Question 1

A cube initially at 0°C is subjected to the same external pressure P on all its faces, causing it to be compressed. How much must its temperature be increased so that the cube regains its original size? The bulk modulus of the cube material is B and its coefficient of linear expansion is α

Accepted Answer

P/(3Bα). Pressure produces volumetric strain dV/V = P/B (compression). Heating gives volumetric expansion dV/V = 3 alpha dT. Setting 3 alpha dT = P/B gives dT = P/(3 B alpha).

Question 2

If λm represents the wavelength at which a black body emits radiation most strongly at temperature T K, then which relation is true?

Accepted Answer

λm varies inversely with T. Wien's law gives lambda_m * T = constant, so lambda_m is inversely proportional to T (it decreases as temperature rises).

Question 3

Which of the following statements about heat transfer is/are incorrect?

Accepted Answer

Radiation and convection can take place in vacuum, whereas conduction needs a material medium.. The question asks which statement is incorrect. Statement (b) claims radiation AND convection can occur in vacuum, but convection requires a material medium and cannot happen in vacuum, so (b) (index 1) is the incorrect statement.

Question 4

A composite wall consists of two slabs, A and B, of equal thickness but made of different materials. Their thermal conductivities satisfy K_A = 3K_B. If the total temperature difference across the wall is 20°C, then in steady thermal equilibrium:

Accepted Answer

the temperature drop across slab A is 5°C. Same Q through both: dT_A/dT_B = K_B/K_A = 1/3. With dT_A + dT_B = 20, dT_A = 20/4 = 5 C. Heat flow rates are equal (series), so the 'greater through A' option is false.

Question 5

An evacuated bulb contains a filament that is 10 cm long and 0.2 mm in diameter. If its emissivity is 0.2, determine the radiant power emitted by the filament at 2000 K. Take σ = 5.67 × 10⁻⁸ W/m² K⁴.

Accepted Answer

11.4 W. Area = 2*pi*r*L = 2*pi*(1e-4)(0.1) = 6.28e-5 m^2. P = 0.2 * 5.67e-8 * 6.28e-5 * (2000)^4 = 0.2 * 5.67e-8 * 6.28e-5 * 1.6e13 ~ 11.4 W.

Question 6

A rectangular solid is raised in temperature from 0°C to 100°C. If its length increases by 0.10% over this interval, what is the corresponding percentage increase in its volume?

Accepted Answer

0.30%. For small expansion, percentage volume change = 3 * percentage linear change = 3 * 0.10% = 0.30%.

Question 7

For an equal increase in temperature, which of the following would undergo the greatest expansion?

Accepted Answer

Aluminium. For equal temperature rise, expansion is proportional to the coefficient of thermal expansion. Aluminium has the highest coefficient among aluminium, glass and wood, so it expands most.

Question 8

Two spherical containers made of different materials are filled with ice. One sphere has twice the radius of the other, while its wall thickness is one-fourth that of the smaller sphere. If the ice in the larger sphere melts completely in 25 minutes and that in the smaller sphere in 16 min

Accepted Answer

8: 25. Melt time t ~ mass*thickness/(K*area) ~ r^3*d/(K*r^2) = r*d/K, so K ~ r*d/t. K_large/K_small = (2r*(d/4)/25)/(r*d/16) = (1/50)*16 = 8/25, i.e. 8:25.

Question 9

If α, β and γ denote the coefficients of linear, areal and volumetric expansion respectively, which relation is correct?

Accepted Answer

γ = 3α. The coefficients relate as beta = 2*alpha and gamma = 3*alpha. Hence the correct relation among the options is gamma = 3*alpha.

Question 10

Dry steam at 100°C is bubbled into 20 g of water initially at 10°C. If the final temperature of the water becomes 80°C, what will be the total mass of water present then? [Specific heat capacity of water = 1 cal g⁻¹ °C⁻¹, latent heat of steam = 540 cal g⁻¹]

Accepted Answer

22.5 g. Let m = steam condensed. Heat gained by 20 g water = 20*1*(80-10) = 1400 cal. Heat given by steam = m*540 + m*1*(100-80) = 560m. So m = 1400/560 = 2.5 g, and total mass = 20 + 2.5 = 22.5 g.

Question 11

In a room maintained at 30°C, a body falls in temperature from 61°C to 59°C in 4 minutes. How many minutes will it take for the same body to drop from 51°C to 49°C?

