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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. |
Mercury boils at `367^(@)C`. However, mercury thermometers are made such that they can measure temperature up to `500^(@)C`. This is done byA. maintaining vacuum above mercury column in the stem of the thermometerB. filling nitrogen gas at high pressure above the mercury columnC. filling nitrogen gas at low pressure above the mercury levelD. filling oxygen gas at high pressure above the mercury column |
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Answer» Correct Answer - B By filling nitrogen gas at high pressure, the boling point of mercury is increased which extend the range upto `500^(@)C`. |
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| 2. |
We consider the radition emitted by the human body which of the following statements is true?A. The radiation is emiited only during the dayB. The radiation is emitted during the summers and absorbed during the wintersC. The radiation emitted lies in the ultraviolet region and hence is not visibleD. The radiation emitted is in the infrared region |
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Answer» Correct Answer - D Everybody at all time, at all temperatures, emits radiation (except at `T=0`). The radiation emitted by the human body is in the infrared region. |
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| 3. |
A body of length `1 m` having cross sectional area `0.75 m^(2)` has heat flow through it at the rate of `6000 "Joule"//sec`. Then find the temperature difference if `K = 200 Jm^(-1) K^(-1)`.A. `20^(@)C`B. `40^(@)C`C. `80^(@)C`D. `100^(@)C` |
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Answer» Correct Answer - B `Q/t = (KADelta theta)/(l)` `rArr 6000=(200xx0.75xxDelta theta)/(1)` `:. Delta theta = (6000xx1)/(200xx0.75) = 40^(@)C`. |
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| 4. |
A body cools down from `45^(@)C` to `40^(@)C` in 5 minutes and to `35^(@)` in another 8 minutes. Find the temperature of the surrounding. |
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Answer» Correct Answer - `29^(@)C` |
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| 5. |
The temperature of the body is increased from `-73^(@)C` to `357^(@)C`, the ratio of energy emitted per second isA. `1 : 3`B. `1 : 81`C. `1 : 27`D. `1 : 9` |
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Answer» Correct Answer - B Energy per second `P=(=(Q)/(t)) prop T^(4)`. |
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| 6. |
The energy supply being cut-off, an electric heater element cools down to the temperature of its surrounding, but it will not cool further becauseA. supply is cut offB. it is made of metalC. surroundings are radiatingD. element and surroundings have the same temperature. |
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Answer» Correct Answer - D When element and surrounding have the same temperature. There will be no temperature difference, hence, heat will not flow from the filaments and its temperature remains constant. |
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| 7. |
The radiation emitted by a star `A` is `1000` times that of the sun. If the surface temperature of the sun and star `A` are `6000 K` and `2000 K` respectively. The ratio of the radii of the star `A` and the sun is:A. `300 : 1`B. `600 : 1`C. `900 : 1`D. `1200 : 1` |
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Answer» Correct Answer - C According to Stefan-Boltzmann law, the energy radiated per second through the surface of area `A` is given by `E = sigma AT^(4)` `:. (E_1)/(E_2) = (A_1)/(A_2) ((T_1)/(T_2))^(4)` or, `10000 = (r_(1)^(2))/(r_(2)^(2))(2000/6000)^(4)` or,`(r_(1)^(2))/(r_(2)^(2))=(30)^(4)` or `r_(1):r_(2) = 900:1`. |
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| 8. |
A rectangular body has maximum wavelength `lambda_(m)` at `2000 K`. Its corresponding wavelength at `3000 K` will beA. `3/2 lambda_(m)`B. `2/3 lambda_(m)`C. `4/9 lambda_(m)`D. `9/4 lambda_(m)` |
