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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.
| 551. |
PV versus T graph of equal masses of `H_(2)`, He and `CO_(2)` is shown in figure. Choose the correct alternative? A. 3 corresponds to `H_(2),2` to He and 1 to `CO_(2)`B. 1 corresponds to He, 2 to `H_(2)` and 3 to `CO_(2)`C. 1 corresponds to He, 3 to `H_(2)` and 2 to `CO_(2)`D. 1 corresponds to `CO_(2),2` to `H_(2)` and 3 to He |
| Answer» Correct Answer - A | |
| 552. |
For an adiabatic process graph between PV & V for a sample of ideal gas will beA. B. C. D. |
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Answer» Correct Answer - B PV a T for adiabatic process, `TV^(gamma-1)` = constant |
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| 553. |
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 - A |
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| 554. |
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 |
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| 555. |
Anideal gas is taken around `ABCA` as shown in the above `P-V` diagram. The work done during a cycle is A. `2 PV`B. `PV`C. `1//2 PV`D. Zero |
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Answer» Correct Answer - A a. Work done = area enclosed by triangle `ABC` `(1)/(2) AC xx BC = (1)/(2) xx (3V - V) xx (3P - P) = 2 PV` |
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| 556. |
A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will beA. v/2B. `Vsqrt(2)`C. 2VD. 4V |
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Answer» Correct Answer - D |
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| 557. |
A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion isA. QB. `(3)/(5) Q`C. `(2)/(5) Q`D. `(2)/(3) Q` |
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Answer» Correct Answer - C For process at constant pressure. `Q=nC_(P)DeltaT=5/2 nR DeltaT` and `W=P DeltaV=nRDeltaT=2/5 Q` |
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| 558. |
A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion isA. `Q`B. `(3)/(5) Q`C. `(2)/(5) Q`D. `(2)/(3) Q` |
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Answer» Correct Answer - C For process at constant pressure. `Q = nC_(P) Delta T = 5/2 nRDelta T and `W = P Delta V = nR Delta T = 2/5 Q` . |
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| 559. |
A close container of volume `0.02 m^(3)` contains a mixture of neon and another gas A of unknown molecular mass. The container contains 4 g of Neon and 24 g of gas A. At a temperature of `27^(@)C` the pressure of the mixture is `105 N//m^(2)`. Is there any possibility of finding gas A in the atmosphere of a planet of radius 600 km and mean density `r = 5 x× 103 kg//m^(3)` ? Temperature of the planet is `2200^(@)C`. Molar molecular mass of neon = 20 g, Gas constant `R = 8.314 "J mol"^(–1) K^(–1)` Gravitational constant `G = 6.67 x× 10^(–11) N m^(2) kg^(–2)` |
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Answer» Correct Answer - No |
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| 560. |
Which of the following statements is correct for any thermodynamic system?A. The internal energy changes in all processesB. internal energy and entropy are state functionsC. The change in entropy can never be zroD. The work done in an adiabatic process is always zero. |
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Answer» Correct Answer - B The internal energy of entropy depend only on the initial and final states of the system and not on the path followed to attain that state. |
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| 561. |
In the previous question let `V_(0)` be the volume of each container. All other details remain the same. The number of moles of gas in the container at temperature `2 T_(0)` will beA. `(p_(0)V_(0))/(2 RT_(0))`B. `(p_(0)V_(0))/(RT_(0))`C. `(2p_(0)V_(0))/(3 RT_(0))`D. `(p_(0)V_(0))/(3 RT_(0))` |
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Answer» Correct Answer - D `(pV_(0))/(R.2T_(0)) = (4p_(0)V_(0))/(3R.2T_(0)) = (2p_(0)V_(0))/(2RT_(0))` |
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| 562. |