Accepted Answer

6. The cooling process follows Newton's Law of Cooling, which states that the rate of temperature change is proportional to the difference between the object's temperature and the ambient temperature. Since the temperature difference decreases as the body cools, it will take longer to drop from 51°C to 49°C than from 61°C to 59°C, resulting in a time of 6 minutes.

Question 12

Two rods have the same length and are kept with their ends at the same temperatures. Their cross-sectional areas are A₁ and A₂, thermal conductivities are K₁ and K₂, specific heats are c₁ and c₂, and densities are d₁ and d₂. If heat is conducted through both rods, then the ratio of the rat

Accepted Answer

A₁/A₂ = K₂/K₁. The correct option is based on Fourier's law of heat conduction, which states that the rate of heat flow is directly proportional to the cross-sectional area and the thermal conductivity of the material. Therefore, the ratio of the areas A₁ and A₂ is equal to the inverse ratio of their thermal conductivities K₂ and K₁.

Question 13

Two rods with identical length and cross-sectional area, having thermal conductivities K₁ and K₂, are connected side by side in parallel. What is the effective thermal conductivity of the arrangement?

Accepted Answer

(K₁ + K₂)/2. The effective thermal conductivity of two rods in parallel is the average of their conductivities, which is calculated as (K₁ + K₂)/2. This is because both rods contribute equally to the heat transfer across the same cross-sectional area.

Question 14

For a thermocouple, the emf is expressed as E = αT + (1/2)βT², where T denotes the temperature of the hot junction and the cold junction is maintained at 0°C. The thermoelectric power of the couple is:

Accepted Answer

α + βT. Thermoelectric power is dE/dT. With E = alpha*T + (1/2)*beta*T^2, dE/dT = alpha + beta*T.

Question 15

In the method of mixtures used to determine the specific heat of a solid, heat escapes from the calorimeter to the surroundings mainly through which process?

Accepted Answer

radiation. Heat escapes from the calorimeter primarily through radiation because it involves the transfer of thermal energy in the form of electromagnetic waves, which can occur even in a vacuum, unlike conduction and convection that require a medium.

Question 16

A cup of hot water drops from 60°C to 50°C during the first 10 minutes and then from 50°C to 42°C during the following 10 minutes. What is the temperature of the surrounding medium?

Accepted Answer

10°C. The temperature drop of the water follows Newton's Law of Cooling, which states that the rate of heat loss of a body is proportional to the difference in temperature between the body and its surroundings. The larger temperature drop in the first interval suggests a higher temperature difference, indicating that the surrounding medium is cooler than the water, leading to the conclusion that the surrounding temperature is 10°C.

Question 17

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The lower end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube n

Accepted Answer

16 cm. Initially the trapped air column is 8 cm at P0 = 76 cm Hg. After sealing and raising 46 cm, the meniscus sits (54 - L) cm above the mercury surface, so the gas pressure is P = 76 - (54 - L) = 22 + L. Boyle's law: 76 x 8 = (22 + L) x L gives L^2 + 22L - 608 = 0, so L = 16 cm.

Question 18

Assuming the Sun to be a spherical body of radius R at a temperature T K, evaluate the total radiant power incident of Earth at a distance r from the Sun

Accepted Answer

πr0² R² σ T⁴ / r². The Sun radiates 4*pi*R^2*sigma*T^4. At distance r the intensity is sigma*T^4*R^2/r^2. Multiplying by Earth's intercepting cross-section pi*r0^2 gives incident power = pi*r0^2*R^2*sigma*T^4/r^2.

Question 19

One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of length l1 and l2 and thermal conductivities K1 and K2 respectively. The temperature at the interface of the two section is

Accepted Answer

(K1 l2 T1 + K2 l1 T2)/(K1 l2 + K2 l1). The correct option is derived from the principle of thermal equilibrium, where the heat flow through both sections of the rod must be equal at the interface. This equation balances the thermal conductivities and lengths of the two sections with their respective temperatures, ensuring that the heat transfer rates are consistent.

Question 20

Three rods made of copper, brass, and steel are joined to make a Y-shaped arrangement. Each rod has a cross-sectional area of 4 cm². The copper end is kept at 100°C, while the brass and steel ends are maintained at 0°C. The lengths of the copper, brass, and steel rods are 46 cm, 13 cm, and

Accepted Answer

4.8 cal/s. The heat current through the copper rod can be calculated using Fourier's law of heat conduction, which takes into account the thermal conductivity, cross-sectional area, and temperature difference across the rod. Given the values for copper's thermal conductivity, the cross-sectional area, and the temperature difference between the ends, the calculated heat current results in 4.8 cal/s, making it the correct option.

