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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.
| 451. |
A body having `110 and70^(@)C` temperatures of surrounding and of itself repectively . What would be the value of coefficient of performance ?A. 1.75B. 2.3C. `1.50`D. 2.55 |
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Answer» Correct Answer - A Given , `T_(1)=110^(@)C,andT_(2)=70^(@)C,beta` = ? Coefficient of performance `beta=(T_(2))/(T_(1)_T_(2))=(70)/(110-70)=(70)/(40)=1.75` |
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| 452. |
Simple behaviour under all conditions of real gas is governed by the equationA. `pV=muRT`B. `(p+(a)/V^(2))(V-b)=muRT`C. `pV="constant"`D. `pV^(gamma)="constant" ` |
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Answer» Correct Answer - B `(P+(a)/(V^(2)))(V-b)=muRT` governs the simple behaviour pf real gas. |
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| 453. |
Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the air temperture is `22^(@)C` During the day , temperature ries to `42^(@)C` and the tube expands by 2% a The pressure of the air in the tube at this temperature, will be approximatelyA. 212 kPaB. 209 kPaC. 206 kPaD. 200 kPa |
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Answer» Correct Answer - B The ideal gas law is the equation of state of an ideal gas . The state of an amount of gas is determined by its pressure , volume and temperature . The equation has the from pV=nRT Where , p is pressure , V the volume , n the number of moles , R the gas constant and T the temperature. `therefore " " (P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` Given , `P_(1)=200kPa,V_(1)=V,T_(1)=273+22=295K,` `V_(2)=V+0.02,T_(2)=273+42=315K` `(200xxV)/(295)=(P_(2)xx1.02V)/(315)rArrP_(2)=(200xx315)/(295xx1.02)` |
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| 454. |
Air is pumped into the tubes of a cycle rickshaw at a pressure of 2atm. The volume of each tube at this pressure is `0.002m^(3). One of the tubes gets punctured and the vloume of the tube reduces to `0.0005m^(3)`.How many moles of air have leaked out? Assume that the temperature remains constant at 300K abd that the air behaves asan ideal gas. |
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Answer» `P_1 = 2 atm = 2 xx (10^5)Pa` ` V_1 = 0.002(m^3), T_1 = 300K` ` (P_1)(V_1) = (n_1)(RT_1)` ` rArr (n_1) = ((P_1)(V_1)/ (RT_1)) = (2 xx (10^5)xx 0.002/ 8.3 300)` ` = (2 xx (10^5) xx 2 xx (10^-3)/ 8.3 xx (10^2) xx 3)` ` = (4/ 8.3 xx 3) = 0.1606.` ` Again, (P_2) = 1 atm = 10^5Pa` ` V_2 = 0.0005 (m^3), (T_2)= 300K` ` ((P_2)(V_2))= (n_2)(RT_2)` ` rArr (n_2) = ((P_2)(V_2)/(RT_2)) = ((10^5) xx 0.0005 / 8.3 xx 300)` ` = ((10^5) xx 5 xx (10^-4)/ 8.3 xx 3 xx (10^2))` ` = (5/ 3 xx 83) = 0.02` ` Delta m = moles leaked out ` ` = 0.16 - 0.02 = 0.14` . |
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| 455. |
A given quantity of an ideal gas at pressure P and absolute temperature T obyes `P alpha T^(3)` during adiabatic process. The adiabatic bulk modulus of the gas isA. `(2)/(3)P`B. `P`C. `(3)/(2)P`D. `2P` |
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Answer» Correct Answer - C `PT^(-3)=` constant, `P^(1-gamma)T^(gamma)=` constant `PT^(gamma//10gamma)=` Constant `-3=(gamma)/(1-gamma)rArrgamma=(3)/(2)` `therefore` Adiabatic bulk modulus `K=gammaP` |
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| 456. |
Three designs are proposed for an engine which is to operate between 300 K and 500 K. design A is claimed to produced 3000 J of work per kcal of heat input, B is claimed to produced 2000 J and C , 1000 J. which design would you choose ? Given 1 kcal =4185 JA. A onlyB. B onlyC. AllD. C only |
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Answer» Correct Answer - D Maximum value of efficiency, `eta=1-(T_(2))/(T_(1))=1-(300)/(500)=(2)/(5)` As, `eta=(W)/(Q_(1))` `W=etaQ_(1)=(2)/(5)xx1000" cal "=400xx4.2J=1680J` As no engine can produce more than 168000 J , designs A and B are not possible . Only design C is possible. |
