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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.
| 1. |
The efficiency of a Carnot engine operating between temperatures of `100^(@)C` and `-23^(@)C` will beA. `(100 - 23)/(273)`B. `(100 + 23)/(373)`C. `(100 + 23)/(100)`D. `(100 - 23)/(100)` |
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Answer» Correct Answer - b |
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| 2. |
Mercury thermometers can be used to measure temperatures uptoA. `260^(@)C`B. `100^(@)C`C. `360^(@)C`D. `500^(@)C` |
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Answer» Correct Answer - c |
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| 3. |
A black body has maximum wavelength `lambda_(m)` at temperature `2000 K`. Its corresponding wavelength at temperature 3000 will beA. `(2)/(3) lambda`B. `(16)/(81) lambda`C. `(81)/(16) lambda`D. `(4)/(3) lambda` |
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Answer» Correct Answer - a |
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| 4. |
Which of the following is close to an ideal black body ?A. Black lampB. Cavity maintained at constant tempratureC. Platinum blackD. A lamp of charcoal heated to high temperature |
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Answer» Correct Answer - b |
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| 5. |
Condider two rods of same length and different specific heats `(s_(1), s_(2))`, thermal conductivities `(K_(1), K_(2))` and areas of cross-section `(A_(1), A_(2))` and both having temperatures `(T_(1), T_(2))` at their ends. If their rate of loss of heat due to conduction are equal, thenA. `K_(1)A_(1) = K_(2)A_(2)`B. `(K_(1)A_(1))/(s_(1)) = (K_(2)A_(2))/(s_(2))`C. `K_(2)A_(1) = K_(1)A_(2)`D. `(K_(2)A_(1))/(s_(2)) = (K_(1)A_(2))/(s_(1))` |
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Answer» Correct Answer - a |
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| 6. |
The efficiency of carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of the sink will be : -A. `600 K`B. `500 K`C. `400 K`D. `100 K` |
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Answer» Correct Answer - c |
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| 7. |
One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is [Given the gas constant R = 8.3 J/ mol. K]A. 198.7 JB. 29 JC. 215.3 JD. 124 J |
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Answer» Correct Answer - d |
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| 8. |
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio `C_P//C_V` for the gas isA. `4/3`B. `2`C. `5/3`D. `3/2` |
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Answer» Correct Answer - D (d) `p prop T^3 rArr PT^-3=constant…(i)` But for an adiabatic process, the pressure temperature relationship is given by `P^(gamma-1)T^(gamma)=constant rArr PT^(gamma/(1-gamma))=constt….(ii)` From (i) and (ii) `gamma/(1-gamma)=-3 rArr gamma=-3+3gamma rArr gamma=3/2` |
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| 9. |
If the temperature of the sun (black body) is doubled, the rate of energy received on earth will be increase by a factor ofA. 2B. 4C. 8D. 16 |
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Answer» Correct Answer - d |
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| 10. |
A gaseous mixture consists of 16g of helium and 16 g of oxygen. The ratio `(C_p)/(C_v)` of the mixture isA. `1.62`B. `1.59`C. `1.54`D. `1.4` |
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Answer» Correct Answer - A (a) `(n_1+n_2)/(r-1)=(n_1)/(r_1-1)+(n_2)/(r_2-1)` `(16/4+16/32)/(r-1)=(16//4)/(5/3-1)+(16//32)/(1.4-1)` `:. gamma=1.62` |
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| 11. |
The temperature of an ideal gas is increased from `27 ^@ C` to `927^(@)C`. The rms speed of its molecules becomes.A. is `sqrt(((927)/(27)))` times the earlier valueB. remains the sameC. gets halvedD. gets doubled |
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Answer» Correct Answer - d |
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| 12. |
110 J of heat is added to a gaseous system, whose internal energy change is 40 j. then the amount of external work done isA. 150 JB. 70 JC. 110 JD. 40 J |