Suppose a container is evacuated to leave just one molecule of a gas in it. Let v_(a) and v_(rms) represents the average speedand the rms speed of the gas.A. (a)`v_(a)gtv_(rms)`B. (b)`v_(a)ltv_(rms)`C. (c)`v_(a)=v_(rms)`D. (d)`v_(rms)` is underfined. |
| Answer» Correct Answer - C | |
| 563. |
The ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobatic process isA. `5/7`B. `3/5`C. `2/7`D. `5/3` |
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Answer» Correct Answer - C `DeltaU = nC_(v)DeltaT = n(5//2)RDeltaT` `DeltaQ = nC_(p)DeltaT = n(7//2)RDeltaT` `W = Delta Q -DeltaU = (n7)/(2) R Delta T -(n5)/(2) Rdelta T = nRDelta T` `W/(DeltaU) = 2/7`. |
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| 564. |
The ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobatic process is |
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Answer» Correct Answer - W/Q=2/7 `triangleU = nC_(1) triangleT= n(5)/(2)R triangle T` `triangleQ = nC_p triangle T = n(7)/(2) R triangleT` `W = triangleQ - triangleU = (n7)/(2) RtriangleT = nR triangle T` `(W)/(Q) = (2)/(7)` |
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| 565. |
Efficiency of a Carnot engine is `50%` when temperature of outlet is `500 K`. In order to increase efficiency up to `60%` keeping temperature of intake the same what is temperature of outlet?A. 200KB. 400KC. 600KD. 800K |
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Answer» Correct Answer - B `eta_(1)=1-(T_(2))/(T_(1)), eta_(2)=10(T_(2)^(1))/(T_(1)` |
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| 566. |
Two vessels have equal volums. One of them contains hydrogen at one atmosphere and the other helium at two atmosphere. If both the samples are at the same temperature, the `rms` velocity of the hydrogen molecules is |
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Answer» According to kinetic theory of gases, `v_(rms)=sqrt((3RT)/(M)), ((v_(rms))_(H))/((v_(rm s))_(He))=sqrt((M_(He))/(M_(H)))` i.e. `v_(H)=sqrt((4//2))v_(He=sqrt(2)(V_He))` |
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| 567. |
Two vessels have equal volums. One of them contains hydrogen at one atmosphere and the other helium at two atmosphere. If both the samples are at the same temperature, the `rms` velocity of the hydrogen molecules isA. equal to that of the helium moleculesB. twice that of the helium moleculesC. half that of the helium moleculesD. `sqrt(2)` times that of the helium molecules |
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Answer» Correct Answer - D `(v_(rms))_(H_(2))/(v_(rms))_(He) = sqrt((M_(He))/((M_(0))_(H_(2)))) = sqrt((4)/(2)) = sqrt(2)` |
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| 568. |
A cornot engine has the same efficiency between (i) 100 K and 500 K and (ii) T and 900 K. Find T. |
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Answer» Correct Answer - `ETA=T/900` `eta_(1) = 1 -(T_1)/(T_2) = 1-(1000)/(500) =(4)/(5)` `eta_2 = 1-(T)/(900) rArr (T)/(900) = (1)/(5) rArr T=(1)/(5) = 180 K` |
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| 569. |
One mole of an ideal gas is contained with in a cylinder by a frictionless piston and is initially at temperature T. The pressure of the gas is kept constant while it is heated and its volume doubles. If R is molar gas constant, the work done by the gas in increasing its volume isA. RT ln2B. 1/2 RTC. RTD. 3/2 RT |
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Answer» Correct Answer - C `W = PV = RT` |
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| 570. |
A U shaped tube has two arms of equal cross section and lengths `l_(1) = 80` cm and `l_(2) = 40` cm. The open ends are sealed with air in the tube at a pressure of 80 cm of mercury. Some mercury is now introduced in the tube through a stopcock connected at the bottom (the air is not allowed to leak out). In steady condition the length of mercury column in the shorter arm was found to be 10 cm. Find the length of the mercury column in the longer arm. Neglect the volume of the part of the tube connecting two arms and assume that the temperature is constant. |
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Answer» Correct Answer - 16.27 cm |
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| 571. |
find the number of molecules of an ideal gas in a volume of`1.000cm^(3)`atSTP. |
