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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.
| 801. |
A cube, which may be regarded as a perfectly black body, radiates heat at the rate of 2770.2 watt when its temperature is `27^(@)C`. The volume of the cube is `(sigma=5.7xx10^(8)J//m^(2)" s "K^(4))`A. `10^(-2)m^(3)`B. `100m^(3)`C. `1m^(3)`D. `10^(3)m^(3)` |
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Answer» Correct Answer - C `(dQ)/(dt)=sigmaAT^(4)` `therefore A=(dQ)/(dt)//rT^(4)` `6l^(2)=6` `l^(2)=1` `V=l^(3)=1m^(3)` |
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| 802. |
Black body at a temperature of `1640K` has the wavelength corresponding to maximum emission equal to `1.75mu m` Assuming the moon to be a perfectly black body the temperature of the moon if the wavelength corresponding to maximum emission is `14.35mum` is .A. 100 KB. 150 KC. 200 KD. 250 K |
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Answer» Correct Answer - C `(T_(2))/(T_(1))=(lamda_(m_(1)))/(lamda_(m_(2)))=(1.75)/(14.35)` `T_(2)=(1.75)/(14.35)xx1640=200K` |
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| 803. |
A stationary cylinder of oxygent used in a hospital has the following characteristics at room temperature `300 K`, gauge pressure `1.38 xx 10^(7)` Pa. volume `16 L`. If the flow area, measured at atmospheric pressure, is constant at `2.4 L//min`, the cylinder will last for nearlyA. `5 h`B. `10 h`C. `15 h`D. `20 h` |
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Answer» Correct Answer - C c. `(P_(1) V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` `((1.38 xx 10^(7) Pa) (16 L))/(300 K)= ((10^(5) Pa)(2.4 (L)/(m) t))/(300 K)` `t = (1.38 xx 10^(7) xx 16)/(10^(5) xx 2.4) = 920 min` `(920)/(60) h = 15 h` (approx.) |
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| 804. |
One mole of a gas is enclosed in a cylinde in a cyclinder and occupies a volume of `1.5 L` at a pressure 1.5 atm. It is subjected to strong heating due to which temperature of the gas increase according to the relation `T = alpha V^(2)`, where `alpha` is a positive constant and `V` is volume of the gas. a. Find the work done by air in increasing the volume of gas to `9 L`. b. Draw the `P - V` diagram of the process. c. Determine the heat supplied to the gas (assuming `gamma = 1.5)`. |
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Answer» `W = int_(V_(1))^(V_(2)) PdV = int_(V_(1))^(V_(2)) (RT)/(V) dV = R int_(V_(1))^(V_(2)) (alpha V^(2))/(V) dV` `alpha R int_(V_(1))^(V_(2)) VdV = (alphaR)/(2) (V_(2)^(2) - V_(1)^(2))``alpha R int_(V_(1))^(V_(2)) VdV = (alphaR)/(2) (V_(2)^(2) - V_(1)^(2))` `W = (R )/(2) (T_(2) - T_(1))` (i) Now, `T_(1) = alpha V_(1)^(2)` `T_(2) = alpha V_(2)^(2)` `T_(2) = T_(1) [(V_(2))/(V_(1))]^(2) = 35 T_(1)` Hence from Eq. (i) we have `W = (R )/(2) [36 T_(1) - T_(2)] = (35 RT_(1))/(2) = (35)/(2) P_(1) V_(1)` `= (35)/(2) xx (1.5 xx 10^(-3)) xx (1.2 xx 10^(5)) J = 31 50 J` b. We know, `PV = RT = R alpha V^(2)` `P prop V` Hence `P -V` graph is a straight line. c. `Delta U = n V_(V) Delta T = n ((R )/(gamma - 1)) (T_(2) - T_(1))` `(R )/(0.5) xx 35 T_(1) = 70 RT_(1) = 70 P_(1) V_(1)` `= 70 xx (1.2 xx 10^(5)) xx (1.5 xx 10^(-3)) J = 12600 J` `Delta Q = Delta U + W 12600 + 3150 = 15750 J` |
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| 805. |
Find the amount of work done to increase the engines ever developed operates between `2100 K and 700 K`. Its actual efficiency is `40%`. What percentage of its maximum possible efficiency is this?A. `40%`B. `60%`C. `66.67%`D. `33.37%` |
