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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. |
At 1 atmospheric pressure, 1.000 g of water having a volume of `1.000 cm^3` becomes 1671 `cm^3` of steam when boiled. The heat of vaporization of water at 1 atmosphere is `539 cal//g` . What is the change in internal energy during the process ? |
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Answer» Heat spent during vaporisation Q = mL = `1.000 xx 539` = 539 cal Work done `W = p(V_(v) - V_(l))` `=1.013 xx 10^(5) xx (1671 - 1.000( xx 10^(-6)` `169.2J = (169.2)/(4.18) cal = 40.5 cal` `therefore` Change in internal energy, U = 539 cal `- 40.5` cal = 498.5 cal |
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| 2. |
A slab consists of two layers of different materials of the same thickness and having thermal conductivities `K_(1)` and `K_(2)`. The equivalent thermal conductivity of the slab isA. K1 + K2B. K1 + K2/2C. 2K1 + K2 /K1 + k2D. K1 + K2/2KK2 |
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Answer» Correct Answer - B `(K_(eq))_(11)= (k_(1)+k_(2))/2` |
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| 3. |
The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along BD . If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L / C is A. 40B. 80C. 100D. 20 |
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Answer» Correct Answer - D Let the quantity of heat supplied per minute be Q, then quantity of heat supplied in 2min = mc(90-80) In 4min, heat supplied = 2mc(90-80) `therefore` 2mc(90-80) = mL `rArr` L/c = 20 |
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| 4. |
If a body cools down from `80^(@) C`to `60^(@) C` in 10 min when the temperature of the surrounding of the is `30^(@) C` . Then, the temperature of the body after next 10 min will beA. `50^(@) C`B. `48^(@) C`C. `30^(@) C`D. None of the above |
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Answer» Correct Answer - D `(theta_(1) - theta_(2))/t = alpha[(theta_(1) + theta_(2))/2 - theta_(0)]` two times we get, Where `alpha` is constant. `(80-60)/(10) = alpha[(80+60)/2 -30]`............(i) `(60-theta)/10 = alpha[(60+theta)/2 -30]`..............(ii) Solving these two equations, we get `theta = 48^(@)`C |
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| 5. |
A geyser heats water flowing at the rate of `3.0` litre per minute from `27^(@)C` to `77^(@)C`. If the geyser operates on a gas burner and its heat of combustion is `4.0 xx 10^(4) J//g`, then what is the rate of combusion of fuel (approx.)? |
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Answer» Given, Volume of water heated, V`=3.0 Lmin^(-1)` = `3 xx10^(-3)m^(3)min^(-1)` As Density, `rho` = `(Mass(m))/(Volume(V))` `therefore` Mass of wate heated, `m=rhoV = 10^(3)xx3xx10^(-3)kg min^(-1)` Rise in temperature of water, `DeltaT = (T_(2)-T_(1)) = 77-21 = 50^(@)C` Specific heat of water, `c = 4.2Jg^(-1)C^(-1)` = `4.2xx10^(3)Jkg^(-2)C^(-1)` Heat required by wate, `DeltaQ = mcDeltaT` `=3xx4.2xx106(3)xx50`Jmin `=63 xx10^(4)Jmin^(-1)`................(i) Heat of combustion of fuel = `4.0 xx 10^(4)Jg^(-1)` = `4.0 xx 10^(7)Jkg^(-1)` Let m kg `min^(-1)` be the rate of combustion of fuedl `= mxx4 xx10^(7Jmin^(-1)`...............(ii) `therefore` Eqs. (i) and (ii) , we get `63 xx10^(4)=mxx4xx10^(7)` or `m=(63xx10^(4))/(4 xx10^(7)` = `15.75 xx 10^(-3)kg min^(-1)` or rate of combution of fuel = `16g min^(-1)` |
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| 6. |
Same quantity of ice is filled in each of the two metal container P and Q having the same size, shape and will thickness but make of different materials. The containers are kept in identical surroundings, The ice in P melts completely in time `t_(1)`, whereas in Q takes a time `t_(2)`. The ratio of thermal conductivities of hte materials of P and Q is:A. `t_(2):t_(1)`B. `t_(1):t_(2)`C. `t_(1)^(2):t_(2)^(2)`D. `t_(2)^(2):t_(1)^(2)` |
