Loss of potential energy \( =mgh \) is converted into the heat that melts \( \frac m5 \) of ice: \( mgh=\frac m5L \).
2. \( h=\frac{L}{5g} \).

Teacher's Note:
1. Minimum distance means all of the potential energy goes into melting.
2. Here L is the heat per unit mass, so the mass cancels from both sides.

Answer: (c) 68 km

Explanation:
1. Heat absorbed \( =\frac12mgh \), and this must equal \( mL \).
2. \( h=\frac{2L}{g}=\frac{2\times3.4\times10^5}{10}=6.8\times10^4 \) m \( =68 \) km.

Teacher's Note:
1. The heat needed to melt unit mass is the latent heat, so the mass cancels out.
2. Such a large height shows that melting ice by a fall is practically impossible.

Answer: (c) 84 kJ kg\( ^{-1} \)

Explanation:
1. Heat absorbed during vaporisation \( =42\ \text{kJ/min}\times10\ \text{min}=420 \) kJ.
2. \( L=\frac{420}{5}=84 \) kJ/kg.

Teacher's Note:
1. The flat part of a heating curve shows a change of state: the heat goes into the latent heat, not into the temperature.
2. The length of the flat part times the heating rate gives the heat absorbed.

Answer: (d) more than \( 50^\circ\)C

Explanation:
1. The hotter body has the larger heat capacity, so it loses a given amount of heat with a smaller fall in temperature than the colder body rises.
2. The final temperature is therefore closer to \( 100^\circ\)C than to \( 0^\circ\)C, that is, more than \( 50^\circ\)C.

Teacher's Note:
1. If the heat capacities were equal and constant, the answer would be exactly 50 degrees Celsius.
2. The one with the larger heat capacity changes its temperature the least.

Answer: (a) 6 kg

Explanation:
1. Heat available if the water cools to \( 0^\circ\)C: \( 5000\times20=100000 \) cal.
2. Heat needed to warm the ice to \( 0^\circ\)C: \( 2000\times0.5\times20=20000 \) cal; heat to melt all the ice: \( 2000\times80=160000 \) cal.
3. Only \( 100000-20000=80000 \) cal remains for melting, so \( \frac{80000}{80}=1000 \) g of ice melts and the mixture stays at \( 0^\circ\)C.
4. Water \( =5+1=6 \) kg.

Teacher's Note:
1. Compare the available heat with the heat needed for the complete melting before assuming a final temperature.
2. If the ice only partly melts, the final temperature is exactly 0 degrees Celsius.

Answer: (a) \( \frac{K_1+3K_2}{4} \)

Explanation:
1. Heat flows along the length through the two materials in parallel.
2. Area of the cylinder \( =\pi R^2 \); area of the shell \( =\pi(4R^2-R^2)=3\pi R^2 \).
3. \( K_{eff}(4\pi R^2)=K_1\pi R^2+K_2\,3\pi R^2\Rightarrow K_{eff}=\frac{K_1+3K_2}{4} \).

Teacher's Note:
1. Paths side by side along the heat flow add their conductances (K times area).
2. The effective conductivity is the area-weighted average of the two conductivities.

Answer: (c) \( \frac{k_1k_2(d_1+d_2)}{d_1k_2+d_2k_1} \)

Explanation:
1. In series, the thermal resistances add: \( \frac{d_1+d_2}{k_{eq}A}=\frac{d_1}{k_1A}+\frac{d_2}{k_2A} \).
2. \( k_{eq}=\frac{d_1+d_2}{\frac{d_1}{k_1}+\frac{d_2}{k_2}}=\frac{k_1k_2(d_1+d_2)}{d_1k_2+d_2k_1} \).

Teacher's Note:
1. Layers in series carry the same heat current, so their resistances add.
2. For equal thicknesses, the result reduces to the harmonic mean 2 k1 k2/(k1 + k2).

Answer: (c) 400 W

Explanation:
1. Let each rod have length \( \ell \) and cross-section A, and let \( x=\frac A\ell \).
2. Series (a): \( R_{total}=\frac1x\left(\frac1{200}+\frac1{400}+\frac1{200}\right)=\frac{1}{x}\cdot\frac{5}{400} \), so \( 40=\frac{100\times400x}{5}\Rightarrow x=0.005 \).
3. Parallel (b): conductance \( =x(200+400+200)=800x=4 \) W/\( ^\circ\)C, so heat flow \( =100\times4=400 \) W.

