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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.
| 251. |
Two solid bodies of equal mass m initially at `T = 0^(@)C` are heated at a uniform and same rate under identical conditions. The temperature of the first object with latent heat `L_(1)` and specific heat capacity in solid state `C_(1)` changes according to graph `1` on the diagram. the temperature of the second object withe latent heat `L_(2)` and specific heat capacity in solid state `C_(2)` changes according to graph `2` on the diagram. based on what is shown on the graph, the latent heat `L_(1) and L_(2)` and the specific heat capacities `C_(1) and C_(2)` in solid state obey which of the following relationships? A. `L_(1) gt L_(2), C_(1) lt C_(2)`B. `L_(1) lt L_(2), C_(1) lt C_(2)`C. `L_(1) gt L_(2), C_(1) gt C_(2)`D. `L_(1) lt L_(2), C_(1) gt C_(2)` |
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Answer» Correct Answer - A If heat is supplied at constant rate `P`, then `Q = P Delta tau` and as during change of state `Q = mL, so,mL = P Delta t` i.e., `L = [P/m] Delta t = P/m` (length of line `AB`) Hence `L_(1) gt L_(2)` i.e., the ratio of latent heat of fusion of the two substances are in the ratio `3:4` In the portion `OA` the substance is in solid state and its temperature is changing. `Delta Q = mCDelta T` and ` Delta Q = P Delta t` so, `(Deltat)/(Delta t) = (P)/(mC)` or slope = `(P)/(mS) = [ "as" (Delta T)/(Delta t)="slope" ]` Hence, `C_(1) lt C_(2)`. |
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| 252. |
A metal ball suspended from the hook of a spring balance is kept immesed in a liquid other than water. On increassing the temperature of this liquid, the reading in the spring balance.A. increasesB. DecreasesC. Remains sameD. May increase or decreases. |
| Answer» Correct Answer - A | |
| 253. |
The coefficient of cubical expansion of liquid and glass are in the ratio of `8:1`. The volume of the liquid to be taken into `800cc` container so that the unoccupied portion remains costant isA. `10cc`B. `100cc`C. `80cc`D. `8cc` |
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Answer» Correct Answer - B `gamma_(l)/(gamma_(g)) = (8)/(1), V_(1)gamma_(g)` |
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| 254. |
On heating a liquid of coefficient of cubical expansion `gamma` in a container having coefficient of linear expansion `gamma//3`. The level of liquid in the container willA. riseB. fallC. reamain sameD. over flows |
| Answer» Correct Answer - C | |
| 255. |
A liquid with coefficient of real volume expansion `(gamma)` is filled in vessel of coefficient of linear expansion `gamma/3`. When the systeam is heated then. a) The volume of spece above liquid remains same. b) The level of liquid relative to vessel reamins same. c) The fraction of volume of liquid in vessel reamains same.A. Only `(a)` is correctB. Only `b` & `c` are correctC. Only `(c)` is trueD. All are true |
| Answer» Correct Answer - D | |
| 256. |
If `alpha, beta` and `gamma` coefficient of linear, superficial and volume expansion respectively, tehnA. `(beta)/(alpha) = 1/2`B. `(beta)/(gamma) = 2/3`C. `(gamma)/(alpha) = 3/2`D. `(beta)/(alpha) = (gamma)/(beta)` |
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Answer» Correct Answer - B As `beta = 2alpha` and `gamma = 3alpha` `:. (beta)/(gamma) = (2alpha)/(3alpha) = (2)/(3)` |
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| 257. |
The ratio among coefficient of volume expansion, superficial expansion and linear expansion i.e., `gamma` : `beta` : `alpha` isA. `1:2:3`B. `3:2:1`C. `4:3:2`D. all of these |
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Answer» Correct Answer - A |
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| 258. |
A liquid with coefficient of volume expansion `gamma` is filled in a container of a material having coefficient of linear expansion `alpha` . If the liquid overflows on heating, thenA. `gamma = 3alpha`B. `gamma gt 3alpha`C. `gamma lt 3alpha`D. `gamma = alpha` |
| Answer» Correct Answer - B | |
| 259. |
The coefficient of linear expansion of crystal in one direction is `alpha_(1)` and that in every direction perpendicular to it is `alpha_(2)`. The coefficient of cubical expansion isA. `alpha_(1) + alpha_(2)`B. `2alpha_(1) + alpha_(2)`C. `alpha_(1) + 2alpha_(2)`D. None of these |