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| 457. |
isobaric modulus of elasticity isA. `oo`B. zeroC. 1D. `(C_(p))/(C_(v))` |
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Answer» Correct Answer - B `DeltaP=0` So, `beta=0` |
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| 458. |
The ratio of r.m.s speed to the r.m.s angular speed of a diatomic gas at certain temperature is (assume m =mass of one molecule, M=molecular mass, I=moment of inertia of the molecules )A. `sqrt(3/2)`B. `sqrt((3I)/(2M))`C. `sqrt((3I)/(2m))`D. 1 |
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Answer» Correct Answer - C `1/2 mV^(2)=3/2 kT`, `1/2 I omega^(2)=2/2 kT` ,`V/(omega)=sqrt((3I)/(2m))` |
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| 459. |
Four particles have velocities `1, 0,2, and 3 m//s`. The root mean square velocity of the particles (definition wise) is.A. 3.5B. `sqrt(3.5)`C. 1.5D. `sqrt(14/3)` |
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Answer» Correct Answer - B `V_(r.m.s)=sqrt((1^(2)+0^(2)+2^(2)+3^(2))/4)=sqrt(3.5)` |
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| 460. |
An ideal heat engine has an efficiency `eta`. The cofficient of performance of the engine when driven backward will beA. `1-((1)/(eta))`B. `eta-((1)/(eta))`C. `((1)/(eta))-1`D. `((1)/(1-eta))` |
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Answer» Correct Answer - C `eta = 1-(T_2)/(T_1)` and `omega = (T_2)/(t_1-T_2) = (T_2//T_1)/(1-(T_2//T_1)) = (1-eta)/(eta) = 1/(eta)-1`. |
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| 461. |
Assertion: Reversible systems are difficult to find in real world. Reason: Most processes are dissipative in nature.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false |
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Answer» Correct Answer - A In a perfectly reversibel system , there is no loss of energy. It is impossible by any means to recover the energy lost in doing work against dissipative forces. Usually all processes are dissipative in nature. |
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| 462. |
A cylinder contains a mixture of helium and argon gas in equilibrium at `150^(@)C`. What is the average kinetic energy for each type of gas molecules?A. `3.26 xx 10^(-21) J`B. `8.76 xx 10^(-21) J`C. `6.28 xx 10^(-21) J`D. `4.14 xx 10^(-21) J` |
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Answer» Correct Answer - B Both kinds of molecules have the same average kinetic as the temperature is same for both the gas molecules The average kinetic energy of a molecules , `bar(K) = 3/2 k_(B) T` `bar(K) = 3/2 (1.38 xx 10^(-23)J//K) (423K) = 8.76xx10^(-21)J` |
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| 463. |
Calculate the total kinetic energy of one kilo mole of Oxygen gas at `27^(@)C` |
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Answer» Since oxygen is a diatomic gas, its molecules have 5 degrees of freedom. Therefore, the total kinetic energy of molecule of Oxygen is `(5)/(2)k_(B)T`. As, 1kg-mole of Oxygen has N molecules, the total kinetic energy of one kg-mole of Oxygen is temperature T is `(5)/(2)xx8.31xx300` `=6.23xx10^(6)` Joule/kg-mole |
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| 464. |
If the temperature of a body is reduced to half of in initial temperature, then radiation power decresed byA. 0.84B. 0.94C. 0.06D. 0.16 |
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Answer» Correct Answer - B `(P_(2))/(P_(1))=((T_(2))/(T_(1)))^(4)=(((T)/(2))/(T))^(4)` `(P_(2))/(P_(1))=(1)/(16)` `(P_(2)-P_(1))/(P_(1))=(1-16)/(16)=-(15)/(16)=93.75%=94%`. |
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| 465. |
The adiabatic elasticity of hydrogen gas `(gamma=1.4)` at `NTP`A. `1xx10^(5)Nm^(-2)`B. `1xx10^(8)Nm^(-2)`C. `1.4Nm^(-2)`D. `1.4xx10^(5)Nm^(-2)` |
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Answer» Correct Answer - D We known that , `E_(phi)=gammap =1.4xx(1xx10^(5))=1.4xx10^(5)Nm^(-2)` |