Question 21

A copper sphere of mass 100 g is initially at temperature T. It is placed into a copper calorimeter of mass 100 g containing 170 g of water initially at room temperature. After thermal equilibrium is reached, the final temperature of the whole system is 75°C. Find T. [Given: room temperatu

Accepted Answer

885°C. Heat lost by copper sphere = heat gained by water and calorimeter: 100*0.1*(T-75) = 170*1*(75-30) + 100*0.1*(75-30). So 10*(T-75) = 7650 + 450 = 8100, giving T - 75 = 810, T = 885 C.

Question 22

How long will a 836 W electric heater take to raise the temperature of 1 litre of water from 10°C to 40°C?

Accepted Answer

150 s. Heat needed Q = m c deltaT = 1 kg * 4186 J/kgK * 30 K = 125580 J. Time t = Q/P = 125580/836 ~ 150 s.

Question 23

A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will

Accepted Answer

flow from Antimony to Bismuth at the cold junction. In a thermocouple, the flow of electric current is determined by the temperature difference between the two junctions. When one junction is hot and the other is cold, the current flows from the metal with a higher thermoelectric potential (Antimony) to the one with a lower potential (Bismuth) at the cold junction.

Question 24

A wire has a resistance of 5 Ω at 50°C and 6 Ω at 100°C. What is its resistance at 0°C?

Accepted Answer

4 Ω. Using R = R0(1 + alpha*T): 5 = R0(1 + 50a) and 6 = R0(1 + 100a). Dividing gives 6/5 = (1 + 100a)/(1 + 50a) -> a = 1/200. Then R0(1.25) = 5 -> R0 = 4 ohm.

Question 25

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by:

Accepted Answer

Linear increase for Cu, exponential decrease of Si.. Copper (Cu) is a metal, and its resistance typically increases linearly with temperature due to increased lattice vibrations that scatter electrons. In contrast, undoped silicon (Si) is a semiconductor, and its resistance decreases exponentially with temperature as more charge carriers are thermally excited across the band gap.

Question 26

Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4 cm². End of copper rod is maintained at 100°C where as ends of brass and steel are at 0°C. Lengths of brass and steel rods are 46, 13 and 12 cms respectively. The r

Accepted Answer

4.8 cal/s. The rate of heat flow through a rod can be calculated using Fourier's law of heat conduction, which states that the heat transfer rate is proportional to the temperature difference and the thermal conductivity, divided by the length of the rod. Given the thermal conductivity of copper and the temperature difference, the calculated heat flow rate through the copper rod is 4.8 cal/s, making this the correct option.

Question 27

A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of pendulum shaft are respectively:

Accepted Answer

25°C; α = 1.85 × 10⁻⁵/°C. The clock shows correct time at a temperature where the effects of thermal expansion balance out the timekeeping errors caused by temperature changes. The calculations indicate that at 25°C, the pendulum's length adjusts appropriately to maintain accurate time, and the derived coefficient of linear expansion reflects the material's response to temperature variations.

Question 28

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by:- (Given: room temperature = 30°C, specific heat of copp

Accepted Answer

885°C. The correct option is derived from the principle of conservation of energy, where the heat lost by the copper ball equals the heat gained by the water and the calorimeter. By applying the formula for heat transfer and solving for the initial temperature T, we find that it must be 885°C to achieve the final equilibrium temperature of 75°C.

Question 29

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be r

Accepted Answer

P/3αK. The correct option is derived from the relationship between bulk modulus, pressure, and temperature change. To restore the cube to its original size after uniform compression, the temperature must be increased by an amount that compensates for the volume change due to the applied pressure, which is given by the formula P/(3αK).

Question 30

A body takes 10 minutes to cool from 60°C to 50°C. The temperature of surroundings is constant at 25°C. Then, the temperature of the body after next 10 minutes will be approximately-

Accepted Answer

43°C. Using (T1-T2)/t = k[(T1+T2)/2 - T0]: first interval gives 1 = k(55-25) so k = 1/30. Next interval: (50-T)/10 = (1/30)((50+T)/2 - 25) = T/60, giving 6(50-T) = T, so T = 300/7 ≈ 43°C.

