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JEE • PHYSICS • THERMAL PROPERTIES OF MATTER

Thermal properties of matter practice · Medium

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271 Medium
271 of 514
Analyzing +4 | -1
During an experiment, an ideal gas is found to obey a condition P2ρ=\frac{P^2}{\rho}= constan tt, ( ρ=\rho= density of the gas). The gas is initially at temperature T, pressure P\mathrm{P} and density ρ\rho. The gas expands such that density changes to ρ/2\rho / 2. Then,
View answer and explanation
Correct option: D
The graph of the above process on P\mathrm{P} - T\mathrm{T} diagram is hyperbola.
Option D is correct because the process follows PT = constant, which represents a hyperbola on the P-T diagram.
272 Medium
272 of 514
Application +4 | -1
Three slabs of thicknesses x1,x2\text{x}_1, \text{x}_2 and x3\text{x}_3 and coefficients of thermal conductivity K1,K2\mathrm{K}_1, \mathrm{K}_2 and K3\mathrm{K}_3 are placed parallel to each other in contact. In steady state the combination behaves as a single slab of material of conductivity K,\mathrm{K}, given by: (Cross section areas are same)
View answer and explanation
Correct option: D
x1 K1+x2 K2+x3 K3=x1+x2+x3 K\frac{\mathrm{x}_1}{\mathrm{~K}_1}+\frac{\mathrm{x}_2}{\mathrm{~K}_2}+\frac{\mathrm{x}_3}{\mathrm{~K}_3}=\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{\mathrm{~K}}
Option D (x1 K1+x2 K2+x3 K3=x1+x2+x3 K\frac{\mathrm{x}_1}{\mathrm{~K}_1}+\frac{\mathrm{x}_2}{\mathrm{~K}_2}+\frac{\mathrm{x}_3}{\mathrm{~K}_3}=\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3}{\mathrm{~K}}) is correct because the thermal resistances add in series, and the equivalent thermal resistance is the sum of individual resistances.
273 Medium
273 of 514
Application +4 | -1
A steel bar 0.5 cm square [cross-sectional area] and 25 cm long at 30∘C30^\circ C fits into the space between two fixed supports. If the bar is now heated to 230∘C230^\circ C, the force it exerts against the supports is [αsteel=11×10−6/∘C  and  Ysteel=12×1011Nm2]\left[ {{\alpha _{\mathrm{steel}}} = 11 \times {{10}^{ - 6}}/^\circ C\,\,\mathrm{and}\,\,{Y_{\mathrm{steel}}} = 12 \times {{10}^{11}}\frac{N}{{{m^2}}}} \right]
View answer and explanation
Correct option: B
22000 N
Option B (22000 N) is correct because the force exerted against the supports is calculated using the formula F=YAαΔtF = YA\alpha \Delta t, where Y is Young's modulus, A is the cross-sectional area, α\alpha is the coefficient of linear expansion, and Δt\Delta t is the change in temperature.
274 Medium
274 of 514
Understanding +4 | -1
Two balloons are filled, one with pure He gas and the other by air, respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
View answer and explanation
Correct option: B
Same in both balloons
Option B (Same in both balloons) is correct because, according to the ideal gas law, at the same pressure and temperature, equal volumes contain the same number of molecules, regardless of the gas.
275 Medium
275 of 514
Understanding +4 | -1
Assertion (A): With increase in temperature, the pressure of given gas increase.
Reason (R): Increase in temperature causes decrease in number of collisions of molecules with the walls of the container.
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 pressure increases with temperature due to increased molecular collisions, not decreased.
276 Medium
276 of 514
Understanding +4 | -1
During the melting of a slab of ice at 273 K273 \mathrm{~K} at atmospheric pressure
View answer and explanation
Correct option: B
Positive work is done on the ice-water system by the atmosphere
Option B (Positive work is done on the ice-water system by the atmosphere) is correct because ice has a lower density than water, so when it melts, the volume decreases, and the atmosphere does positive work on the system.
277 Medium
277 of 514
Understanding +4 | -1
A black body is at a temperature 300 K. It emits energy at a rate, which is proportional to
View answer and explanation
Correct option: D
(300)4{\left( {300} \right)^4}
Option D ((300)4{\left( {300} \right)^4}) is correct because the energy emitted by a black body is proportional to the fourth power of its absolute temperature (Stefan-Boltzmann Law).