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Answer» Correct Answer - B `eta=1-(T_L)/(T_H) = 1-700/2100=2/3` `^ eta = 2/3xx100=66%` Actual efficiency is `40%` which is `60%` of the theoretical effienency`. |
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| 806. |
A carnot engine absorbs `1000J` of heat energy from a reservoir at `127^(@)C` and rejecs `600J` of heat energy during each cycle. Calculate (i) efficiency of the engine, (ii) temperature of sink, (iii) amount of useful work done per cycle.A. 20% and `-43^(@)C`B. 40% and `-33^(@)C`C. 50% and `-20^(@)C`D. 70% and `-10^(@)C` |
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Answer» Correct Answer - B `eta=1-(Q_(2))/(Q_(1))=1-(T_("sin k"))/(T_("source"))` |
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| 807. |
if R is the molar gas constant and `gamma=C_(P)//C_(V),` then `C_(V)` is equal toA. `R//gamma`B. `gammaR`C. `(2R)/(gamma-1)`D. `(R)/(gamma-1)` |
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Answer» Correct Answer - D `C_(p)_C_(V)=R` `(C_(p)-R)/(C_(V))=(C_(V))/(C_(V))+1` `(C_(p))/(C_(V))-(R)/(C_(V))=1` `gamma-(R)/(C_(V))=1` `gamma-1=(R)/(C_(V))" "thereforeC_(V)=(R)/(gamma-1)` |
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| 808. |
The root mean square velocity of the molecules in a sample of helium is `5//7th` that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is `0^(@)C`, then the temperature of the helium sample is aboutA. `0^(@)C`B. `0 K`C. `273^(@)C`D. `100^(@)C` |
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Answer» Correct Answer - A `((v_(rms))_(He))/((v_(rms))_(H_(2))) = sqrt(T_(He)/T_(H_(2)).((M_(0))_(H_(2)))/((M_(0))_(He)))` `5/7=sqrt(T_(He)/((0+273)).2/4)=sqrt(T_(He)/546)` `T_(He) = ((5)/(7))^(2) xx 546 = 278.6 K` `= 278.6 - 273 = 5.6^(@)C cong 0^(@)C` |
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| 809. |
The rate of emission of heat energy of an iron ball of radius 5 cm is 10 J/s, then rate of emission of heat energy by a copperr ball of radius 1 cm at same temperature will be (If emissivity of both the balls is same)A. 2 J/sB. 0.4 J/sC. 2 cal/sD. 250 J |
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Answer» Correct Answer - B `(R_(2))/(R_(1))=(A_(2))/(A_(1))((T_(2))/(T_(1)))^(4)` as `T_(1)=T_(2)` `(R_(2))/(R_(1))=(1)/(25)` `therefore R_(2)=(10)/(25)=0.4J//s` |
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| 810. |
The amount of thermal radiations emitted from one square metre area of a black body in one second when at a temperature of 100 K isA. 5.67 JB. 56.7 JC. 567 JD. 5670 J |
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Answer» Correct Answer - A `(dQ)/(dtA)=sigmaT^(4)` |
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| 811. |
Logarithms of readings of pressure and volume for an ideal gas were plotted on a graph as shown in Fig. By measuring the gradient, it can be shwon that the gas may be A. monatomic and undergoing an adiabatic changeB. monatomic and undergoing an isothermal changeC. diatomic and undergoing an adiabatic changeD. triatomic and undergoing an isothermal change |
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Answer» Correct Answer - C c. `log P = m log V + C_(1)`, where `C_(1)` is positive, `m` is slope `m = (2.38 - 2.10)/(1.1 - 1.3) = 1.4` `log P = 1.4 log V + C_(1)` `log PV^(1.4) = C_(1)` `PV^(1.4) = k` Thus, it represent an ideal diatomic gas undergoing adiabatic change. |
