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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.
| 51. |
`1kg` ice at `0^(@)C` is mixed with `1kg` of steam at `100^(@)C` what will be the composition of the system when thermal equilibrium is reached ? Latent beat of fusion of ice `= 3.36xx 10^(5)J kg^(-1)` and latent head of vaporization of water `= 2.26 xx 10^(6)J kg^(-1)` |
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Answer» Correct Answer - A::C::D Heat obserbed by the ice to raise the temperature `100^(@)C` `Q_(1) = 1 xx 3.36 xx 10^(5) + 1 xx 4200 xx 100` ` = 3.36 xx 10^(5) + 4.2 xx 10^(5)` ` = (3.36 + 4.2) xx 10^(5)` ` = 7.56 xx 10^(5) = 0.756 xx 10^(6)` `Q_(2) ` heat released by steam ` = 1 xx 2.26 xx 10^(6)J` ` = 3.26 xx 10^(6)J` THe extra heat `= Q_(2) - Q_(1)` ` = (2.26 - 0.756 xx 10^(6)` `= 1.506 xx 10^(6)` THe arrount of steam condensed ` = (1.506 xx 10^(6))/(2.26 xx 10^(6))` `= 0.665 kg = 665 gm` The extra ice` = (1000 - 665) = 335gm` The amount of ice formed `= 1000 +335 = 1335g` |
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| 52. |
What is the temperature of the steel copper junction in the steady state of the system shown if Fig. Length of the steel rod `=25cm`, length of the copper rod`=50cm`, temperature of the furnace`=300^@C`, temperature of the other end `=0^@C`. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel `=50J//s//m` and of copper `=400Js//m//K`). |
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Answer» `(k_1A_1(T_1-T))/(L_1)=(k_2A_2(T-T_2))/(L_2)` `300-T=((L_1)/(L_2))((k_2)/(k_1))((A_2)/(A_1))(T-0)` `300-T=2T` `T=100^@C` |
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| 53. |
Power of a man who can chew `0.3 kg` ice in one minute is (in cal/s).A. 400B. 4C. 24D. 240 |
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Answer» Correct Answer - A `P = (mL_(f))/(t)`. |
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| 54. |
A 0.60 kg sample of water and a sample of ice are placed in two compartmetnts A and B separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the compartments remain homogeneous during whole heat transfer process. Given specific heat of ice `=2100 J//kg-K`, specific heat of water `=4200 J//kg-K`, and latent heat of fusion of ice `=3.3xx10^5 J//kg`.A. The value of rate P isB. 42.0 WC. 36.0 WD. 21.0 W |
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Answer» Correct Answer - A In 40 min. Temperature of water has water has come down by `40^@C` . Therefore rate `P=(mSDeltaT)/(t)=(0.60xx4200xx40)/(40xx60)=42W` |
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| 55. |
A piece of ice of mass of `100g` and at temperature `0^(@)C` is put in `200g` of water of `25^(@)C` .How much ice will melt as the temperature of the water reaches `0^(@)C`? The specific heat capacity of water`= 2400 J kg ^(-1)K^(-1)` and the specific latent heat of ice `= 3.4 xx 10^(3)J kg^(-1)` |
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Answer» Heat released by 0.15 kg of water in being cooled to `0^@C=0.15xx1xx20=3kcal` Heat absorbed by ice from `-10^@`C to `0^@C=0.1xx0.5xx10=0.5kcal` The balance heat is available for malting ice. Let m kg of ice melt. Then `mxx80=2.5` or `m=0.03kg` Thus the final temperature is `0^@C` with `0.07kg` of ice and `0.18kg` of water |
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| 56. |
A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C. |
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Answer» mass of Iron = 100g water Eq of caloriemeter = 10g mass of water = 240g Let the Temp. of surface = 0ºC Siron = 470J/kg°C Total heat gained = Total heat lost. So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20) |
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| 57. |
A vessel is completely filled with `500g` of water and `1000g` of mercury. When 21,200 calories of heat are given to it `3.52g` of water overflows. Calculate the volume expansion of mercury. Expansion of the vessel may be neglected. Given that coefficient of volume expansion of water `=1.5xx10^(-4)//C^(@)`, density of mercury `=13.6g//cm^(3)`, density of water `e=1g//cm^(3)`, and specific heat of mercury `=0.03"cal"//gC^(@)` |
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Answer» Correct Answer - `1.7xx10^(-4)//C^(@)` |
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| 58. |