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Answer» Correct Answer - B `lambda_(m2) = (T_1)/(T_2) xx lambda_(m1)=2000/3000xxlambda_(m1) = 2/3 lambda_(m1) = 2/3 lambda_(m)`. |
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| 9. |
Assertion : Coefficient of absorption of radiation of an ideal black body is `1`. Reason : An ideal black body emits radiation of all wavelengths.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B We know that for ideal black body reflectance `(r) =0` transmittance `(t)=0` absorptance `(a)=1`. It is clear that when heated the black body will radiated all energy. Which is absorbed by it. |
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| 10. |
By virtue of some internal mechanism, temperature of spherical shell is maintained in steady state which emits radiation like black body while kept in vaccum and isolated from all radiation. If this shell is enveloped by another shell `B` of same radius, still emitted like a black body, find the ratio of temperatures of shell `B` and shell `A` when it was alone:A. `(2)^(1//4)`B. `(3)^(1//4)`C. `(4)^(1//4)`D. None of these |
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Answer» Correct Answer - A When shell `A` alone is there `P_(S) = sigma AT_(0)^(4)` When there are both the shells `sigma AT_(1)^(4) = P_(S) = sigma A T_(0)^(4)` `sigma AT_(2)^(4) -sigma A T_(1)^(4) = P_(S) = sigmaA T_(0)^(4)` `T_(2)^(4) = 2T_(0)^(4)` `T_(2)//T_(0) = (2)^(1//4)`. |
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| 11. |
The molar specific heats of an ideal gas at constant pressure and volume are denotes by `C_(P)` and `C_(upsilon)` respectively. If `gamma = (C_(P))/(C_(upsilon))` and `R` is the universal gas constant, then `C_(upsilon)` is equal toA. `(1 + gamma)/(1 - gamma)`B. `(R)/((gamma - 1))`C. `((gamma - 1))/(R)`D. `gammaR` |
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Answer» Correct Answer - B We have `C_(p)-C_(v)=R`...(i) and `(C_p)/(C_v) = gamma` ...(ii) From (i) and (ii), we get `C_(v) = (R)/((gamma-1))`. |
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| 12. |
Given below are observations on molar specific heats at room temperature of some common gases The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ? |
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Answer» The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion). Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases. If only rotational mode of motion is considered, then the molar specific heat of a diatomic `gas=(5)/(2)R` `=(5)/(2)xx1.98=4.95 "cal mol"^(1)K^(-1)` With the exception of chlorine, all the observations in the given table agree with`((5)/(2)R)`.This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion. |
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| 13. |
Two rods of the same length, have radii in the ratio `3:4` Their densities are respectively `8000` and `9000 kg//m^(3)`. Their specific heats are in the ratio of `2:3`. When the same amount of heat is supplied to both, the changes in their lengths are in the ratio. (If their linear coefficients are in the ratio `5:6`)A. `1:1`B. `5:2`C. `5:12`D. `12:5` |
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Answer» Correct Answer - B `Q = msDeltatheta, m = rhopir^(2)l` `Delta theta = (Deltal)/(lalpha) rArr (Deltal_(1))/(Deltal_(2)) = (rho_(2))/(rho_(1)) xx (r_(2)^(2))/(r_(1)^(2)) xx (s_(2))/(s_(1)) xx (alpha_(1))/(alpha_(2))` |
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| 14. |
If the temperature of a gas is increased by `1K` at constant pressure its volume increase by `0.0035` of the initial volume. The temperature of the gas isA. `100K`B. `150K`C. `300K`D. `285.7K` |
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Answer» Correct Answer - D `V prop T , (DeltaV)/(V) = (DeltaT)/(T)` |
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| 15. |
Select the correct graphs A) the `P - 1//V` graph at constant temperature is a rectangular hyperbola. B) the `PV-V` graph is a straight line parallel to the `Y-` axis. C) `V-T` graph at constant pressure is a straight line passing through the origin.A. `A`B. `B`C. `C`D. `A,B,C` |