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Answer» Correct Answer - A a) We know that, relation between the temperature gradient (TG) and thermal conductivity (K), `(dQ)/(dt) = -KA(d(theta))/(dx)=-KAxx(TG)` i.e., `TG alpha 1/K` `(dQ)/(dt)` = constant or `Kalpha1/t rArrK_(1)alpha1/(t_(1))`...........(i) `K_(2)alpha1/t_(2)`..............(ii) From the Eqs. (i) and (ii), we get `K_(1)/K_(2)=(1//t_(1))/(1//t_(2)) rArrK_(1):K_(2)=t_(2):t_(1)` |
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| 7. |
The calories of heat developed in 200 W heater in 7 min is estimatedA. 15000B. 100C. 1000D. 20000 |
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Answer» Correct Answer - D `H = (200 xx 7 xx 60)/(4.18)cal = 20095.7 ~~20000 cal` |
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| 8. |
The thickness of a metallic plate is 0.4 cm. The temperature between its two surfaces is `20^(@)C`. The quantity of heat flowing per second is 50 calories from `5cm^(2)` area. In CGS system, the coefficient of thermal conductivity will beA. 0.4B. 0.6C. 0.2D. 0.5 |
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Answer» Correct Answer - C Heat flowing per second, `Q/t=(KA(DeltaT))/t rArr 50 = (5 xx20K)/0.4rArrK = 1/5=0.2` |
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| 9. |
Equal masses of three liquids A, B and C have temperature `10^(@)C, 25^(@)C` and `40^(@)c` respectively. If A and B are mixed, the mixture has a temperature of `15^(@)C`. If B and C are mixed, the mixture has a temperature of `30^(@)C`, if A and C are mixed will have a temperature ofA. `16^(@)C`B. `20^(@)C`C. `25^(@)C`D. `29^(@)C` |
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Answer» Correct Answer - A When A and B are mixed. `mc_(A)(15-10) = mc_(B)(25-15)` or `c_(A)/c_(B) = 2`...............(i) When B and C are mixed `mc_(B)(30-25) = mc_(cC(40-30)` or `c_(B)/c_(C) = 2`...........(ii) From Eqs. (i) and (ii), we have `c_(A)/c_(C) = 4` Now let `theta` be the equilibrium temperature when A and C are mixed. Then `mc_(A)(theta-10) = mc_(C)(40-theta)` or `4(theta-10)=40-theta` (Using `c_(A)/c_(C) = 4`) or `theta=16^(@)`C |
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| 10. |
Two identical conducting rods AB and CD are connected to a circular conducting ring at two diametrically opposite points B and C, the radius of the ring is equal to the length of rods AB and CD. The area of cross-section, thermal conductivity of the rod and ring are equal. points A and D are maintained at temperatures of `100^(@)C` and `0^(@)C`. temperature at poimt C will be A. `62^(@)C`B. `37^(@)C`C. `28^(@)C`D. `45^(@)C` |
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Answer» Correct Answer - C `R_(AB) = R_(CD)=R`(say) Length of semicircle = `pil` Resistance of semicircle = `piR` as `Ralphal` `R_(BC) = (piR)/2`(in parallel) Heat current, H = `(100-0)/(R+R+(pi)/2.R)=28/R` units Now, `theta_(C) - 0^(@)` = HR = `28^(@)`C `theta_(C)` = `28^(@)`C |
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| 11. |
In a room containing air, heat can go from one place to anotherA. conductionB. convectionC. radiationD. All the three |
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Answer» Correct Answer - B In gases conductions, is most prominent mode. |
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| 12. |
Ice formed over lakes hasA. has vey high thermal conductivity and helpsinfurther ice formationB. has very low conductivity and retards further formation of iceC. permits quick convection and retards further formation of iceD. is very good radiator |
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Answer» Correct Answer - B Ice is non-conducting. |
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| 13. |
