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JEE • PHYSICS • THERMAL PROPERTIES OF MATTER

Thermal properties of matter practice · Mixed difficulty

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Showing 241-270 of 553 questions

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241 Medium
241 of 553
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.
242 Medium
242 of 553
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}.
243 Medium
243 of 553
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.
244 Medium
244 of 553
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.
245 Medium
245 of 553
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.
246 Medium
246 of 553
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.
247 Medium
247 of 553
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.
248 Medium
248 of 553
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.
249 Medium
249 of 553
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.
250 Medium
250 of 553
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.
251 Medium
251 of 553
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.
252 Medium
252 of 553
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.
253 Medium
253 of 553
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.
254 Medium
254 of 553
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.
255 Easy
255 of 553
Application +4 | -1
What is the percentage change in length of 1m iron rod if its temperature changes by 100∘C {100^ \circ }C\, [α  for iron = 2×10−5 / ∘C]\left[ {\alpha \,\,\mathrm{for}\,\mathrm{iron}\, = \,2 \times {{10}^{ - 5}}\,/{\,^ \circ }C} \right]
View answer and explanation
Correct option: A
0.002
Option A (0.20%) is correct because the percentage change in length is calculated using the formula αΔt×100\alpha \Delta t \times 100.
256 Medium
256 of 553
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.
257 Medium
257 of 553
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.
258 Medium
258 of 553
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.
259 Easy
259 of 553
Application +4 | -1
Calculate radiation power for sphere whose temperature is 227∘C227^{\circ} \mathrm{C} and radius 0.20.2 and emissivity 0.8.0.8.
View answer and explanation
Correct option: A
1425 W1425 \mathrm{~W}
Option A (1425 W1425 \mathrm{~W}) is correct because it is calculated using the Stefan-Boltzmann law: P=σAeT4P = \sigma A e T^4.
260 Medium
260 of 553
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.
261 Medium
261 of 553
Application +4 | -1
A body cools down from 50∘C50^{\circ} \mathrm{C} to 45∘C45^{\circ} \mathrm{C} in 5 min5 \mathrm{~min} and 45∘C45^{\circ} \mathrm{C} to 40∘C40^{\circ} \mathrm{C} is 8 min8 \mathrm{~min}. Find the temperature of the surrounding. Assume Newton's law of cooling is valid.
View answer and explanation
Correct option: A
34∘C34^{\circ} \mathrm{C}
Option A (34∘C34^{\circ} \mathrm{C}) is correct because it is the solution to the system of equations derived from Newton's Law of Cooling.
262 Medium
262 of 553
Analyzing +4 | -1
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100oC{100^\mathrm{o}}C, while the other one is at 0oC{\mathrm{0}^\mathrm{o}}C, If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is
View answer and explanation
Correct option: C
more than 50oC{50^\mathrm{o}}C
Option C (more than 50oC{50^\mathrm{o}}C) is correct because the body at 100oC{100^\mathrm{o}}C has a greater heat capacity, leading to a final temperature closer to 100oC{100^\mathrm{o}}C.
263 Medium
263 of 553
Application +4 | -1
If the temperature of radiating body increased by 50%50 \%. The percentage increase in radiation energy is closest to
View answer and explanation
Correct option: D
400%400 \%
Option D (400400 \\%) is correct because the radiation energy is proportional to T4T^4, and a 50% increase in temperature leads to approximately a 400% increase in radiation energy.
264 Medium
264 of 553
Understanding +4 | -1
Statement-I: Invar steel is used to prepare clock pendulum.
Statement-II: The coefficient of linear expansion of invar steel is infinity.
View answer and explanation
Correct option: B
Statement-I is true and statement-II is false
Option B is correct because Invar steel is used for clock pendulums due to its very low coefficient of linear expansion, not infinity.
265 Medium
265 of 553
Understanding +4 | -1
The coefficient of cubical expansion of water is negative between
View answer and explanation
Correct option: A
273 K and 277 K273 \mathrm{~K} \text{ and }277 \mathrm{~K}
Option A (273 K and 277 K273 \mathrm{~K} \text{ and }277 \mathrm{~K}) is correct because water has a negative coefficient of cubical expansion in this temperature range.
266 Medium
266 of 553
Application +4 | -1
A pendulum clock having copper rod keeps correct time at 20∘C20^{\circ} \mathrm{C}. It gains 15 seconds per day if cooled to 0∘C0^{\circ} \mathrm{C}. The coefficient of linear expansion of copper is
View answer and explanation
Correct option: B
1.7×10−5∘C−11.7 \times 10^{-5}{}^{\circ} \mathrm{C}^{-1}
Option B (1.7×10−5∘C−11.7 \times 10^{-5}{}^{\circ} \mathrm{C}^{-1}) is correct because the coefficient of linear expansion is calculated based on the time gained by the pendulum clock.
267 Medium
267 of 553
Understanding +4 | -1
A solid ball of metal has a concentric spherical cavity within it. If the ball is heated, the volume of the cavity will
View answer and explanation
Correct option: B
Decrease
Option A (Increase) is correct because when the ball is heated, both the outer and inner surfaces expand, leading to an increase in the cavity's volume.
268 Medium
268 of 553
Understanding +4 | -1
The coefficient of volume expansion of an ideal gas at constant pressure and at temperature TT. K is equal to
View answer and explanation
Correct option: D
1T\frac{1}{T}
Option D (1T\frac{1}{T}) is correct because the coefficient of volume expansion for an ideal gas at constant pressure is the inverse of the temperature.
269 Medium
269 of 553
Application +4 | -1
An iron sphere has a radius of 10 cm10 \mathrm{~cm} at a temperature of 0∘C0^{\circ} \mathrm{C}. Calculate the change in the volume of the sphere, if it is heated to 100∘C100^{\circ} \mathrm{C}. Coefficient of linear expansion of iron =11×10−6∘C−1=11 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}.
View answer and explanation
Correct option: A
13.8 cm313.8 \mathrm{~cm}^3
Option A (13.8 cm313.8 \mathrm{~cm}^3) is correct because it is the change in volume calculated using the formula for volumetric expansion with the given linear expansion coefficient and temperature change.
270 Medium
270 of 553
Application +4 | -1
A faulty thermometer reads the melting point and boiling point of water as −1∘C-1^{\circ} \mathrm{C} and 104∘C104^{\circ} \mathrm{C} respectively. The temperature of the object is measured as 40∘C40^{\circ} \mathrm{C} using that thermometer. The true temperature of the object is:
View answer and explanation
Correct option: B
39∘C39^{\circ} \mathrm{C}
Option B (39∘C39^{\circ} \mathrm{C}) is correct because it is the true temperature calculated using the faulty thermometer readings and the given formula.
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