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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. |
Two metal rods of lengths `L_(1)` and `L_(2)` and coefficients of linear expansion `alpha_(1)` and `alpha_(2)` respectively are welded together to make a composite rod of length `(L_(1)+L_(2))` at `0^(@)C.` Find the effective coefficient of linear expansion of the composite rod.A. `(L_(1)alpha_(1)^(2)-L_(2)alpha_(2)^(2))/(L_(1)^(2)+L_(2)^(2))`B. `(L_(1)^(2)alpha_(1)-L_(2)^(2)alpha_(2))/(L_(1)^(2)+L_(2)^(2))`C. `(L_(1)alpha_(1)+L_(2)alpha_(2))/(L_(1)-L_(2))`D. `(L_(1)alpha_(1)+L_(2)alpha_(2))/(L_(1)+L_(2))` |
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Answer» Correct Answer - D Given, the length of rods are `L_(1) and L_(2)` Coefficient of linear expansions are `alpha_(1) and alpha_(2)` As we know that `DeltaL_(1)=alpha_(1)L_(1)DeltatrArrDeltaL_(2)=alpha_(2)L_(2)Deltat` and for combined rod `DeltaL=alpha(L_(1)+L_(2))Deltat` Now, `DeltaL=DeltaL_(1)+DeltaL_(2)` `rArr" "alpha(L_(1)+L_(2))Deltat=alpha_(1)L_(1)Deltat+alpha_(2)L_(2)Deltat` `rArr" "alpha(L_(1)+L_(2))Deltat=alpha_(1)L_(1)Deltat+alpha_(1)L_(2)` `rArralpha=(alpha_(1)L_(1)+alpha_(2)L_(2))/(L_(1)+L_(2))` |
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| 2. |
If at same temperature and pressure, the densities for two diatomic gases are respectively `d_(1) and d_(2)` , then the ratio of velocities of sound in these gases will beA. `d_(1)d_(2)`B. `sqrt(d_(2)//d_(1))`C. `sqrt(d_(1)//d_(2))`D. `sqrt(d_(1) d_(2))` |
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Answer» Correct Answer - B Speed of sound in gases is given by `v_("sound")=sqrt((gammap)/(p)) rArr (v_(1))/(v_(2))=sqrt((p_(2))/(p_(1)))=sqrt((d_(2))/(d_(1)))` |
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| 3. |
Find the coefficient of colume oexpansion of for an idela gas at consant pressure. |
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Answer» For an ideal gas, pV= nRT As P is constant, we have `p.dV=nRT` `therefore" " (dV)/(dT)=(nR)/(p)` Accordign to volume expansion, dV=`Vgammadt` `or" " gamma=(1)/(V).(dV)/(dT)=(nR)/(pV)=(nR)/(nRT)=(1)/(T)` `therefore" " gamma=(1)/(T)` |
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| 4. |
A vessel contains 1 mole of `O_(2)` gas (molar mass 32) at a temperature T. The preesure of the gas is p. An identical vessel containing one mole of He gas (molar mass 4) at temperatuer 2T has a pressure ofA. p/8B. pC. 2pD. 8p |
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Answer» Correct Answer - C `pV=muRT rArrp prop muT " "(thereforeV and R ="constant")` `rArr" " (p_(2))/(p_(1))=(mu_(2))/(mu_(1))xx(T_(2))/(T_(1))rArr(P_(He))/P=1/1xx(2T)/TrArrP_(He)=2p` |
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| 5. |
What is the degree of freedom in case of a monoatomic gas?A. 1B. 3C. 5D. None of these |
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Answer» Correct Answer - B f = 3N - k For monomatomic gas, `" "N=1, k =0` where, N=numbe of particales in the sysetem K= number of independent ralations between the particles. |
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| 6. |
Two vessels A and B having equal volume contain equal masses of hydrogen in A and helium in B at 300 K. Then, mark the correct statement?A. The pressure exerted by hydrogen is half that exerted by heliumB. The pressure exerted by hydrogen is equal to that exacted by heliumC. Average KE of the molecule of hydrogen is half the average KE of the molecules of heliumD. The pressure exeted by hydrogen is twice theat exerted by helium |
