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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies `E_(1)` and `E_(2)` respectively. ThenA. `E_(1) = E_(2)`B. `E_(1) gt E_(2)`C. `E_(1) lt E_(2)`D. `E_(1)` and `E_(2)` cannot be compared |
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Answer» Correct Answer - a |
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| 52. |
An engine takes heat from a reservior and converts its `1//6` part into work. By decreasing temperature of sink by `62^(@)C`, its efficiency becomes double. The temperatures of source and sink must beA. `90^(@)C, 37^(@)C`B. `99^(@)C, 37^(@)C`C. `372^(@)C, 37^(@)C`D. `206^(@), 37^(@)C` |
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Answer» Correct Answer - b |
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| 53. |
The molar specific heat at constant pressure of an ideal gas is `(7//2 R)`. The ratio of specific heat at constant pressure to that at constant volume isA. `7//5`B. `8//7`C. `5//7`D. `9//7` |
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Answer» Correct Answer - a |
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| 54. |
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in (figure). The change in internal energy of the gas during the transition is `(gamma=3//5)` A. `20 kJ`B. `- 20 kJ`C. `20 J`D. `- 12J` |
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Answer» Correct Answer - B |
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| 55. |
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will beA. `pV = ((5)/(32)) RT`B. `pV = 5RT`C. `pV = ((5)/(32)) RT`D. `pV = ((5)/(16)) RT` |
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Answer» Correct Answer - a |
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| 56. |
2kg of ice at `20^[email protected]` is mixed with 5kg of water at `20^[email protected]` in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are `1kcal//kg//^@C 0.5` `kcal//kg//^@C` while the latent heat of fusion of ice is `80kcal//kg`A. `7kg`B. `5kg`C. `4kg`D. `2kg` |
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Answer» Correct Answer - B (b) Heat required to convert 5 kg of water at `20^@C` to 5kg of water at `0^@C` `=mC_(omega)DeltaT=5xx1xx20=100kcal` Heat released by 2kg. Ice at `-20^@C` to convert into 2kg of ice at `0^@C` `=mC_(ice)DeltaT=2xx0.5xx20=20kcal.` How much ice at `0^@C` will convert into water at `0^@C` for giving another 80 kcal of heat `Q=mL rArr 80=mxx80` `rArr m=1kg` Therefore the amount of water at `0^@C` `=5kg+1kg=6kg` Thus, at equilibrium, we have, [6 kg water at `0^@C`+1kg ice at `0^@C`]. |
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| 57. |
300 grams of water at `25^@ C` is added to 100 grams of ice at `0^@ C.` The final temperature of the mixture is …..^@C |
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Answer» Correct Answer - C The heat required for 100g of ice at `0^@C` to change into Water at `0^@C=mL=100xx80xx4.2=33,600J…(i)` The heat released by 300g of water at `25^@C` to change its temperature to `0^@C=mcDeltaT=300xx4.2xx25=31,500J….(ii)` Since the energy in eq.(ii) is less than of eq(i) therefore the final temperature will be `0^@C` |
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| 58. |
2kg of ice at `20^[email protected]` is mixed with 5kg of water at `20^[email protected]` in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are `1kcal//kg//^@C 0.5` `kcal//kg//^@C` while the latent heat of fusion of ice is `80kcal//kg`A. `7kg`B. `5kg`C. `4kg`D. `2kg` |
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Answer» Correct Answer - B (b) Heat required to convert 5 kg of water at `20^@C` to 5kg of water at `0^@C` `=mC_(omega)DeltaT=5xx1xx20=100kcal` Heat released by 2kg. Ice at `-20^@C` to convert into 2kg of ice at `0^@C` `=mC_(ice)DeltaT=2xx0.5xx20=20kcal.` How much ice at `0^@C` will convert into water at `0^@C` for giving another 80 kcal of heat `Q=mL rArr 80=mxx80` `rArr m=1kg` Therefore the amount of water at `0^@C` `=5kg+1kg=6kg` Thus, at equilibrium, we have, [6 kg water at `0^@C`+1kg ice at `0^@C`]. |
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| 59. |
if 1 g of system is mixed with 1 g of ice, then the resultant temperature of the mixture isA. `270^(@)C`B. `230^(@)C`C. `100^(@)C`D. `50^(@)C` |
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Answer» Correct Answer - c |
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| 60. |
If mass-energy equilivalence is taken into account, when water is cooled to from ice, the mass of water shouldA. increaseB. remains unchangedC. decreaseD. first increase then decrease |
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Answer» Correct Answer - C When water is cooled to form ice, energy is released from water in the form of heat. As energy is equilivalent to mass therefore when water is cooled to ice, its mass decreases. |