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Answer» We know, ` n=(PV)/(RT) =(1 xx 1 xx 10^-3)/(0.082 xx 273)` ` = (10^-3)/(22.4) =(1)/(22400)` No. of molecules `=0.023 xx 10^23 xx 1/22400` `=2.688 xx 10^-4 xx 10^23` ` =2.688 xx 10^19` |
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| 572. |
The height of Niagra falls is 50 m. Calculate the difference in temperature of water at the top and at the bottom of fall, if `J = 4.2 J cal^(-1)`. |
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Answer» P.E. is converted into heat, mgh=`JmSDeltat` `Deltat=(gh)/(JS)=(980xx5000)(4.2xx10^(7))=0.117^(@)C`. |
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| 573. |
One mole of an ideal gas at temperature `T_` expands slowly according to the law `p/V=` constant. Its final temperature is `T_2`. The work done by the gas isA. `R(T_(2) - T_(1))`B. `2R(T_(2) - T_(1))`C. `R/2(T_(2) - T_(1))`D. `2R/3(T_(2) - T_(1))` |
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Answer» Correct Answer - C `PV^(-1)=C` Comparing `PV^(N) = C` `N+-1` `W=(nR(T_1-T_2))/(N-1) = (1xxR(T_1-T_2))/(-2)` `W = (R(T_2-T_1))/(2)` . |
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| 574. |
If the amount of heat given to a system is 40 J and the amount of work done on the system is 15 J then the change in internal energy isA. 50 JB. `-50J`C. `30J`D. `55J` |
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Answer» Correct Answer - D `dQ=du+domega` `du=dQ-(domega)` `=40+15=55J`. |
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| 575. |
If the amount of heat given to a system is 50 J and work done on the system is 15 J, then change in internal energy of the system will beA. 35 JB. 50 JC. 65 JD. 15 J |
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Answer» Correct Answer - C `dQ=50J` `d u=dQ-dw` `=50-(-15)=50+15=65J` |
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| 576. |
When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphre, the volume increases from 1cc to 1671cc. The heat of vaporisation at his pressure is 540 cal/gm. Increase in internal energy of water is (1 atmosphre =1.01x `10^(6)` dyne/`cm^(2)`)A. 4200JB. 8200JC. 1200JD. 2100J |
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Answer» Correct Answer - D `dQ=mL,dW=P(V_(2)-V_(1)),dU=dQ-dW` |
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| 577. |
The time taken by an electric heater to rise the temperature of 100cc of water through `10^(@)C` is 7s. If there is no loss in energy. Power of that motor is (J=4.2J/cal)A. 420WB. 42WC. 4.2WD. 0.6KW |
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Answer» Correct Answer - D `P=(W)/(t)=(JmSDeltatheta)/(t)` |
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| 578. |
A steel drill is making 180 revolution per minute, under a constant forque of 5N-m. If it drills a hole in 7sec in a steel block of mass 600gm, rise in temperature of the block is `(S=0.1 gm^(-1).^(@)C^(-1))`A. `2.6^(@)C`B. `1.3^(@)C`C. `5.2^(@)C`D. `3^(@)C` |
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Answer» Correct Answer - A `tau omegat=JmSDeltatheta(omega=2pin)` |
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| 579. |
A lead ball moving with a velocity v strikes a wall and stops. If 50% of its energy is onverted into heat. The increase in temperature is (Specific heat of lead is S)A. `2v^(2)//JS`B. `v^(2)//4JS`C. `v^(2)S//J`D. `v^(2)S//2J` |
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Answer» Correct Answer - B `(1)/(2)((1)/(2)mv^(2))=JmSDeltatheta` |
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| 580. |
Which type of motion of the molecules is responsible for internal energy of a monoatomic gas?A. translationalB. rotationalC. vibrationalD. Isothermal |
| Answer» Correct Answer - A | |
| 581. |
How much will the temperature of 100g of water be rised by doing 4200 J of work in stirring the water?A. `0.01^(@)C`B. `0.1^(@)C`C. `1^(@)C`D. `10^(@)C` |
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Answer» Correct Answer - D `W=JHrArrW=JmSDeltatheta` |
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| 582. |
The internal energy of a perfect monoatomic gas isA. complete kineticB. complete potentialC. sum of potential and kinetic energy of the moleculesD. difference of kinetic and potential energies of the molecules. |