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Answer» Correct Answer - C `V=V_(0)(1+gamma Delta theta)` `L^(3) = L_(0)(1+alpha_(1) Delta theta)L_(0)^(2) (1+alpha_(2)Delta theta)^(2)` `L_(0)^(3)(1+alpha_(1)Delta theta)(1+alpha_(2)Delta theta)^(2)` Since `L_(0)^(3) = V_(0)` and `L^(3) =V` Hence `1+gamma Delta theta = (1+alpha_(1)Delta theta)(1+alpha_(2)Delta theta)^(2)` `~=(1+alpha_(1)Delta theta) (1+2 alpha_(2)Delta theta)` `~=(1+alpha_(1)Delta theta+2 alpha_(2)Delta theta)` `rArr gamma = alpha_(1)+ 2 alpha_(2)`. |
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| 260. |
The coefficient of linear expansion of an in homogeneous rod change linearly from `alpha_(1)` to `alpha_(2)` from one end to the other end of the rod. The effective coefficient of linear expansion of rod isA. `alpha_(1) + alpha_(2)`B. `(alpha_(1) + alpha_(2))/(2)`C. `sqrt(alpha_(1)alpha_(2))`D. `alpha_(1) - alpha_(2)` |
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Answer» Correct Answer - B `alpha_(n)=alpha_(1)+((alpha_(2)-alpha_(1))/(L))x,DeltaL=overset(L)underset(0)intalpha_(n)dxDeltat` `L=((alpha_(1)+alpha_(2))/(2))LDeltaT , alpha_(eff) = ((alpha_(1)+alpha_(2))/(2))` |
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| 261. |
Statement-1: On a cold winter day, an iron railing appears much colder than a wooden fence post. Statement-2: heat capacity of iron is different from wood.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
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Answer» Correct Answer - B |
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| 262. |
Water is used as a collent becauseA. it has lower densityB. it has low specific heat.C. it has high specific heat.D. It is earily available. |
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Answer» Correct Answer - C Owing to its high specific heat, water is used as a coolant in automobile radiators as well as a heater in hot water begs. |
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| 263. |
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. Find the rise in temperature of the block in `2.5` minutes, assuming `50%` power is used up in heating the machine itself or lost to the surropundings. (Specific heat of aluminium `= 0.91 J g^(-1).^(@)C^(-1)`)A. `100 .^(@)C`B. `103 .^(@)C`C. `150 .^(@)C`D. `155 .^(@)C` |
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Answer» Correct Answer - B Here, `P = 10 kW = 10^(4) W, m = 8 kg` time, `t = 2.5` minute `= 2.5 xx 60 = 150 s` Specific heat, `s = 0.91 j g^(-1) C^(-1)` Total energy `= P xx t = 10^(4) xx 150 = 15 xx 10^(5) J` As `50%` of energy is lost, `:.` Energy available, `DeltaQ = 1/2 xx 15 xx 10^(5) = 7.5 xx 10^(5) J` As `DeltaQ = msDeltaT` `DeltaT = (DeltaQ)/(ms) = (7.5 xx 10^(5))/(8 xx 10^(3) xx 0.91) ~~ 103 .^(@)C` |
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| 264. |
A copper block of mass `2.5 kg` is heated in a furnace to a temperature of `500^(@)C` and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat copper `= 0.39 J//g^(@)C` heat of fusion of water `= 335 J//g`). |
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Answer» Mass of the copper block, m = 2.5 kg = 2500 g Rise in the temperature of the copper block, `Delta theta = 500^(@)C` Specific heat of copper, `C = 0.39 J g^(–1)"@C^(–1)` Heat of fusion of water, `L = 335 J g^(–1)` The maximum heat the copper block can lose, Q = mCΔθ = 2500 × 0.39 × 500 = 2500 × 0.39 × 500 Let `m^(1) g` be the amount of ice that melts when the copper block is placed on the ice block. The heat gained by the melted ice, `Q = m^(1)L` `:.m_(1)=(Q)/(L)=(487500)/(335)=1455.22 g` Hence, the maximum amount of ice that can melt is 1.45 kg |
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| 265. |
As the temperature is increased, the period of a pendulumA. Increases as its effective length increases even through its centre of mass still remains at the centre of the bob.B. decreases as its effective length increases even though its centre of mass still remains at the centre of thed bob.C. Increases as its effective length increases due to shifting of centre of mass below the centre of the bob.D. decreases as its effective length remains same but the centre of mass shifts above the centre of the bob. |
