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JEE • PHYSICS • THERMAL PROPERTIES OF MATTER

Thermal properties of matter practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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211 Medium
211 of 514
Application +4 | -1
The length of a metal rod at 00C{0^0}C is 0.5 m0.5\,m. When it is heated, its length increases by 2.7 mm2.7\,mm. The final temperature of rod is [α=90×10−610C]\left[ {\alpha = 90 \times {{10}^{ - 61}}^0C} \right]
View answer and explanation
Correct option: D
600C{60^0}C
Option D (600C{60^0}C) is correct because it is calculated using the formula for linear thermal expansion.
212 Medium
212 of 514
Application +4 | -1
Two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio 2:3. These two cylinders are combined to make a cylinder. One end of P is kept of 1000C{100^0}C and another end of Q at 00C{0^0}C. The temperature at the interface of P and Q is
View answer and explanation
Correct option: B
400C{40^0}C
Option B (400C{40^0}C) is correct because the temperature at the interface is calculated using the thermal conductivities and temperatures of the two cylinders.
213 Medium
213 of 514
Understanding +4 | -1
The wavelength of the radiation emitted by a body depends upon
View answer and explanation
Correct option: C
The temperature of the surface
Option C (The temperature of the surface) is correct because Wien's displacement law states that the wavelength of maximum emission is inversely proportional to the temperature of the black body.
214 Medium
214 of 514
Application +4 | -1
The tungsten filament of an electric lamp has a surface AA area and a power rating PP. If the emissivity of the filament is ε\varepsilon and σ\sigma is Stefan’s constant, the steady temperature of the filament will be
View answer and explanation
Correct option: D
T=(PAεσ)14T = {\left( {\frac{P}{{A\varepsilon \sigma }}} \right)^{\frac{1}{4}}}
Option D (T=(PAεσ)14T = {\left( {\frac{P}{{A\varepsilon \sigma }}} \right)^{\frac{1}{4}}}) is correct because the steady temperature of the filament is found by equating the power rating to the power radiated by the filament using Stefan's law: P=AεσT4P = A\varepsilon \sigma T^4.
215 Medium
215 of 514
Application +4 | -1
A particular star (assuming it as a block body has a surface temperature of about 5×104K.5 \times {10^4}K. The wavelength in nanometers at which its radiation becomes maximum is (b=0.0029  mk)(b = 0.0029\,\,mk)
View answer and explanation
Correct option: B
58
Option B (58) is correct because it is calculated using Wien's displacement law, where the product of wavelength and temperature is constant.
216 Medium
216 of 514
Analyzing +4 | -1
Assertion (A): At constant temperature, when pressure of a gas is increased by 10%10 \%, then volume decreases by less than 10%10 \%
Reason (R): At constant temperature the volume of gas is inversely proportional to the pressure acting on it.
View answer and explanation
Correct option: A
If both (A) and (R) are true and (R) is the correct explanation of (A).
Option A (If both (A) and (R) are true and (R) is the correct explanation of (A).) is correct because Boyle's law states that volume is inversely proportional to pressure, and the percentage decrease in volume is less than the percentage increase in pressure due to the inverse relationship.
217 Medium
217 of 514
Analyzing +4 | -1
(A): Two bodies at different temperatures, if brought in thermal contact do not necessary settle to the mean temperature.
(R): The two bodies may have different thermal capacities.
View answer and explanation
Correct option: A
If both (A) and (R) are true and (R) is the correct explanation of (A)
The correct answer is A because two bodies at different temperatures do not necessarily settle to the mean temperature due to different thermal capacities, making the reason a correct explanation.
218 Medium
218 of 514
Understanding +4 | -1
Heat is associated with
View answer and explanation
Correct option: A
Kinetic energy of random motion of molecules
Option A (Kinetic energy of random motion of molecules) is correct because heat is associated with the disordered motion of molecules.
219 Medium
219 of 514
Application +4 | -1
In an isochoric process if T1=27oC   and   T2=127oC{T_1} = {27^\mathrm{o}}C\mathrm{ }\,\,\,\mathrm{and}\,\,\,\mathrm{ }{T_2} = {127^\mathrm{o}}C, then P1/P2{P_1}/{P_2} will be equal to
View answer and explanation
Correct option: C
45750
Option C (03-Apr) is correct because at constant volume, the ratio of pressures equals the ratio of absolute temperatures, resulting in 3/4.
220 Medium
220 of 514
Understanding +4 | -1
Assertion (A): Sometimes water pipes burst in severe winter.
Reason (R): On freezing water expands due to anamolous expansion.
View answer and explanation
Correct option: A
If both (A) and (R) are true and (R) is the correct explanation of (A).
Option A (If both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the anomalous expansion of water upon freezing causes increased pressure, leading to burst pipes.
221 Medium
221 of 514
Application +4 | -1
Estimate the rate at which ice would melt in a wooden box 2.0 cm2.0 \mathrm{~cm} thick and of inside measurements 200 cm×120 cm×120 cm200 \mathrm{~cm} \times 120 \mathrm{~cm} \times 120 \mathrm{~cm} assuming that the external temperature is 30∘C30^{\circ} \mathrm{C} and coefficient of thermal conductivity of wood is 0.0004 cal s−1 cm−1∘C0.0004 \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~cm}^{-1}{ }^{\circ} \mathrm{C}.