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Answer» Correct Answer - b |
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| 13. |
If `c_(0)` and `c` denote the sound velocity and the rms velocity of the molecules in a gas, thenA. `c_(s) lt c`B. `c_(s) = c`C. `c_(s) = c ((gamma)/(3))^(1//2)`D. None of these |
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Answer» Correct Answer - c |
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| 14. |
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cycle process, thenA. `W = 0`B. `Q = W = 0`C. `E = 0`D. `Q = 0` |
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Answer» Correct Answer - c |
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| 15. |
A gas is taken through the cycle `A rarrB rarr C rarr A`, as shown in figure, what is the net work done by the gas? A. 2000 JB. 1000 JC. ZeroD. `-2000 J` |
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Answer» Correct Answer - b |
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| 16. |
The molar specific heats of an ideal gas at constant pressure and volume are denotes by `C_(P)` and `C_(upsilon)` respectively. If `gamma = (C_(P))/(C_(upsilon))` and `R` is the universal gas constant, then `C_(upsilon)` is equal toA. `(1 + gamma)/(1 - gamma)`B. `(R)/((gamma - 1))`C. `((gamma - 1))/(R)`D. `gamma R` |
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Answer» Correct Answer - c |
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| 17. |
A gas mixture consists of 2 moles of oxygen and 4 of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. 4RTB. 15 RTC. 9RTD. 11RT |
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Answer» Correct Answer - D |
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| 18. |
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by `1^@C` abd it is defined under which of the following conditions?A. From `14.5^@C to 15.5^@C` at 760mm of Hg.B. From `98.5^@C to 99.5^@C` at 760mm of Hg.C. From `13.5^@C to 14.5^@C` at 76mm of Hg.D. From `3.5^@C to 4.5^@C` at 76mm of Hg. |
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Answer» Correct Answer - A (a) 1 Calorie is the amount of heat required to raise temperature of 1gm of water from `14.5^@C to 15.5^@C` at 760mm of Hg. |
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| 19. |
The thermal capacity of 40 g of aluminium (specific heat `=0.2 cal//gm^(@)C`)A. `168 J//.^(@)C`B. `672J//.^(@)C`C. `840 J//.^(@)C`D. `33.6 J//.^(@)C` |
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Answer» Correct Answer - d |
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| 20. |
The internal energy change in a system that has absorbed `2 kcal` of heat and done `500J` of work isA. 8900 JB. 6400 JC. 5400 JD. 7900 J |
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Answer» Correct Answer - d |
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| 21. |
The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, A. the process during the path `AtoB` is isothermalB. work done during the path `BtoCtoD`C. heat flows out of the gas during the path `BtoCtoD`D. the process during the path `AtoB` is isothermal |
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Answer» Correct Answer - B::D (b,d) In case of an isothermal process we get a rectangular hyperbola in a P-V diagram. Therefore option (a) is wrong. `T_DltT_B.` Therefore in process `BtoCtoD, DeltaU` is negative. PV decreases and volume also decrease, therefore W is negative. From first law os thermodynamic, Q is negative i.e. There is a heat loss option (b) is correct. `W_(AB)gtW_(BC).` Therefore work done during path `AtoBtoC` is positive, option (c) is wrong Work done is clockwise cycle in a PV diagram is positive. Option (d) is correct. |
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| 22. |
Two containers A and B are partly filled with water and closed. The volume of A is twice that of B and it contains half the amount of water in B. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio ofA. `1 : 2`B. `1 : 1`C. `2 : 1`D. `4 : 1` |
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Answer» Correct Answer - b |
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| 23. |
