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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. |
Real gases behave as ideal gases most closely at low pressure and high temperature. Intermolecular force between ideal gas molecules is assumed to be zero.A. If both Assertion and Reason are true and the reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - B | |
| 2. |
Vibrational kinetic energy is insignificant at low temperatures. Interatomic forces are responsible for vibrational kinetic energy.A. If both Assertion and Reason are true and the reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - B | |
| 3. |
Change each of the given temperature to the Fahrenheit and Reaumur scale : `30^@ C, 5^@ C and -20^@ C`. |
| Answer» Correct Answer - A::D | |
| 4. |
The temperature Ton thermodynamic scale is defined in terms of a property `p` by the relation `T=a lnp+b` where `a` and `b` are constants.The termperature of the ice point and steam point are assigned the numbers 32 and 212 respectively and the values of `p` at these temperatures are 1.86 and 6.81 respectively. Calculate the temperature on this scale when `p=2.50` |
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Answer» Correct Answer - `73.0` |
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| 5. |
Mercury thermometers can be used to measure temperatures uptoA. `100^(@)C`B. `212^(@)C`C. `360^(@)C`D. `500^(@)C` |
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Answer» Correct Answer - C |
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| 6. |
An ideal gas exerts a pressure of `1.52 MP`a when its temperature is `298.15 K` and its volume is `10^-2m^3`. (a) How many moles of gas are there ? (b) What is the mass dencity if the gas is molecular hydrogen ? ( c) What is the mass density if the gas is oxygen ? |
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Answer» Correct Answer - A::B::C::D (a) `n = (p V)/(R T)= ((1.52 xx 10^6)(10^-2))/((8.31)(298.15)` = `6.135` (b) `rho = (PM)/(RT) =((1.52 xx 10^6)(2 xx 10^-3))/(8.31 xx 298.15)` = `1.24 kg//m^3` ( c) `p prop M` (if (p) and (T) are same) `:. rho_(O 2)/(P_(H 2))= (M_(O_2))/(M_(H_2))` `:. rho _(O 2) = ((M_(O 2))/(M_(H 2))) rho_(H 2)` = `((32)/2)(1.23)` = `19.68 kg//m^3`. |
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| 7. |
Three moles of an ideal gas having `gamma = 1.67` are mixed with 2 moles of another ideal gas having `gamma = 1.4`. Find the equivalent value of `gamma` for the mixture. |
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Answer» Correct Answer - A::C `(n_1 + n_2)/(gamma - 1)= (n_1)/(gamma_1 - 1)+(n_2)/(gamma_2 - 1)` `:. (3 + 2)/(gamma -1) = 3/((1.67 - 1)) + 2/(1.4 - 1)` `5/(gamma - 1) = 9.48` Solving, we get `gamma = 1.53`. |
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| 8. |
Two moles of helium (He) are mixed with four moles of hydrogen `(H_2)`. Find (a) `(C_(V)` of the mixture (b) `(C_(P)` of the mixture and ( c) `(gamma) `of the mixture. |
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Answer» Correct Answer - A::B::C Helium is a monoatomic gas. :. `C_V = 3/2 R, C_P = 5/2 R and gamma = 5/3` Hydrogen is a diatomic gas. `:. C_V = 5/2 R` and `C_V = 7/2 R and gamma = 7/5` (a) `(C_V)` of the mixture `C_V = (n_1 C_V 1 + n_2 C_(V 2))/ (n_1 + n_2) = ((2)(3/2 R) + 4(5/2 R))/(2 + 4)` = `(13)/6 R` (b) `(C_P)` of the mixture `C_P = C_V + R = (13)/6 R + R` = `(19)/6 R`. (c) `gamma` of the mixture `gamma = (C_p)/(C_V) = ((19//6) R)/((13//6)R)` = `(19)/(13)`. |
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| 9. |
Two gases occupy two containers (A) and (B). The gas in (A) of volume `0.11 m^3` experts a pressure of `1.38 Mpa`. The gas in (B) of volume `0.16 m^3` experts a pressure of `0.69 Mpa. Two containers are united by a tube of negligible volume and the gases are allowed to intermingle. What is the final pressure in the container if the temperature remains constant ? |
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Answer» Correct Answer - A `n = n_1 + n_2` `(p V)/(R T) = (p_1 V_1)/(R T) + (P_2 V_2)/( R T)` `:. p = (p_1 V_1 + p_2 V_2)/(V)` = `((0.11)(1.38) + (0.16)(0.69))/(0.11 +0.16)` = `0.97 Mpa`. |
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| 10. |