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| 466. |
A 3000 - mL flask is initailly open while in a room containing air at 1.00 atm and `20^(@)C`. The flask is the closed and immersed in a bath of boiling water. When the air in the flask has reach thermodynamic equilibrium, the flask is opend and air is allowed to escap.The flask is then closed and cooled back `20^(@)C` (A) What is the maximum pressure reached in the flask? (b) How many moles escape when air is released from the flask? (c) What is the final pressure in the flask? |
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Answer» Correct Answer - (a)1.27 atm (b)(0.0268) (0.786 atm) (a) P = 1 atm `T = 20^(@)C` V=3000ml isothoric process (b) `P =? T=100^(@)C` V=3000ml (a) `(P_(2))/(P_(1)) = T_(2)/T_(1) rArr P_(2) =1 xx (373)/(293)` atm= 1.27 atm (b) when flask opened (Temperature constant) change in no. of moles `= |(P_(f)V_(f)-P_(i)V_(i))/(RT)|` `=|(1xx10^(5)xx3000xx10^(-6))/(Rxx373)-(373)/(293)xx(3000xx10^(-6)xx105)/(Rxx373)|` `|(300)/(Rxx373)-(300xx1.27)/(Rxx373)|`=0.026 moles (c) final pressure (ischoric processs) `P_(1)/T_(1) =P_(2)/T_(2) rArr (latm)/(373) = p_(2)/(293)` `P_(2) = (293)/(373) tam =0.786 atm` |
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| 467. |
An ideal Black-body at room temperature is thrown into a furnace. It is observed thatA. it is the darkest body at all timesB. it cannot be distinguished at all timesC. initially it is the darkest body and later it become brightestD. initially it is the darkest body and later it cannot be distinguished |
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Answer» Correct Answer - C When the temperature of black body becomes equal to the temperature of the furnace , the black body will radiate maximum energy and it will be bightest of all initially it will absorb all the radiant energy incident in it. So , it the darkest one. |
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| 468. |
The maximum rarefaction produced by up-to-date laboratory methods is `10^(-11)` mm of mercury. What is the density of the rarest air at `17^(@)C`? Molecular weight of air `=28` and `760mm` of mercury `=1.013xx10^(5)Pa`. |
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Answer» Correct Answer - `1.5x10^(-14)kgm^(-3)` |
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| 469. |
Temperature of 1 mole of an ideal gas is increased from 300K to 310K under isochoric process. Heat supplied to the gas in this process is Q=25R, where R=universal gas constant. What amount of work has to be done by the gas if temperature of the gas decreases from 310K to 300K adiabatically? |
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Answer» `DeltaQ=nC_(V)DeltaT` `therefore25R=(1)(C_(V))(310-300)` or `C_(V)=(5)/(2)R` As the gas is diatomic `gamma=1.4` Now work done in adiabatic process `W=(nR(T_(1)-T_(2)))/(-1)=((1)(R)(310-300))/(1.4-1)=25R` |
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| 470. |
Two cylinders fitted with pistons and placed as shown, connected with string through a small tube of negligible volume, are filled with gas at pressure `P_(0)` and temperature `T_(0)`. The radius of smaller cylinder is half of the other. If the temperature is increased to `2 T_(0)`, find the pressure if the piston of bigger cylinder moves towards left by 1 metre ? A. `(4)/(5)P_(0)`B. `(3)/(5)P_(0)`C. `(2)/(5)P_(0)`D. `(5)/(4)P_(0)` |
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Answer» Correct Answer - D Initial volume `=1xx[piR^(2)+(piR^(2))/(4)]=(5)/(4)piR^(2)` Piston moves towards left by one metre on heating Additional volume available due to heating. `=1xxpiR^(2)-(1xxpiR^(2))/(4)=(3)/(4)piR^(2)` Total volume after heating `=2piR^(2)` `P_(0)xx(5)/(4)piR^(2)=nRT_(0)` After heating, `P_(2)xx2piR^(2)=nRxx2T_(0)` `(P_(0)//P_(2))xx(5)/(4)xx(1)/(4)=(1)/(2)(P_(0)//P_(2))xx(5)/(4)xx(1)/(2)=(1)/(2)` `P_(2)=5p_(0)//4` |
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| 471. |