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Answer» Correct Answer - D As, pV=nRT `rArr" "P_(H_(2))=(m)/(M_(H_(2))).(RT)/(V)andP_(He)=(m)/(M_(He)).(RT)/(V)` `therefore" "(P_(H_2))/(P_(He))=(M_(He))/(M_(H_2))=(4 xx 10^(-3))/(2xx10^(-3))=2rArrP_(H_2)=2p_(He)` |
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| 7. |
The `p -V` diagram of two different masses `m_(1)` and `m_(2)` are drawn (as shown) at constant temperature `(T)`. State whether `m_(1) gt m_(2) "or" m_(2) gt m_(2)`. |
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Answer» The ideal gas equation is `pV=nRT=m/MRT` `therefore" " m=(pV)((M)/(RT))` or `" " mproppV` if `T=` constant From the graph we can see that `p_(2)Vgtp_(1)V_(1)` (for same p or V). Therefroe, `m_(2)gtm_(1)` |
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| 8. |
Oxygen is filled in a closed metal jar of volume `1.0xx10^(-3)m^(3)` at a pressure of `1.5xx10^(5)Pa.` and temperature `400K`. The jar has a small leak in it. The atmospheric pressure is `1.0xx10^(5)Pa` and the atmospheric temperature is `300K`. Find the mass of the gas that leaks out by time the pressure and the temperature inside the jar equalise with the surrounding. |
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Answer» Given, volume `V =10^(-3) m^(3)`, Pressure `P_(1)=1.5xx10^(5)Pa,P_(2)=10^(5)Pa` Temperature `T_(1)=400 K, T_(2)=300K,` From ideal gas equation, `p_(1)V=n_(1)RT_(1)` `Rightarrow" " n_(1)=p_(1)/(RT_(1))` Similarly,`Similarly" " p_(2)V=n_(2)RT_(2)Rightarrown_(2)=(P_(2)V)/(RT_(2))` `therefore` Number of mole leaked `Delta=n_(1)n_(2)=(P_(1)/T_(1)-P_(2)/T_(2))(V)/(R)=((1.5)/(400)-(1)/(300))xx(10^(5)xx10^(-3))/(8.3)=5.02xx10^(-3)` Mass of gas leaked =`Deltan M_(o)=5.02xx 10^(-3)xx32=0.16 g` |
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| 9. |
One mole of an ideal monoatomic gas is taken at a temperature of `300 K`. Its volume is doubled keeping its pressure constant. Find the change in internal energy. |
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Answer» Since, pressure is constant `therefore" " VpropTtherefore(V_(i))/T_(i)=(V_(f))/T_(f)` `therefore" " T_(f)=(V _(f))/V_(i)TirArrT_(f)=2T_(i)=600K` `therefore" " DeltaU=f/2n.RDeltaT=3/2R(600-300)=450R` |
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| 10. |
The rms speed of oxygen molecules in a gas is `v`. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will becomeA. vB. `sqrt2v`C. 2vD. 4v |
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Answer» Correct Answer - C As the rms speed is given by `v_(rms)=sqrt((3RT)/(M))rArrv_(rms)propsqrt((T)/(M))` When temperature is double and molecules dissociates into atoms, then `(v_(rms1))/(v_(rms2))=(sqrt((T)/(M)))/(sqrt((2T)/(M//2)))=(sqrt((T)/(M)))/(sqrt((4T)/(M)))=(sqrtT)/(sqrt4T)=1/2` If `v_(rms1) is v, then v_(rms2)` will be 2v. |
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| 11. |
Assertion Any straight line on V-T diagram represents iisobaric process. Reason In isobaric process, if V is doubled, then T will also become two times.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explation os Assertion.C. If Assertionis true bur Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D Only that straight line paassing through origin represents isobaric process. |
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| 12. |
A gas has volume `V` and pressure `p`. The total translational kinetic energy of all the molecules of the gas isA. 3/2 pV only if th gas in monoatomicB. 3/2 pV only if the gas is diatomicC. 3/2 pV in all casesD. None of above |