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| 812. |
dU+dW=0 is valid forA. adiabatic processB. isothermal processC. isobaric processD. isochoric process |
| Answer» Correct Answer - A | |
| 813. |
Isothermal curves for a given mass of gas are shown at two different temperture `T_(1) and T_(2)` in Fig. State whether `T_(1) gt T_(2) or T_(2)gt T_(1)`. Justify your answer. A. `T_(1)=T_(2)`B. `T_(1)gtT_(2)`C. `T_(1)lt^(2)`D. `T_(1)geT_(2)` |
| Answer» Correct Answer - C | |
| 814. |
The pressure `p` for a gas is plotted against its absolute temperature `T` for two different volumes `V_(1)` and `V_(2)`. If `p` is plotted on `y-` axis and `T` on `x-`axis, thenA. the curve for `V_(1)` has greater slope than the curve for `V_(2)`B. the curve for `V_(2)` has greater slope than the curve for `V_(1)`C. the curve must intersect at some point other than T=0D. the curves have the same slope and do not intersect |
| Answer» Correct Answer - B | |
| 815. |
The temperature range in the definition of standard calorie isA. `14.5^(0)C` to `15.5^(0)C`B. `15.5^(0)C` to `16.5^(0)C`C. `1^(0)C` to `2^(0)C`D. `13.5^(0)C` to `14.5^(0)C` |
| Answer» Correct Answer - A | |
| 816. |
If the rms speed of nitrogen molecules is `490 m s^(-1)`at 273 K, find the rms speed of hydrogem molecules at the same temperature. |
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Answer» The molecular weight of nitrogen=n is `28 g mol^(-1) and that of hydrogen is `2 g mol^(-1)`. Let `m_(1), m_(2)` be the masses and `v_(1)`,`v_(2)`be the rms speeds of a nitrogen molecule and a hydrogen molecule respectively. Then `m_(1) = 14m_(2)`.Using equation`(24*6)`, `1/2m_(1) v_(1)^(2)=1/2m_(2)v_(2)^(2)` or, `v_(2) = v_(1)sqrt(m_(1)/m_(2))=490m s^(-1)xx sqrt14=1830ms^(-1)`. |
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| 817. |
Certain amount of heat supplied to an ideal gas under isothermal conditions will result inA. rise in temperature qB. doing external work and a change in temperatureC. doing external workD. an increase in the internal energy of the gas |
| Answer» Correct Answer - C | |
| 818. |
A very tall vertical cylinder is filled with a gas of molar mass M under isothermal conditions temperature T. the density and pressure of the gas at the base of the container is `rho_(0)` and `P_(0)`, respectively Select the incorrect statementA. Pressure decreases with heightB. The rate of decrease of pressure with height is a constantC. `(dP)/(dh)=-rhog` where `rho` is density of the gas at a height hD. `P=rho(RT)/(M)` |
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Answer» Correct Answer - B Considering the equilibrium of differential layer of a thickness dh and height h, we have `(P+dP)A+dmg=PA` or `A dP=-g dm=0grhoA dh` From ideal gas equation, we have `P=(rhoRT)/(M)` or `rho=(PM)/(RT)` From eqns. (1) and (2), we get `dp=-(PM)/(RT)g dh` or `In (P)/(P_(0))=-(Mg)/(RT)h` or `P=P_(0)e^(-Mgh//RT)` In a similar manner if we eliminate P with the of ideal gas equation, we get `dP=(drhoRT)/(M)` The eqn. becomes `drho(RT)/(M)=-rhog dh` or `int_(P_(0))^(rho) (drho)/(rho)=-(Mg)/(RT)int_(0)^(h)dh` or `rho=rho_(0)e^(-Mgh//RT)` |
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| 819. |
Calculate the rms speed of nitrogen at STP (pressure `=1` atm and temperature `= 0^0 C`. The density of nitrgoen in these conditions is `1.25 kg m^(-3)` |
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Answer» At STP, the pressure is `1.0 xx 10^(5) N m^(-2)`. The rms speed is `v_(rms) = sqrt(3p/rho)` `= sqrt((3 xx 10^(5) N m^(-2))/((1.25 kg m^(-3))` `=490 m s^(-1)`. |