The inernal energy of a solid also increases when heat is transferred to it from its surroundings. A 5 kg solid bar is heated at atmospheric pressure. Its temperature increases from `20^@C` to `70^@C`. The linear expansion coefficient of solid bar is `1xx10^(-3)//^(@)C`. The density of solid bar is `50 kg//m^3`. The specific heat capacity of solid bar is `200 J//kg C^@`. The atmospheric pressure is `1xx10 N//m^2`. The work done by the solid bar due to thermal expansion, under atmospheric pressure isA. 500 JB. 1000 JC. 1500 JD. 2000 J |
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Answer» Correct Answer - C `DeltaV=gammaVDeltaT` `=3alphaVDeltaT=3xx1xx10^-3xx((5kg)/((50kg)/(m^2)))xx50` `=15xx10^-3m^3` `W=PDeltaV` `=(1xx10^5(N)/(m^2))xx(15xx10^-3m^3)=15xx10^2` `=1500J` |
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| 59. |
One end of a copper rod of uniform cross section and length 1.5 m is kept in contact with ice and the other end with water at `100^@C`. At what point along its length should a temperature of `200^@C` be maintained so that in the steady state, the mass of ice melting be equal to that of the steam produced in same interval of time. Assume that the whole system is insulated from surroundings: `[L_("ice")=80 cal//g,L_("steam")=540 cal//g]`A. 8.59 cm from ice andB. 10.34 cm from water endC. 10.34 cm from ice endD. 8.76 cm from water end |
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Answer» Correct Answer - B If the point is at a distance x from water at `100^@C`, heat conducted to ice in time t, `Q_(ice)=KA((200-0))/((1.5-x))xxt` So ice melted by this heat `m_(ice)=(Q_(ice))/(L_F)=(KA)/(80)((200-0))/((1.5-x))xxt` Similarly heat conducted by the rod to the water at `100^@C` in time t, `Q_(water)=KA((200-100))/(x)t` Steam formed by this heat `m_(steam)=(Q_(water))/(L_V)=KA((200-100))/(540xxx)t` According to given problem `m_(ice)=m_(steam)`, i.e., `(200)/(80(1.5-x))=(100)/(540xx x)x=(6)/(58)m=10.34cm` i.e., `200^@C` temeprature must be maintained at a distance 10.34 cm from water at `100^@C`. |
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| 60. |
One end of a copper rod of uniform cross section and length 1.5 m is kept in contact with ice and the other end with water at `100^@C`. At what point along its length should a temperature of `200^@C` be maintained so that in the steady state, the mass of ice melting be equal to that of the steam produced in same interval of time. Assume that the whole system is insulated from surroundings: `[L_("ice")=80 cal//g,L_("steam")=540 cal//g]`A. 10.34 cm from the end at `100^@C`B. 10.34 mm from the end at `100^@C`C. 1.034 cm from the end at `100^@C`D. 1.034 m from the end at `100^@C` |
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Answer» Correct Answer - A If the point is at a distance x from water at `100^@C`, heat transferred to ice in time t to melt it is `m_1L_1=(KA(200-0)t)/((1.5-x))` or `m_1=(KAxx200t)/(80(1.5-x))` similarly, heat conducted by the rod to water at `100^@C` in time t is `Q=(KA(200-100)t)/(x)=m_SL_S` `m_S=(KA(200-100)t)/(xL_S)=(KAxx100t)/(x xx540)` According to problem, `m_1=m_S` i.e., i.e., `(KAxx200t)/(80(1.5-x))=(KAxx100t)/(x xx540)` or `(2)/(8(1.5-x))=(1)/(54x)` solving it, we get `x=0.1034m` or `10.34cm` |
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| 61. |
The temperature of a room heated by heater is `20^@C` when outside temperature is `-20^@C` and it is `10^@C` when the outside temperature is `-40^@C`. The temperature of the heater isA. `80^@C`B. `100^@C`C. `40^@C`D. `60^@C` |
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Answer» Correct Answer - D Under steady state condition heat released to the room`=`heat dissipated out of the room Let `theta` be the temperature of heater Then `theta-20=alpha[20-(-20)]` .(i) and `theta-10=alpha[10-(-40)]` .(ii) Solving Eqs. (i) and (ii) we get we get `theta=60^@C` |
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| 62. |
`1 kg` ice at `-20^(@)C` is mixed with `1 kg` steam at `200^(@)C`. The equilibrium temperature and mixture content is |