suppose you want to cool 0.25 kg of cola (mostly water ), at `25^(@)C` by adding ice initially at `-20^(@)C`. How much ice should you add so that the final temperature will be `0^(@)C` with all the ice melt? Neglect the heat capacity of the container. specific heat of ice is` 2000 Jkg^(-1)K^(-1)`. [take specific heat of cola `4160 Jkg^(-1)K^(-1)`.] |
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Answer» Heat lost by cola is `Q_(cola)=m_(cola)c_(cola)DeltaT_(cola)` `(0.25kg)(4160 Jkg^(-1)K^(-1)25^(@)C - 0^(@)`C ltbrge26000J Let the mass of the required ice be `m_(ice)`, then the heat needed to warm it from `-20^(@)`C to `0^(@)`C is `Q_(1) = m_(ice)c_(ice)DeltaT_(ice)` `m_(ice)(2000)Jkg^(-1)K^(-1))(0^(@)C-(-20^(@)C))` `m_(ice)(3.34 xx 10^(5)Jkg^(-1)` `therefore` Total energy gained by ice is `Q_(1)+Q_(2)=m_(ice)(3.74 xx 10^(5)Jkg^(-1)` Using principle of calorimetry, heat lost = heat gained, we get 26000 J = `m_(ice)(3.74 xx 10^(5)Jkg^(-1)` or, `m_(ice)`=0.07kg = 70g |
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| 14. |
Three bars of equal lengths and equal area of cross-section are connected in series fig. their thermal conducitives are in the ratio `2:3:4`. If at the steady state the open ends of the first and the last bars are at temperature `200^(@)C` and `20^(@)C` respectively, find the temperature of both the junctions. . |
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Answer» Let the temperature of the junction B and C be `theta_(1)` and `theta_(2)`, respectively. In the steady state, the rate of flow of heat through all rods will be the same. Therefore, Heat transfer `H = (2K) A(200-theta_(1))/L` (4K)A(theta_(1)-theta_(2))/L= (3K) A(theta_(2) - 18)/L)` or `2(200-theta_(1)) = 4(theta_(1)-theta_(2)) = 3(theta_(2)-18)` Solving, `theta_(1)=116^(@)`C, `theta_(2)=74^(@)`C |
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| 15. |
If the rate of emission of radiation by a body at temperature `TK` is `E` then graph between log `E` and log `T` will be .A. B. C. D. |
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Answer» Correct Answer - C According to Stefan-Boltzmann law, the emissive power is directly proportional to the fourth powerof temperature. So, it will give a straight line. |
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| 16. |
A block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a room free from air currents. Which of the following curves correctly represents the rate of coolingA. B. C. D. |
| Answer» Correct Answer - B | |
| 17. |
When 400 J of heat are added to a 0.1 kg sample of metal, its temperature increase by `20(@)`C. What isthe specific heat of the metal? |
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Answer» Given, Heat `DeltaQ` = 400 J, Mass m = 0.1kg Temperature `DeltaT` = `20^(@)C` `therefore` Specific heat, `c=1/m.(DeltaQ)/(DeltaT) = (1/0.1)(400/3)= 200 Jkg^(-1)^(@)C^(-1)` |
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| 18. |
A black body with surface area 0.001 `m^(2)` is heated upto a temperature 400 K and is suspended in a room temperature 300K. The intitial rate of loss of heat from the body to room isA. 10 WB. 1WC. 0.1WD. 0.5 W |
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Answer» Correct Answer - B b) The rate of heat emission, u = `sigmaAT^(4)`. At room temperatures to the rate of absorption is `u_(0)=sigmaAT_(0)^(4)` The net rate of heat loss. `u-u_(0)=sigmaA(T^(4)-T_(0)^(4))` Given that A=0.001`m^(2)` T=400K and `T_(0)`=300K Thus, `u-u_(0)` = `5.67 xx 10^(-8) xx 0.001 xx [(400)^(4)-(300)^(4)]`=0.99 W |
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| 19. |
Assertion : For higher temperature, the peak emission wavelength of a black body shifts to lower wavelengths. Reason : Peak emission wavelength of a black body is proportional to the fourth power of temperature.A. If both Assertion and Reason are correct and Reason is also a good absorber of radiation at a given wavelength.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B b) `lambda_(m) alpha 1//T` |