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| 820. |
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). The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)A. `rho_(1)nRgT_(0)((P_(0)+rho_(l)gH)^(2//5))/((P_(0)+rho_(l)gy)^(7//5))`B. `(rho_(l)nRgT_(0))/((P_(0)+rho_(l)gH)^(2//5){(P_(0)+rho_(l)g)(H-y)}^(3//5))`C. `rho_(l)nRgT_(0)((P_(0)+rho_(l)gH)^(3//5))/((P_(0)+rho_(l)gy)^(8//5))`D. `(rho_(l)nRgT_(0))/((P_(0)+rho_(l)gH)^(3//5){(P_(0)+rho_(l)g)(H-y)}^(2//5))` |
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Answer» Correct Answer - B `F_(b)`=Volume of bubble `rho_(l)g` `=(nRT_(2))/(P_(2))rho_(1)g` `F_(b)=(rho_(l)nRgT_(0))/((P_(0)+rho_(1)gH)^((2)/(5))[P_(0)+rho_(l)g(H-Y)]^((3)/(5)))` |
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| 821. |
Find the mean free path and collision frequency of a hydrogen molecule in a cylinder containing hydrogen at 3atm and temperature `27^(@)C`. Take the radius of a hydrogten molecule to be `1overset(0)A`. Given, one atmospheric pressure `=1.013xx10^(5)N m^(-2)` and molecular mass of hydrogen=2.A. `7.66xx10^(-8)m, 2.52xx10^(5)s^(-1)`B. `7.66xx10^(-5)m,2.52xx10^(5)s^(-1)`C. `7.66xx10^(-8)m,2.52xx10^(10)`D. `7.66xx10^(-10)m, 2.52xx10^(8)s^(-1)` |
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Answer» Correct Answer - C `lamda=(m)/(sqrt(2)pid^(2)e)` `lambda=7.66xx10^(-8)m` `f=(V_(rms))/(lambda)` `f=2.25xx10^(10)s^(-1)` |
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| 822. |
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. Only the force of gravityB. The force due to gravity and the force due pressure of the liquidC. The force due to gravity, the force due pressure of the liquid and the force due to visco of the liquid.D. The force due to gravity and the force due viscosity of the liquid. |
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Answer» Correct Answer - D Gravity and viscous force |
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| 823. |
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)+P_(l)gh)/(P_(0)+rho_(l)gy))^(2//5)`B. `T_(0)((P_(0)+rho_(l)g(H-y))/(P_*(0)+rho_(l)gH))^(2//5)`C. `T_(0)((P_(0)+rho_(l)gH)/(P_(0)+rho_(l)gy))^(3//5)`D. `T_(0)((P_(0)+rho_(l)g(H-y))/(P_(0)+rho_(l)gH))^(3//5)` |
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Answer» Correct Answer - B `T_(1)P_(1)^((1-gamma)/(gamma))=T_(2)P_(2)^((1-gamma)/(gamma))` `T_(2)=T_(0)[((P_(0)+rho_(l)g(H-y)))/(P_(0)+rho_(l)gH)]^((2)/(5))` |
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| 824. |
The density of a polyatomic gas in stantard conditions is `0.795 kg m^(-3)`. The specific heat of the gas at constantA. `930 J-kg^(-1) K^(-1)`B. `1400 J-kg^(-1) K^(-1)`C. `1120 J-kg^(-1) K^(-1)`D. `1600 J-kg^(-1) K^(-1)` |
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Answer» Correct Answer - B b. Ideal gas equation for `m` grams of gas is `PV = mr T` (where `r=` specific gas constant) or `P = (m)/(V) rT = rho rT` `implies r = (P)/(rho T) = (1.013 xx 10^(5))/(0.795 xx 273) = 466.7` Specific heat at constant volume `c_(v) = (r )/(gamma - 1) = (466.7)/((4)/(3) -1) = 1400 J//kg-K` `(gamma = 4//3` for polyatomic gas) |
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| 825. |
A thermodynamic system is taken through the cyclic `PQRSP` process. The net work done by the system is A. `20 J`B. `- 20 J`C. `400 J`D. `- 374 J` |