A block of ice at `-10^@C` is slowly heated and converted to steam at `100^@C.` Which of the following curves represents the phenomenon qualitatively?A. B. C. D. |
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Answer» Correct Answer - A |
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| 11. |
Compared to a burn due to water at `100^(@)C` , a burn due to steam at `100^(@)C` isA. More dangerousB. Less dangerousC. Equally dangerousD. None of these |
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Answer» Correct Answer - A |
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| 12. |
The temperature of an ideal gas is increased from `27 ^@ C` to `927^(@)C`. The rms speed of its molecules becomes.A. twiceB. halfC. four timesD. one - fourth |
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Answer» Correct Answer - A `T_1 = 273 + 27 = 300 K` `T_2 = 273 + 927 = 1200 K` `v_(rms) prop sqrt(T)` (T) has become four times. Therefore, v_(rms) will become two times. |
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| 13. |
In a crude model of a rotating diatomic molecule of chlorine `(Cl_2)`, the two `(C l)` atoms are `2.0 xx 10^-10 m` apart and rotate about their centre of mass withb angular speed `omega = 2.0 xx 10^12 "rad" //s`. What is the rotational kinetic energy of one molecule of `C l_(2)` , Which has a molar mass of `70.0 g // mol` ? . |
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Answer» Correct Answer - A::B::C Moment of inertia, `I = 2 (mr^2) = 2mr^2` Here, `m = (70 xx 10^-3)/(2 xx 6.02 xx 10^23)` = `5.81 xx 10^-26 kg` and `r = (2.0 xx 10 ^-10)/2` `1.0 xx 10^-10 m` `:. I = 2 (5.81 xx 10^-26) ( 1.0 xx 10^-10)^2` `= 1.16 xx 10^-45 kg - m^2` `:. K_R = 1/2 I omega^2` `= 1/2 xx (1.16 xx 10^-45) xx (2.0 xx 10^12)^2` `= 2.32 xx 10^-21 J`. |
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| 14. |
Calculate the change in internal energy of `3 .0` mol of helium gas when its temperature is increased by `2. 0 K`. |
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Answer» Correct Answer - D Helium is a monatomic gas. Internal energy of (n) moles of the gas is `U = 3/2 nRT` `:. Delta U = 3/2 nR (Delta T)` Substituting the values, `Delta U = (3/2)(3) (8.31)(2.0)` = `74.8 J`. |
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| 15. |
A tank used for filling helium balloons has a volume of `0 .3 m^3` and contains (2.0) mol of helium gas at `20 .0^@ C`. Assuming that the helium behaves like an ideal gas. (a) What is the total translational kinetic energy of the molecules of the gas ? (b) What is the average kinetic energy per molecule ? |
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Answer» Correct Answer - A::B::C (a) Using `(KE)_(Trans) = 3/2 nRT` with `n = 2.0 mol and T = 273 + 20 = 293 K,` we find that `(KE) _(Trans) = 3/2 (2.0) (8.31)(293)` = `7.3 xx 10^3 J` (b) The average kinetic energy per molecule is `3/2 kT` or `1/2 m bar v^2 = 1/2 m bar (v^2)_(rms) = 3/2 kT` = `3/2 (1.38 xx 10^-23) (293)` = `6.07 xx 10^-21 J`. |
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| 16. |
At what temperature will the particles in a sample of helium gas have an rms speed of `1.0 km//s` ? |
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Answer» Correct Answer - A `v_(rms) = sqrt (3 R T)/M` `:. T = (mv_("rms")^(2))/(3 R)= ((4 xx 10^-3)(10^3)^2)/((3 xx 8.31))` = `160 K`. |
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| 17. |
Four particles have velocities `1, 0,2, and 3 m//s`. The root mean square velocity of the particles (definition wise) is.A. `3.5 m//s`.B. `sqrt (3.5 m//s)`.C. `1.5 m//s`D. `sqrt((14)/(3)) m//s`. |
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Answer» Correct Answer - B `v_(rms) = sqrt(((1)^(2) + (0)^(2) + (2)^(2) + (3)^(4))/(4))` = `sqrt(3.5) m//s`. |
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| 18. |
Consider an `1100` particels gas system with speeds distribution as follows : 1000 particles each with speed `100 m//s` 2000 particles each wityh speed `200 m//s` 4000 particles each with speed `300 m//s` 3000 particles each with speed `400 m//s` and 1000 particles each with speed `500 m//s` Find the average speed, and rms speed. |