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Answer» Correct Answer - C Translational degree of freedom for any type of gas is three. `therefore` Total translational kinetic energy `=3(1/2nRT)=3/2pV` |
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| 13. |
Prove that the pressure of an ideal gas is numerically equal to two third of the mean translational kinetic energy per unit volume of the gas. |
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Answer» Translational `kE` per unit volume `E=(1)/(2)` ( mass perunit volume) `(v^(2))=(1)/(2)(p)((3p)/(p)=(3)/(2)p)` `or" " p=(2)/(3)E` |
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| 14. |
The molecules of a given mass of a gas have root mean square speeds of `100 ms^(-1) "at " 27^(@)C` and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at `127^(@)C` and 2.0 atmospheric pressure? |
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Answer» Given, At `27^(@)C,` root meann square speed `v_(rms)=100ms^(-1)` `(v_(rms_(1)))=sqrt((3p_(1))/(d_(1)))=sqrt((3p_(1)V_(1))/(M))` According to ideal gas equation `(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))` `Rightarrow" " (V_(1))/(V_(2))=(p_(2)T_(1))/(p_(1)T_(2))=(2xx300)/(400)=(3)/(2)` |
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| 15. |
Which one of the following is a wrong statement in kinetic theory of gases?A. The gas molecules are in random motion.B. The gas molecules are perfect elastic spheres.C. The volum occupied by the molecules of a gas is negligible.D. The collision between molecules is inelastic. |
| Answer» The kinetic therory of gases assume the colision between molecules as perfectly elastic. | |
| 16. |
The temperature at which the root mean squres speed of a gas will be half its value at `0^(@)C` is (assume the pressure remains constant)A. `-86.4^(@)C`B. `-204.75^(@)C`C. `-104.75^(@)C`D. `-6825^(@)C` |
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Answer» Correct Answer - A `v_(rms)propsqrtT` rms speed will remain half if temperature become (1/4) th or `((273)/(4)-273)"^(@)C` or `-204.75^(@)C` |
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| 17. |
Two identical containers joned by a small pipe initially contain the same gas at pressue `p_(o)` and abosolute temperature `T_(o^.)` One container is now mantained at the same temperature while the other is heated to `2T_(0^.)` The commmon pressure of the gases will beA. `3/2p_(o)`B. `4/3p_(o)`C. `3/5p_(o)`D. `2p_(o)` |
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Answer» Correct Answer - B `(n_(1)+n_(2))=(n_(1)+n_(2))_(f)` `(p_(o)V)/(RT_(o))+(p_(o)V)/(RT_(o))=(pV)/(RT_(o))+(pV)/(2RT_(o))` `" "p=(4//3)p_(o)` |
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| 18. |
For a molecule of an ideal gas `n=3xx10^(8)cm^(-3)` and mean free path is `10^(-2)` cm. Calculate the diameter of the lomecule. |
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Answer» Given, `n=3xx10^(8)cm^(-1),lambda=10^(-2)` cm Mean free path is given by `lambda=1/(sqrt2pind_(2))` `rArr" "d^(2)=1/(sqrt2pinlambda` `rArr" " d^(2)=1/(sqrt2xx3.14xx3xx10^(8)xx10^(-2))` `=10^(-6)/(1.414xx3.14xx3)` `rArr" "d=sqrt(7.5xx10^(-6))=sqrt7.5xx10^(-4)` `rArr" " d=2.7xx10^(-4)`cm Hence, the diameter of the molecule of the gas is `2.7xx10^(-4)`cm |
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| 19. |