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuousely form 0 to 500K at a constant rate. Ignorign any volume change, the following statement (s) is (are) correct to a reasonable approximation. A. The rate at which heat is absorbed in the range 0-100K varies linearly with temperature T.B. Heat absorbed in increasing the temperature from 0-100K is less than the heat required for increasing the temperature from 400-500K.C. There is no change in the rate of heat absorption in the range 400-500K.D. The rate of heat absorption increases in the range 200-300K. |
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Answer» Correct Answer - A::B::C::D (a,b,c,d) We know that `dQ=mCdT` in the range 0 to 100K From the graph, C increases linearly with temperature therefore the rate at which heat is absorbed varies linearly with temperature. Option (a) is correct As the value of C is greater in the temperature range `400- 500K,` the heat absorbed in increasing the temperature from `0-100K` is less than the heat required for increasing the temperature from `400-500K` option (b) is correct. From the graph it is clear that the value of C does not change in the temperature range `400-500K,` therefore there is no change in the rate of heat absorption in this range. Option (c) is correct. As the value of C increasing from `200-300K,` the rate of heat absorption increase in the range `200-300K.` Option (d) is also correct. |
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| 24. |
During an experiment, an ideal gas is found to obey an additional law `VP^2=constant,` The gas is initially at a temprature T, and volume V. When it expands to a volume `2V,` the temperature becomes……. |
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Answer» Correct Answer - B `PV=RT(Ideal gas equation)` `rArrP=(RT)/V….(i)` Given that `VP^2=const …..(ii)` From (i) and (ii) `:. (T^2)/V=const.` `:. (T_1^2)/(V_1)=(T_2^2)/(V_2) rArr T_2=T_1sqrt((V_2)/(V_1))=Tsqrt((2V)/V)=sqrt2T` |
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| 25. |
The variation of temprature of a material as heat is given to it at a constant rate is shown in the figure. The material is in solid state at the point O. The state of the material at the point P is….. |
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Answer» Correct Answer - A::D AB represent a process when physical state change from solid to liquid and temperature remains unchnged. Since P is a point between A and B, therefore the material is partly solid and partly liquid. |
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| 26. |
The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300K isA. `(sqrt(2//7))`B. `(sqrt(1//7))`C. `(sqrt3)//5`D. `(sqrt6)//5` |
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Answer» Correct Answer - C (c)`(V_N)/(V_He)=sqrt(gamma_(N_2)M_(He))/(gamma_(He)M_(N_2))=sqrt((7//5xx4)/(5//3xx28))=sqrt3/5` |
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| 27. |
A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be:A. efficiency of Carnot engine cannot be made larger than `50%`B. `1200K`C. `750K`D. `600K` |
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Answer» Correct Answer - C (c) `0.4=1-(T_2)/500 and 0.6=1-(T_2)/(T_1)` on solving we get `T_2=750K` |
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| 28. |
Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas) A. `15.4%`B. `9.1%`C. `10.5%`D. `12.5%` |
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Answer» Correct Answer - A (a) Heat given to system =`(nC_VDeltaT)_(AtoB)+(bC_PDeltaT)_(BtoC)` `=[3/2(nRDeltaT)]_(AtoB)+[5/2(nRDeltaT)]_(BtoC)` `[3/2xxV_0DeltaP]_(AtoB)+[5/2xx2P_0xxV_0]` `=13/2P_0V_0` and `W_0=P_0V_0` `eta=Work/(heat given)=(P-0V_0)/(13/2P_0V_0)xx100=15.4%` |
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| 29. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `p_(0)v_(0)`B. `2p_(0)v_(0)`C. `(p_(0)v_(0))/(2)`D. zero |
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Answer» Correct Answer - d |
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| 30. |
A Carnot engine, having an efficiency of `eta=1//10` as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature isA. `100J`B. `99J`C. `90J`D. `1J` |