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Answer» Correct Answer - A::B::C The average speed is `v_(av) = ((1000) (100) + (2000) (200) + (4000) (300) +(3000) (400) + (1000) (500))/(1100)` `309 m//s` The rms speed is `v_(rms) = sqrt ((1000)(100)^2 + (2000) (200)^2 + (4000) (300)^2 + (3000) (400)^2 + (1000) (500)^2)/(1100)))` `328 m//s`. |
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| 19. |
For any distribution of speeds `v_(rms) ge v_(av)` Is this statement true or false ? |
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Answer» Correct Answer - A Suppose `n_(1)` molecules have `v_(1)` velocity and `n_(2)` molecules have `v_(2) velocity. Then, `v_(av) = (n_1 v_1 + n_2 v_2)/(n_2 + n_2)` But, `v_(rms) = sqrt((n_1 v_1^2 + n_2 v_2^2)/(n_1 +n_2)` Now, `v_(rms) ge v_(av)` because `v_1" and" v_2` may be in opposite directions also. |
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| 20. |
A hammer of mass `1 kg` having speed of `50 m//s`, hit a iron nail of mass `200 gm`. If specific heat of iron is `0.105 cal//gm^(@)C` and half the energy is converted into heat, the raise in temperature of nail isA. `7.1^(@)C`B. `9.2^(@)C`C. `10.5^(@)C`D. `12.1^(@)C` |
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Answer» Correct Answer - A |
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| 21. |
The thermal capacity of 40 g of aluminium (specific heat `=0.2 cal//gm^(@)C`)A. `40cal//^(@)C`B. `160 cal//^(@)C`C. `200 cal//^(@)C`D. `8 cal//^(@)C` |
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Answer» Correct Answer - D |
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| 22. |
Which of the curves in figure represents the telation between Celsius and Fahrenheit temperatures? ` `A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A |
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| 23. |
Assertion : Two bodies at different temperatures if brought in thermal contact do not necessary settle to the mean temperature Reason : The two bodies may have different thermal capacities.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 24. |
Statement-1: Latent heat of fusion of ice is `336000Jkg^(-1)` Statement-2: Latent heat refers to change of state without any change in temperature.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - B |
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| 25. |
When an ideal gas undergoes an adiabatic change causing a temperature change `DeltaT` (i) there is no heat ganied or lost by the gas (ii) the work done by the gas is equal to change in internal eenrgy (iii) the change in internal energy per mole of the gas is `C_(V)DeltaT`, where `C_(V)` is the molar heat capacity at constant volume.A. WeightB. Specific heatC. Relative densityD. Temperature change |
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Answer» Correct Answer - C |
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| 26. |
The latent heat of fusion of a substance is always less than the latent heat vapourization or latent heat of sublimation of the same substace. Explain.A. Greater than its latent heat of fusionB. Greater than its latent heat of sublimationC. Equal to its latent heat of sublimationD. Less than its latent heat of fusion |
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Answer» Correct Answer - A |
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| 27. |
`A` lead bullet of `10g` travelling at `300m//s` strikes against a block of wood and comes to rest. Assuming `50%` heat is absorbed by the bullet, the increase in its temperature is (sp-heat of lead is `150J//Kg-K`)A. `100^(@)C`B. `125^(@)C`C. `150^(@)C`D. `200^(@)C` |
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Answer» Correct Answer - C |
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| 28. |
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by `1^@C` and it is defined under which of the following conditions?A. From `14.5^(@)C` to `15.5^(@)C` at 760 mm of HgB. From `98.5^(@)C` to `99.5^(@)C` at 760 mm of HgC. From `13.5^(@)C` to `14.5^(@)C` at 76 mm of HgD. From `3.5^(@)C` to `4.5^(@)C` at 76 mm of Hg |
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Answer» Correct Answer - A |
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| 29. |