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Answer» Correct Answer - B b. Work done by the system = area of shade protion on `P - V` diagram. `= (300 - 100) 10^(-6) xx (200 - 100) xx 10^(3) = 20 J` And direction of process is anticlock wise so work done will be negative, i.e., `Delta W = - 20 J`. |
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| 826. |
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. `15RT`B. `9RT`C. `11RT`D. `4RT` |
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Answer» Correct Answer - C `U=f/2 nRT` `U_(t otal) = 5/2(2)RT+3/2(4)RT` `U_(T otal) = 11RT`. |
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| 827. |
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. `4 RT`B. `15 RT`C. `9 RT`D. `11 RT` |
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Answer» Correct Answer - D d. Total internal energy of systme `U_(o x y g e n) + U_(argo n) = mu_(1) (f_(1))/(2) RT + mu_(2) (f_(2))/(2) RT` [As `f_(1) = 5` (for oxygen) and `f_(2) = 3` (for argon)] |
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| 828. |
A vessel containing one gram -mole of oxygen is enclosed in a thermally insulated vessel. The vessel is next moved with a constant speed `v_(0)` and then suddenly stopped. The process results in a rise in the temperature of the gas by `1^(@)c`. Calculate the speed `v_(0)`. |
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Answer» Correct Answer - `36.0,ms^(-1)` |
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| 829. |
It is known that pressure of a gas increases if– (A) You increase the temperature of the gas while holding its volume constant. (B) You compress the gas holding its temperature constant. (i) In which of the two cases [A or B] the average impulse imparted by a gas molecule to the container wall during a collision increases? (ii) In which of the two cases the frequency of collisions of gas molecules with the container wall increases. |
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Answer» Correct Answer - (a) A (ii) A and B |
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| 830. |
An insulated cylindrical vessel is divided into three identical parts by two partitions 1 and 2. The left part contains `O_(2)` gas, the middle part has `N_(2)` and the third chamber has vacuum. The average molecular speed in oxygen chamber is `V_(0)` and that in nitrogen chamber is `sqrt((8)/(7)) V_(0)`. Pressure of the gases in two chambers is same. Partition 1 is removed and the gases are allowed to mix. Now the stopper holding the partition 2 is removed and it slides to the right wall of the container, so that the mixture of gases occupy the entire volume of the container. Find the average speed of `O_(2)` molecules now. |
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Answer» Correct Answer - `V_(0)` |
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| 831. |
Select the incorrect statement about ideal gas.A. Molecules of gas are in incessant random motion collinding against one another and with the walls of the container.B. The gas is not isotropic and the constant `(1/3)` in equation `P = (1/3)rho v_rms^2` is result of this propertyC. The time during which a collision lasts is negligible compared to the time of free path between collisions.D. There is no force of interaction between molecules among themseleves or between molecules and the wall except during collision |
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Answer» Correct Answer - B |
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| 832. |