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Answer» Let equilibrium temperature be `100^@`C, heat required to canvert 1 kg ice at `-20^@`C to 1 kg water at `100^@`C is equal to `H_1=1xx(1)/(2)xx20+1xx80+1xx1xx100=190kcal` Heat released by steam to convert 1 kg steam at `200^@` C to 1 kg water at `100^@` C is equal to `H_2=1xx(1)/(2)xx100+1xx540=590kcal` 1 kg ice at `-20^@C=H_1+1kg` water at `100^@`C ..(i) 1 kg steam at `200^@`C`=H_2+1kg` water at `100^@`C ..(ii) By adding Eqs. (i) and (ii) 1kg ice at `-20^@C+1kg` steam at `200^@C=H_1+H_2+2kg` water at `100^@`C. Here heat required to ice is less than heat supplied by steam, so mixture equilibrium temperature is `100^@`C. THen steam is not completely converted into water. So mixture has water and steam, which is possible only at `100^@` C. Mass of steam which converted into water is equal to `m=(190-1xx(1)/(2)xx100)/(540)=(7)/(27)kg` So mixture content Mass of steam`=1-(7)/(27)=(20)/(27)kg` Mass of water`=1+(7)/(27)=(34)/(27)kg` |
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| 63. |
Can heat be considered to be a form of stored energy? |
| Answer» No, heat cannot be considered to be a form of stored energy. Had it been stored energy it would have been possible to remove all of it. | |
| 64. |
A thermally isulated piece of metal is heated under atmospheric pressure by an electric current so that it receives electric energy at a constant power P. This leads to an increase of absolute temperature T of the metal with time t as follows: `T(t)=T_0[1+a(t-t_0)]^(1//4)`. Here, a, `t_0` and `T_0` are constants. The heat capacity `C_p(T)` of the metal isA. `(4P)/(aT_0)`B. `(4PT)/(aT_0^4)`C. `(2PT)/(aT_0^4)`D. `(2P)/(aT_0)` |
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Answer» Correct Answer - B Heat given to the metal `dQ=Pdt=C_P(t)dT` ..(i) At constant pressure in time interval at Given `T=T_0[1+a(t-t_0)]^(1//4)` `(dT)/(dt)=(T_0)/(4)[1+a(t-t_0)^(-3//4)xxa` ..(ii) From Eqs. (i) and (ii) `C_P(T)=(P)/(((dT)/(dt)))=(4P[1+a(t-t_0)]^(3//4))/(T_0a)=(4PT^3)/(aT_0^4)` |
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| 65. |
A planet is at an average distance d from the sun and its average surface temeperature is T. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If `Tpropd^(-n)`, the value of n isA. 2B. 1C. `(1)/(2)`D. `(1)/(4)` |
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Answer» Correct Answer - C Let P be the power radiated by the sun and R be the radiys planet. Energy radiated by planet `=4piR^2xx(sigmaT^4)` For thermal equiliibrium `(P)/(4pid^2)xxpiR^2=4piR^2(sigma^4)` `:. T^4prop(1)/(d^2)` or `Tpropd^((-1)/(2))` Hence `n=(1)/(2)` |
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| 66. |
A metallic ball and highly stretched spring are made of the same material and have the same mass. They are heated so that they melt. The latent heat requiredA. Are the same for theB. is greater fore the ballC. is greater for the springD. For the two may or may not be the same depending upon the metal |
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Answer» Correct Answer - A Latent heat is independent of configuration. Ordered energy spent in stretching the spring will not contribute to heat which is disorder kinetic energy of molecules of substance. |
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| 67. |
When `20xx10^(-3)` kg of water at `15^(@)C` is placed in the tube of an ice calorimeer, it si found that the mercury thread mvoes through `29cm`. If a metal of mass `12g` and `100^(@)C` is placed in the tube, the mercury thread contracts by `12cm`. Find the specific heat capacity of metal. (Specific heat capacity of water `=4200J kg^(-1)K^(-1)`) |
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Answer» Correct Answer - `4543.5Jkg^(-1)K^(-1)` |
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| 68. |
A mercury in glass thermometer has a stem of internal diameter 0.06 cm and contains 43 g of mercury. The mercury thread expands by 10 cm when the temperature changes from `0^@C` to `50^@C`. Find the coefficient of cubical expansion of mercury. Relative density of mercury `=13.6` and `alpha_("glass")=9xx10^(-6)//K`. |
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Answer» Apparent expansion of mercury `=(pi)/(4)(6xx10^-4)^2xx10xx10^-2` `=9pixx10^-9m^3` volume of mercury `=(43xx10^-3)/(13.6xx1000)` `=3.16xx10^-6m^3` `:. 9pixx10^-9=3.16xx10^-6xxgamma_(alpha)xx50` `(DeltaV=VgammaxxDeltaT)` or `gamma_(alpha)=18xx10^-5` `gamma_r=gamma_(alpha)+gamma_g=gamma_(alpha)+3alpha_(g) (gamma_g=3alpha_g)` `:. gamma_r=18xx10^-5+3xx9xx10^-6` `=20.7xx10^-5K^-1` |
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| 69. |