The length of a metallic rod is 5 m at `0^(@)C` and becomes `5.01 m` , on heating upto `100^(@)C` . The linear expansion of the metal will beA. `2.33xx10^(-5)//^(@)C`B. `6.0xx10^(-5)//^(@)C`C. `4.0xx10^(-5)//^(@)C`D. `2.0xx10^(-5)//^(@)C` |
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Answer» Correct Answer - D |
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| 30. |
Ten moles of `(O_2)` gas are kept at temperature `T`. At some higher temperature `2T`, fourty percent of molecular oxygen breaks into atomic oxygen. Find change in internal energy of the gas. |
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Answer» Correct Answer - A::B::D Initial energy Using the relation, `U = (nf)/2 RT` we have, `n = 10, f = 5` for diatomic `O_(2)` gas `U_i = (10 xx 5)/2(RT) = 25 RT` Final energy fourty percent means 4 moles `O_(2)` breaks into `O`. So, it will become `8 moles` of monoatomic gas (O). remaining 6 moles are of diatomic gas `(O_(2)`. But now the new tempreature is (2 T). `:. Uf =(8 xx 3)/2(R) (2T) +(6 +5)/2(R)(2T) =54 RT` So, change in internal energy, `DeltaU = Uf -U_i = 29RT`. |
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| 31. |
At what temperature is the "effective" speed of gaseous hydrogen molecules (molecular weight = 2) equal to that of oxygen molecules (molecular weight = 32) at `47^@ C`? |
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Answer» Correct Answer - B::C `v_(rms)= sqrt((3 R T)/(M)) prop sqrt((T)/(M))` `v_(rms)` is same. Hence, `((T)/M)_(H 2) = ((T)/(M))_(O 2)` or `T_(H 2) = (M_(H 2)/(M_(O 2))) T_(O2)` `= (2/(32)) (273 + 47)` `= 20 K = - 253^@ C`. |
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| 32. |
A partially inflated balloon contains `500 m^3` of helium at `27^@ C` and `1 atm` pressure. What is the volume of the helium at an altitude of `18000 ft`, where the pressure is `0.5 atm` and the temperature is `-3^@ C` ? |
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Answer» Correct Answer - C `n_1 = n_2` `:. (p_1 V_1)/(R T_1) = (P_2 V_2)/(RT_2)` `:. V_2 = ((p_1)/(p_2))((T_2)/(T_1))V_1 =(1/(0.5))((270)/(300))(500)` =`900 m^3`. |
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| 33. |
Latent heat of ice 80 cal/gm . A man melts 60 g of ice by chewing in 1 minute . His power isA. 4800 WB. 336 WC. `1.33 W`D. `0.75 W` |
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Answer» Correct Answer - b |
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| 34. |
find the minimum attainable pressure of an ideal gs in the process `T = t_0 + prop V^2`, where `T_(0)n` and `alpha` are positive constants and (V) is the volume of one mole of gas. |
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Answer» Correct Answer - B `p = (R T)/(V)` `(n = 1)` or `p = (R) /(V) (T_0 + alpha V^2)` For minimum attainable pressure `(d p)/(d V) = 0 or (- R T_0)/(V^2) + alpha R = 0` or `V = sqrt((T_0)/(alpha))` At this volume we can see that `(d^2 p)/(d V^2)` is positive or (p) is minimum. From Eq. (i) `p_(min) = (R T_0)/(sqrt(T_0//alpha)) + alpha R sqrt(T _0 // prop)` = `2 R sqrt (alpha T_0)`. |
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| 35. |
A perfectly conducting vessel of volume ` V = 0.4m^3` contains an ideal gas at constant temperature `T = 273 K`. A portion of the gas is let out and the pressure of the falls by `Delta p = 0.24 atm`. (Density of the gas at (STP) is `rho = 1.2 kg//m^3`). Find the mass of the gas which escapes from the vessel. |
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Answer» Correct Answer - A::B `p_1 V = n_1 R T` `p_2 V = n_2 R T` `(T = 273 K)` `:. (p_1 -p_2) V = (n_1 - n_2)R T = ((m_1 - m_2)/(M)) R T` `(Delta p)V = (Delta m)/(M) R T`…(i) In the initial condition (atv STP) `(R T)/(M) = (p_0)/(rho)`….(ii) `(T = 273 K)` From Eqs. (i) and (ii), we get `Delta m (rho V Deltap)/(p_0) = (1.2 xx 0.4 xx 0.24)/(1.0)` = `0.1152 kg = 115.2 g`. |
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| 36. |
During an experiment, an ideal gas is found to obey a condition `Vp^2 =` constant. The gas is initially at a temperature (T), pressure (p) and volume (V). The gas expands to volume (4V).A. The pressure of gas changes to `(p)/(2)`B. The temperature of the gas changes to `4 T`C. The graph of the above process on (p - T) diagram is parabola.D. The graph of the above process on (p - T) diagram is hyperbola. |