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| 20. |
These question consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five responses. Assertion: Absorptive power of any substance is temperature independent. Buit emissive power depends on the temperature. Reason: Emissive power `alpha T^(4)`A. If both Assertion and Reason are correct and Reason is also a good absorber of radiation at a given wavelength.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A a) `E= intE_(lambda)dlambdaalphaT^(4)` |
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| 21. |
Assertion : A normal body can radiate energy more than a perfectly black body. Reason : A perfectly black body is always black in colour.A. If both Assertion and Reason are correct and Reason is also a good absorber of radiation at a given wavelength.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is falseD. If both Assertion and Reason are false. |
| Answer» e) A normal body can also radiate energy more than a perfectly black body if its temperature is very high. | |
| 22. |
Assertion : Snow is better insulator than ice. Reason : Snow contain air packet and air is good insulator of heat.A. If both Assertion and Reason are correct and Reason is also a good absorber of radiation at a given wavelength.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A a) Both Assertion and Reason are correct and reason explaains Assertion. |
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| 23. |
A piece of ice falls from a height `h` so that it melts completely. Only one-quarter of the heat produced is absobed by the ice and all energy of ice gets converted into heat during its fall. The value of `h` is [Latent heat of ice is `3.4 xx 10^(5) J//kg` and `g = 10 N//kg`]A. 544 kmB. 136kmC. 68 kmD. 34km |
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Answer» Correct Answer - B b) According to quesion as conservation of energy, energy grained by the ice during its fall from height h is given by E =mgh So, `(mgh)/4 = mL_(f)` As given, only one quarter of its energy is absorbed by the ice. So, `(mgh)/4 = mL_(f)` `rArr h=(mL_(f) xx 4)/(mg) = (L_(f) xx 4)/g = (3.4 xx 10^(5)xx4)/10 = 13.6 xx 10^(4)` =13600 m = 136 km |
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| 24. |
Snow is more heat insulating than ice, becauseA. Air is filled in pores of snowB. Ice is more bad conductor than snowC. Air is filled in pores of iceD. Density of ice is more |
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Answer» Correct Answer - A Snow contains air pockets. |
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| 25. |
One likes to sit under sunshine in winter seasons, becauseA. we get heat by radiation from SunB. we get heat by conduction by SunC. we get heat by conduction from Sun.D. we get heat by conduction from Sun. |
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Answer» Correct Answer - B Heat is received from sun by radiation. |
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| 26. |
The thermal conductivity of a rod depends onA. lengthB. massC. area of x-sectionD. material of the rod |
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Answer» Correct Answer - D Thermal conducting is a material property. |
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| 27. |
The unit of thermal conductivity is :A. ` Js^(-1) K`B. `Js^(-1) m^(2)k`C. `Jm^(-1)K`D. `Js^(-1) m^(-1) K^(-1)` |
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Answer» Correct Answer - D `Q= (KA(theta_(1)-theta_(2)t)/l) ` `therefore` `K= (Q xx l)/(A(theta_(1) - theta_(2))t` Unit of K = (Jm)/(m^(2)K-s)` = `J/(s-mK)` |
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| 28. |