Statement I: The expanded length l of a rod of original length `l_0` is not correctly given by (assuming `alpha` to be constant with T) `l=l_0(1+alphaDeltaT)` if `alphaDeltaT` is large. Statement II: It is given by `l=l_0e^(alphaDeltaT)`, which cannot be treated as being approximately equal to `l=l_0(1+alphaDeltaT)` for large value a `DeltaT`.A. Statement I is true, Statement II is true and Statement II is the correct explanation for statement I.B. Statement I is true, statement II is true and statement II NOT the correct explanation for Statement IC. Statement I is true, Statement II is false.D. Statement I is false, statement II is true. |
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Answer» Correct Answer - A `alpha=(1)/(l)(dl)/(dT)` `impliesalphaint_(T0)^(T)dT=int_(l0)^(l)(dl)/(l)` `impliesl=l_0e^(alphaDeltaT)` |
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| 70. |
In natural convection, the fluid motion is caused due to density difference produced by temperature gradient. Statement II: In forced convection, the fluid is forced to flow along the solid surface by means of fans or pumps.A. Statement I is true, Statement II is true and Statement II is the correct explanation for statement I.B. Statement I is true, statement II is true and statement II NOT the correct explanation for Statement IC. Statement I is true, Statement II is false.D. Statement I is false, statement II is true. |
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Answer» Correct Answer - B Statement I and II are true but Statement II is not correct explanation for Statement I. |
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| 71. |
Two rods having length `l_1` and `l_2`, made of materials with the linear expansion coefficient `alpha_1` and `alpha_2`, were soldered together. The equivalent coefficient of linear expansion for the obtained rod isA. `(l_1alpha_2+l_2alpha_1)/(l_1+l_2)`B. `(l_1alpha_1+l_2alpha_2)/(alpha_1+alpha_2)`C. `(l_1alpha_1+l_2alpha_2)/(l_1+l_2)`D. `(l_2alpha_1+l_1alpha_2)/(alpha_1+alpha_2)` |
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Answer» Correct Answer - C `L_1(1+alpha_1Deltat)+l_2(1+alpha_2Deltat)=l_f` `l_f=l_1+l_2+(l_1alpha_1+l_2alpha_2)Deltat` `l_f=l_1+l_2(1+(l_1alpha_1+l_2alpha_2)/(l_1+l_2))` |
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| 72. |
On a Celsius thermometer the distance between the readings `0^@C` and `100^@C` is 30cm and the area of cross section of the narrow tube containing mercury is `15xx10^-4cm^2`. Find the total volume of mercuty in the thermometer at `0^@C`. `alpha` of glass`=9xx10^-6//K` and the coefficient of real expansion of mercury`=18xx10^(-5)//K`. |
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Answer» Apparent increase in volume `=Vxx(18xx10^-5-27xx10^-6)xx100` Also apparent increase`=15xx10^-4xx30` `:. 15xx10^-4xx30` `=Vxx(18xx10^-5-27xx10^-6)xx100` or `V=2.94cc` |
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| 73. |
Two insulated metal bars each of length 5 cm and rectangular cross section with sides 2 cm and 3 cm are wedged between two walls, one held at `100^@C` and the other at `0^@C`. The bars are made of lead and silver. `K_(pb)=350 W//mK,K_(Ag)=425 W//mK`. Equevalent thermal resistence of the two bar system isA. `0.1(K)/(W)`B. `0.23(K)/(W)`C. `0.19(W)/(W)`D. `0.42(K)/(W)` |
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Answer» Correct Answer - A `(1)/(R_(eq))=(1)/(R_(pb))+(1)/(R_(Ag))=(21)/(5)+(51)/(10)` `=(42+51)/(10)=(93)/(10)` `R_(eq)=(10)/(93)=0.107Wcong0.1(K)/(W)` |
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| 74. |
Two insulated metal bars each of length 5 cm and rectangular cross section with sides 2 cm and 3 cm are wedged between two walls, one held at `100^@C` and the other at `0^@C`. The bars are made of lead and silver. `K_(pb)=350 W//mK,K_(Ag)=425 W//mK`. Total thermal current through the two bar system isA. 210 WB. 420 WC. 510 WD. 930 W |
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Answer» Correct Answer - D Resistance of silver bar `R_(Ag)=(l_Ag))/(K_(Ag)A_(Ag))=(5xx10^-2)/(425xx6xx10^-4)` `=(100)/(85xx6)=(20)/(17xx6)=(10)/(51)(K)/(W)` Thermal current through silver bar `I_(Ag)=(DeltaT)/(R_(Ag))=(100xx51)/(10)=510W` Total thermal current `I_(pb)+I_(Ag)=930W` |
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| 75. |