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Answer» Correct Answer - A::D `p prop (1)/sqrt(V)` If volume becomes 4 times than (p) will remain half `V prop (T)/(p)` `:. Vp^2 =` constant `:. (T/P) p^2 =` constant or `p T =` constant or `p prop (1)/(T)` i.e. (p - T) graph is a rectangular hyperbola. If (p) is halved than (T) will becomes two times. |
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| 37. |
Three closed vessels A, B and C are at the same temperature T and contain gasses which obey the Maxwellian distribution of velocities. Vessel A contain only `O_2 and N_2`. If the average speed of the `O_2` molecules in vessel A is `v_1` that of the `N_2` molecules in vessel B us `v_2,` the average speed of the `O_2` molecules in vessel C isA. `((v_1 + v_2))/(2)`B. `V_1`C. `sqrt(v_1 v_2)`D. None of these |
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Answer» Correct Answer - B `v_(av) = sqrt((8 R T) /(pi M))` (T) and (M) are same. Hence, average speed of `O_(2) molecules in both vessels will remain same or `v_(1)`. |
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| 38. |
The resistances of a platinum resistance thermometer at the ice point, the steam point and the boiling point of sulphur are `2.50, 3.50 and 6.50 (Omega)` respectively. Find the boiling point of sulphur on the platinum scale. The ice point and the steam point measure `0^@ and 100^@`, respectively. |
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Answer» Correct Answer - D `t = ((R_t - T_0)/(T_(100) - R_0)) xx 100 = ((6.5 - 2.5)/(3.5 -2.5)) xx 100` = `400^@`. |
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| 39. |
A platinum resistance thermometer reads `0^@ C` when its resistance is `80 Omega and 100^@ C` when its resistance is `90 Omega`. Find the temperature at which the resistance is `86 Omega`. |
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Answer» Correct Answer - C The temperature on the platinum scale is `t = (R_t - R_0)/(R_100 - R_0) xx 100^@` = `(86 - 80)/(90 - 80) xx 100^@ C` = `60^@ C`. |
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| 40. |
The sream poinr and rht ice point of a mercuty thermometer are marked as `80^0 and 20^0`. What will be the temperature in centigrade mercury scale when this thermometer reads `32^0`? |
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Answer» Correct Answer - B::C `(T_C)/(32 - 20) = (100)/(80 - 20) or T_C = 20^@ C`. |
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| 41. |
You work in a materials testing lab and your boss tells you to increase the temperature of a sample by `40.0^@ C`. The only thermometer you can find at your workbench reads in .^@ F`. If the initial temperature of the sample is `68.2^@ F`. What is its temperature in .^@ F` when the desired temperature increase has been achieved ? |
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Answer» Correct Answer - A::B::D Initially `T_C = ((T_F - 32)/(180)) (100)` =`((68.2 - 32)/(180)) (100)` = `20.11^@ C` Finally `T_C = 20.11 + 40 = 60.11^@ C` `:. T_F = 32 + ((180)/(100))T_C` = `32 + ((180)/(100)) (60.11)` = `140.2^@ F`. |
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| 42. |
A metal rod (A) of `25 cm` length expands by `0.050 cm` when its temperature is raised from `0^@ C` to `100^@ C`. Another rod (B) of a different metal of length `40 cm` expamds by `0.040 cm` for the same rise in temperature. A third rod ( C) of `50 cm` length is made up of pieces of rods (A) and (B) placed end to end expands by `0.03 cm` on heating from `0^@ C`. Find the lengths of each portion of the composite rod. |
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Answer» Correct Answer - A::C::D From `Delta l = l prop Delta theta` we have, `0.05 = 25 alpha_A (100)` `:. alpha_A = 0.00002 per^@ C` `0.04 = 40 alpha_B(100)` `alpha_B = 0.00001 per^@ C` ,brgt In third case let (l) is the length of rod (A). Then length of rod (B) will be`(50 - l) cm`. `Delta l = Delta l_1 + Delta l_2` or `0.03 = l (0.00002)(50)+(50 - l)(0.00001)(50)` Solving we get, `l = 10 cm and 50 - l = 40 cm`. |