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Answer» Correct Answer - A Time period of the sample pendulum is `T = 2pi sqrt((L)/(g))` Where L is effiective length of the pendulum . With increases in temperature , the effective (L) of simple pendulum increase even through its centre of mass still remains at the centre of the bob. From ` (i), T prop sqrt(L)` So T increases as temperature increases. |
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| 266. |
A woman working in the field drinks her morning coffee out of an aluminium cup. The cup has a mass of 0.120 kg and is initially at `20.0^(@)C` when she pours in 0.300 kg of coffee initially at `70.0^(@)C`. What is te final temperature after the coffee and the cup attain thermal equilibrium? (Assume that coffee has the same specific heat as water and that there is no heat exchange with the surroundings). |
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Answer» `Q_("coffee")=m_("coffee")DeltaT_("coffee")` `Q_("coffee")=(0.300kg)(4190J//kg.K)(T-70.0^(@)C)` The positive heat is gained by the aluminium cup is ltBrgt `Q_("aluminium")=M_("aluminium").C_("aluminium")DeltaT_("aluminium")` ltBrgt `=(0.120kg)(910J//kg.K)(T-20.0^(@)C)` We equate the sum of these two quantities of heat to zero, obtaining an algebraic equation for T. `Q_("coffee")=Q_("aluminium")=0` `(0.300kg)(4190J//kg.K)(T-70.0^(@)C)+(0.120kg)(910J//kg.K)(T-20.0^(@)C)=0` `T=66.0^(@)C`. |
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| 267. |
Assertion : Water kept in an open vessel will quickly evaporate on the surface of the moon. Reason : The temperature at the surface of the moon is much higher than boiling point of the water.A. If both the assertion and reason are true and reason is the correct explanation of the assertion.B. If both the assertion and reason are true but the reason is not correct explanation of the assertion.C. If the assertion is true but the reason is falseD. If the both the assertion and reason are false. |
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Answer» Correct Answer - A water would evaporate quickly because there is no atmosphere on moon, due to which surface temperature of moon is much higher than earth (maximum surface temperature of moon is `123^(@)C`). |
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| 268. |
`20 gm` ice at `-10^(@)C` is mixed with `m gm` steam at `100^(@)C`. The minimum value of `m` so that finally all ice and steam converts into water is: `("Use " s_("ice") = 0.5 "cal gm"^(@)C, S_("water") = 1 cal//gm^(@)C, L`) (melting) `= 80 cal// gm` and `L ("vaporization") = 540 cal//gm`)A. `(185)/(27) gm`B. `(135)/(17) gm`C. `(85)/(32) gm`D. `(113)/(17) gm` |
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Answer» Correct Answer - C For minimum value of `m`, the final temperature of the mixture must be `0^(@)C` `:. 20xx1/2xx10+20xx80=m 540+m*1*100` `:. M=1700/640 = 85/32 gm`. |
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| 269. |
The coefficient of real expansion of liquid is `gamma_(R)` and the coefficient of appareent expansion of the liquid is `gamma_(A)`. The coefficient of cubical expansion of the vessel is `gamma`. If `gamma_(R) : gamma_(A) = 4:1` then `gamma_(A) : gamma` isA. `3 : 1`B. `1 : 3`C. `4 : 1`D. `1 : 4` |
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Answer» Correct Answer - B `gamma_(R) = gamma_(A) + gamma_(g)` |
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| 270. |
A cube of coefficient of linear expansion `alpah` is floating in a bath containing a liquid of coefficeient of volume expansion `gamma_(l)`. When the temperature is raised by `DeltaT`, the depth upto which the cube is submerged in the liquid remains the same. Find relation between `alpha` and `gamma_(l)` |
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Answer» Befor heating `Mg = F_(b), Mg = V_(1)rho_(1)g = Axrho_(1)g` ……..(1) After heating `Mg = F_(b)^(1), Mg = V^(1)rho_(1)^(1)g` ……(2) Equating `(1)` & `(2) Axrho_(1)^(1)g` `Arho_(1)g = A(l + betaDelta t)(rho_(1))/(1 + gammaDetla t)g` `1 + gammaDelta t = 1 + betaDeltat, gamma = beta :. gamma = 2alpha` |
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| 271. |
A spherical black body of radius n radiates power p and its rate of cooling is R. then.A. `p prop n`B. `p prop n^(2)`C. `R prop n^(2)`D. `R prop (1)/(n)` |
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Answer» Correct Answer - B::D |