A constant volume gas thermometer show pressure reading of 50cm and 99 cm of mercury at `0^(@)C` and `100^(@)C` respectively. When the pressure reading is 60 cm of mercury, the temperature isA. `25^(@)`B. `40^(@)`C. `15^(@)`D. `12.5^(@)` |
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Answer» Correct Answer - A `t=(P_(t)-P_(o))/(P_(100)-P_(o))xx100^(@)C=((60-50))/((90-50))=25^(@)C` |
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| 20. |
A perfect gas at `27^(@)C` is heated at constant pressure so as to triple its volume. The tmemperature of th gas will beA. `84^(@)C`B. `900^(@)C`C. `627^(@)C`D. `450^(@)C` |
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Answer» Correct Answer - C `V prop T rArr (V_(1))/(V_(2))=(T_(1))/(T_(2)) rArr (V)/(3V)=((273+27))/(T_(2))` `rArr" "T_(2)= 900K = 627^(@)C` |
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| 21. |
Mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it .At any given time. A. the pressure on EFGH would be zeroB. the pressur all the faces will the equalC. the pressure pf EGGH would be double the pressure on ABCDD. the pressure on EFGH would be half that on ABCD |
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Answer» Correct Answer - A In an ideal gas, when a molecule colides elasitically with a wall, the momentum transferred to each molecule will be twice the manitude of its normal momentum. For the face EFGH, it transfere only half of that on other faces. |
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| 22. |
A sealed container with negiligible coefficient of volumetric expansion contains helium (a monatomic gas). When it is heated from 300 K to 600 K, the average KE of helium atoms isA. halvedB. unchangedC. doubledD. increased by factor `sqrt2` |
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Answer» Correct Answer - C As, kinetic enregy of gas is given as `" "K=1/2mv_(rms)^(2)` Also,`" "v_(rms)=sqrt((3KT)/M)` Now, `v_(rms)prop sqrtT rArr(K_(2))/(K_(1))=[((v_(rms))_(2))/((v_(rms))_(1))]=[sqrt(T2)/sqrt(T_(1))]^(2)` `rArr" "K_(2)=K_(1)((T_(2))/(T_(1)))rArrK_(2)=K_(1)((600)/(300))` `therefore" "K_(2)=2K_(1)` |
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| 23. |
A ballon has 5.0 g mole of helium at `7^(@)C` Calculate (a) the number of atoms of helium in the balloon, (b) the total internal energy of the system. |
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Answer» Given, `n=5.0,T=7^(@)C=7+273=280K` Number of atoms `=nN_(A)=6.02xx10^(23)=30xx10^(23)` Average kinetic energy per molecule `=(3)/(2)kT` `therefore` Total internal energy `=(3)/(2)kTxxN` `=(3)/(2)xx1.3.8xx10^(23)xx28030xx10^(23)` `=(3)/(2)xx30xx280xx1.38=1.74xx10^(4)J` |
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| 24. |
A metal ball having a diameter of `0.4` m is heated from `273` to `360 K.` If the coefficient of areal expansion of the material of the ball is `0.000034 K^(-1)`, then determine the increase in surface area of the ball. |
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Answer» Given, dimeter =`0.4` m and radius, `r=(0.4)/(2)=0.2` m `therefore " " ` Area of ball `A_(o)=4pir^(2)=4xxpixx(0.2)^(2)=0.5024 m^(2)` Temperature, `DeltaT=T_(2)-T_(1)-T_(1)=360K-273K=87K` Coefficient of area expansion, `beta=0.000034K^(-1)` Apply `DeltaA=betaA_(o)DeltaT` `Rightarrow" " DeltaA=0.000034xx0.5024xx0.5024xx87=0.001486=1.486xx10^(3)m^(2)` |
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| 25. |
A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through 1K, then the percentage increase in its pressure will beA. `0.4%`B. `0.2%`C. `0.1%`D. `0.8%` |