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| 61. |
At what temperature is the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at `24^@C`?A. `80K`B. `-73K`C. `3K`D. `20K` |
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Answer» Correct Answer - D (d) `v_(rms)=sqrt((8RT)/(piM))` For `v_(rms) to be equal `(T_(H_2))/(M_(H_2))=(T_(O_2))/(M_(O_2))` Here, `M_(H_2)=2, M_(O_2)=32,` `T_(O_2)=47+273=320K` `:. (T_(H_2))/2=320/32 rArr T_(H_2)=20K` |
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| 62. |
Even Carnot engine cannot give `100%` efficiency because we cannotA. prevent radiationB. find ideal sourcesC. reach absolute zero temperatureD. eliminate friction. |
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Answer» Correct Answer - C (c) `eta=1-(T_2)/(T_1)` For `eta=1 or 100%, T_2=0K` The temperature of 0K (absolute zero) can not be obtained. |
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| 63. |
The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding PV diagram. A. B. C. D. |
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Answer» Correct Answer - B (b) If we study the P-T graph we find AB to be a isothermal process, AC is adiabatic process given. Also for an expansion process, the slope of adiabatic curve is more (or we can say that the area under the P-V graph for isothermal process is more than adiabatic process for same increase in volume). only graph (b) fits the above criteria. |
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| 64. |
Cooking gas container are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside willA. increaseB. decreaseC. remain sameD. decrease for some, while increase for others |
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Answer» Correct Answer - A (c) Since pressure and volume are not changing, so temperature remains same. |
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| 65. |
A refrigerator works between `4^(@)C` and `30^(@)C`. It is required to remove `600 calories` of heat every second in order to keep the temperature of the refrigerator space constant.The power required is (Take `1calorie= 4.2 J`)A. 23.65 WB. 236.5 WC. 2365 WD. 2.365 W |
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Answer» Correct Answer - B |
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| 66. |
A water cooler of storages capacity 120 liters can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3kW of heat (thermal load). The temperature of water fed into the device cannot exceed `30^@C` and the entire stored 120 liters of water is initially cooled to `10^@C.` The entire system is thermally insulated. The minimum value of P ( in watts) for which the device can be operated for 3hours is (Specific heat of water is `4.2kJkg^-1K^-1` and the density of water is `1000kgm^-3`)A. `1600`B. `2067`C. `2533`D. `3933` |
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Answer» Correct Answer - B (b) `P_(heater)-P_(cooler)=(mcDeltaT)/t=(VrhocDeltaT)/t` `:. (3000-P_(cooler))=(0.12xx1000xx4.2xx10^3xx20)/(3xx60xx60)` `:. P_(cooler)=2067W` |
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| 67. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at ita bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then beA. `P_0`B. `(P_0)/2`C. `(P_0)/2+(Mg)/(piR^2)`D. `(P_0)/2-(Mg)/(piR^2)` |
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Answer» Correct Answer - A (a) When the piston is pulled out slowly, the pressure drop produced inside the cylinder is almost instantaneously neutralised by the air entering from outside into the cylinder (through the small hole at the top). Therefore , the pressure inside the cylinder is `P_0` throughout the slow pulling process. |
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| 68. |
Parallel rays of light of intensity `I=912 WM^-2` are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan-Boltzmann constant `sigma=5.7xx10^-8` `Wm^-2K^-4` and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toA. `330K`B. `660K`C. `990K`D. `1550K` |
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Answer» Correct Answer - A (a) In steady state `Energy lost =Energy gained` `sigma(T^4-T_0^4)xx4piR^2=I(piR^2)` `:. 5.7xx10^-8[T^4-(300)^4]xx4=912` `:. T=330K` |
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| 69. |
A black body of temperature T is inside chamber of `T_0` temperature initially. Sun rays are allowed to fall from a hole in the top of chamber. If the temperature of black body (T) and chamber `(T_0)` remains constant, then A. Black body will absorb more radiationB. Black body will absorb less radiationC. Black body emit more energyD. Black body emit energy equal to enrgy absorbed by it |