| Answer» Correct Answer - B | |
| 16. |
`PV = n RT` holds good for A) Isobaric process B) Isochoric process C) Isothermal process D) Adiabatic processA. `A` & `B `B. `A, B` & `C`C. `A, B` & `D`D. All |
| Answer» Correct Answer - D | |
| 17. |
Certain volume of a liquid is taken in a long glass tube and its temperature is increased at a uniform rate, the rate of increase in the length of the liquid depends on a) length of the liquid b) area of cross section of the glass tube c) coefficient of expansion of glassA. Only `(a)` is correctB. `(a)` & `(b)` are correctC. `(b)` & `(c)` are correctD. `(a), (b)` & `(c)` are correct |
| Answer» Correct Answer - D | |
| 18. |
For a constant volume gas thermometer, one should fill the gas atA. high temperature and high pressureB. high temperature and low pressureC. low tempreature and low pressureD. low temperature and high pressure |
| Answer» Correct Answer - B | |
| 19. |
An ideal gas is that whichA. cannot be liquefiedB. can be easily liqefiedC. has strong inter molecular forcesD. has a large size of molecules. |
| Answer» Correct Answer - A | |
| 20. |
In a gas equation, `PV = RT, V` refers to the volume ofA. any amount of a gasB. 1 gram mass of a gasC. `1` gram mole of a gasD. `1` litre of a gas |
| Answer» Correct Answer - C | |
| 21. |
A mercury thermomeeter is transferred from melting ice to a hot liquid. The mercury rises to `9//10` of the distance between the two fixed points. Find the temperature of the liquid in Fahrenheit scaleA. `194^(@)F`B. `162^(@)F`C. `112^(@)F`D. `113^(@)F` |
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Answer» Correct Answer - A `((F - 32)/(180) = (9)/(10))` |
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| 22. |
If an air bubbles rises from the bottom of a mercury tank to the top its volume becomes `1(1)/(2)` times. When normal pressure is `76 cm` of `Hg` then the depth of the `Hg` tank isA. `38 cm`B. `123 cm`C. `76 cm`D. `49 cm` |
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Answer» Correct Answer - A `P_(1)V_(1) = P_(2)V_(2) , (H + h)V_(1) = HV_(2)` |
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| 23. |
A quill tube contains a mercury column of length `19cm`. The length of air column is `24cm` when it is held veritcally. On inverting it with its open end downwards the length of air column will be (atmospheric pressure `= 76cm` of `Hg`)A. `20cm`B. `30cm`C. `40cm`D. `35cm` |
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Answer» Correct Answer - C `(H + h)l_(1) = (H - h)l_(2)` |
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| 24. |
The length of a metal rod at `0^(@)C` is `0.5m` When it sis heated, its length incerases by `2.7mm`. The final temperature of rod is (coeff. Of linear excpansion of metal `= 90 xx 10^(-6//@)C`)A. `20^(@)C`B. `30^(@)C`C. `40^(@)C`D. `60^(@)C` |
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Answer» Correct Answer - D `t_(2) - t_(1) = (l_(2) - l_(1))/(l_(1)alpha)` |
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| 25. |
State the equation corresponding to `8g` of `O_(2)` isA. `PV = 8RT`B. `PV = RT//4`C. `PV = RT`D. `PV = RT//2` |
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Answer» Correct Answer - B `PV = (m)/(M)RT` |
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| 26. |
A metal sheet having size of `0.6 xx 0.5m^(2)` is heated from `293 K` to `520^(@)C`. The final area of the hot shee is {`alpha` of metal `= 2 xx 10^(-5 //@)C`]A. `0.306 m^(2)`B. `0.0306 m^(2)`C. `3.06 m^(2)`D. `1.02m^(2)` |
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Answer» Correct Answer - A `A_(2) = A_(1)(1 + betaDeltat)` |
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| 27. |
A given amoount of gas is heated until both its pressure and volume are doubled. If initial temperature is `27^(0)C`, its final temperature isA. `300 K`B. `600 K`C. `1200 K`D. `900 K` |
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Answer» Correct Answer - C `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` |