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Answer» Correct Answer - A `p prop T rArr(p_(1))/(p_(2))=(T_(1))/(T_(2)) rArr(p_(2)-p_(1))/(p_(1))=(T_(2)-T_(1))/(T_(1))` `rArr" "((Deltap)/(p))%=((251-250)/(250))xx100=0.4%` |
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| 26. |
On heating a glass block of `10,000 cm^(3)` from `25^(@)C` to `40^(@)C`, its volume increase by `4cm^(3)`. Calculate coefficient of linear expansion of glass. |
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Answer» Given Volume V=`10000 cm^(3)` Temperature, `DeltaT=(40-25)=15^(@)C,DeltaV=4cm^(3)` `therefore ` Ceefficeient of cubical expansion is given by `gamma=(DeltaV)/(V.DeltaT)=(4)/(10000xx5)` `=26.67xx10^(-6)"^(@)C^(-1)` `therefore` Coeffcient of linear expansion, `alpha=(gamma)/(3)=(26.67xx10^(-6))/(3)=8.89xx10^(-6)" ^(@)C^(-1)` |
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| 27. |
When a liquid is heated in a glass vessel, its coefficient of apparent expension is `1.03xx10^(-3)//^(@)C.` When the same liquid is heated in a copper vessel, its coefficient of apparent expansion is `1.006xx10^(-3)//^(@)C.` If the coefficient of linear expension of copper is `17xx10^(-6)//^(@)C,` then the coefficient of linear expansion of glassA. `8.5xx10^(-4)//^(@)C`B. `9xx10^(-6)//^(@)C`C. `27xx10^(-6)//^(@)C`D. `10xx10^(-4)//^(@)C` |
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Answer» Correct Answer - B Here, `gamma_(r)=gamma_(ag)+3alpha_(g)=alpha_(ac)+3alpha_(c)` `rArr1.03xx10^(-3)+3alpha_(g)=1.006xx10^(-3)+3xx(17xx10^(-6))` `alpha_(g)=9xx10^(-6)//""^(@)C` |
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| 28. |
A glass vessel of volume `V_(o)` is comopleated filled with volume of the liquid will overflow? Cofficient of linear expansion of gass =`alpha` and coefficient of volume expension of the liquid =`gamma_(l)`A. `2V_(o)AT(gamma_(l)-3alpha_(g))`B. `V_(o)DeltaT(gamma_(l)-3alpha_(g))`C. `V_(o)DeltaT(gamma_(l)-alpha_(g))`D. `(V_(o)DeltaT)/2(gamma_(l)-3alpha_(g))` |
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Answer» Correct Answer - B Volume of the liquid overflow = Increase in the volume of the liquid -Increase in the volume of the container `=[V_(o)(1+gamma_(t)DeltaT)-V_(o)]-[V_(o)(1+gamma_(g)DeltaT)-V_(o)]` `=V_(o)DeltaT(gamma_(l)-gamma_(g))=V_(o)DeltaY(gamma_(l)-3alpha_(g))" "(therefore gamma_(g)~~ 3alpha_(g))` |
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| 29. |
Figure shows two flasks connected to each other. The volume of the flask `1` is twice that of flask `2`. The system is filled with an ideal gas at temperature `100 K` and `200 K` respectively. If the mass of the gas in `1` be `m` then what is the mass of the gas in flask `2` A. mB. m/2C. m/4D. m/8 |
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Answer» Correct Answer - C `pV=m/MRTrArrVpropmT rArr(V_(1))/(V_(2))=(m_(1))/(m_(2)).(T_(1))/(T_(2))` `rArr" "(2V)/V=m/(m_(2))xx100/200 rArrm_(2)=m/4.` |
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| 30. |
A glass flask of volume `200cm^(3)` is just filled with mercury at `20^(@)C`. The amount of mercury that will overflow when the temperature of the system is raised to `100^(@)C` is `(gamma_(glass)=1.2xx10^(-5)//C^(@),gamma_(mercury)=1.8xx10^(-4)//C^(@))`A. 2.15 `cm^(3)`B. 2.69 `cm^(3)`C. 2.52 `cm^(3)`D. 2.25 cm^(3)` |
| Answer» Correct Answer - B | |
| 31. |
Assertion In summers, a metallic scale will read more than the actual. Reason In summers, length of methallic scale will increase.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explation os Assertion.C. If Assertionis true bur Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D It will read less than the actual. |