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Answer» Correct Answer - C (c) The efficiency `(eta)` of a Carnot engine and the coefficient of performance `(beta)` of a refrigerator are related as `beta=(1-eta)/eta Here, eta=1/10 :. Beta=(1-1/10)/(1/10)=9`. Also, Coefficient of performance `(beta)` is given by `beta=(Q_2)/W,` where `Q_2` is the energy absorbed from the reservoir. or, `9=(Q_2)/10 :. Q_2=90J` |
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| 31. |
When a system is taken from state i to state f along the path iaf, it is found that `Q=50 cal and W=20 cal`. Along the path ibf `Q=36cal`. W along the path ibf is A. `14 cal`B. `6 cal`C. `16 cal`D. `66 cal` |
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Answer» Correct Answer - B (b) For path iaf, `DeltaU=Q-W=50-20=30cal` For path ibf `W=Q-DeltaU=36-30=6cal`. |
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| 32. |
A thermodynamic system is taken from an initial state I with internal energy `U_i=-100J` to the final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the pat af, ib and bf are `W_(af)=200J, W_(ib)=50J and W_(bf)=100J` respectively. The heat supplied to the system along the path iaf, ib and bf are `Q_(iaf), Q_(ib),Q_(bf)` respectively. If the internal energy of the system in the state b is `U_b=200J and Q_(iaf)=500J`, The ratio `(Q_(bf))/(Q_(ib))` is |
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Answer» Correct Answer - B Applying first law of thermodynamics to path iaf `Q_(iaf)=DeltaU_(iaf)+W_(iaf)` `500=DeltaU_(iaf)+200 :. DeltaU_(iaf)=300J` Now, `Q_(ibf)=DeltaU_(ibf)+W_(ib)+W_(bf)` `=300+50+100` `Q_(ib)+Q_(bf)=450J….(1)` Also `Q_(ib)=DeltaU_(ib)+DeltaW_(ib)` `:. Q_(ib)=100+50=150J....(2)` From (1) & (2) `(Q_(bf))/(Q_(ib))=300/150=2` |
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| 33. |
A diatomic ideal gas is compressed adaibatically to 1/32 of its initial volume. If the initial temperature of the gas is `T_i` (in Kelvin) and the final temperature is a `T_i`, the value of a is |
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Answer» Correct Answer - D For an adiabatic process, the temperature-volume relationship is `T_1V_1^(gamma-1)=T_2V_2^(gamma-1) rArr T_1=T_2((V_2)/(V_1))^(gamma-1)` Here `gamma=1.4 (for diatomic gas). V_2=(V_1)/32, T_1=T_i, T_2=aT_i` `:. T_i=aT_i[1/32]^(1.4-1) =aT_i[1/(2^5)]^0.4=(aT_i)/4 :. a=4` |
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| 34. |
An ideal gas at `27^(@)C` is compressed adiabatically to `8//27` of its original volume. If `gamma = 5//3`, then the rise in temperature isA. `475^(@)C`B. `402^(@)C`C. `275^(@)C`D. `375^(@)C` |
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Answer» Correct Answer - d |
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| 35. |
`300K` a gas `(gamma = 5//3)` is compressed adiabatically so that its pressure becomes `1//8` of the original pressure. The final temperature of the gas is :A. `420 K`B. `300 K`C. `-142^(@)C`D. `327 K` |
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Answer» Correct Answer - c |
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| 36. |
A diatomic gas initially at `18^(@)` is compressed adiabatically to one- eighth of its original volume. The temperature after compression will bA. `18^(@)C`B. `668.4 K`C. `395.4^(@) C`D. `144^(@)C` |
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Answer» Correct Answer - b |
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| 37. |
A jar contains a gas and a few drops of water at absolute temperture `T_(1)`. The pressure in the jar is `830mm` of mercury. The temperature of the jar is reduced by `1%`. The saturation vapour pressures of water at the two temperatures are `30mm` of mercury and `25mm` of mercury. Calculate the new pressure in the jar. |
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Answer» Correct Answer - A `P_1=830-30=800mm Hg, P_2 ?` `V_1=V , V_2=V , T_1=T , T_2=T-0.01T=0.99T` `(P_1V_1)/(T_1)=(P_2V_2)/(T_2) :. P_2=(P_1V_1)/(T_1)=(800xx0.099T)/T=792 mmHg` `:. Total pressure in the jar =792+25=817mm Hg` |
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| 38. |
A given quantity of a ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas isA. `2/3P`B. `P`C. `3/2P`D. `2P` |