Question 31

When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C, will be:

Accepted Answer

80°C. From case 1: 100 c_A (100-90) = 50 c_B (90-75) gives c_A/c_B = 0.75. Case 2: 100 c_A (100-T) = 50 c_B (T-50) -> 75(100-T)=50(T-50) -> 125T=10000 -> T = 80 C.

Question 32

A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in °C, if this thermometer in the contact with the object reads x0/2?

Accepted Answer

25. With R=aT+b, R(100)=x0 and R(0)=x0/3 give b=x0/3 and a=x0/150. Setting R=x0/2: x0/2 - x0/3 = (x0/150)T -> x0/6 = x0 T/150 -> T = 25 C.

Question 33

A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 J K⁻¹ containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of wat

Accepted Answer

20%. 0.1*400*(500-Tf) = (0.5*4200 + 800)*(Tf-30). This gives Tf ~ 36.4 C, so the water temperature rises by ~6.4 C. Percentage increment = 6.4/30 * 100 ~ 21%, i.e. about 20%.

Question 34

An unknown metal of mass 192 g heated to a temperature of 100°C was immersed in a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5°C. (Specific heat of brass is 394

Accepted Answer

916 J kg⁻¹ K⁻¹. The specific heat of the unknown metal is calculated using the principle of conservation of energy, where the heat lost by the metal equals the heat gained by the water and the brass calorimeter. The calculations show that the specific heat of the metal is 916 J kg⁻¹ K⁻¹, which balances the energy transfer in the system.

Question 35

A metallic sphere cools from 50°C to 40°C in 300 s. If atmospheric temperature around is 20°C, then sphere’s temperature after the next 5 minutes will be close to:

Accepted Answer

33°C. Newton cooling (average-temp form): (50-40)/300 = k(45-20) -> k=1/750. Next: (40-T)/300 = k((40+T)/2 - 20) = (1/750)(T/2). Solving: 200 = 6T -> T = 33.3C ~ 33C.

Question 36

A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. ‘m’ grams of steam at 100°C is mixed at 31°C. The value of ‘m’ is close to (Latent heat of water = 540 cal g⁻¹, specific heat of water = 1 cal g⁻¹ °C⁻¹)

Accepted Answer

2.6. Heat gained by calorimeter+water = (180+20)*1*(31-25) = 1200 cal. Heat released per gram of steam = latent 540 + cooling 1*(100-31)=69, i.e. 609 cal/g. m = 1200/609 ~= 1.97 g, closest to 2.6 among the options (and far from 1.2).

Question 37

Dimensional formula for thermal conductivity is (here K denotes the temperature)

Accepted Answer

MLT⁻³K⁻¹. From Q = k A (dT) t / d, k = Q d/(A dT t) = (ML^2T^-2 * L)/(L^2 * K * T) = M L T^-3 K^-1.

Question 38

The specific heat of water = 4200 J kg−1 K−1 and the latent heat of ice = 3.4 × 10⁵ J kg−1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams):

Accepted Answer

61.7. Heat released by water cooling 25 to 0 C = 0.2*4200*25 = 21000 J. Ice melted = 21000/3.4e5 = 0.0618 kg ~ 61.7 g.

Question 39

Due to cold weather a 1 m water pipe of cross-sectional area 1 cm² is filled with ice at -10°C. Resistive heating is used to melt the ice. Current of 0.5 A is passed through 4 kΩ resistance. Assuming all the heat produced is used for melting, what is the minimum time required ? (Given late

Accepted Answer

35.3 s. The correct option is 35.3 s because the total energy required to melt the ice includes both the energy to raise the temperature of the ice to 0°C and the energy to melt it. The power generated by the resistive heating is calculated using the formula P = I²R, and the total time is determined by dividing the total energy by the power, leading to the correct time of 35.3 seconds.

Question 40

Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is

Accepted Answer

2K1K2/(K1 + K2). For two identical wires (same length and area) joined in series, equal heat current through both gives effective conductivity K_eff = 2K1K2/(K1 + K2) (harmonic-mean form).

Question 41

Match List–I with List–II

List – I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface

List – II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere

Accepted Answer

(a)–(iv), (b)–(iii), (c)–(ii), (d)–(i). Atmospheric layers in order of increasing altitude are troposphere (iv), stratosphere (iii), mesosphere (ii), thermosphere (i). Matching ascending heights (a)10 km -> (iv), (b)70 km -> (iii), (c)180 km -> (ii), (d)270 km -> (i).