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| 32. |
A faulty thermometer has its fixed points marded 5 and 95. if the temperature of a body as shown on the Celsius scale is 40, then its temperature shown on this faulty thermometer isA. `39^(@)`B. `40^(@)`C. `41^(@)`D. `44.4^(@)` |
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Answer» Correct Answer - C Temperature shown of faulty thermometer will be `" "t=5+((95-5)/(100))xx40=41^(@)` |
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| 33. |
Assume that one early morning when the temperatures in `10^(@)C,` a driver of an automobile gets his gasoline tank which is made of steel, tilled with `75` L of gasoline, which is also at `10^(@)C.` During the day, the tempaerature rises to `30^(@)C,` how much gasoline will overflow ? (Given, `alpha ` for steel `=1.2xx 10^(-5)"^(@)C^(^1), gamma ` for gasoline `=9.5xx10^(-4)"^(@)C^(-1))` |
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Answer» Change in volume of gasoline, ` DeltaV_(g)= gamma_(g)VDeltaT` `=9.5xx10^(-4)xxVxx20=190xx10^(-4)V` Change in volume of steel tanks, `DeltaV_(s)=gamma_(s)VDeltaT` `=3alpha_(s)VDeltaT=3xx1.2xx10^(-5)xxVxx20=7.2xx10^(-4))V` `therefore` Volume of gasoline that overflows `=DeltaV_(g)-DeltaV_(s)=(190xx10^(-4)-7.2xx10^(-4))V` `=182.8xx10^(-4)xx75=1.37L.` |
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| 34. |
At `27^(@)C` temperature, the kinetic energy of an ideal gas is `E_(1^.)` If the temperature is increassed to `327^(@)C,` then the kinetic energy will beA. `(E_(1))/sqrt2`B. `sqrt2 E`C. `2E_(1)`D. `(E_(1))/2` |
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Answer» Correct Answer - C `E=3/2kTrArr E prop T` `therefore" "(E_(2))/(E_(1))=(T_(2))/(T_(1))=(600)/(300)=2rArr E_(2)=2E_(1)` |
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| 35. |
A bar of iron is 10 cm at `20^(@)C`. At `19^(@)C` it will be (`alpha` of iron `=11xx10.//^(@)C`)A. `11xx10^(-6)` cm longerB. `11xx10^(-6)` cm shorterC. `11xx10^(-5)` cm shorteD. `11xx10^(-5)` cm longer |
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Answer» Correct Answer - C Chane is length, `Deltal=l_(o)alphaDeltaT=10xx11xx10^(-6)(19-20)` `" "=-11xx10^(-5)` cm |
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| 36. |
Heat given to a system can be associated withA. kinetic energy of amotion of moleculesB. kinetic energy of orderly motio of moleculesC. total kinetic energy of random and orderly motion of moleculesD. dinetic energy of random motion in some csases and kinetic energy of orderly motion in other |
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Answer» Correct Answer - A We know that as temperature increases vibrationof molecules about their mean position increases hence, kinetic associated with random motion of molecules increase. |
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| 37. |
ABCD is thin iron shet. A hole with radius r is made in it in the middle. If we heat up the sheet, what will happen to the holw.A. Radius will increaseB. Radius will decreasesC. It will remains constantD. It can increase or decrease depending upn size |
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Answer» Correct Answer - A The hole will increase in size. |
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| 38. |