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| 28. |
A vessel contains `110 g` of water. The heat capacity of the vessel is equal to `10 g` of water. The initial temperature of water in vessel is `10^(@)C`. If `220 g` of ho, water at `70^(@)C` is poured in the vessel, the final temperature neglecting radiation loss, will beA. `70^(@)C`B. `80^(@)C`C. `60^(@)C`D. `50^(@)C` |
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Answer» Correct Answer - D Let final temperature of water be `theta` Heat taken = Heat given `110xx1(theta-10)+10(theta-10)` `=220 xx 1(70-theta)` `rArr theta = 48.8^(@)C = 50^(@)C`. |
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| 29. |
A gas is kept at `13^(0)C` in a vessel, if the volume of the gas is kept constant and is heated, the pressure will be doubled to its initial pressure at a temperatureA. `572 K`B. `286 K`C. `143 K`D. `73 K` |
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Answer» Correct Answer - A `P prop T , (P_(1))/(P_(2)) = (T_(1))/(T_(2))` |
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| 30. |
Water of mass `m_(2) = 1 kg` is contained in a copper calorimeter of mass `m_(1) = 1 kg`. Their common temperature `t = 10^(@)C`. Now a piece of ice of mass `m_(2) = 2 kg` and temperature is `-11^(@)C` dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper `=0.1 Kcal//kg^(@)C`, specific heat of water `= 1 Kcal//kg^(@)C`, specific heat of ice `= 0.5 Kcal//kg^(@)C`, latent heat of fusion of ice `= 78.7 Kcal//kg`]A. `0^(@)C`B. `4^(@)C`C. `-4^(@)C`D. `-2^(@)C` |
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Answer» Correct Answer - A Loss in heat from calorimeter+water as temperature change form `10^(@)C` to `0^(@)C` `=m_(1)C_(1)10+m_(2)C_(2)10` `=1xx1xx10+1xx0.1xx10=11kcal` Gain in heat of ice as its temperature changes from `-11^(@)C` to `0^(@)C` `=m_(3)C_(3)xx2xx0.5xx11 = 11kcal` Hence ice and water will coexist at `0^(@)C` without any phase change. |
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| 31. |
When `0.400 kg` of water at `30^(@)C` is mixed with `0.150 kg` of water at `25^(@)C` contained in a calorimeter, the final temperature is found to be `27^(@)C`, find the water equivalent of the calorimeter.A. `0.450 kg`B. `1 kg`C. `0.50 kg`D. `1.5 kg` |
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Answer» Correct Answer - A Water of mass `m_(1)=0.150 kg` is taken in the calorimeter at temperature, `T_(1) = 25^(@)C` is mixed with `N` another known mass of pure water `m_(2) =0.400 kg` at a temperature `T_(2) = 30^(@)C` and final temperature is found to be `T=27^(@)C`. Let `s_(w)` and `W` be the heat capacity of water and the water equivalent of the calorimeter. Heat gained by (water+calorimeter) at temperature `T_(1)`= Heat lost by water at temperature `T_(2)` i.e., `m_(1)s_(w)(T-T_(1))+Ws_(w) (T-T_(1)) = m_(2)s_(w)(T_(2)-T)` or, `W = m_(2)((T_(2)-T)/(T-T_(1)))-m_(1)` Hence, `W = 0.400kg[(30^(@)C-27^(@)C)/(27^(@)C-25^(@)C)]-0.150 kg` `=0.450 kg`. |
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| 32. |
If the pressure of a gas contained in a closed vessel increases by `X%` when heated bu `1^(0)C`, it initial temperature isA. `(100//x)` KelvinB. `(100//x)` CelsiusC. `((x + 100)/(x))` KelvinD. `((100 - x)/(x))` Kelvin |
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Answer» Correct Answer - A `(P_(2))/(P_(1)) = (T_(1))/(T_(2))` and `(P_(2) - P_(1))/(P_(1)) = (T_(2) - T_(1))/(T_(1))` |
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| 33. |
A current of `2.50` A passing through a heating coil immersed in `180 g` of paraffin (specific heat capacity `2.00 J g^(-1) K^(-1)`) contained in a `100 g` calorimeter (specific heat capacity `0.400 J g^(-1) K^(-1)`) raise the temperature from `5^(@)C` below room temperature to `5^(@)C` above room temperature in `100 s`. the reading of the voltmeter connected across the heating coil isA. `8.0 V`B. `16.0 V`C. `24.0 V`D. `32.0 V` |