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Answer» Correct Answer - B (b) For an isothermal process, `PV=constant` On differentiating, we get , `PdV+VdP=0` `rArrP=(dP)/(dV//V)=K (Bulk modulus)` |
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| 39. |
The curves A and B in the figure shown P-V graphs for an isothermal and an adiabatic process for an idea gas. The isothermal process is represented by the curve A. |
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Answer» The slope of P-V curve is more for adiabatic process than for isothermal process. From the graph it is clear that slope for B is greater than the slope for A. |
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| 40. |
A small spherical monoatomic ideal gas bubble `(gamma=5//3)` is trapped inside a liquid of density `rho` (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is `T_0`, the height of the liquid is H and the atmospheric pressure `P_0` (Neglect surface tension). As the bubble moves upwards, besides the buyoncy force the following forces are acting on itA. Only the force of gravityB. The force due to gravity and the force due to the pressure of the liquidC. The force due to gravity, the force due to the pressure of the liquid and the force due to viscocity of the liquidD. The force due to gravity and the force due to viscocity of the liquid |
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Answer» Correct Answer - D (d) The forces acting besides buoyancy force are (a) Force of gravity (vertically downwards) (b) Viscous force (vertically downwards) Foce due to pressure of the liquid is the buoyant force. |
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| 41. |
A small spherical monoatomic ideal gas bubble `(gamma=5//3)` is trapped inside a liquid of density `rho` (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is `T_0`, the height of the liquid is H and the atmospheric pressure `P_0` (Neglect surface tension). When the gas bubble is at a height y from the bottom, its temperature is-A. `T_0((P_0+rho_lgH)/(P_0+rho_lgy))^(2//5)`B. `T_0((P_0+rho_lg(H-y))/(P_0+rho_lgH))^(2//5)`C. `T_0((P_0+rho_lgH)/(P_0+rho_lgy))^(3//5)`D. `T_0((P_0+rho_lg(H-y))/(P_0+rho_lgH))^(3//5)` |
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Answer» Correct Answer - B (b) It is given that the bubble does not exchange any heat with the liquid. This means that while the bubble moves up and expand, the process is adiabatic. For adiabatic expansion the pressure -temperature relationship is `T_2=T_1[(P_1)/(P_2)]^((1-gamma)/gamma)` Here `T_1=T_0, P-1=P_0+Hrho_lg,` `P_2=P_0+(H-y)rho_lg, gamma=5/3` `:. T_2=T_0[(P_0+Hrho_lg)/(P_0+(H-y)rho_lg)]^(1-5/3//5//3)` `T_0[(P_0+Hrho_lg)/(P_0+(H-y)rho_lg)]^((-2)/3xx3/5)` `T_2=T_0[(P_0+(H-y)rho_lg)/(P_0+Hrho_lg)]^(2/5) |
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| 42. |
"Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement or consequence ofA. second law of thermodynamicsB. conservation of momentumC. conservation of massD. first law of thermodynamics |
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Answer» Correct Answer - A (a) This is a statement of second law of thermodynamics |
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| 43. |
An ice cube of mass 0.1kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat S of the container varies with temperature T according to the empirical relation `S=A+BT`, where `A=100 cal//kg-K and B=2xx10^-2cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion of water =`8xx10^4cal//kg`, Specific heat of water=`10^3cal//kg-K`). |
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Answer» Correct Answer - D Here the equilibrium temperature is `273+27=300K` Also according to the principle of calorimetry Heat lost by container=Heat gained by ice. Heat lost by container: Since specific heat is variable, we need to take the help of calculus to find the heat lost by the container. Let dQ be the heat lost when the temperature of the container is T. `:. dQ=mcdT` Where m is the mass of the container and `C=A+BT` is specific heat at that temperature `:. dQ=m(A+BT)dT` On integrating, we get `Q=int_500^300m(A+BT)dT=m[AT+(BT)^2/2]_500^300` `=-21600m` calorie(heat lost) Heat gained by ice This heat is to be divided into two parts (i) `0^@ iceto0^@ water` `0^@water to 27^@ water` `Q_1=mL Q_2=mcDeltaT` `=0.1xx80,000 =0.1xx10^3xx27` `=8000cal =2700cal` `:. Q_1+Q_2=8000+2700=10,700cal...(i)` Heat lost =heat gained `21600m=10,700` `rArrm=0.495kg` |