Out of 10J of radiant energy incident on a surface, the energy absorbed by the surface is 2 J and the energy reflected is 7 J. Then, coefficient of transmission of the body isA. 0.2B. 0.7C. 0.1D. zero |
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Answer» Correct Answer - C `Q=Q_(a)+Q_(r)+Q_(t)` `Q_(t)=10-(7+2)=1` `t=(Q_(t))/(Q)=(1)/(10)=0.1`. |
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| 833. |
A Carnot engine workds between `200^(@)C` and `0^(@)C` and `.-200^(@)C`. In both caes the working substance absorbes 4 kilocalories of heat from the source. The efficiency of first engine will beA. `(100)/(173)`B. `(200)/(473)`C. `(173)/(273)`D. `(273)/(373)` |
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Answer» Correct Answer - B `eta=1-(T_(2))/(T_(1))` |
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| 834. |
An ideal heat engine works between source at `127^(@)C` and sink `27^(@)C`. If 800 J heat is taken from reservoir, the amount of heat rejected to sink isA. 300 JB. 400 JC. 500 JD. 600 J |
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Answer» Correct Answer - D `eta=1-(T_(2))/(T_(1))` `eta=1-(Q_(2))/(Q_(1))=1-(T_(2))/(T_(1))=(300)/(600)=(1)/(2)` `therefore (Q_(2))/(Q_(1))=(T_(2))/(T_(1))` `Q_(2)=(T_(2))/(T_(1))xxQ_(1)=(300)/(400)xx800=600J` |
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| 835. |
A sink, that is a system where heat is rejected, is essential for the conversion of heat into work. From which law the above inference follows?A. zerothB. FirstC. SecondD. Third |
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Answer» Correct Answer - C This is the statement of the second law of thermodynamics. |
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| 836. |
Two moles of an ideal gas at temperature `T_(0) = 300 K` was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. Find the total amount of heat absorbed by the gas in the processsA. `150 R "joules"`B. `300 R "joules"`C. `75 R "joules"`D. `100 R joules` |
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Answer» Correct Answer - B For one mole of gas `Delta Q = C_v DeltaT + P DeltaT` At constant volume `DeltaT = 0` for two moes of gas, `DeltaT = 0` From `PV = nRT = 2R xx 100` and `P/2 V = 2RT_(f)` `:. T_(f) = 150K` `:. DeltaQ = 2C_(v)(T_f-T_i) = 2C_v(150-300)` `=-300 C_v` joules In the next process, `Delta = 2C_p DeltaT=2C_p(300-150)` `=300C_p` joules. `:.` Net heat absorbed = `-300C_v+300C_p` `=300(C_p-C_v)` `=300R "joules"`. |
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| 837. |
If `300 ml` of a gas at `27^(@)` is cooled to `7^(@)` at constant pressure, then its final volume will beA. `540 ml`B. `350 ml`C. `280 ml`D. `135 ml` |
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Answer» Correct Answer - C `V prop T` at constant pressure `implies (V_1)/(V_2)=(T_1)/(T_1) implies V_(2)=(V_(1)T_(2))/(T_1)=(300xx280)/(300) = 280ml`. |
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| 838. |
The ratio of the number of moles of a monoatomic to a polyatomic gas in a mixture of the two, behaving as an diatomic gas is : (vibrational modes of freedom is to be ignored)A. `2:1`B. `1:2`C. `2:3`D. `3:2` |
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Answer» Correct Answer - B Let the required ratio be `n:1` `Q_(1) = n((f_(1)R)/(2)) Delta T` and `Q_(2) = 1((f_(3)R)/(2)) Delta T` `and Q = (n+1)((f_(2)R Delta T)/(2))` `Q = Q_(1)+Q_(2)`, so `(n+1)f_(2)=nf_(1)+f_(3)` or, `(n+1)5=n(3)+6impliesn=1//2`. |