The pressure exterted on the walls of the container by a gas is due to the fact that the gas moleculesA. lose their kinetic energyB. stick to the wallsC. are accelerated towards the wallsD. change their momenta due to collision with the walls |
| Answer» Correct Answer - D | |
| 472. |
A vessel `A` has volume `V` and a vessel `B` has volume `2V`. Both contain some water which has constant volume. The pressure in the space above water is `p_(a)` for vessel `A` and `p_(b)` for vessel `B`.A. `P_(a) = P_(b)`B. `P_(a) = 2P_(b)`C. `P_(b) = 2P_(a)`D. `P_(b) = 4P_(a)` |
| Answer» Correct Answer - A | |
| 473. |
The molecules of a given mass of gas have a rms velocity of `200 m//sec at 127^(@)C and 1.0 xx 10^(5) N//m^(2)` pressure. When the temperature is `127^(@)C` and pressure is `0.5 xx 10^(5) N//m^(2)` the rms velocity in `m//sec` will beA. `(100sqrt(2))/(3)`B. `100sqrt(2)`C. `(400)/sqrt(3)`D. 400 |
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Answer» Correct Answer - C `v_(rms)=sqrt((3RT)/(M))rArrv_(rms)propsqrt(T)` |
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| 474. |
At `127^(@)C` and `1.00 xx 10^(-2)` atm pressure, the density of a gas is `1.24 xx 10^(-2) kg m^(-3)`. a. Find `v_(rms)` for the gas molecules. b. Find the molecular weight of the gas and identify it. |
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Answer» `p = (1)/(3) rho v_(rms)^(2) implies v_(rms) = sqrt((3 P)/(rho))` `:. V_(rms) = sqrt((3 xx 1.00 xx 10^(-2) xx 1.013 xx 10^(5))/(1.24 xx 10^(-2)))` `= 495 m//s` Again, `v_(rms) = sqrt((3 RT)/(M))` or `M = (3RT)/(v_(rms)^(2))` `= (3 xx 8.3 xx 400)/((495)^(2)) = 40.6 g//mol` This is the molecular weight of argon and so the gas is argon. |
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| 475. |
At what temperature the molecule of nitrogen will have same rms velocity as the molecule of oxygen at `127^(@)C` ?A. `457^(@)C`B. `273^(@)C`C. `350^(@)C`D. `77^(@)C` |
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Answer» Correct Answer - D The root mean square velocity , `v_" rms"sqrt((3RT)/(M))` where , R is gas constant , T the absolute temperature ans M the molecular weight. Given , `M_(N_(2))=28,M_(O_(2))=32` , `T_(O_(2))=127^(@)C=127+273=400K` `therefore" " (v_(O_(2)))/(v_(N_(2)))=sqrt((T_(O_(2)))/(M_(O_(2)))xx(M_(N_(2)))/(T_(N_(2))))=sqrt((400)/(32)xx(28)/T_(N_(2)))=1` `rArr" " T_(N_(2))=350K=77^(@)C` |
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| 476. |
Gas at a pressure `P_(0)` in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubles. The resulting pressure P will be equal toA. `4 P_(0)`B. `2 P_(0)`C. `P_(0)`D. `(P_(0))/(2)` |
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Answer» Correct Answer - B `P_(0) = (1)/(3)(MN)/(V) bar(v^2)` `P = (1)/(3)((m//2)N)/(V)(4 bar(v^(2)))` `(P)/(P_(0)) = 2 rArr = P = 2P_(0)` |
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| 477. |
The molecules of a given mass of gas have a rms velocity of `200 m//sec at 127^(@)C and 1.0 xx 10^(5) N//m^(2)` pressure. When the temperature is `127^(@)C` and pressure is `0.5 xx 10^(5) N//m^(2)` the rms velocity in `m//sec` will beA. `100(sqrt(2))/3`B. `100(sqrt(2))`C. `400/(sqrt(3))`D. None of these |
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Answer» Correct Answer - C Change in pressure will not affect the rms velocity of molecules. So we will calculate only the effect of temperature. As, `v_(rms) prop sqrt(T)` `:. (v_(300^@))/(v_(400^@)) = sqrt((300)/(400)) = sqrt((3)/(4)) implies (200)/(v_(400)) = sqrt(3/4)` `implies v_(400) = (200xx2)/(sqrt(3)) = (400)/(sqrt(3)) m//s`. |
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| 478. |
The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociates into two atoms. The new root mean square speed of the atom isA. `sqrt(2)v`B. yC. 2vD. 4v |
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Answer» Correct Answer - C |
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| 479. |
The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociates into two atoms. The new root mean square speed of the atom isA. `100 (sqrt(2))/3`B. `100 (sqrt(2))`C. `400/(sqrt(3))`D. None of these |