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Answer» Correct Answer - A::C::D (a,c,d) Since sun rays fall on the black body, it will absorb more radiation and since, its temperature will emit more radiation. The temperature will remain same only when energy emitted is equal to energy absorbed. |
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| 70. |
The volume V versus temperature T graphs for a cetain amount of a perfect gas at two pressure `p_1 and p_2` are as shown in Fig. It follows from the graphs that `p_1` is greater than `p_2.` |
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Answer» For a particle temperature T,`Vprop1/p` Volume is greater for pressure `P_1` `:. P_1ltP_2` |
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| 71. |
An insulated container containing monoatomic gas of molar mass s is moving with a velocity `v_0`. If the container is suddenly stopped, find the change in temperature. |
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Answer» Correct Answer - B::C When container is stopped, velocity decrease by `v_0.` `Therefore change in kinetic energy=1/2(nm)v_0^2…..(i)` Here n=number of moles of gas present in the container. The kinetic energy at a given temperature for a monatomic gas is `=3/2xxnRT` `:. Change in kinetic energy =3/2xxnR(DeltaT).....(ii)` where `DeltaT=Change in temperature` From (i) and (ii) `3/2nR(DeltaT)=1/2(nm)v_0^2 :. DeltaT=(mv_0^2)/(3R)` |
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| 72. |
Consider a compound slab consisting of two different material having equal thickness and thermal conductivities `K` and `2K` respectively. The equivalent thermal conductivity of the slab isA. `3 K`B. `(4)/(3) K`C. `(2)/(3) K`D. `sqrt2 K` |
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Answer» Correct Answer - b |
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| 73. |
We consider the radition emitted by the human body which of the following statements is true?A. The radiation is emitted during the summers and absorbed during the wintersB. The radiation emitted lies in the ultraviolet region and hence is not visibleC. The radiation emitted is in the infrared ragionD. The ratiation is emitted only during the day |
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Answer» Correct Answer - c |
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| 74. |
An ideal gas heat engine operates in Carnot cycle between `227^@C` and `127^@C`. It absorbs `6.0 xx 10^4 cal` of heat at high temperature. Amount of heat converted to work is :A. `2.4 xx 10^(4)` calB. `6 xx 10^(4)` calC. `1.2 xx 10^(4)` calD. `4.8 xx 10^(4)` cal |
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Answer» Correct Answer - c |
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| 75. |
An ideal gas heat engine operates in a Carnot cycle between `227^(@)C and 127^(@)C`. It absorbs `6K cal.` of heat at higher temperature. The amount of heat in `k cal` rejected to sink isA. 1.6B. 1.2C. 4.8D. 3.5 |
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Answer» Correct Answer - b |
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| 76. |
In the figure, a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulated material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700K and the upper compartment is filled with 2 moles of an ideal diatmoic gas at 400K. The heat capacities per mole of an ideal monatomic gas are `C_V=3/2R, C_P=5/2R,` and those for an ideal diatomic gas are `C_V=5/2R, C_P=7/2R,` Now consider the partition to be free to move without friction so that the pressure of gasses in both compartments is the same. The total work done by the gases till the time they achives equilibrium will beA. `-200K`B. `200R`C. `100R`D. `-100R` |
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Answer» Correct Answer - D (d) In this case both the gases are at constant pressure `:. nC_(p_1)(700-T)=nC_(p_2)(T-400)` `5/2R(700-T)=7/2R(T-400)` `3500-5T=7T-2800` `rArr 12T=6300` `:. T=525K` Applying first law of thermodynamics `DeltaW_1+DeltaU_1=DeltaQ_1` and `DeltaW_2+DeltaU_2=DeltaQ_2` As the gas two system in thermally insulated, therefore `DeltaQ_1+DeltaQ_2=0` `-(DeltaW_1+DeltaW_2)=DeltaU_1+DeltaU_2` `=nC_(v_1)(525-700)+n_2C_(v_2)(525-400)` `=-2xx(3R)/2xx175+2xx(5R)/2xx125` `=-525R+625R=-100R` Therefore, total work done=-100R |
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| 77. |
0.05 kg steam at 373K and 0.45kg of ice at 253K are mixed in an insultated vessel. Find the equilibrium temperature of the mixture. Given, `L_(fusion)=80cal//g=336J//g, L_(vaporization)=540 cal//g=2268J//g,C_(ice)=2100J//KgK=0.5calg//gK and S_(water)=4200J//KgK=1cal//gK` |