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| 272. |
A gas is enclosed in a vessel at a pressure of `2.5 atm`. Due to leak in the vessel., after some time the pressure is reduced to `2 atm`, temperature remaining unchanged. The percentage of gas that has leaked out isA. `40`B. `15`C. `20`D. `25` |
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Answer» Correct Answer - C `P prop m` when `V` & `T` are constant `rArr (DeltaP)/(P) = (Deltam)/(M)` |
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| 273. |
Liquid oxygen at `50 K` is heated to `300 K` at constant pressure of `1 atm`. The rate of heating is constant. Which of the following graphs represents the variation of temperature with time?A. B. C. D. |
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Answer» Correct Answer - B Iniitially liquid oxygen will gain the temperature up to its bolling temperature, then it changes its state of gas. After this again its tmperature will increase, so coresponding graph will be option (b). |
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| 274. |
A solids floats submerged in a liquid. When the liquid is heated, which of the following is most likely to happen?A. Solid may sinkB. solid may float with a part outside the surfaceC. Solid may first sink and then rise upwardsD. Solid may oscillate vertically |
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Answer» Correct Answer - A The density of the liquid will decrease on heating. However, the relative change in volume of the solid will be negligible as compared to that in density of liquid. Hence, upthrust action on the solid will decrease and the solid will sink. |
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| 275. |
A sphere of diameter `8cm` and mass `275g` floats in a bath of liquid. As the temperature is raised, the sphere begins to sink at a temperature of `40^(0)C`. If the density of the liquid is `1.5g//cm^(3)` at `0^(0)C`, find the coefficient of cubical expansion of the liquid. Neglect the expansion of the sphereA. `12 xx 10^(-4 //^(0))C`B. `25 xx 10^(-4 //^(0))C`C. `15 xx 10^(-4 //^(0))C`D. `115 xx 10^(-4 //^(0))C` |
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Answer» Correct Answer - A `gamma = (d_(0) - d_(t))/(d_(t) xx t),d_(t) = (m)/(V)` |
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| 276. |
A non-conducting body floats in a liquid at `20^(@)C`. with `2//3` of its volume immersed in the liquid When liquid temperature is increased to `100^(@)C, 3//4` of body volume is immersed in the liquid. Then the coefficient of real expansion of the liquid is... (neglecting the expansion of container of the liquid)A. `1.56 xx 10^(-4 //@)C`B. `15.6 xx 10^(-4 //@)C`C. `1.56 xx 10^(-5 //@)C`D. `15.6 xx 10^(-5 //@)C` |
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Answer» Correct Answer - B `V_(1) = 2V//3, V_(2) = 3V//4, gamma = (V_(2) - V_(1))/(V_(1)Deltat)` |
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| 277. |
An enclosure of volume `3` litre contains `16g` of oxygen, `7 g` of nitrogen and `11 g` of carbon - di-oxide at `27^(@)C`. The pressure exerted by the mixture is approximately `[R = 0.0821 lit "atm mole"^(-1) K^(-1)]`A. `1` atmosphereB. `3` atmosphereC. `9` atmosphereD. `8.3` atmosphere |
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Answer» Correct Answer - D `PV = (m)/(M)RT , P = (RT)/(V)((m_(1))/(M_(1)) + (m_(2))/(M_(2)) + (m_(3))/(M_(3)))` |
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| 278. |
Consider a compound slab consisting of two different material having equal thickness and thermal conductivities `K` and `2K` respectively. The equivalent thermal conductivity of the slab isA. `2/3 K`B. `sqrt(2) K`C. `3 K`D. `4/3 K` |
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Answer» Correct Answer - D The equivalent thermal conductivity when the slabs are connected in series is `K_(eq) = (sumx_(n))/((sumx_(n))/(sumK_(n)))= (x+x)/(x/K+x/(2K)) = (2x(2K^(2)))/((2Kx+Kx))=4/3 K` |
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| 279. |
Consider a compound slab consisting of two different material having equal thickness and thermal conductivities `K` and `2K` respectively. The equivalent thermal conductivity of the slab isA. `sqrt(2K)`B. `3K`C. `4/3K`D. `2/3 K` |