| Answer» Correct Answer - A | |
| 583. |
An ice block is projected vertically up with a velocity 20 `ms^(-1)`. The amount of ice that melt when it reaches the ground and if the loss of P.E. is converted into heat energy if the mass of ice block is 4.2 kgA. 2.5gmB. 2.5kgC. 0.25kgD. 0.25gm |
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Answer» Correct Answer - A Maximum height attained, `h=(u^(2))/(2g)` `W=JHrArrmgh=J(xL_("ice"))` |
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| 584. |
Which of the following is constant in an isochoric process?A. PressureB. VolumeC. TemperatureD. Mass |
| Answer» Correct Answer - B | |
| 585. |
In the free expansion of a gas, its internal energyA. remains constantB. increasesC. Decreases both for ice and waxD. sometimes increases, sometimes decreases |
| Answer» Correct Answer - A | |
| 586. |
Consider the melting of 1g of ice at `0^(@)C` to was at `0^(@)C` at atmospheric pressure. Then the change in internal energy of the system (density of ice is 920kg/`m^(3)`)?A. 334JB. 420JC. 540JD. 680J |
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Answer» Correct Answer - A Heat required to change the phase of a solid `DeltaQ=mL_(f)` Work done by system at constant pressure, `DeltaW=P(V_("liquied")-V_("solid"))` From first law of thermodynamics, `DeltaU=DeltaQ-DeltaW` `=mL_(f)-P(V_("liquid")-V_("solid"))` Latent heat of fusion of water, `L_(f)=3.335xx10^(5) J//Kg` `DeltaQ=(1xx10^(-3))(3.335xx10^(5))=334J` The density of ice is `920 kg//m^(3)`. `V_("solid")=(1xx10^(-3))/(920)=1.09xx10^(-6)m^(3)` `V_("liquid")=1xx10^(-6)m^(3)` `=1.013xx10^(5))(1xx10^(-6)-1.09xx10^(-6))` `=-9xx10^(-3)J` Work done by the system is negative because the system is negative because the system decreased in volume. `DeltaU=DeltaQ-DeltaW=334-(-9xx10^(-3))` `=334J` |
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| 587. |
How does the internal energy change when the ice and wax melt at their normal melting points?A. Increases for ice and decreases for waxB. Decreases for ice and increases for waxC. Decreases both for ice and waxD. Increases both for ice and wax |
| Answer» Correct Answer - A | |
| 588. |
The average speed of air molecules is `485 "ms"^(-1)` . At STP the number density is `2.7xx10^(25)m^(-3)` and diameter of the air molecule is `2xx10^(-10)` m . The value of mean free path for the air molecule isA. `2.5xx10^(-7)m`B. `2.9xx10^(-7)m`C. `3.5xx10^(-7)m`D. `3.9xx10^(-7)m` |
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Answer» Correct Answer - B Given , number density of the air molecule, `n=2.7xx10^(25)//m^(3)` diameter of the air molecule, `d=2xx10^(-10)m` mean free path , `lambda` = ? we know that , mean free path, `lambda=(1)/(sqrt2npid^(2))=(1)/(sqrt2pixx2.7xx10^(25)(2xx10^(-10))^(2))=2.9xx10^(-7)m` |
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| 589. |
The adiabatic Bulk modulus of a diatomic gas at atmosheric pressure isA. `0 Nm^-2`B. `1 Nm^-2`C. `1.4 xx10^4 Nm^-2`D. `1.4 xx 10^5 Nm^-2` |
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Answer» Correct Answer - D adiabatic bulk modulus = rp |
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| 590. |
N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the velocity of the molecules of B type is denoted by `v^(2)` and the mean square of the x-component of the velocity of a tye is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`A. 2B. 1C. `(1)/(3)`D. `(2)/(3)` |
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Answer» Correct Answer - D Mean kinetic energy of two types of molecules should be equal. so, `(1)/(2)m(3omega^(2))=(1)/(2)(2m)v^(2)rArr(omega^(2))/(v^(2))=(2)/(3)` |
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| 591. |
If temperature of black body increases from `300K` to `900K`, then the rate of energy radiation increases byA. 81B. 3C. 9D. 2 |
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Answer» Correct Answer - A We can write , `((T_(2))/T_(1))^(4)=(E_(2))/(E_(1))` `(E_(2))/(E_(1))=((900)/(300))^(4)rArr(E_(2))/(E_(1))=(3)^(4)` , `E_(2)=81E_(1)rArr(E_(2))/(E_(1))=81` |
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| 592. |