Two insulated metal bars each of length 5 cm and rectangular cross section with sides 2 cm and 3 cm are wedged between two walls, one held at `100^@C` and the other at `0^@C`. The bars are made of lead and silver. `K_(pb)=350 W//mK,K_(Ag)=425W//mK`. Thermal corrent through lead bar isA. 210 WB. 420 WC. 510 WD. 930 W |
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Answer» Correct Answer - B Resistance of lead bar `R_(pb)=(l_(pb))/(K_(pb)A_(pb))=(5xx10^-2)/(350xx6xx10^-4)` `=(10)/(21)=(5)/(21)(K)/(W)` Thermal current through lead bar `I_(pb)=(DeltaT)/(R_pb))=(100xx21)/(5)=420W` |
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| 76. |
The specific head capicity of a body depends onA. the heat givenB. the temperature releasedC. the mass of the bodyD. the material of the body |
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Answer» Correct Answer - D the material of the body |
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| 77. |
The head capacity of a body depends onA. the heat givenB. the temperature releasedC. tha mass of the bodyD. the material of the body |
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Answer» Correct Answer - C::D tha mass of the body ,the material of the body |
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| 78. |
In following equiation calculate value of H: 1kg ice at `-20^@C=H+1`kg water at `100^@C`, here H means heat required to change the state of substance. |
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Answer» Heat required to convert 1 kg ice at `-20^@C` into 1 kg water at `100^@C` `=1kg` ice at `-20^@C` to 1 kg ice at `0^@C` ice at `0^@C+1kg` water at `0^@C+1kg` water at `0^@C` to 1 kg water at `100^@C` `=1xx(1)/(2)xx20+1xx80+1xx100=190kcal`. So `H=-190kcal` Negative sign indicates that 190 kcal heat is with drawn from 1 kg water at `100^@C` to convert it into 1 kg ice at `-20^@C` |
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| 79. |
Name two factors on which the heat absorbed or given out by a body depends. |
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Answer» The heat absorbed or given out by a substance depends upon (i) mass of the body, (ii) rise or fall of temperature |
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| 80. |
A block of wood is floating in water at `0^@ C`. The temperature of water is slowly raised from `0^@ C` to `10^@ C`. How will the precentage of volume of block above water level change with rise in temperature?A. increaseB. decreaseC. first increase and then decreaseD. first decrease and then increase |
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Answer» Correct Answer - C Fraction of wooden block immersed at `0^@C` `(V_1)/(V_0)=((rho_("wood"))_(0^@C))/((rho_(H)_(2)O)_(0^@C))` `f_1=(V_0-V_1)/(V_0)=((rho_(H2O))_(0^@C)-(rho_("wood"))_(0^@C))/((rho_(H_2O))_(0^@C))` `V_1`- volume of wood immersed in water at `0^@C` `V_0`- Volume of wood `(rho_("wood"))_(0^@C)`- Density of wood at `0^@C` When the temperature is raised to `10^@C` the volume of wood immersed in water changes to `V_2`. `(V_2)/(V_0)=((rho_("wood"))_(0^@C))/((rho_(H_2O))_(0^@C))` `f_2=(V_0-V_2)/(V_0)=((rho_(H_2O))_(10^@C)-(rho_("wood"))_(10^@C))/((rho_(H_2O))_(10^@C))` From `0^@C` to `4^@C`, the density of water increases, and from `4^@C` to `10^@C` the density of water decreases. But for wood density decreases as temperature increases. The volume of block above water level will first increase and then decrease. |
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| 81. |
When the temperature of a copper coin is raised by `80^@C`, its diameter increases by `0.2%`.A. percentage rise in the area of a face is `0.4%`B. percentage rise in the thickness is `0.4%`C. percentage rise in the volume is `0.6%`D. corfficient of linear expansion of copper is `0.25xx10^(-4)//^(@)C` |
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Answer» Correct Answer - A::C::D `(DeltaA)/(A)xx100=2((Deltal)/(A))xx100` `%` increase in area`=2xx0.2=0.4` `(DeltaV)/(V)xx100=3xx0.2=0.6%` Since `DeltaI=IalphaDeltaT` `(Deltal)/(l)xx100=alphaDeltaTxx100=0.2` `alpha=0.25xx(10^-4^@)/(C )` |
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| 82. |
State the effect of presence of impurity on the melting point of ice. Give one use of it. |
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Answer» The melting point of ice decreases by the presence of impurity in it. Use: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams. |