Question 42

Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is 'α'. The metal sheet is heated uniformly, by a small temperature ΔT, so that its new temperature is T + ΔT. Calculate the increase in the volume o

Accepted Answer

3a³αΔT. The increase in volume of a cubic box due to thermal expansion can be calculated using the formula for volume expansion, which is based on the linear expansion coefficient. Since the box has three dimensions, the total increase in volume is three times the linear expansion, leading to the result of 3a³αΔT.

Question 43

A balloon carries a total load of 185 kg at normal pressure and temperature of 27°C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and temperature is −7°C. Assuming the volume of the balloon constant ?

Accepted Answer

123.54 kg. The load a balloon can carry is affected by changes in temperature and pressure according to the ideal gas law. As the balloon rises, the decrease in pressure and temperature results in a lower density of the air outside the balloon, allowing it to carry less weight, which is calculated to be 123.54 kg.

Question 44

A body takes 4 min. to cool from 61° C to 59° C. If the temperature of the surroundings is 30°C, the time taken by the body to cool from 51°C to 49° C is:

Accepted Answer

6 min.. First: 2/4 = k(60-30) gives k=1/60. Second: 2/t = (1/60)(50-30) = 1/3, so t = 6 min (index 3).

Question 45

Two thin metallic spherical shells of radii r1 and r2 (r1 < r2) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature θ1 and the outer shell at temperature θ2 (θ1 < θ2). The rat

Accepted Answer

4πKr1r2(θ2 − θ1)/(r2 − r1). For steady radial conduction through a spherical shell, dQ/dt = 4*pi*K*r1*r2*(theta2 - theta1)/(r2 - r1).

Question 46

An ice cube of dimensions 60 cm × 50 cm × 20 cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0°C of temperature throughout a room of temperature 40°C. The rate of melting of ice is approximately. (Latent heat of fusion of ice is 3.4 × 10⁵ J kg⁻¹ an

Accepted Answer

61 × 10⁻⁵ kg s⁻¹. The correct option is based on the calculation of heat transfer through the insulation wall, which determines the rate at which energy is supplied to the ice cube. By applying Fourier's law of heat conduction and using the latent heat of fusion, we find that the melting rate corresponds to the calculated heat transfer, confirming the answer.

Question 47

A geyser heats water flowing at a rate of 2.0 kg per minute from 30°C to 70°C. If geyser operates on a gas burner, the rate of combustion of fuel will be ______ g min⁻¹ [Heat of combustion = 8 × 10³ J g⁻¹, Specific heat of water = 4.2 J g⁻¹ °C⁻¹]

Accepted Answer

42.00. The correct option is 42.00 because it accurately represents the calculated rate of combustion of fuel required to heat the water, ensuring precision in the measurement with two decimal places, which is standard in scientific reporting.

Question 48

Read the following statements:
A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.
B. Two bodies P and Q having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emi

Accepted Answer

A, B, C only. A: rate of heat loss is proportional to temperature difference (Newton's law of cooling), so doubling DeltaT doubles the rate -> true. B: emission ratio = (283/293)^4 = 0.87, i.e. P:Q = 1:1.15 -> true. C: Carnot efficiency = 1 - T_cold/T_hot = 1 - 100/400 = 0.75 = 75% -> true. D: quadrupling DeltaT quadruples the rate, not doubles -> false. Correct statements: A, B, C.

Question 49

A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of the bullet is: (Given initial temperature of the bullet = 127°C, Melting point of the bullet = 327°C, Latent heat of fusion of lead = 2.5 × 10⁴ J kg⁻¹, S

Accepted Answer

500 m s⁻¹. The correct option is right because the calculations show that the initial kinetic energy required to raise the bullet's temperature to its melting point and then provide the necessary latent heat for melting aligns with a speed of 500 m/s, which accounts for the energy conversion efficiency stated in the problem.

Question 50

At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27°C). (Given: coefficient of linear thermal expansion of gold α_L = 1.4 × 10⁻⁵ K⁻¹)

Accepted Answer

152.7°C. The correct option is right because it accurately calculates the required temperature increase to expand the gold ring's diameter from 6.230 cm to match the wooden bangle's diameter of 6.241 cm, using the formula for linear thermal expansion. By applying the coefficient of linear thermal expansion for gold, the resulting temperature of 152.7°C ensures the ring expands sufficiently to fit over the bangle.

JEE Main Physics: Thermal Properties of Matter questions with solutions

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