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| 839. |
n moles of a gas are filled in a container at temperature T. If the gas is slowly and isothermally compressed to half its initial volume, the work done by the atmosphere on the gas is A. `(nRT)/2`B. `-(nRT)/2`C. `nRT(ln 2-1/2)`D. `-nRT//n 2` |
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Answer» Correct Answer - A Work done by atmosphere `P_(atm) DeltaV` `=P_(aim)V/2.....(i)` As, Initially gas in container is in thermodynamic equilibrium with its surroundings. `:. ` pressure inside cylinder `=P_(atm)` & PV=nRT `rArr P_(atm) V=nRT ` or `V=(nRT)/(P_(atm))` Putting in (i) `W=(nRT)/2` |
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| 840. |
`4.0 g` of a gas occupies `22.4` litres at NTP. The specific heat capacity of the gas at constant volume is `5.0 JK^(-1)mol^(-1)`. If the speed of sound in this gas at NTP is `952 ms^(-1)`. Then the heat capacity at constant pressure isA. `8.5 JK^(-1) mol^(-1)`B. `8.0 JK^(-1) mol^(-1)`C. `7.5 JK^(-1) mol^(-1)`D. `7.0 JK^(-1) mol^(-1)` |
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Answer» Correct Answer - B Speed of sound `V = sqrt((gammaRT)/(M))` `V = sqrt((C_P)/(C_v)(RT)/(M))` `952=sqrt((C_(P)xx8.3xx273)/(5xx4xx10^(-3))` `C_(p)=8JK^(-1) mol^(-1)`. |
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| 841. |
A Carnot engine operates between `327^(@)C` and `27^(@)C` How much heat (in joules) does it take from the `327^(@)C` reservoir for every `100 J` of work done?A. `100J`B. `200J`C. `300J`D. `400J` |
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Answer» Correct Answer - B `W=Q_(1)-Q_(2)` As, `eta=(Q_(1)-Q_(2))/(Q_1) = (T_1-T_2)/(T_1)` `:. Q_(1)((T_1)/(T_1-T_2))W = (600)/(600-300)xx100` `=200J`. |
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| 842. |
Ratio of isothermal elasticity of gas to the adiabatic elasticity isA. `gamma`B. `(1)/(gamma)`C. `1-gamma`D. `(1)/(1-gamma)` |
| Answer» Correct Answer - B | |
| 843. |
An ideal gas expands isothermally from volume `V_(1)` to `V_(2)` and is then compressed to original volume `V_(1)` adiabatically. Initialy pressure is `P_(1)` and final pressure is `P_(3)`. The total work done is `W`. ThenA. `P_(3)gtP_(1),Wgt0`B. `P_(3)ltP_(1),Wlt0`C. `P_(3)gtP_(1),Wlt0`D. `P_(3)=P_(1),W=0` |
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Answer» Correct Answer - C |
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| 844. |
A monoatomic gas at pressure `P_(1)` and volume `V_(1)` is compressed adiabatically to `1/8th` of its original volume. What is the final pressure of gas.A. `64P_(1)`B. `P_(1)`C. `16 P_(1)`D. `32 P_(1)` |
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Answer» Correct Answer - D Ideal gas equation, for an adiabatic process is |
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| 845. |
Your are given the following group of particles, `n_(i)` represents the number of molecules with speed `v_(i)` `(P_(2))/(P_(1))=(m_(2))/(m_(1)) : (p_(2))/(76)=(m_(1)+(50)/(100)m_(1))/(m_(1))=(3)/(2)` calculate (i) average speed (ii) rms speed (iii) most probable speed. |
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Answer» `v_(ang)=(n_(1)v_(1)+n_(2)v_(2)+n_(3)v_(3)+n_(4)v_(4)+n_(5)v_(5))/((n_(1)+n_(2)+n_(3)+n_(4)+n_(5)))` `=((2xx1)+(4xx2)+(8xx3)+(6xx4)+(3xx5))/((2+4+8+6+3))` `=(73)/(23)=3.17ms^(-1)` (ii) root mean square speed is `v_(rms)=((n_(1)v_(1)^(2)+n_(2)v_(2)^(2)+n_(3)v_(3)^(2)+n_(5)v_(5)^(2))/(2+4+8+6+3))^(1//2` `=3.36 ms^(-1)` (iii) By definition, most probable speed `=3.0ms(-1)`. |