Teacher's Note:
1. In series the resistances add; in parallel the conductances add.
2. The rate in the parallel arrangement is far larger than in the series one.

Answer: (a) 1 minute

Explanation:
1. (a) Series: length \( 2\ell \), area A, so rate \( \propto\frac{KA}{2\ell} \).
2. (b) Parallel: length \( \ell \), total area \( 2A \), so rate \( \propto\frac{K\,2A}{\ell} \).
3. The rate in (b) is 4 times that in (a), so the time is \( \frac44=1 \) minute.

Teacher's Note:
1. Doubling the length halves the rate, doubling the area doubles it.
2. Compare the two rates through the ratio of area to length.

Answer: (c) its density is less than that of the surrounding fluid

Explanation:
1. On heating, the liquid expands and its density falls.
2. The lighter heated portion is pushed up by the denser surrounding fluid (buoyancy), and cooler fluid takes its place.

Teacher's Note:
1. Natural convection is driven by density differences and gravity.
2. In forced convection a pump or fan drives the flow instead.

Answer: (c) \( \frac{625}{81} \)

Explanation:
1. Wien's law: \( \lambda_mT=\text{constant}\Rightarrow\frac{T'}{T}=\frac{\lambda_0}{\frac35\lambda_0}=\frac53 \).
2. Stefan's law: \( P\propto T^4\Rightarrow n=\left(\frac53\right)^4=\frac{625}{81} \).

Teacher's Note:
1. Use Wien's law to find the temperature ratio and Stefan's law for the power.
2. The wavelength of maximum emission falls when the temperature rises.

Answer: (c) All glow with equal brightness

Explanation:
1. Inside a furnace at uniform temperature, each object absorbs and emits radiation in equilibrium with the cavity radiation.
2. By Kirchhoff's law, a good absorber is a good emitter; the extra emission of the black object is exactly balanced by its extra absorption, and the white object reflects the cavity radiation to make up the difference.
3. Hence all of them glow with equal brightness.

Teacher's Note:
1. In thermal equilibrium, the radiation leaving every object (emitted plus reflected) equals the cavity radiation.
2. The colours only show up when the objects are taken out and cooled in air.

Answer: (c) Wien's displacement law

Explanation:
1. As the temperature rises, the wavelength of maximum emission shifts from the infrared towards shorter wavelengths: \( \lambda_mT=b \).
2. This shift in colour from red to yellow to white is explained by Wien's displacement law.

Teacher's Note:
1. Stefan's law explains how much energy is radiated, Wien's law explains the colour.
2. The peak wavelength is inversely proportional to the absolute temperature.

Answer: (c) 3240

Explanation:
1. \( P=\sigma AT^4\propto r^2T^4 \).
2. \( \frac{P'}{P}=\left(\frac12\right)^2(2)^4=\frac{16}{4}=4 \).
3. \( P'=4\times810=3240 \) W.

Teacher's Note:
1. Surface area goes as the square of the radius.
2. Combine the changes in area and temperature as a product of ratios.

Answer: (d) \( \frac32T \)

Explanation:
1. Excess over the room temperature: from \( 3T-T=2T \) to \( 2T-T=T \) in 15 min (a halving).
2. In the next 15 min the excess again halves: from \( T \) to \( \frac T2 \).
3. Temperature \( =T+\frac T2=\frac32T \).

Teacher's Note:
1. Newton's law means the excess temperature decays exponentially, so equal times give equal ratios.
2. The body never cools below the room temperature.

Answer: (a) \( E\propto T^4;\ E_{max}\propto T^5 \)

Explanation:
1. Stefan-Boltzmann law: total energy \( E\propto T^4 \).
2. The peak spectral emissive power \( E_{\lambda,max}\propto T^5 \) (the peak height grows as the fifth power of temperature, while the peak wavelength shifts as \( T^{-1} \)).

Teacher's Note:
1. Total emission goes as T to the fourth, but the height of the peak grows even faster.
2. Both results follow from Planck's radiation law.