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| 43. |
A faulty thermometer reads `5^@ C` melting ice and `99^@ C` in steam. Find the correct temperature in .^(@) F` when this faulty thermometer reads `52^@ C`. |
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Answer» Correct Answer - A::B `(T_C)/(52 - 5)= (100)/(99 - 5)` `:. T_C = 50^@ C` Now, using the equation, `(T_C -0)/(100) = (T_F - 32)/(180)` Putting `T_C = 50^@ C`, we get `T_F = 122^@ F`. |
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| 44. |
An air bubble of `20 cm^3` volume is at the bottom of a lake `40 m` deep where the temperature is `4^@ C`. The bubble rises to the surface which is at a temperature of `20 ^@ C`. Take the temperature to be the same as that of the surrounding water and find its volume just before it reaches the surface. |
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Answer» Correct Answer - A::C `n_1 = n_2` `:. (p_1 V_1)/(R T_1) = (p_2 V_2)/(R T_2)` `:. V_2 = (p_1 V_1 T_2)/(p_2 T_1) = ((p_0 + rho g h)V_1 T_2)/(p_0 T_1)` = `((1.01 xx 10^5 + 10^3 xx 10 xx 40)(20)(273 +20))/(1.01 xx 10^5 (273 +4)` =`105 cm^3`. |
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| 45. |
A cylinder whose inside diameter is `4.00 cm` contains air compressed by a piston of mass `m = 13.0 kg` which can slide freely in the cylinder. The entire arrangement is immersed in a water bath temperature can be controlled. The system is initially in equilibrium at temperature `t_i = 20^@ C`. The initial height of the piston above the bottom of the cylinder is `h_i = 4.00 cm`. The temperature of the water bath is gradually increased to afinal temperature `t_f = 100^@ C`. Calculate the final height `h_f` of the piston. . |
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Answer» Correct Answer - C `p = constant` `:. V prop T` or `h prop T` `:. (h_f)/(T_f) = (h_i)/(T_i) h_i` or `h_(f) = ((T_(f))/(T_(i))) h_(i)` = `((273 + 100)/(273 + 20)) (4) = 5.09 cm`. |
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| 46. |
Change each of the given temperature to the Celsius and Kelvin scales : `68^@ F, 5^@ F and 176 ^@ F`. |
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Answer» Correct Answer - A::B::C For (Q). Nos. 1 and 2 Apply the formula, `(T_C - 0)/(100) = (T_K - 273)/(100) = (T_F -32)/(32) = (T_R -0)/(80)`. |
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| 47. |
What is the value of. (a) `0^@ F` in Celsius scale ? (b) (0 K) on fahrenheit scale ? |
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Answer» Correct Answer - A::B::C::D (a) `(T_C - 0)/(100) = (T_F - 32)/(180)` Putting `T_F = 0`, we get `T_C = -17.8^@ C` (b) `(T_C - 0)/(100) = (T_F - 32)/(180)` or `(T_K -273)/(100) = (T_K - 32)/(180)` putting `T_K = 0`, we get `T_F = -459.67^@ F`. |
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| 48. |
At what temperature the Fahrenheit and Celsius scales of temperature give the same reading ?. |
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Answer» Correct Answer - C::D `(T_C - 0)/(100) = (T_F - 32)/(180)` Putting `T_C = T_F`, we get `T_C or T_F = -40^@ C or -40^@ F`. |
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| 49. |
An electric bulb of volume `250`cc was sealed during manufacturing at a pressure of `10^(-3)mm` of mercury at `27^(@)C`. Compute the number of air molecules contained in the bulb. Avogadro constant `=6xx10^(23)mol^(-1),` density of mercury`=13600kgm^(-3)` and `g=10ms^(-2)`. |
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Answer» Correct Answer - A `pV = nRT` `:. n = (p V)/(RT) = ((rho g h)V)/(RT)` `= ((13.6 xx 10^3)(9.8)(10^-6)(250 xx 10^-6))/(8.31 xx 300)` `= 1.33 xx 10^-8` `:.` Number of molecules = `(n) N_A` `= (1.33 xx 10^-8)(6.02 xx 10^23)` `= 8 xx 10^15`. |
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| 50. |
At what temperature the Fahrenheit and kelvin acales of temperature give the same reading ? |
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Answer» Correct Answer - B::D `(T_C - 0)/(100) = (T_F -32)/(180) or (T_K - 273)/(100) = (T_F - 32)/(180)` Putting `T_F = T_K`, We get, `T_F or T_K = 574.25`. |
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