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| 846. |
Two cylinder having `m_(1)g` and `m_(2)g` of a gas at pressure `P_(1)` and `P_(2)` respectively are put in cummunication with each other, temperature remaining constant. The common pressure reached will beA. `(m_(1)m_(2)P_(2))/(P_(2)m_(1)+P_(1)m_(2))`B. `(P_(1)P_(2)m_(1))/(P_(2)m_(1)+P_(1)m_(2))`C. `(m_(1)m_(2)(P_(1)+P_(2)))/(P_(2)m_(1)+P_(1)m_(2))`D. `(P_(1)P_(2)(m_(1)+m_(2)))/(P_(2)m_(1)+P_(1)m_(2))` |
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Answer» Correct Answer - D As temperature is constant `P_(1)V_(1) = ((m_1)/(M))RT`..(i) `P_(2)V_(2) = ((m_2)/(M))RT` ..(ii) When both cylinder are connected `P(V_(1)+V_(2)) = ((m_(1)+m_(2)))/(M) RT` ..(ii) From (i), (ii) and (iii) `P[(m_1)/(M)(RT)/(p_1)+(m_2)/(M)(RT)/(p_2)]=((m_(1)+m_(2))/(m) RT` `implies P[(m_1)/(P_1)+(m_2))/(P_2)] = (m_(1)+m_(2))` Hence, `P = (P_(1)P_(2)(m_(1)+m_(2)))/((m_(1)P_(2)+m_(2)P_(1))`. |
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| 847. |
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 monoatomic gas at absolute temperature T. The mass of a molecule is m. ThenA. no molecule can have speed greater than `sqrt(2)^(v_(ms))`B. no molecule can have speed less than `underset(p)("V/")sqrt2`C. `V_(p)lt V_(av) ltV_(ms)`D. the average kinetic energy of a molecule is `3//4 mv_(p)^2` |
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Answer» Correct Answer - C::D Refer to maxvell dismintion curve |
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| 848. |
A body radiates heat at the rate of 50 J/s at 300 K. when the same body is at 600 K then its rate of radiation of heat will beA. 100 J/sB. 200 J/sC. 400 J/sD. 800 J/s |
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Answer» Correct Answer - D `(E_(2))/(E_(1))=((T_(2))/(T_(1)))^(4)=((600)/(300))^(4)=2^(4)` `E_(2)=16xx50=800J//s`. |
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| 849. |
The variation of magnetic susceptibility `(chi)` with temperature for a diamagnetic substance is best represented byA. B. C. D. |
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Answer» Correct Answer - D |
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| 850. |
A monatomic gas undergoes a cycle consisting of two isothermals and two isobarics. The minimum and maximum temperatures of gas during the cycle are `T_(1)= 400 K` and `T_(2)=800 K`, respectively, and the ratio of maximum to minimum volume is 4. The volume at `B` isA. `1.5 V_(0)`B. `2 V_(0)`C. `3 V_(0)`D. `2.5 V_(0)` |
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Answer» Correct Answer - B Process `CB` is done at minimum temperature, while process `DA` is done at maximum temperature. Process `A rarr b(` isobaric `)` `(V_(A))/(V_(B))=(T_(A))/(T_(B))=(800)/(400)=2` hence` V_(B)=2V_(0)` `B rarr C(` isothermal `)` `C rarr D (` isobaric `)` `(V_(D))/(V_(C))=(T_(D))/(T_(C))implies V_(D)=2V_(0)` `D rarr(` isothermal `)` `W_(AB)=P(V_(B)-V_(A))=nRDT` `=nR(-400)=-400nR` `W_(BC)=nRT1n(V_(C))/(V_(B))=nR4001n(V_(0))/(2V_(0))` `=-400nR 1n2` `W_(CD)=P(V_(D)-V_(C))=nR(400)=400nR` `W_(DA)=nRT1n(V_(A))/(V_(D))` `=nRxx800 xx 1 n (2V_(0))/(V_(0))=800nR1n2` `Delta W=400nR1n2` `DeltaQ=` heat is extracted `Q_(CD)=nC_(p) DeltaT=n(5)/(4)R400=1000nR` `=800 n R 1 n 2= 1000 n R+800 nR 1 n 2` Efficiency `=(DeltaW)/( Delta Q)` `eta =(2 1n 2)/(5 +4 1n 2 ) xx 100%` |
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