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| 83. |
Explain why the weather becomes very cold after a hailstorm. |
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Answer» As the ice starts melting after a hailstorm, it absorbs latent heat of fusion from the surrounding air. This leads to the cooling of atmosphere. |
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| 84. |
What happens to the heat supplied to a substance when the heat supplied causes no change in the temperature of the substance? |
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Answer» Heat supplied to a substance during its change of state is called latent heat. It is used up in increasing the potential energy of the molecules of the substance and in doing work against external pressure if there is an increase in volume. Hence there is no change of temperature. |
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| 85. |
A solid floats in a liquid at `20^@`C with 75 % of it immersed. When the liquid is heeated to `100^@`C, the same solid floats with `80%` of it immersed in the liquid. Calculate the coefficient of expansion of the liquid. Assume the volume of the solid to be constant. |
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Answer» Let m be the mass of the solid and V its volume. By the law of floatation Weight of floating object = Buoyant force In Case I: `mg=((3)/(4)V)rho_(20)g` Where `rho_(20)=` density of liquid at `20^@`C In Case II : `mg=((8)/(100)V)rho_(100)g`, Where `rho_(100)=`density of liquid at `100^@`C Considering both the cases `implies(3)/(4)rho_(20)=(4)/(5)rho_(100)implies(3)/(4)(rho_0)/(1+gammaxx20)=(4)/(5)(rho_0)/(1+gammaxx100)` After solving we get `gamma=(1)/(1180)=8.47xx10^(-4)//^(@)C` |
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| 86. |
A glass flask whose volume is exactly 1000`cm^3` at `0^@C` is filled level full of mercury at this temperature. When the flask and mercury are heated to `100^@C`, `15.2cm^3` of mercury overflows. The coefficient of cubical expansion of Hg is `1.82xx10^(-4)//^@ C`. Compute the coefficient of linear expansion of glass. |
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Answer» As `15.2cm^3` of Hg overflows at `100^@`C (final volume of Hg)`-`(final volume off glass flask)`=15.2cm^3` `implies1000(1+gamma_(l) theta)-1000(1+gamma_(g) theta)=15.2`, where `theta=` rise in temperature `=100-0=100^@C` `implies gamma_g=gamma_i-(15.2)/(1000theta)=-.000182-0.000152` `implies gamma_g =3 xx 10^(-5)//^@C implies alpha=(gamma_g)/(3)=1xx10^(-5)//C` |
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| 87. |
A `250cm^3` glass bottle is completely filled with water at `50^@C`. The bottle and water are heated to `60^@C`. How much water runs over if: a. the expansion of the bottle is neglected: b. the expansion of the bottle is included? Given the coefficient of areal expansion of glass `beta=1.2xx10^(-5)//K` and `gamma_("water")=60xx10^(-5)//^@C`. |
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Answer» Water overflow`=`(final volume of water)`-`(final volume of bottle) a. If the expansion of bottle is neglected: water overflow`=250(1+gamma_ltheta)-250` `=250xx60xx10^-5xx10` `implies`water overflow`=1.5cm^3` b. If the bottle (glass) expands: Water overflow. `=("final volume of water")-("final volume of glass")` `=250(1+gamma_(l) theta)-250(1+gamma_(g) theta)` `=250(gamma_l-gamma_g)theta`, where `gamma_g=3//2beta=1.8xx10^(-5)//.^@C` `=250(58.2xx10^-5)xx(60-50)` Water overflow`=1.455cm^3` |
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| 88. |
Why does ice-cream feel more colder than water at 0°C? Explain. |
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Answer» Reasons used : 1kg of ice on meting absorbs 336000J of heat energy and 1kg of water to freeze will absorb 336000J of heat energy Sp. latent heat of ice is very high and it is 336000J Hence ice will absorb more heat from mouth and temp, of mouth will fall considerably and ice cream feels more colder than water. |
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| 89. |
Why does the weather become warm, when it snows? Explain. |
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Answer» Reasons used : 1kg of ice on meting absorbs 336000J of heat energy and 1kg of water to freeze will absorb 336000J of heat energy. When it snows, water evolves heat i.e. it gives out 336000J for every 1kg, in the surrounding and it becomes warm. |