A cylindrical steel plug is inserted into a circular hole of diameter 2.60 cm in a brass plate. When the plug and the plates are at a temperature of `20^(@)C,` the diameter of the plug is 0.010 cm cmaller than that of the hole. The temperature at which the plug will just fit in it is `("Given",alpha_(steel)=(11xx10^(-6))/("^(@)C)and alpha_(bress)=(19xx10^(-6))/C)`A. `-48^(@)C`B. `-20^(@)C`C. `-10^(@)C`D. `-485^(@)C` |
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Answer» Correct Answer - D Diameter of brass plate = 2.6 cm and diameter of steel plate `" "=(2.6-0.01)cm=2.59cm` Now, `" "(d+Deltad)_(b)=(d+Deltad)_(s)` or `" "d_(b)(1+alphaDeltatheta)_(b)=d_(s)(1+alphaDeltatheta)_(s)` `therefore" "2.6 (1+19xx10^(-6)Delta theta )=2.59 (1+11 xx10^(-6)Deltatheta)` `therefore" "Deltatheta=-478^(@)C or theta_(f)=-458^(@)C` |
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| 39. |
A steel rod of diamerer 1 am is clamped firmly at each end when its temperature is `25^(@)C` so that it cannot contract on cooling The tension in the rod at `0^(@)C` is approximately `(alpha=10^(-5)//^(@)C,Y=2xx10^(11)Nm^(-2))`A. 4000 NB. 7000 NC. 7400 ND. 4700 N |
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Answer» Correct Answer - A Strain `=(Deltal)/l=alphaDelta theta` Strss = Y xx strain = Y `alpha Delta theta` `therefore` Force or tension, T = stress xx are = YA `alpha Delta theta` `" " =(piYalphad^(2)Deltatheta)/4(A=(pid^(2))/4)` or `" " T=(pi xx 2 xx 10^(11) xx 10^(-5) xx 10^(-4) xx25)/(=3926N~~400N)` |
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| 40. |
Bimetal strips are used forA. metal thermometerB. opening or closing electrical circuitsC. thermomsatsD. All of the above |
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Answer» Correct Answer - D When bimetallic strip is heated, the bress expands more than the steel the strip curves with the brass on the outside. If strip is cooled, it curves with the steel on outside and can be used in thermostat, as a switch or in metal thermometers. |
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| 41. |
Some gas at 300K is enclosed in a container. Now the container is placed on a fast moving train. While the train is in motion, the temperature of the gasA. rises above 300 KB. falls below 300 KC. remains unchangedD. becomes unsteady |
| Answer» Correct Answer - C | |
| 42. |
Match the following columns. `{:(,"ColumnI",, "ColumnII"),((A),Inp=2/3E","Eis,(p),"isochoric"),((B),In U =3RT "for and monotomic gas"U is,(q),"Translational kinetic energy of unit volume"),((C),V/T= "constant is valid for",(r), "Internal energy of one mole"),((D), p/T= "constant is associated with",(s),"isobaric process"):}` |
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Answer» Correct Answer - `(Atos,Btor,Ctoq,Dto"p")` Internal enerfy of moles of an ideal gas, `" "U=n(f/2RT)" "(f="degree of freedom")` For monotomic gas, f=3 `" "U=(3nRT)/(2)` `or U=3RT ` for n = 2 is valid for any gas |
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| 43. |
A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowerd until the Fahrenheit thermometer registenrs `140^(@).` What is the fall in temperature as registered by the centigrade thermometer?A. `30^(@)`B. `40^(@)`C. `60^(@)`D. `80^(@)` |
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Answer» Correct Answer - B `(DeltaT_(c))/(100)=(DeltaT_(F))/(180)=(212-140)/(180)` i.e., `" "DeltaT_(c)=100xx(72)/(180)=40^(@)C` `therefore` Fall temperature `= 40^(@)C.` |
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| 44. |