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Answer» Correct Answer - C `v_(rms) = sqrt((3RT)/(M))`. According to problem `T` will `T//2` and `M` will becomes `M//2` so the value of `v_(rms)` will increase by `sqrt(4) = 2` times i.e., new root mean square velocity will be `2v`. |
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| 480. |
The molecular weighs of oxygen and hydrogen are 32 and 2 respectively. The root mean square velocities of oxygen and hydrogen at `NTP` are in the ratioA. `4:1`B. `1:16`C. `16:1`D. `1:4` |
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Answer» Correct Answer - C `(C_(H_2))/(C_(O_2)) = sqrt((2)/(32)) = 1/4`. |
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| 481. |
Calculate the root mean square velocity of a nitrogen molecule at NTP if the density of hydrogen under the same conditiion is `9xx10^(-2)kgm^(-3)` |
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Answer» Correct Answer - `4.91xx10^(2)ms^(-1)` |
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| 482. |
Calculate the number of molecules per unit volume of a perfect gas at `27^(@)C` and `10mm` of mercury. Density of mercury`=13.6xx10^(3) kgm^(-3)` and Boltzmann constant `=1.38xx10^(-23)JK^(-1)` |
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Answer» Correct Answer - `3.22xx10^(23)` |
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| 483. |
Ten small planes are flying at a speed of `150 km//h` in total darkness in an air space that is `20 xx 20 xx 1.5 km^(3)` in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are, On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximately by a sphere of radius 10 m.A. 250 hB. 225 hC. 330 hD. 360 h |
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Answer» Correct Answer - B Given , `v=150kmh^(-1)` ,`V=20xx20xx1.5km^(3)` , N = 10,`d=2xx10=20m` time taken by plane , `t=lamda//V` and mean free path, `lambda=(1)/(sqrt2pi^(2))n` ,where d= diameter and n = number of density , So ,`n=(N)/(V)=(10)/(20xx20xx1.5)= 0.0167km^(-3)` and time elapse between collision, `t=(1)/(sqrt2d^(2)(N//V)xxv)` `=(1)/(1.414xx3.14xx(20)^(2)(xx10^(-6))xx0.0167xx10^(-3)xx150)` `t=225h` |
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| 484. |
One mole of an ideal gas undergoes a process `P = P_(0) [1 + ((2 V_(0))/(V))^(2)]^(-1)`, where `P_(0) V_(0)` are constants. Change in temperature of the gas when volume is changed from `V = V_(0) to V = 2 V_(0)` is:A. `(4)/(5) (P_(0) V_(0))/(nR)`B. `(3)/(4) (P_(0) V_(0))/(nR)`C. `(2)/(3) (P_(0) V_(0))/(nR)`D. `(9)/(7) (P_(0) V_(0))/(nR)` |
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Answer» Correct Answer - A `P=P_(0) [1+(2V_(0)//V)^(2)T^(-1)` at `V =V_(0), P = P_(0)//5` `T_(i) = (PV)/(nR) = ((P_(0)//5)V_(0))/(nR) = (P_(0)V_(0))/(5nR)` At `V = 2V_(0), P=P_(0)//2` `T_(f) = (PV)/(nR) = ((P_(0)//2)(2V_(0)))/(nR) = (P_(0)V_(0))/(nR)` `Delta T = T_(f)-T_(i) = (4P_(0)V_(0))/(5nR)`. |
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| 485. |
The given curve represents the variation of temperature as a function of volume for one mole of an ideal gas. Which of the following curves best represents the variation of pressure as a function of volume? A. B. C. D. |
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Answer» Correct Answer - A Here `T = V tan 45^(@) +T_(0) implies T = V +T_(0)` And `P = (nRT)/(V)` `:. (PV)/(R) = V +T_(0)` (since , n=1) `(P-R)V = RT_(0)`. |
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| 486. |
At what temperature will the R.M.S. velocity of a gas be double its value at N.T.P.?A. `273^(@)C`B. `546^(@)C`C. `819^(@)C`D. `1092^(@)C` |
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Answer» Correct Answer - C `C_(2)=2C_(1)` `(C_(2))/(C_(1))=sqrt((T_(2))/(T_(1)))` `(2C_(1))/(C_(1))=sqrt((T_(2))/(T_(1)))` `4=(T_(2))/(T_(1))` `T_(2)=4xx273=1092K` `T_(2)=1092-273` `=819C`. |
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| 487. |