5 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter Find the final temperature of the mixure. Assume water equivalent of calorimeter to be negligible, sp. Heats of ice and water are 0.5 and `1 cal//g C^@`, and latent heat of ice is `80 cal//g` |
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Answer» In this case heat is given by water and taken by ice. Heat available with water to cool from `30^(@)`C to `0^(@)`C `=mcDeltaT=5xx1xx30`=150cal ltbrlt Heat required by 5g ice to increase its temperature up to `0^(@)`C `mcDeltaT` = `5xx0.5xx20`=50cal Out of 150 cal heat available, 50cal is used for increasing tempeature of ice from `-20^(@)`C to `0^(@)`C. The remaining heat 100 cal is used for melting the ice. If mass of ice melted is mg then `mxx80 =100 rArr` m=1.25g Thus, 1.25g ice out of 5g melts and mixture of ice and water is at `0^(@)`C. |
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| 29. |
50 g of copper is heated to increase its temperature by `10^@C`. If the same quantity of heat is given to `10g` of water, the rise in its temperature is (specific heat of copper`=420J//kg^(@)//C`)A. `5^(@)C`B. `6^(@)C`C. `7^(@)C`D. `8^(@)C` |
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Answer» Correct Answer - A Same amount of heat is supplied to copper and water. So, `m_(C)c_(C)DeltaT_(c)= m_(W)c_(W)DeltaT` `rArr` `DeltaT_(W) = (m_(C)c_(C)(DeltaT)_(c))/(m_(W)c_(W))` =`(50 xx 10^(-3) xx 420 xx 10)/(10 xx 10^(-3) xx 4200)`= 5^(@)C` |
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| 30. |
A liquid of mass m and specific heat c is heated to a temperature 2T. Another liquid of mass m/2 and specific heat 2 c is heated to a temperature T. If these two liquids are mixed, the resulting temperature of the mixture isA. (2/3) TB. (8/5)TC. (3/5)TD. `(3/2)`T |
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Answer» Correct Answer - D Temperature of mixture, `theta(mix) = ((m_(1)c_(1)T_(1)+ m_(2)c_(2)T_(2))/(m_(1)c_(1)+ m_(2)c_(2))` = `(m xx cxx 2T+(m/2)(2c)T)/(mxxc+m/2(2c)) = 3/2 T` |
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| 31. |
`5g` ice at `0^(@)C` is mixed with `5g` of steam at `100^(@)C`. What is the final temperature? |
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Answer» Given, mass of the steam, m=1g Temperature `DeltaT = T_(2)-T_(1)` = 100-0=`100^(@)`C Heat required to melt ice at `0^([email protected])C=Q_(1)=m_(i)c_(w)DeltaT = 5xx1xx100`=500cal This heat rejected can melt the ice completely but cannot raise the temperature of water from 0 to `100^(@)`C as there is a deficiency of heat = `Q_(1) + Q_(2)-Q_(3)`=(900-540)cal = 360 cal `therefore` Resulting temperature `=100^(@)`C-(Heat deficient)/(Thermal capacity of system i.e., 6 g water) `=100^(@)C-(360 cal)/(6gxx1calg^(-1^(@))C^(-1))=100^(@)C -(360^(@)C)/6 =100^(@)C-60^(@)C=40^(@)`C |
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| 32. |
`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3m^(3)` of water at `60^(@)C`. The finial temparature of the mixture isA. `70^(@)C`B. `65^(@)C`C. `60^(@)C`D. `75^(@)C` |
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Answer» Correct Answer - B Density of water = `10^(3)kg m^(-3)` Let the final temperature of the mixture be t. Assuming no heat transfer to or from container. Heat lost by water = `0.1 xx 10^(3) xx S_(water) xx (80-t)` Heat gained by water = `0.3 xx 10^(3) xx S_(water) (t-60)` According to principle of calorimetry Heat lost = heat gain `0.1 xx 10^(3) xx S_(water) xx (80-t) = 0.3 xx 10^(3) xx S_(water) xx (t-60)` `rArr 1 xx(80-t)= 3xx (t-60) rArr t=65^(@)`C |
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| 33. |
How much heat energy is gained when 5 kg of water at `20^(@)C` is brought to its boiling point (Specific heat of water `= 4.2` kj kg c)A. 1680 kJB. 1700 kJC. 1720 kJD. 1740 kJ |
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Answer» Correct Answer - A Heat energy gained Q = `m_(1)c_(1)DeltaT + m_(2)c_(2)DeltaT + m_(3)c_(3)DeltaT` `= (2.4 xx 0.216 + 1.6 x 0.0917 + 0.8 xx 0.0931)`(60) `44.376 cal = 44.44 cal |