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Answer» Correct Answer - C `K=(2K_(1)K_(2))/(K_(1)+K_(2)) = (2.K.2K)/(K+2K) = 4/3 K`. |
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| 280. |
Which of the following statement are true (a) Rubber contracts on heating (b) Water expands on freezing (c) Waterr expands on heating from `4^(@)C` to `40^(@)C`A. `(a)` is correctB. `(b)` and `(c)` are correctC. `(a)` & `(c)` correctD. all are correct |
| Answer» Correct Answer - D | |
| 281. |
When a metal ring having some gap is heated (a) length of gap increases (b) radius of the ring decreases (c) the angle substanded by the gap at the centre remains same (d) length of gap decreasesA. Only `d` is correctB. `a` and `b` are correctC. `a` & `c` are correctD. all are correct |
| Answer» Correct Answer - C | |
| 282. |
Temperature of gas is a measure ofA. (a) the average translational kinetic energy of the gas moleculesB. (b) the averge potential energy of the gas moleculesC. (c) the average distance of the gas moleculesD. (c) the size of the molecules of the gas |
| Answer» Correct Answer - A | |
| 283. |
The standard scale of temperature isA. the mercury scaleB. the gas scaleC. the platinum resistance scaleD. liquid scale |
| Answer» Correct Answer - B | |
| 284. |
Celsius is the unit ofA. (a) TemperatureB. (b) HeatC. (c) Specific heatD. (d) Latent heat |
| Answer» Correct Answer - A | |
| 285. |
The absolute zero is the temperature at whichA. water freezesB. all substances exist in solidC. molecular motion ceasesD. all of these |
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Answer» Correct Answer - D |
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| 286. |
The correct value of `0^(@)C` on the Kelvin scale isA. (a) `273.15K`B. (b) `273.16K`C. (c) `273K`D. (d) `273.2K` |
| Answer» Correct Answer - A | |
| 287. |
On the celsius scale the absolute zero of temperature is atA. `0^(@)C`B. `-32^(@)C`C. `100^(@)C`D. `-273.15^(@)C` |
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Answer» Correct Answer - D |
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| 288. |
On the Celsius scale the absolute zero of temperature is atA. (a) `0^(@)C`B. (b) `-32^(@)C`C. (c) `100^(@)C`D. (d) `.-273.15^(@)C` |
| Answer» Correct Answer - D | |
| 289. |
Heat given to a system can be associated withA. kinetic energy of random motion of moleculesB. kinetic energy of orderly motion of moleculesC. total kinetic energy of random and orderly motion of moleculesD. kinetic energy of random motion in some case and kinetic energy of orderly motion in other |
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Answer» Correct Answer - A We know that as temperature increases vibration of molecules about their mean position increases hence, kinetic energy associated with random motion of molecules increases. |
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| 290. |
Water of volume 2 litre in a container is heated with a coil of `1kW at 27^@C.` The lid of the container is open and energy dissipates at rate of `160J//s.` In how much time temperature will rise from `27^@C to 77^@C` Given specific heat of water is `[4.2kJ//kg`]A. `8 min 20 s`B. `6 min 2 s`C. `7 min`D. `14 min` |
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Answer» Correct Answer - A Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to enviroment) `rArr mc Delta theta = P_("Coil") t - P_("Loss")t` `rArr 2 xx 4.2 xx 10^(3) xx (77-27) = 1000t = 160t` `rArr t = (4.2xx10^(5))/(840) = 500 sec = 8 min 20 sec`. |
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| 291. |
Six identical conducting rods are connected as shown in figure. In steady state temperature of point `phi_(1)` is fixed at `100^(@)C` and temperature of `phi_(6)` at `-80^(@)C` Q. What is the temperature of `phi_(5)`?A. `10^(@)C`B. `30^(@)C`C. `-20^(@)C`D. `-10^(@)C` |
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Answer» Correct Answer - C |
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| 292. |
Six identical conducting rods are connected as shown in figure. In steady state temperature of point `phi_(1)` is fixed at `100^(@)C` and temperature of `phi_(6)` at `-80^(@)C` Q. What is the temperature of `phi_(2)`?A. `10^(@)C`B. `40^(@)C`C. `-20^(@)C`D. `-10^(@)C` |
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Answer» Correct Answer - B |
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| 293. |