278 Medium
278 of 514
Understanding +4 | -1
The two ends of a rod of length LL and a uniform cross-sectional area AA are kept at two temperatures T1T_1 and T2T_2 (T1>T2)\left(T_1 > T_2\right). The rate of heat transfer, dQdt\frac{d Q}{d t} through the rod in a steady state is given by:
View answer and explanation
Correct option: C
kA(T1−T2)L\frac{{kA}\left({T}_1-{T}_2\right)}{{L}}
Option C (kA(T1−T2)L\frac{{kA}\left({T}_1-{T}_2\right)}{{L}}) is correct because it represents the rate of heat transfer through a rod in steady state according to Fourier's law of heat conduction.
279 Medium
279 of 514
Remembering +4 | -1
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by 10C{1^0}C and it is defined under which of the following conditions?
View answer and explanation
Correct option: A
From 14.50C{14.5^0}C to 15.50C{15.5^0}C at 760 mm of Hg
Option A is correct because 1 calorie is defined as the heat required to raise the temperature of 1 g of water from 14.50C{14.5^0}C to 15.50C{15.5^0}C at 760 mm of Hg.
280 Medium
280 of 514
Application +4 | -1
If the temperature of the Sun were to increase from TT to 3T3 T and its radius from RR to 1.5R1.5 R. The ratio of power, radiated by it would become:
View answer and explanation
Correct option: D
36 times
Option D (36 times) is correct because the power radiated is proportional to the area and the fourth power of the temperature, resulting in a 36-fold increase.
281 Medium
281 of 514
Understanding +4 | -1
Assertion: Perspiration from human body helps in cooling the body.
Reason: A thin layer of water on the skin enhances its emissivity.
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 perspiration cools the body through evaporation, not by enhancing emissivity.
282 Medium
282 of 514
Analyzing +4 | -1
Assertion (A): For a constant volume gas thermometer one should fill the gas at high temperature and low pressure
Reason (R): Many of the properties of gases are common at low pressures and high temperatures.
View answer and explanation
Correct option: B
If both (A) and (R) are true but (R) is not the correct explanation of (A).
Option B (If both (A) and (R) are true but (R) is not the correct explanation of (A)) is correct because while both the assertion and reason are true, the reason doesn't directly explain why a constant volume gas thermometer should be filled at high temperature and low pressure.
283 Medium
283 of 514
Understanding +4 | -1
At atmospheric pressure, the water boils at 100∘C100^{\circ} \mathrm{C}. If pressure is reduced, it will boil at
View answer and explanation
Correct option: B
Lower temperature
Option B (Lower temperature) is correct because the boiling point of water decreases as the pressure is reduced.
284 Medium
284 of 514
Understanding +4 | -1
Statement I: Water expands both when heated or cooled
from 4°C.
Statement II: Density of water is minimum at 4°C.
View answer and explanation
Correct option: C
Statement I is true; Statement II is false
Option D (Statement I is false; Statement II is true) is correct because water expands when cooled or heated from 4°C, and its density is maximum at 4°C.
285 Medium
285 of 514
Understanding +4 | -1
Statement-I: A metallic plate containing circular hole is heated then the size of the hole increases.
Statement-II: The expansion of the solid always takes place radially outwards.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A is correct because when a metallic plate is heated, it expands radially outwards, causing the hole to increase in size.
286 Medium
286 of 514
Application +4 | -1
A copper rod of 132 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is:

(αCu=1.7×10−5 K−1\left(\alpha_{Cu}=1.7 \times 10^{-5} \mathrm{~K}^{-1}\right. and αAl=2.2×10−5 K−1)\alpha_{Al}=2.2 \times 10^{-5} \mathrm{~K}^{-1})
View answer and explanation
Correct option: B
102 cm
Option B (102 cm) is correct because the length of the aluminum rod is calculated by equating the increase in length of both rods, considering their respective coefficients of linear expansion.
287 Medium
287 of 514
Remembering +4 | -1
At temperature TT, the emissive power and absorption power of a body for certain wavelength are eλe_\lambda and aλa_\lambda respectively, then:
View answer and explanation
Correct option: C
eλ=aλe_\lambda=a_\lambda
Option C (eλ=aλe_\lambda=a_\lambda) is correct because, according to Kirchhoff's law, at a given temperature, the emissive power and absorptive power of a body are equal.
288 Medium
288 of 514
Application +4 | -1
A pendulum clock loses 12 s a day if the temperature is 40∘C40^\circ C and gains 4 s in a day if the temperature is 20∘C20^\circ C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α)\left( \alpha \right) of the metal of the pendulum shaft are, respectively
View answer and explanation
Correct option: A
25∘C,  α=1.85×10−5/∘C25^\circ C,\,\,\alpha = 1.85 \times {10^{ - 5}}/^\circ C
Option A (25∘C,  α=1.85×10−5/∘C25^\circ C,\,\,\alpha = 1.85 \times {10^{ - 5}}/^\circ C) is correct because it satisfies the conditions where the clock shows correct time and the calculated coefficient of linear expansion.