View answer and explanation
Correct option: A
9.36 g s−19.36 \mathrm{~g~s}^{-1}
Option A (9.36 g s−19.36 \mathrm{~g~s}^{-1}) is correct because the rate of melting is calculated using the formula for heat transfer through conduction.
222 Medium
222 of 514
Understanding +4 | -1
The latent heat of vaporization of a substance is always
View answer and explanation
Correct option: A
Greater than its latent heat of fusion
Option A is correct because the latent heat of vaporization is always greater than the latent heat of fusion due to the larger energy requirement for phase change from liquid to gas.
223 Medium
223 of 514
Application +4 | -1
The normal body-temperature of a person is 97∘F97^{\circ} \mathrm{F}. Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature =47∘F=47^{\circ} \mathrm{F}, surface of the body under clothes =1.6 m2,=1.6 \mathrm{~m}^2, conductivity of the cloth =0.04 J s−1 m−1∘C−1=0.04 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}, thickness of the cloth =0.5 cm=0.5 \mathrm{~cm}.
View answer and explanation
Correct option: C
356 J/s
Option C (356 J/s) is correct because it applies the formula for heat transfer through conduction, converting Fahrenheit to Celsius correctly.
224 Medium
224 of 514
Application +4 | -1
The temperature of body is increased from 27∘C27^{\circ} \mathrm{C} to 127∘C127^{\circ} \mathrm{C}. The radiation emitted by it increases by a factor of
View answer and explanation
Correct option: A
25681\frac{256}{81}
Option A (25681\frac{256}{81}) is correct because the radiation emitted is proportional to the fourth power of the absolute temperature, and the ratio of the fourth powers of the temperatures is 25681\frac{256}{81}.
225 Medium
225 of 514
Understanding +4 | -1
Two thermometers are used to record the temperature of a room. If the bulb of one is wrapped in wet hanky
View answer and explanation
Correct option: C
The temperature recorded by dry-bulb thermometer will be greater than that recorded by the other
Option C (The temperature recorded by dry-bulb thermometer will be greater than that recorded by the other) is correct because the wet hanky causes evaporative cooling, lowering the temperature recorded by that thermometer.
226 Medium
226 of 514
Application +4 | -1
Two litre glass flask contains some mercury. It is found that at all temperatures the volumes of the air inside the flask remains the same. The volume of the mercury inside the flask is
(α  for  glass = 9×10−6/∘C,  γ  for  mercury = 1.8×10−4/∘C)\left( {\alpha \,\,\mathrm{for}\,\,\mathrm{glass}\, = \,9 \times {{10}^{ - 6}}/^\circ C,\,\,\gamma \,\,\mathrm{for}\,\,\mathrm{mercury}\, = \,1.8 \times {{10}^{ - 4}}/^\circ C} \right)
View answer and explanation
Correct option: D
300 CC
Option D (300 CC) is correct because it is the volume of mercury that keeps the air volume constant, calculated using the coefficients of volume expansion.
227 Medium
227 of 514
Application +4 | -1
A metal block of density 6000 kg m−36000 \mathrm{~kg} \mathrm{~m}^{-3} and mass 1⋅2 kg1 \cdot 2 \mathrm{~kg} is suspended through a spring of spring constant 200 N m−1200 \mathrm{~N} \mathrm{~m}^{-1}. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260 g260 \mathrm{~g} and the block is at a height 40 cm40 \mathrm{~cm} above the bottom of the vessel. If the support to the spring is broken, what will be the rise in the temperature of the water. Specific heat capacity of the block is 250 J kg−1 K−1250 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} and that of water is 4200 J kg−1 K−14200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}. Heat capacities of the vessel and the spring are negligible.
View answer and explanation
Correct option: B
3×10−30C3 \times 10^{-30} \mathrm{C}
Option B (3×10−30C3 \times 10^{-30} \mathrm{C}) is correct because it is the temperature rise calculated from the potential energy conversion and heat transfer to the water and block.
228 Medium
228 of 514
Application +4 | -1
If a body coated black at 600 K600 \mathrm{~K} surrounded by atmosphere at 300 K300 \mathrm{~K} has cooling rate r0r_0, the same body at 900 K900 \mathrm{~K}, surrounded by the same atmosphere, will have cooling rate equal to
View answer and explanation
Correct option: A
163r0\frac{16}{3} r_0
Option A (163r0\frac{16}{3} r_0) is correct because the cooling rate is proportional to the difference of the fourth power of the temperatures.
229 Medium
229 of 514
Application +4 | -1
A black body emits heat at the rate of 20 W20 \mathrm{~W}. When its temperature is 727∘C727^{\circ} \mathrm{C}. Another black body emits heat at the rate of 15 W15 \mathrm{~W}, when its temperature is 227∘C227^{\circ} \mathrm{C}. Compare the area of the surface of the two bodies, if the surrounding is at NTP.