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| 90. |
An iron ball (coefficient of linear expansion`=1.2xx(10^(-5)//^(@)C`) has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate (coefficient of linear expansion`=1.9xx10^-5//^(@)C`) when the ball and the plate are both at a temperature of `30^@C` At what common temperature of the ball and the plate will the ball just pass through the hole in the plate ?A. `23.8^@C`B. `53.8^@C`C. `42.5^@C`D. `63.5^@C` |
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Answer» Correct Answer - B Using the suffixes I and B for the iron ball and the brass plate we have `L_I=6cm`,`L_I-L_B=0.001cm` at `t=30^@C` Heating both the ball and the plate increases the diameters of the ball as well as the hole in the plane, with the hole diameter increasing at a faster rate, since `alpha_Bgtalpha_L`. Now we require that `DeltaL_B-DeltaL_I=0.001cm` at the desired temperature t, with `DeltaL_B=L_Balpha_BDeltat` and `DeltaL_I=L_Ialpha_IDeltat`. Then `DeltaL_B-DeltaL_I=(L_Balpha_B-L_Ialpha_I)DeltaT=0.001cm` `approxL_I(alpha_B-alpha_I)Deltat`, Approximating by putting `LcongL_I` or `(6cm)[1.9xx10^-5-1.2xx10^-5]Deltat=0.001cm` Hence `Deltat=(0.001)/(6[1.9xx10^-5-1.2xx10^-5J]).^@C` `Deltat=23.8^@C` Hence final temperature `=30+23.8^@C=53.8^@C` |
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| 91. |
A table top is made of aluminium and has a hole of diameter 2 cm. An iron sphere of diameter 2.004 m is resting on this hole. Below the hole, an insulated container has 2 kg of water in it. Everything is at ambient temperature of `25^(@)C`. The table top along with the iron sphere is heated till the ball falls through the hole into the water. Find the equilibrium temperature of the ball and water system Neglect any heat loss from ball–water system to the surrounding and assume the heat capacity of the container to be negligible. Relevant data: Coefficient of linear expansion for aluminium and iron are `2.4 xx 10^(-5) .^(@)C^(-1)` and `1.2 xx 10^(-5) .^(@)C^(-1)` respectively. Specific heat capacity of water and iron are `4200 J .^(@)C^(-1) g^(-1)` and `450 J .^(@)C^(-1) g^(-1)` respectively. Density of iron at `25^(@)C` is `8000 kg//m^(3)`. |
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Answer» Correct Answer - `27.4^(@)C` |
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| 92. |
A tube leads from a flask in which water is boiling under atmospheric pressure to a calorimeter. The mass of the calorimeter is 150 g, its specific heat capacity is `0.1 cal//g//^@C`, and it contains originally 340g of water at `15^@C`. Steam is allowed to condense in the colorimeter until its temperature increases to `71^@C`, after which total mass of calorimeter and contents are found to be 525g. Compute the heat of condensation of steam. |
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Answer» Mass of calorimeter and contents before passing steam `=(150+340)=490g` mass after passing steam`=525g` `implies`mass of steam which condenses`=(525-490)g=35g` Let L`=` latent heat of steam. Heat lost by steam`=`heat gained by water`+`heat gained by calorimeter `35L+35xx1(100-71)=340xx1xx(71-15)+150` `xx0.1xx(71-15)` `implies=539 cal//g` |
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| 93. |
Why do bottled soft drinks get cooled, more quickly by the ice cubes than by the iced water, both at 0℃? |
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Answer» This is because 1 g of ice at 0oC takes 336 J of heat energy from the bottle to melt into water at 0oC. Thus, bottle loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1 g iced water at 0oC. Therefore, bottled soft drinks get cooled, more quickly by the ice cubes than by iced water. |
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| 94. |
Determine the lengths of an iron rod and copper ruler at `0^@` C if the difference in their lengths at `50^@`C and `450^@`C is the same and is equal to 2 cm. the coefficient of linear expansion of iron`=12xx10^(-6)//K` and that of copper`=17xx10^(-6)//K`. |