A fixed mass of gas is taken through a process `A rarr B rarr C rarrA`. Here `A rarr B` is isobaric, `B rarr C` is adiabatic and `C rarr A `is isothermal. Find pressure at `C`.A. `(10^(5))/( 64) N //m^(2)`B. `(10^(5))/( 32) N //m^(2)`C. `(10^(5))/( 12) N //m^(2)`D. `(10^(5))/( 6) N //m^(2)` |
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Answer» Correct Answer - A a For adiabatic process `BC`, `P_(B)=P_(C)` ...(i) For isothermal process `CA` `P_(A)V_(A)=P_(C)V_(C)` ...(ii) From Eqs. (i) and (ii) `V_(C)=[(V_(B)^(gamma))/(V_(A))]^((1)/( gamma-1))=64m^(3)` `P_(C)=(P_(A)V_(A))/(V_(C))=(10^(5))/( 64)N//m^(2)` b Work done, `W=W_(AB)+W_(BC)+W_(CA)` `=P(V_(B)-V_(A))+(1)/(gamma-1)[PV_(B)-P_(C)V_(C)+PV_(A)1n(V_(A)//V_(C))]` Putting the values, `W=4.9 xx 10^(5)J` |
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| 488. |
Statement I: A quasi -static process is so called bacause it is a sudden and large change of the system. Statement II: An adiabatic process is not quasi- static because it is a sudden and large change of the system.A. Statement I: is true, Statement II is true and Statement II is the correct explanation for Statement I.B. Statement I: is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
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Answer» Correct Answer - B It is true that for real gases, and for the real procedures in the laboratory, an adaibatic process is a sudden and large change of the system obtained in a small time. However, for an ideal gas, and for an ideal procedure, we assume for such a gas, an adiabatic process is considered quasi`-` static, in which the system passes through a continuous succession of equilibrium states, which we can plot on a `P-V` diagram. Such a process is also assumed to be reversible through, of course, a quasi `-` static process may or may not be reversible. Such ideal consitions as we assume may not be obtained in particle but almose all real systems deviate little from the results of our ideal procedures, which may therefore be considered approximations to real processes. |
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| 489. |
A cylindrical container has cross sectional area of `A = 0.05 m^(2)` and length L = 0.775 m. Thickness of the wall of the container as well as mass of the container is negligible. The container is pushed into a water tank with its open end down. It is held in a position where its closed end is h = 5.0 m below the water surface. What force is required to hold the container in this position? Assume temperature of air to remain constant. Atmospheric pressure `P_(0) = 1 x× 10^(5)` Pa, Acceleration due to gravity `g = 10 m//s^(2)` Density of water = `10^(3) kg//m^(3)` |
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Answer» Correct Answer - 250 N |
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| 490. |
Assertion: Air quickly leaking out of a balloon becomes coolers. Reason: The leaking air undergoes adiabatic expansion.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false |
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Answer» Correct Answer - A The leaking air of balloon undergoes adiabatic expansion . In this expansion , due to work done against external pressure, the internal energy of air reduces. Thus, it becomes cooler. Adiabatic expansion produces cooling. |
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| 491. |
In a certain process, 400 cal of heat are supplied to a system and the same time 105 J of mechanical work done on the system . The increase in its internal energy isA. 20 calB. 303 calC. 404 calD. 425 cal |
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Answer» Correct Answer - D Here, `dQ=400"cal",dW=-105 J` = `-105//4.2"cal"=-25"cal",dU` =? `"Now"" " du=dQ-dW=400-(-25)=425"cal"` Note dW is negative because ,work done on the system . |
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| 492. |