View answer and explanation
Correct option: D
1:121: 12
Option D (1:121: 12) is correct because it is calculated by comparing the power emitted by two black bodies using Stefan-Boltzmann law, considering their temperatures and areas.
230 Medium
230 of 514
Analyzing +4 | -1
(A): Cooking food is difficult on hills.
(R): The boiling point decreases with increases in pressure
View answer and explanation
Correct option: C
If (A) is true but (R) is false
Option C (If (A) is true but (R) is false) is correct because cooking is difficult on hills due to lower atmospheric pressure, which *decreases* the boiling point, not increases it.
231 Medium
231 of 514
Analyzing +4 | -1
A cube of a substance of coefficient of cubical expansion γ1\gamma_1 is floating in a liquid of coefficient of cubical expansion γ2(γ2<γ1)\gamma_2\left(\gamma_2 < \gamma_1\right), such that 50%50 \% of its volume is inside the liquid. If temperature of the system is now increased, the percentage of volume of cube inside the liquid may be
View answer and explanation
Correct option: C
45%45 \%
Option C (45 %) is correct because the volume expansion of the cube is greater than that of the liquid, causing a decrease in the submerged percentage.
232 Medium
232 of 514
Application +4 | -1
A large steel wheel is to be fitted on to a shaft of the same material. At 27∘C27^{\circ} \mathrm{C}, the outer diameter of the shaft is 8.70 cm8.70 \mathrm{~cm} and the diameter of the central hole in the wheel is 8.69 cm8.69 \mathrm{~cm}. The shaft is cooled using 'dry ice' (solid carbon dioxide). At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range. αsteel =1.20×10−5 K−1.\alpha_{\text {steel }}=1.20 \times 10^{-5} \mathrm{~K}^{-1} .
View answer and explanation
Correct option: A
−68.8∘C-68.8^{\circ} \mathrm{C}
Option A (−68.8∘C-68.8^{\circ} \mathrm{C}) is correct because it is the temperature at which the shaft's diameter shrinks enough for the wheel to slip on.
233 Medium
233 of 514
Analyzing +4 | -1
A solid sphere of radius R\mathrm{R} and a hollow sphere of inner radius r\mathrm{r} and outer radius R\mathrm{R} made of copper are heated to the same temperature and are allowed to cool in the same environment. Then, choose the CORRECT statement :-
View answer and explanation
Correct option: A
Hollow sphere cools faster
Option A (Hollow sphere cools faster) is correct because the hollow sphere has a smaller mass for the same surface area, leading to a faster cooling rate.
234 Medium
234 of 514
Understanding +4 | -1
Assertion: Radiation is the speediest mode of heat transfer.
Reason: Radiation can be transmitted in zig-zag motion.
View answer and explanation
Correct option: C
If assertion is true, but reason is false
Option C (If assertion is true, but reason is false) is correct because radiation is the fastest mode of heat transfer, but it travels in straight lines, not zig-zag motion.
235 Medium
235 of 514
Understanding +4 | -1
Statement A: Greater the mass of radiating body, slower will be cooling
Statement B: Greater the temperature of the surroundings, lower will be cooling
View answer and explanation
Correct option: A
Statement A is true and statement B is true
Statement A is true because the rate of cooling is inversely proportional to the mass, and Statement B is true because the rate of cooling is proportional to the temperature difference between the body and surroundings.
236 Medium
236 of 514
Remembering +4 | -1
An ideal gas equation can be written as P=ρRTM0P=\frac{\rho R T}{M_0} where ρ\rho and M0M_0 are respectively
View answer and explanation
Correct option: C
Mass density, molar mass
Option C (Mass density, molar mass) is correct because ρ\rho represents mass density (mass/volume) and M0M_0 represents molar mass in the ideal gas equation.
237 Medium
237 of 514
Understanding +4 | -1
A real gas can be approximated to an ideal gas at
View answer and explanation
Correct option: A
low density
Option B (Metal has high thermal conductivity) is correct because metals conduct heat away from the body more quickly than wood, resulting in a colder sensation.
238 Medium
238 of 514
Understanding +4 | -1
Volume of gas becomes four times if
View answer and explanation
Correct option: A
Temperature becomes four times at constant pressure
Option A (Temperature becomes four times at constant pressure) is correct because, according to Charles's Law, volume is directly proportional to temperature at constant pressure.
239 Medium
239 of 514
Understanding +4 | -1
Assertion: The amount of radiation from sun's surface varies as the fourth power of its absolute temperature.
Reason: The sun is black body.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because the Stefan-Boltzmann law states that the energy radiated by a black body is proportional to the fourth power of its absolute temperature, and the sun approximates a black body.
240 Medium
240 of 514
Understanding +4 | -1
Assertion (A): Solids expands on heating
Reason (R): Atoms of the solid are an harmonic oscillators
View answer and explanation
Correct option: A
If both (A) and (R) are true and (R) is the correct explanation of (A).
Option A is correct because solids expand due to the anharmonic nature of atomic vibrations, which increases with temperature.
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