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| 34. |
If specific heat of a substance is infinite, it meansA. heat is given outB. heat is taken inC. no change in temperature takes place whether heat is taken in a or given outD. All the above |
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Answer» Correct Answer - C Q =mcQT `rArr c=(Q)/(mDeltaT)` When, `Delta`=0 rArr c=infty` |
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| 35. |
Water is used in car radiators as coolant becauseA. of its lower densityB. it is easily availableC. it is cheapD. it has high specific heat |
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Answer» Correct Answer - D Due to large specific heat of water, it releases large heat with very small temperature change. |
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| 36. |
The spectrum of a black body at two temperatures `27^(@)C` and `327^(@)C` is shown in the figure. Let `A_(1)` and `A_(2)` be the areas under the two curves respectively. Find the value of `(A_(2))/(A_(1))` A. `1 : 16`B. `4 : 1`C. `2 : 1`D. `16 : 1` |
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Answer» Correct Answer - D Area `alpha` Total energy radiated `alphaT^(4)` `T_(1) = 300K` and `T_(2)` = 600K |
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| 37. |
On heating one end of a rod the temperature of the whole rod will be uniform when .A. K = 1B. K = 0C. K = 100D. `K = oo |
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Answer» Correct Answer - D `(dQ)/(dT) = -KA(dtheta)/(dx)`, When `K = infty`, `(dtheta)/(dt) = 0` i.e. `theta` is independent of x, i.e., constnat or uniform |
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| 38. |
80 g of water at `30^(@) C` is mixed with 50 g of water at `60^(@) C`, final temperature of mixture will beA. 30 gB. 80 gC. 1600 gD. 150 g |
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Answer» Correct Answer - A If m g ice melts, then Heat lost = Heat gain `80 xx 1 xx (30-0) = m xx 80 rArr m = 30kg` |
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| 39. |
50 gram of ice at `0^(@)`C is mixed with 50 gram of water at `60^(@)C` , final temperature of mixture will be :-A. `0^(@)C`B. `40^(@)C`C. `10^(@)C`D. `15^(@)C` |
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Answer» Correct Answer - A Heat liberated when 50 g of water at `60^(@)`C converts into water at `0^(@)`C Q = ms`DeltaQ`= 50 xx 1 x 60 = 3000 cal From this heat, mass of ice which can be melted is `m= Q/L = 3000/80` = 37.5g i.e, whole ice is not melted and temperature of mixture will be `0^(@)`C. |
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| 40. |
A `10 kW` drilling machine is used to drill a bore in a small aluminium block of mass `8.0 kg`. How much is the rise in temperature of the block in `2.5` minutes, assuming `50%` of power is used up in heating the machine itself or lost to the surrounding? Specific heat of aluminium `= 0.91 J//g^(@)C`. |
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Answer» Given, Power P = 10kW = `10 xx 10^(3)`W Time, t = 2.5min = `2.5 xx 60s` As power is defined as rate at which energy is consumed, i.e. `P=E/t)` `therefore` Total energy used E = `Pt xx 10^(3) xx 2.5 xx 60` = `1.5 xx10^(6)`J Energy absorbed by aluminium block, Q = 50% of the total energy = 50% pf `1.5 xx 10^(6)` = `0.75 xx 10^(6)`J Also, m = 8.0Kg = `8.0 xx 10^(3)`g, c = 0.91 `Jg^(-1^(@)C^(-1)` As, `Q = mc DeltaT` `therefore` `DeltaT = Q/(mc)=(0.75xx10^(6)(8.0 xx10^(3)xx 0.91)` = `(103.02^(@)C` |
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| 41. |
A cylinder of radius R made of a material of thermal conductivity `K_1` is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity `K_2`. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system isA. `K = K_(1)+K_(2)`B. `K = (K_(1)K_(2))/(K_(1)+K_(2))`C. `K=(K_(1)+3K_(2))/4`D. `K = (3K_(1)+K_(2))/4` |