We need mechanical equivalent of heat becauseA. in C.G.S system, heat is not measured in the units of workB. in SI system, heat is measured in the units of workC. of some reason other than those mentioned in the units of workD. of some reason other than those mentioned above. |
| Answer» Correct Answer - B | |
| 493. |
If an electric fan be switched on in a closed room, will the air of the room be cooled? If not, why do we feel cold?A. increasesB. decreasesC. remains unchangedD. may increase or decrease depending on the speed of rotation of the fan. |
| Answer» Correct Answer - A | |
| 494. |
A scientist says that the efficiency of his heat engine which operates at source temperature `127^(@)C` and sink temperature `27^(@)C is 26%`, thenA. it is impossible.B. it is possible but less probableC. it is quite probableD. data is incomplete |
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Answer» Correct Answer - A `eta=1-(T_(2))/(T_(1))` |
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| 495. |
Calculate the volume of 1 mole of an ideal gas at STP. |
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Answer» Volume of 1 mole gas `PV= nRT` ` rArr V= (RT)/P= 0.082 xx 273( as P= 1)` ` V= 22.386 =22.4L ` ` =22.4 xx (10^-3)m^3` `=2.24 xx (10^-2)m^3` |
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| 496. |
A lead bullet of mass 21g travelling at a speed of 100 `ms^(-1)` comes to rest in a wooden block. If no heat is taken away by the wood, the rise in temperature of the bullet in the wood nearly is (Sp. Heat of lead 80cal/kg `.^(@)C`) |
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Answer» Kinetic energy of the bullet=heat gained by the bullet, `(1)/(2)mv^(2)=mS Deltat` `Deltat=(v^(2))/(2S)=((100)^(2))/(2xx0.03xx4.2xx1000)=39.68^(@)C`. |
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| 497. |
Which of the following is incorrect regarding the first law of thermodynamics?A. It introduce the concept of the internal energyB. it introduce the concept of the entropyC. it is not applicable to any cyclic processD. None of the above |
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Answer» Correct Answer - B Entropy is related to second law of thermodynamics |
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| 498. |
A system undergoes a change of state during which 100 kJ of heat is transferred to it and it does 50 kJ of fwork. The system is brought back to its original state through a process during which 120 kJ of heat is transferred to it. Find the work done by the system in the second process. |
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Answer» Correct Answer - 170KJ `triangleQ_(1) = 100 Kj traingle W_(1) = 50 kj` `triangleQ_(2) = 120 kj W_(2) = ?` given process is cyclic process Hence `triangleU = 0` ` rArr triangleQ_(1) + triangleQ_(2) = triangleW_(1) + triangleW_(2)` `100+120 = 50+ triangleW_(2)` `triangleW_(2) = 170 kj` |
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| 499. |
The change in internal energy of two moles of a gas during adiabatic expansion is found to be -100 Joule. The work done during the process isA. 100 jouleB. `-100` jouleC. zeroD. 200 joule |
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Answer» Correct Answer - A `DeltaW=-DeltaU` |
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| 500. |
As a result of heating a mole of an ideal gas at constant pressure by `72^(@)C`, a heat flow by an amount 1600 joules takes place. Find the work performed by the gas, the increment of its internal energy, and the value of `gamma`. [Hint:`triangleQ = triangleU + triangleW` or `C_(p) triangleT = C_(v)triangleT + triangleW` or `triangleW = (C_p-C_v)triangleT=RtriangleT] h` |
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Answer» Correct Answer - `597.6J,0.65^(@)C` `triangleQ = nC_(p)triangleT` `triangleU = nC_(v) triangleT` `triangleW = triangleQ - triangleU = n(C_(p)-C_(v))triangleT` `1(R) triangleT` `=8.3xx72 = 597.6 J` `triangleU = 1600 - 597.6 = 1002.4 J` `r=(C_(p))/(C_(v)) =(1600)/(1002.4)=5/3` |
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