A large cylindrical tower is kept vertical with its ends closed. An ideal gas having molar mass M fills the tower. Assume that temperature is constant throughout, acceleration due to gravity (g) is constant throughout and the pressure at the top of the tower is zero. (a) Calculate the fraction of total weight of the gas inside the tower that lies above certain height h. (b) At what height `h_(0)` the quantity of gas above and below is same. How does the value of h0 change with temperature of the gas in the tower? |
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Answer» Correct Answer - `(a) e^((-Mgh)/(RT)) " " (b) ((RT)/(Mg))ln2` (c ) `h_(0)` increase with T. |
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| 593. |
A physics book was found on a newly discovered island in `18^(th)` century. A problem in the book was as follows. “1 pinch of an ideal gas is kept in a container of volume 1.5 volka. When the temperature is 40 tapu, the gas pressure is 25 phatka. When the temperatue is reduced to – 20 tapu the gas pressure becomes 10 phatka. Find the temperature of absolute zero in tapu. |
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Answer» Correct Answer - `-60` tapu |
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| 594. |
The efficiency of the reversible heat engine is `eta_(r)` and that of irreversible heat engine is `eta_(l)`. Which of the following relations is correct?A. `eta_(1) gt eta_(2)`B. `eta_(1) lt eta_(2)`C. `eta_(1)=eta_(2)`D. `eta_(1) le eta_(2)` |
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Answer» Correct Answer - A In a irreversible heat engine a part of energy is lost due to friction. |
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| 595. |
The efficiency of the reversible heat engine is `eta_(r)` and that of irreversible heat engine is `eta_(l)`. Which of the following relations is correct?A. `eta_(r)gteta_(1)`B. `eta_(r)ltn_(1)`C. `eta_(r)geeta_(1)`D. `eta_(r)gt1` and `eta_(1)ltl` |
| Answer» Correct Answer - C | |
| 596. |
Two spherical vessel of equal volume are connected by a n arrow tube. The apparatus contains an ideal gas at one atmosphere and `300 K`. Now if one vessel is immersed in a bath of constant temperature `600 K` and the other in a bath of constant temperature `300 K`. then the common pressure will be A. 1 atmB. `(4)/(5)` atmC. `(4)/(3)` atmD. `(3)/(4)` atm |
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Answer» Correct Answer - C |
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| 597. |
Water is used in car radiators as coolant becauseA. its density is moreB. high specific heatC. high thermal conductivityD. free availability |
| Answer» Correct Answer - B | |
| 598. |
In the following pressure-volume diagram, the isochoric, isothermal and isobaric parts, respectively, are A. `BA, AD, DC, CB`B. `DC, CB, BA, AD`C. `AB, BC, CD, DA`D. `CD, DA, AB, BC` |
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Answer» Correct Answer - D Process `AB` is isobaric, process `CD` is isochoric . The slope of BC gt slope of `AD` , therefore `BC` is adiabatic and `AD` is isothermal. |
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| 599. |
`2` moles of a monoatomic gas are expanded to double its initial volume, through a process `(P)//(V) =` constant. If its initial temperature is `300 K`, then which of the following is not true.A. `triangleT = 900 K`B. `triangleQ = 3200R`C. `triangleQ = 3600 R`D. `W = 900 R` |
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Answer» Correct Answer - B `(P)/(V) constant = K` PV = nRT `rArr KV^2 = nRT` `KV^2 = nRT` `rArr (V_(1)^2)/(V_(2)^2) = (T_1)/(T_2)` `V_1= V,V_2 = 2V,T_1 =300k` `triangleW = intpdv` `=overset(2v)underset(v)(int)kvdv =(3kv^2)/(2)` `triangleQ = triangleu + trianglew` |
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| 600. |
Suppose a container is evacuated to leave just one molecule of a gas in it. Let `v_(a)` and `v_(rms)` represents the average speed and the rms speed of the gas.A. `v_(a) gt v_(rms)`B. `v_(a) lt v_(rms)`C. `v_(a) = v_(rms)`D. `v_(rms)` is undefined |
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Answer» Correct Answer - C Since there is only one molecule `bar(v) = v_(rms)` |
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