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Answer» Let x be the length of the iron rod at `0^@`C y that of the copper rod at `0^@`C, and `Deltal` the difference in lengths at `t_1` and `t_2C^@`. Then `Deltal=x(1+alpha_1t_1)-y(1+alpha_2t_1)` ..(i) and `+-Deltal=x(1+alpha_1t_2)-y(1+alpha_2t_2)` ..(ii) Taking positive sign in equation (ii) `Deltal=x(1+alpha_1t_2)-y(1+alpha_2t_2)` ..(iii) From Eqs. (i) and (iii) we get `xalpha_1=yalpha_2` .(iv) From Eqs. (i) and (iv), we get `y=(lalpha_1)/(alpha_2-alpha_1)=(2xx12xx10^-6)/((17-12)xx10^-6)=4.8cm` and `x=(2xx17xx10^-6)/((17-12)xx10^-6)cm=6.8cm` Taking negative sign in equaiton (ii) `y=(2I+Ialpha_1(t_1+t_2))/((t_2-t_1)(a_2-a_1))`, `x=(2I+Ialpha_2(t_1+t_2))/((t_2-t_1)(alpha_2-alpha_1))` `:. y=(2xx2+2xx12xx10^-6(450+50))/((450-50)(17-12)xx10^-6)cm=2006cm=20.06m` `x=(2xx2+2xx17xx10^-6(450+50))/((450-50)(17-12)xx10^-6)cm=2008.5cm=20.08m` |
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| 95. |
Two iron spheres of the same diameter are heated to the same temperature. One is soled, and the other is hollow which will expand more? |
| Answer» They will show same expansion because expansion of a solid does not depend on its internal construction. | |
| 96. |
Two bodies have the same heat capacity. If they are combined to form a fsingle composite body, show that the equivalent specific heat of this composite body is independent of the masses of the individual bodies. |
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Answer» Let the two bodies have masses `m_1`,`m_2` and specific heats `S_1`,`S_2`. Then `m_1S_1=m_2S_2` Let s`=`specific heat of the composite body. Then `(m_1+m_2)s=m_1S_1+m_2S_2=2m_1S_1` `S=(2m_1S_1)/(m_1+m_2)=(2m_1S_1)/(m_1+m_1(S_1//S_2))=(2S_1S_2)/(S_2+S_1)`. |
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| 97. |
Two bodies of equal masses are heated at a uniform rate under identical conditions. The change in temperature in the two cases in shown graphically. What are their melting points? Find the ratio of their specific heats and latent heats. |
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Answer» The melting points liquids I and II are `60^@` C and `40^@`C, respectively. Let R be the rate of supply of heat. We note from the graph that liquid I is heated through `60^@`C in 2 units of time and that liquid II is heated through `40^@`C in 4 units of time. `2R=mxxc_1xx60` and `4R=mxxc_2xx40` Hence, `(c_1)/(c_2)=(1)/(3)` We note further that the temperature of I remains constant for 4 units of time and that of II for 2 units of time. `4R=mL_1` and `2R=mL_2implies(L_1)/(L_2)=2` |
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| 98. |
What is wrong with following statement Given any two bodies, the one with the higher temperature contains more heat. |
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Answer» The statement shows a misunderstanding of the concept of heat. Heat is a process by which energy is transferre, not a form of energy that is held or contained. If you wish to speak of energy that is contained, you speak of internal energy not heat. Further, even, if the statement used the term internal energy, it would still be incorrect, since the effects of specific heat and mass are both ignore. A 1 kg mass of water at `20^@C` has more internal energy than a 1 kg mass of air at `30^@C`. Similarly, the earth has far more internal energy than a drop of molten titanium metal. correct statements would be: 1. Given any two bod-ies in thermal contact, the one with the higher product of absolute temperature and specific heat contains more internal energy. All to say is that internal energy depends not only on temperature but also on mass and nature of body. |
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| 99. |
An electric heater supplies 1.8 kW of power in the form of heat to a tank of water. How long will it take to heat the 200 kg of water in the tank from `10^@` to `70^@`C ? Assume heat losses to the surroundings to be negligible. |
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Answer» The heat added to tank `DeltaQ=` power`xx`time `DeltaQ=1.8xx10^3xxt(J)` .(i) The heat absorbed in water `DeltaQ=mcDeltaQ=mcDeltaT=200xx10^3xx1xx60=12xx10^6cal` `=12xx10^66xx4.2J` Hence `t=(12xx10^6xx4.2)/(1.8xx10^3)=2.78xx10^4s=7.75h` |
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| 100. |
`M g` of ice at `0^@C` is mixed with `M g` of water at `10^@ c`. The final temperature isA. `8^@ C`B. `6^@C`C. `4^@ C`D. `0^@ C` |
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Answer» Correct Answer - D `(M xx 80) gt (M xx 10)` because Final Temp. is `0^@ C`. |
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