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Answer» Correct Answer - B Power `= ("heat supplied")/("Time")` `=(m_(1)s_(1)Delta theta+m_(2)s_(2)Delta theta)/(T)` `(180g xx 2 Jg^(-1)K^(-1)xx10K)+` `=((100gxx0.4Jg^(-1)K^(-1)xx10K))/(100s)` `=40W` `=VI = 40 W rArr V = (4.0)/(2.5) = 16.0V`. |
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| 34. |
Statement-1: Ventilators are provided at the top of the room. Statement-2: Their purpose is to bring oxygen for breathing.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 - C |
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| 35. |
Statement-1: Steam burns are more painful than that caused by boiling water at same temperature. Statement-2: Internal energy of steam at `100^(@)C` is greater than that of water at `100^(@)C`.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 - A |
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| 36. |
A meter scale made of steel is calibrated at `20^@C` to give correct reading. Find the distance between 50 cm mark and 51 cm mark if the scale is used at `10^@C`. Coefficient of linear expansion of steel is `1.1 xx 10^(-5 @)C^(-1)`A. `1.00011 cm`B. `1.0011 cm`C. `1.011 cm`D. `1.000011 cm` |
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Answer» Correct Answer - A `l_(c)=l_(m)(1+(alpha_(S)-alpha_(b))(t_(2)-t_(1)))` |
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| 37. |
A polished metallic piece and a black painted wooden pieceare kept in open in bright sun for a long time.A. The wooden piece will absorb less heat than the metallic pieceB. the wooden piece will hae a lower temperature than the metallic pieceC. If touched, the metallic piece will be felt hotter than the wooden pieceD. When the two pieces are removed from the open to a cold room, the wooden piece will loose heat at a faster rate than the metallic piece. |
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Answer» Correct Answer - C::D |
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| 38. |
When m gm of water at `10^(@)C` is mixed with m gm of ice at `0^(@)C`, which of the following statements are false?A. The temperature of the system will be given by the equation `mxx80+mxx1xx(T-0)=mxx1xx(10-T)`B. Whole of ice will melt and temperature will be more than `0^(@)C` but lesser than `10^(@)C`C. Whole of ice will melt and temperature will be `0^(@)C`D. Whole of ice will not melt and temperature will be `0^(@)C`. |
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Answer» Correct Answer - A::B::C |
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| 39. |
How many grams of liquid of specific heat `0.2` at temperature `40^(@)C` must be mixed with `100 gm` of a liquid of specific heat of `0.5` at temperature `20^(@)C`, so that the final temperature of the mixture becomes `32^(@)C`A. `175 gm`B. `300 g`C. `295 gm`D. `375 g` |
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Answer» Correct Answer - D Temperature of mixture `theta = (m_(1)c_(1)theta_(1)+m_(2)c_(2)theta_(2))/(m_(1)c_(1)+m_(2)theta_(2))` `rArr 32 = (m_(1)xx 0.2 xx 40 + 100 xx 0.5 xx20)/(m_(1)xx0.2+100xx0.5)` `rArr m_(1) =375 gm`. |
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| 40. |
A pendulum clock runs fast by `5` seconds per day at `20^(0)c` and goes slow by `10` seconds per day at `35^(0)C`. It shows correct time at a temperature ofA. `27.5^(0)C`B. `25.^(0)C`C. `30.^(0)C`D. `33.^(0)C` |
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Answer» Correct Answer - B `(1)/(2)alpha(35-t)xx86400=10----(1)` `(1)/(2)alpha(t-20)xx86400=5----(2)` Solving `1` and `2` |
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| 41. |
The coefficient of linear expansion of a metal rod is `12 xx 10^(-6//0)C`, its value in per ``^(0)F`A. `(20)/(3) xx 10^(-6//0)F`B. `(15)/(4) xx 10^(-6//0)F`C. `21.6 xx 10^(-6//0)F`D. `12 xx 10^(-6//0)F` |
| Answer» Correct Answer - A | |
| 42. |
The co-efficient of thermal expansion of a rod is temperature dependent and is given by the formula `alpha = aT`, where `a` is a positive constant at T `"in"^(@)C`. if the length of the rod is l at temperature `0^(@)C`, then the temperature at which the length will be `2l` isA. `sqrt((1n2)/alpha)`B. `sqrt((1n4)/alpha)`C. `1/(alpha)`D. `2/alpha` |