Six identical conducting rods are connected as shown in figure. In steady state temperature of point `phi_(1)` is fixed at `100^(@)C` and temperature of `phi_(6)` at `-80^(@)C` Q. What is the temperature of `phi_(3)`?A. `10^(@)C`B. `40^(@)C`C. `-10^(@)C`D. `-15^(@)C` |
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Answer» Correct Answer - A |
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| 294. |
Always platinum is fused into glass, becauseA. Plaltinum is good conductor of heatB. melting point of platinum is very highC. they have equal specific heatsD. their coefficients of linear expansion are equal |
| Answer» Correct Answer - D | |
| 295. |
The resistance of a platinum wire is `15 Omega` at `20^(@)C`. This wire is put in hot furnace and lthe resistance of the wire is found to be `40 Omega`. Find the temperature of the hot furnace if temperature coefficient of resistance of platinum is `3.6 xx 10^(-3)^(0)C^(-1)` |
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Answer» `R_(1) = R_(0)(l + alpha t) rArr (R_(2))/(R_(1)) = ((l + alphat))/((l + alphat_(1))` `(40)/(15)` `= ((l + alphat_(2))/(l + alphat_(1))) rArr 40 - 15 = alpha(15t_(2) - 40alpha_(1))` `15t_(2) = (25)/(3.6 xx 10^(-3)) + 40 xx 20 approx` `Rightarrowt_(2)=516^(@)C` |
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| 296. |
A constant volume gas thermometer shows pressure readings of `50 cm` and `90 cm` of mercury at `0^(0)C` and `100^(0)C` respectively. What is the temperature on gas scale when the pressure reading is `60 cm` of mercury ? |
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Answer» Given that `P_(0) = 50cm` of `Hg,P_(100) = 90 cm` of `Hg` `P_(t) = 60 cm` of `Hg` `t = (P_(1) - P_(2))/(P_( 100) - P_(0)) xx 100 = (60 - 50)/(90 - 50) xx 100 = 25^(@)C` |
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| 297. |
Two rods are joined between fixed supports as shown in the figure. Condition for no change in the length of individual rods with the increase of temperature will be `"(" alpha_(1), alpha_(2)=` linear expansion coefficient `A_(1), A_(2)=` Area of rods `Y_(1), Y_(2)=` Young modulus `")"` A. `(A_(1))/(A_(2)) = (alpha_(1) Y_(1))/(alpha_(2) Y_(2))`B. `(A_(1))/(A_(2)) = (L_(1) alpha_(1) Y_(1))/(L_(2) alpha_(2) Y_(2))`C. `(A_(1))/(A_(2)) = (L_(2) alpha_(2) Y_(2))/(L_(1) alpha_(1) Y_(1))`D. `(A_(1))/(A_(2)) = (alpha_(2) Y_(2))/(alpha_(1) Y_(1))` |
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Answer» Correct Answer - D `Y=("Stress")/("strain") = (T//A)/(Delta l//l)` `T = (Y*Deltal)/(l) A = Y*A alpha Delta T` In both the rods tension will be same so `T_(1)=T_(2) ,` Hence, `Y_(1)A_(1)alpha_(1)Delta T = Y_(2)A_(2)alpha_(2)Delta T` `(A_1)/(A_2) = (Y_(2)alpha_(2))/(Y_(1)alpha_(1))`. |
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| 298. |
A vessel contains a gas under a pressure of `5 xx 10^(5)pa`. If `3//5` of the mass of the gas is flown out, What will be the gas pressure if the temperature being maintained constant,A. `50 MPa`B. `2 MPa`C. `0.2 MPa`D. `0.5 MPa` |
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Answer» Correct Answer - C `P prop d, d prop m , P prop m , (P_(1))/(P_(2)) = (m_(1))/(m_(2))` mass of gas taken out of cylinder is `(m_(1) - m_(2))` |
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| 299. |
Two thermally insulated vessel 1 and 2 are filled with air at temperature `(T_1T_2), volume (V_1V_2)` and pressure `(P_1P_2)` respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will beA. `T_(1) + T_(2)`B. `T_(1)T_(2) (P_(1)V_(1) + P_(2)V_(2)) // (P_(1)V_(1)T_(1) + P_(2)V_(2)T_(2))`C. `T_(1)T_(2) (P_(1)V_(1) + P_(2)V_(2)) // (P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))`D. `(T_(1) + T_(2)) // 2` |
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Answer» Correct Answer - C `n = (PV)/(RT) = Const` |
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| 300. |
Two identical beakers with negligible thermal expansion are filled with water to the same level at `4^(@)C`. If one says `A` is heated while the other says `B` is cooled, then:A. water level in A will riseB. Water level in B will riseC. Water level in A will fallD. Water level in B will fall |
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Answer» Correct Answer - A::B |
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