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| 44. |
A sample of gas expands from volume `V_(1)` to `V_(2)`. The amount of work done by the gas is greatest when the expansion isA. adiabaticB. isobaricC. isothermalD. Equal in all above cases |
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Answer» Correct Answer - b |
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| 45. |
On observing light from three different stars `P`, `Q` and `R`, it was found that intensity of violet colour is maximum in the spectrum of `P`, the intensity of green colour is maximum in the spectrum of `R` and the intensity of red colour is maximum in the spectrum of `Q`. if `T_(P)`, `T_(Q)` and `T_(R)` are respective absolute temperature of `P`, `Q` and `R`. then it can be concluded from the above observation thatA. `T_(p) gt T_(Q) gt T_(R)`B. `T_(P) gt T_(R) gt T_(Q)`C. `T_(P) lt T_(R) lt T_(Q)`D. `T_(P) lt T_(Q) lt T_(R)` |
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Answer» Correct Answer - B |
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| 46. |
A fiven sample of an ideal gas occupise a volume V at a pressure p and sbsoulte temperature T.The mass of each molecule of the gas is m. Which of the following fives the dinsity of the gas ?A. `p//(kT)`B. `p m//(kT)`C. `p//(kTV)`D. `mkT` |
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Answer» Correct Answer - B |
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| 47. |
One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperature at A,B and C are 400K, 800K and 600K respectively. Choose the correct statement: A. The change in internal energy in whole cyclic process is 250R.B. The change in internal process CA is 700R.C. The change in internal energy in the process AB is -350R.D. The chang in internal energy in the process BC is -500R. |
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Answer» Correct Answer - D (d) In cyclic process, change in total internal energy is zero. `DeltaU_(cyclic)=0` `DeltaU_(BC)=nC_vDeltaT=1xx(5R)/2DeltaT` Where, `C_V` =molar specific heat at constant volume. For BC, `DeltaT=-200K` `:. DeltaU_(BC)=-500R` |
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| 48. |
A system goes from A and B via two processes. I and II as shown in figure. If `DeltaU_1 and DeltaU_2` are the changes in internal enregies in the processes I and II respectively, then A. relation between `DeltaU_1`and `Deltau_2` can not be determinedB. `DeltaU_1=DeltaU_2`C. `DeltaU_2ltDeltaU_1`D. `DeltaU_2gtDeltaU_1` |
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Answer» Correct Answer - B (b) Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e., `DeltaU_1=DeltaU_2` |
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| 49. |
Which of the following is incorrect regarding the first law of thermodynamics?A. It is a restatement of the principle of conservation of energyB. It is not applicable to any cyclic processC. It introduces the concept of the internal energyD. It introduces the concept of the internal energy |
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Answer» Correct Answer - B::C (b,c) First law is applicable to a cyclic process. Concept of entropy is introduced by the second law. |
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| 50. |
During the melting of a slab of ice at 273K at atmospheric pressure,A. positive work is done by the ice-water system on the atmosphere.B. positive work is done on the ice-water system by the atmosphere.C. the internal energy of the ice-water system increases.D. the internal energy of the ice-water system decreases. |
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Answer» Correct Answer - B::C (b,c) There is a decrease in volume during melting of an ice slab at 273K. Therefore, negative work is done by ice-water system on the atmosphere or positive work is done on the ice-water system by the atmosphere . Hence, option (b) is correct. |
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