From the p-T graph want conclusion can be drawn? A. `V_(2)= V_(1)`B. `V_(2) lt V_(1)`C. `V_(2) gt V_(1)`D. None of these |
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Answer» Correct Answer - C As `theta_(2)gttheta_(1)rArrtantheta_(2)gttan theta_(1)` `rArr" "((T)/(p))_(2)gt((T)/(p))_(1) rArr T/p propVrArrv_(2)gtv_(1)` |
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| 45. |
The `p-T` graph for the given mass of an ideal gas is shown in figure. What inference can be drawn regarding the change in Volume (whtherit is constant, increasing or decreasing) ? |
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Answer» Form th given graph we can write the `p-T` equation as, `" " p/T=a+b/T` now, `" " T_(B)gtT_(A)` `therefore" " b/T_(B)ltb/T_(A)` or `" " (p/T)_(B)lt(p/T)_(A)` or `" " (T/p)_(B)gt(p/T)_(A)` or `" " V_(B)gtV_(A)` Thus, us we move from A to B, volume of the is increasing. |
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| 46. |
The rms speed of oxygen molecule in a gas at `27^(@)C` would be given byA. 483 `ms^(-1)`B. 966 `ms^(-1)`C. 4.83 `ms^(-1)`D. 9.66 `ms^(-1)` |
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Answer» Correct Answer - A The rmsspeed is given by `v_(rms)=sqrt((3RT)/(Mol.wt))=sqrt((3xx8.3xx(273+27))/(2xx16xx10^(-3)))=sqrt((3xx8.3xx300)/(32xx10^(-3)))` `rArr" "v_(rms)=483ms^(-1)` |
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| 47. |
Calculate the ratio of KE of molecule of oxygen and neonj gas at `27^(@)C.` |
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Answer» For oxygen gas number of degree of freedonis `5` `therefore" "(KE)_("oxygen")=5/2kT` None is monatomic, thrrefore it has `3` degrees of freedom `" "(KE)_("neon")=3/2kT` Hence, the ratio of their `KE=((KE)_("oxygen"))/((KE)_("neon"))=(5//2kT)/(3//2kT)=5/3` |
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| 48. |
The mass of hydrogen molecule is `3.32xx10^(-27)` kg. If `10^(23)` hydrogen molecules strick per second at 2 `cm^(2)` area of a rigid wall at an angle of `45^(@)` from the normal and rebound back with a speed of 1000 `ms^(-1)`, then the pressure ezxeted on the wass is A. `2.34xx10^(3)` PaB. `0.23xx10^(6)`PaC. `0.23xx10^(3)`PaD. `23.4xx10^(3)`Pa |
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Answer» Correct Answer - A `F=(Deltap)/(Deltat)=(nm)(2vcos theta)` Pressure, `p=F/A=((nm)(2vcos theta))/A` `" "=(10^(23)xx3.32xx10^(-27)xx2xx100xxcos45^(@))/(2xx10^(-4))` `" "=2.34xx10^(3)Nm^(-2)` |
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| 49. |
A cylinder containe 20 kg of `N_(2)` gas (M= 28 kg `K^(-1) mol^(-1))` at a [ressire pf 5 atm. The mass of hydrogen (M = 28 kg `K^(-1) mol^(-1))` at a pressure of 3 atm contained in the same cylinder at same temperature isA. 1.08 kgB. 0.86 kgC. 0.68 kgD. 1.68 kg |
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Answer» Correct Answer - B `V/T=(nR)/p` V and T for both cases in same. Hence `(n_(1))/(p_(1))=(n_(2))/(p_(2))or (m_(1))/(p_(1)M_(1))=(m_(2))/(p_(2)M_(2))` or `" "m_(2)(p_(2)M_(2))/(p_(1)M_(1)).m_(1)=((3)(2))/((5)(28)).20=0.86kg` |
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| 50. |
Calculate the average kinetic enrgy of oxygen molecule at `0^(@)`C. (R=8.314 J `mol^(-1)` `K^(-1),N_(A)`=`6.02xx10^(23))` |
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Answer» Given, T = `0^(@)C` = 273 K Oxygen is diatomic molecule, therefore it has 5 degrees of freedom, 3 translationl 2 rotational. `therefore" "KE=5/2(RT)/N_(A)=5/2xx(8.314)/(6.023xx10^(23))xx273=9.4xx10^(-21)J` |
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