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Answer» Correct Answer - B As, `dl=alpha l dT :. underset(l)overset(2l)(int) (dl)/(l) = a underset(0)overset(T)(int) T dT` `1n 2 = a(T^2)/(2) :. T = [ (1n 4)/(a)]^(1//2)`. |
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| 43. |
A metal rod having a linear coefficient of expansion `2 xx 10^(-5 //@) C` has a length `1m` at `25^(@)C`, the temperature at which it is shortened by `1mm` is `A. `50^(@)C`B. `-50^(@)C`C. `-25^(@)C`D. `-12.5^(@)C` |
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Answer» Correct Answer - C `t_(2) - t_(1) = (l_(2) - l_(1))/(l_(1)Deltat)` |
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| 44. |
The co-efficient of linear expansion of iron is `11//180` of volume coefficient of expansion of mercury which is `18 xx 10^(-5) // ^(0)C`. An iron rod is `10m` long at `27^(@)C`. The length of the rod will be decreased by `1.1mm` then the termperature the rod changes byA. `0^(@)C`B. `10^(@)C`C. `20^(@)C`D. `170^(@)C` |
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Answer» Correct Answer - B `alpha_(iron) = (11)/(800)gamma_(Hg) , Deltal = lalphaDeltat` |
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| 45. |
The coefficient of real expansion of liquid is `7 xx 10^(-4//0)C`. The co efficient of linear ex/pansion of the vessel is `1 xx 10^(-5)//^(0)C`. The coefficient of apparent expansion of the liquid isA. `7 xx 10^(-4) // ^(0)C`B. `6 xx // ^(0)C`C. `67 xx // ^(0)C`D. `73 xx // ^(0)C` |
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Answer» Correct Answer - C `gamma_(R) = gamma_(A) + gamma_(g)` |
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| 46. |
Which of the following processes will quadruple the pressureA. Reduce `V` to half and double `T`B. Reduce `V` to `1//8th` and reduce `T` to halfC. Double `V` and half `T`D. Increase both `V` and `T` to double the values. |
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Answer» Correct Answer - A::B `pv = nRT` |
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| 47. |
In heat transfer, which method is based on gravitationA. Natural convectionB. ConductionC. RadiationD. All of these |
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Answer» Correct Answer - A |
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| 48. |
Two rods `A` and `B` of different materials are welded together as shown in figure. Their thermal conductivities are `K_(1)` and `K_(2)`. The thermal conductivity of the composite rod will be A. `(3 (K_(1) + K_(2)))/(2)`B. `K_(1) + K_(2)`C. `2 (K_(1) + K_(2))`D. `(K_(1) + K_(2))/(2)` |
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Answer» Correct Answer - D In parallel `(1)/(R_(eq)) = (1)/(R_1) +(1)/(R_2)` `(K_(eq)(2A))/(l) = (k_(l)A)/(l)+(K_(2)A)/(l)` `K_(eq) = (K_(1)+K_(2))/(2)`. |
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| 49. |
Five rods of same dimensions are arranged as shown in the figure. They have thermal conductivities `k_(1),k_(2),k_(3) and k_(5)`. When points A and C are maintained at different temperature, no heat flows through the central rod if A. `K_(1)=K_(4) and K_(2)=K_(3)`B. `K_(1)K_(4)=K_(2)K_(3)`C. `K_(1)K_(2)=K_(3)K_(4)`D. `(K_(1))/(K_(4))=(K_(2))/(K_(3))` |
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Answer» Correct Answer - B |
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| 50. |
The graph between two temperature scales `A` and `B` is shown in Fig. Between upper fixed point and lower fixed point there are `150` equal divisions on scales `A` and `100` on scale `B`. The relation between the temperature in two scales is given by_ A. `(t_(A) - 180)/(100) = (t_(B))/(150)`B. `(t_(A) - 30)/(150) = (t_(B))/(100)`C. `(t_(B) - 180)/(150) = (t_(A))/(100)`D. `(t_(B ) - 40)/(100) = (t_(A))/(180)` |
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Answer» Correct Answer - B `("Reading on any scale" - LFP)/(UFP - LFP) ="Constant"` |
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