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Answer» Correct Answer - B::C (1) Heat lost by steam at `100^@C` to change to `100^@C` water `mL_(vap)=0.05xx2268xx1000=1,13400J` (2) Heat lost by `100^@C` water to change to `0^@C` water `=0.05xx4200xx100=21,000J` (3) Heat required by o.45kg of ice to change its temperature from 253K to 273K `=mxxS_(ice)xxDeltaT=0.45xx2100xx20=18,900J (4) Heat required by `0.45kg` of ice at 273K to convert into `0.45kg` water at 273K `=mL_(fusion)=0.45xx336xx1000=151,200J` From the above data it is clear that the amount of heat required by 0.45kg of ice at 253K to convert into 0.45kg of water at `273K(1,70,100J)` cannot be provided by heat lost by 0.05 kg of steam at 373K to convert into water at 273K. Therefore the final temperature will be 273K or `0^@C.` |
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| 78. |
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process untill its volume is again reduced to half. ThenA. compressing the gas through adiabatic process will require more work to be done.B. compressing the gas isothermally or adiabatically will require the same amount of work.C. which of the case (whether compression through isothermal or through adiabatic process) reequires more work will depend upon the atomicity of the gas.D. compressing the gas isothermally will require more work to be done |
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Answer» Correct Answer - A |
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| 79. |
Two vessel separately contains two ideal gases A and B at the same temperature, the pressure of A being twice that of B. under such conditions, the density of A is found to be 1.5 times the density of B. the ratio of molecular weight of A and B isA. `(2)/(3)`B. `(3)/(4)`C. 2D. `(1)/(2)` |
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Answer» Correct Answer - B |
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| 80. |
A black body is at `727^(@)C`. It emits energy at a rate which is proportional toA. `(727)^(2)`B. `(1000)^(4)`C. `(1000)^(2)`D. `(727)^(4)` |
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Answer» Correct Answer - b |
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| 81. |
The temperature of the two outer surfaces of a composite slab, consisting of two materails having coefficients of two materails having coefficients of termal conductivity K and 2K and thickness x and 4x, respectively, are `T_2 and T_1(T_1gtT_1)`. The rate of heat transfer through the slab, in a steady state is `((A(T_2-T_1)K)/2)f`, with f equal to A. `2/3`B. `1/2`C. `1`D. `1/3` |
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Answer» Correct Answer - D (d) The thermal resistance `x/(KA)+(4x)/(2KA)=(3x)/(KA)` `:. (dQ)/(dt)=(DeltaT)/((3x)/(KA))=((T_2-T_1)KA)/(3x)=1/3{(A(T_2-T_1)K)/x}` `:. f=1/3` |
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| 82. |
If the radius of a star is `R` and it acts as a black body, what would b the temperature of the star, in which the rate of energy production is `Q`?A. `Q//4pi R^(2) sigma`B. `(Q//4 pi R^(2) sigma)^(-1//2)`C. `(4pi R^(2)Q//sigma)^(1//4)`D. `(Q// 4pi R^(2) sigma)^(1//4)` |
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Answer» Correct Answer - d |
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| 83. |
A thermodynamic process is shown in the figure. The pressure and volumes corresponding to some points in the figure are A. `p_(A) = 3 xx 10^(4)` pa,B. `V_(A) = 2 xx 10^(-3) m^(3)`C. `P_(B) = 8 xx 10^(4) pa`,D. `V_(B) = 5 xx 10^(-3) m^(3)` |
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Answer» Correct Answer - a |
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| 84. |
In thermodynamic processes which of the following statement is not true?A. In an adiabatic process the system is insulated from the surroundingB. In an isochoric process remains constantC. In an isothermal process the temperature remains constantD. In an adiabatic process `pV^(gamma)` = constant |
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Answer» Correct Answer - b |
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| 85. |
Relation between pressure (p) and energy (E) of a gas isA. `p = (2)/(3) E`B. `p = (1)/(3) E`C. `p = (3)/(2) E`D. `p = 3 E` |
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Answer» Correct Answer - a |
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| 86. |
In the given (V-T) diagram, what is the relation between pressure `P_(1) and P_(2)` ? A. `p_(2) = p_(1)`B. `p_(2) gt p_(1)`C. `p_(2) lt p_(1)`D. Cannot be predicated |
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Answer» Correct Answer - c |
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| 87. |
A thermodynamic system is taken through the cycle `ABCD` as shown in the figure. Heat rejected by the gas during the cycle is A. 2 pVB. 4 pVC. `(1)/(2) pV`D. `pV` |