289 Medium
289 of 514
Understanding +4 | -1
Assertion (A): The lowest attainable temperature is absolute zero, i.e., 0 K=−273.15∘C0 \mathrm{~K}=-273.15^{\circ} \mathrm{C}.
Reason (R): Size of each degree on Kelvin scale is same as that on Celsius scale.
View answer and explanation
Correct option: B
If both (A) and (R) are true but (R) is not the correct explanation of (A).
Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion. Absolute zero is indeed -273.15°C, and the Kelvin and Celsius scales have the same degree size, but the latter doesn't explain the former.
290 Medium
290 of 514
Understanding +4 | -1
When the specific heat of a solid is measured by the method of mixture, the heat is lost to surroundings from the calorimeter by the process of
View answer and explanation
Correct option: A
radiation
Option A (radiation) is correct because the calorimeter loses heat to the surroundings primarily through radiation due to its surface exposure.
291 Medium
291 of 514
Application +4 | -1
In the spectrum of a back body at two temperature T{T} and 2T2 {T}, let A1{A}_1 and A2{A}_2 be area under the two curves respectively. The value of A1A2\frac{A_1}{A_2} will be
View answer and explanation
Correct option: A
1:161: 16
Option A (1:161: 16) is correct because the area under the black body spectrum curve is proportional to the fourth power of the temperature, according to Stefan-Boltzmann law.
292 Medium
292 of 514
Analyzing +4 | -1
(A): It is hotter over the top of a fire than at the same distance on the sides.
(R): Air surrounding the fire conducts more heat upwards.
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 it is hotter over the top of a fire due to convection, not conduction.
293 Medium
293 of 514
Understanding +4 | -1
According to Ingen-Hauz's experiment the thermal conductivity KK and length LL of the metal rod upto which the wax melts are related as
View answer and explanation
Correct option: C
KL2=\frac{K}{L^2}= constant
Option C (KL2=\frac{K}{L^2}= constant) is correct because, according to Ingen-Hauz's experiment, the thermal conductivity is proportional to the square of the length up to which the wax melts.
294 Medium
294 of 514
Understanding +4 | -1
Colour of shinning bright star is an indication of its
View answer and explanation
Correct option: C
Temperature
Option C (Temperature) is correct because the color of a star is directly related to its surface temperature, as described by Wien's displacement law.
295 Medium
295 of 514
Understanding +4 | -1
A heated piece of green glass seen in darkness will
View answer and explanation
Correct option: D
Not be visible
Option D is correct. A heated piece of green glass in darkness will not be visible because it will primarily emit infrared radiation, which is invisible to the human eye.
296 Medium
296 of 514
Application +4 | -1
In Mks system, stefan’s consant is denoted by σ\sigma. In cGs system multiplying factory of σ\sigma will be
View answer and explanation
Correct option: B
103{10^3}
Option B (103{10^3}) is correct because it represents the conversion factor from the MKS unit of Stefan's constant to the CGS unit.
297 Medium
297 of 514
Application +4 | -1
The temperature gradient in the earth’s crust is 32∘C^ \circ C km−1{\mathrm{m}^{ - 1}} and the mean conductivity of earth is 0.008 cals−1cm−1∘C−1{\mathrm{s}^{ - 1}}\mathrm{c}{\mathrm{m}^{ - 1}}^ \circ {C^{ - 1}}. Considering earth to be a sphere of radius 6000 km loss of heat by earth everyday is about
View answer and explanation
Correct option: D
1018{10^{18}} cal
Option D (1018{10^{18}} cal) is correct because it is calculated using the formula for heat loss considering the temperature gradient, conductivity, and earth's surface area.
298 Medium
298 of 514
Understanding +4 | -1
(A): A body that is a good radiator is also a good absorber of radiation at a given wavelength.
(R): According Kirchoff's law the absorptivity of a body is equal to its emissivity at a given wavelength.
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 Kirchhoff's law states that a good radiator is also a good absorber at a given wavelength.
299 Medium
299 of 514
Remembering +4 | -1
A perfect black body is one whose emissive power is
View answer and explanation
Correct option: C
Unity
Option C (Unity) is correct because a perfect black body absorbs all incident radiation, and its emissivity is defined as 1.
300 Medium
300 of 514
Remembering +4 | -1
The surface temperature of the stars is determined using
View answer and explanation
Correct option: B
Wien’s displacement law
Option B (Wien’s displacement law) is correct because it relates the peak wavelength of emitted radiation to the temperature of a black body, like a star.
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