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Answer» Correct Answer - a |
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| 88. |
The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from `T_(1)` K to `T_(2)` K is :A. `(3)/(8) N_(a)K_(B) (T_(2) - T_(1))`B. `(3)/(2) N_(a) K_(B) (T_(2) - T_(1))`C. `(3)/(4) N_(a) K_(B) (T_(2) - T_(1))`D. `(3)/(4) N_(a)K_(B) ((T_(2))/(T_(1)))` |
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Answer» Correct Answer - a |
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| 89. |
The ratio of the velocity of sound in Hydrogen gas `(gamma=7/5)` to that in Helium gas `(gamma=5/3)` at the same temperature is sqrt(21/3). |
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Answer» `((C_(H_2))_1)/((C_(H_e))_2)=sqrt((gamma_1RT)/M_1)/sqrt((gamma_2RT)/M_2)=sqrt((gamma_1)/(gamma_2)xx(M_2)/(M_1))` `=sqrt((7//5)/(5//3)xx4/2)=sqrt(7/5xx3/5xx2)=sqrt(42/25)` |
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| 90. |
If liquefied oxygen at 1 atmospheric pressure is heated from 50K to 300k by supplying heat at constant rate. The graph of temperature vs time will beA. B. C. D. |
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Answer» Correct Answer - C (c) `Q=mcDeltaT` `rArrQ=mc(T-t_0)…..(i)` `:.` From 50K to boiling temperature, T increases linearly. During boiling, equation is `Q=mL` Temperature remains constant till boiling is complete After that, again eqn. (i) is followed and temperature increases linearly. |
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| 91. |
An ideal gas is expanding such that `PT^@=constant.` The coefficient of colume expansion of the gas is-A. `1//T`B. `1//T`C. `3//T`D. `4//T` |
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Answer» Correct Answer - C (C) `PT^2=constant (given)` Also for an ideal gas `(PV)/T=constt` From the above two equations, after eliminatig P. `V/T^3=constt rArr V=kT^3 where k=constant` `rArr(dV)/V=3(dT)/T` `rArrdv=(3/T)VdT……(i)` We know that change in volume due to thermal expansion is given by `dV=VgammadT......(ii)` Where gamma=coefficient of volume expansion. From (i) and (ii) `VgammadT=(3/T)VdT rArr gamma=3/T` |
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| 92. |
The work of 142kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3Jmol^-1K^-1)`A. diatomicB. triatomicC. a mixture of monoatomic and diatomicD. monoatomic |
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Answer» Correct Answer - A (a) `W=(nRDeltaT)/(1-gamma) rArr -146000=(1000xx8.3xx7)/(1-gamma)` or `1-gamma=-58.1/146 rArr gamma=1+58.1/146=1.4` Hence the gas is diatomic. |
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| 93. |
Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. ThenA. no molecules can have a speed greater than `sqrt2v_(rms)`B. no molecule can have a speed less than `v_p//sqrt2`C. `v_p lt barv lt v_(rms)`D. the average kinetic energy of a molecules is `3/4mv_p^2`. |
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Answer» Correct Answer - C::D (c,d) We know that `barv=sqrt((8RT)/(piM)),v_(rms)=sqrt((3RT)/M) and `v_p=sqrt((2RT)/M)` From these expressions, we can conclude that `v_pltbarvltv_(rms)` Also the average kinetic energy of gaseous molecule is `barE=1/2mv_(rms)^2=1/2m(3/2v_p^2)=3/4mv_p^2` |
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| 94. |
The above p-v diagram represents the thermodymic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle isA. `p_0v_0`B. `(13/2)p_0v_0`C. `(11/2)p_0v_0`D. `4p_0v_0` |
| Answer» Correct Answer - B | |
| 95. |
One mole of an ideal monatomic gas undergoes a process described by the equation `PV^(3)`= constant. The heat capacity of the gas during this process isA. `(3)/(2)R`B. `(5)/(2) R`C. `2R`D. `R` |
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Answer» Correct Answer - D |
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| 96. |
70 calories of heat required to raise the temperature of 2 moles of an ideal gas at constant pressure from `30^@Cto 35^@C.` The amount of heat required (in calories) to raise the temperature of the same gas through the same range `(30^@C to 35^@C)` at constant volume is:A. `30`B. `50`C. `70`D. `90` |
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Answer» Correct Answer - B (b) `(Q_2)/(Q_1)=(nC_vDeltaT)/(nC_pDeltaT)=1/gamma rArr Q_2=(Q_1)/gamma` |
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| 97. |
One mole of an ideal gas at an initial temperature true of `TK` does `6R` joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is `5//3`, the final temperature of the gas will beA. `(T + 2.4)K`B. `(T - 2.4)K`C. `(T + 4) K`D. `(T - 4) K` |
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Answer» Correct Answer - d |
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