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Answer» Correct Answer - C c) At steady state, `(dQ)/(dt) = (dQ_(1))/(dt) +(dQ_(2))/(dt)` or `(K(4piR^(2))(T_(2)-T_(1)))/L=(K_(1)(piR^(2))(T_(2)-T_(1)))/L + ((K_2)(3piR^(2))(T_(2)-T_(1)))/L` or `4K = K_(1)+3K_(2)` `therefore` `K=(K_(1)+3K_(2))/4` |
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| 42. |
A cylinder of radius `R` made of a material of thermal conductivity `K_(1)` is surrounded by cylindrical shell of inner radius `R` and outer radius `2R` made of a material of thermal con-ductivity `K_(2)` The two ends of the combined system are maintained at two differnet tem-peratures There is no loss of heat across the cylindrical surface and system is in steady state What is the effective thermal conductivity of the system . |
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Answer» In this situation, a rod of length L and area of cross-section `piR^(2)` and another of same length L and area of cross-section `pi[(3R)^(2)-R^(2)]=8pi(R^(2)` will coduct heat simultaneously, so total heat flowing per second will be `(dQ)/(dt) = (dQ_(1)/(dt) + (dQ_(2))/(dt)` `=K_(1)pir^(2)(T_(1)-T_(2))/(L) +(K_(2)8piR^(2)(T_(1)-T_(2))/(L)`..............(i) Now , if the equivalent conductivity is K. Then, `(dQ)/(dt) = K(9piR^(2)(T-(1)-T_(2))/(L)` [As A `=pi(3R)^(2)]`.......(ii) So, from Eqs, (i) amnd (ii), we have 9K = `K_(1) + 8K_(2)` i.e., `K = (K_(1)+8K_(2))/9 = (K+8xx2K)/(9) = (17K)/(9)` |
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| 43. |
A sphere, a cube and a thin circular plate are heated to the same temperature. All are made of the same material and have the equal masses. Tf `t_(1),t_(2) and t_(3)` are the repective time taken by the sphere, cube and the circular plate in cooling down to common temperature, thenA. `t_(1) gt t_(2) gt t_(3)`B. `t_(1) lt t_(2) lt t_(3)`C. `t_(2) gt t_(1) gt t_(3)`D. `t_(1) = t_(2) = t_(3)` |
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Answer» Correct Answer - A A body having maximum surface are will cool at fastest rate. |
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| 44. |
A solid at temperature `T_(1)` is kept in an evacuated chamber at temperature `T_(2)gtT_(1)` . The rate of increase of temperature of the body is proportional toA. `t_(2)^(4)-t_(1)^(4)`B. `(t_(4)^(2)+273)-(t_(1)^(4)+273)`C. `t_(2)-t_(1)`D. `t_(2)^(2)-t_(1)^(2)` |
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Answer» Correct Answer - C c) As we know that, the rate of cooling is `(dtheta)/(dt) = bA(theta-theta_(0))` where, bA=constant `rArr (dtheta)/(dt) alpha (theta-theta_(0))` Also `-(dT)/(dt) alpha T-T_(0)` where, T=temperature and t=time So, the rate of heat absorbed by cooled body is proportional to `t_(2)-t_(1)`. |
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| 45. |
A cubical block of mass `1.0kg` and edge `5.0cm` is heated to `227^(@)C` . It is kept in an evacuated chamber maintained at `27^(@)C` . Assming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decreases. Specific heat capacity of the material of the block is `400Jkg^(-1)K^(-1)` . |
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Answer» Rate fo cooling `(dT)/(dt)` = `(esigmaA)/(ms)(T^(4)-T_(0)^(4))` Given, `sigma=1`, Area, `A = 6a^(2)= 6 xx (5xx10^(-2)^(2)=150 xx 10^(-4)m^(2)` Temperature, T = 227 + 273 = 500K `T_(0) (27 + 273) = 300K Given, m = 1kg, s=400J/kg.K `=`6 xx10^(-8) xx 150 xx 10^(-4)[(500)^(4)-(300)^(4)]` `=2.25 xx 10^(-12)xx(625 -8)xx10^(8) = 0.12^(@)C//s` |
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| 46. |
A student takes 50 g wax (specific heat `=0.6 kcal//kg^(@)C`) and heats it till it boils. The graph between temperature and time is as follows. Heat supplied to the wax per minute and boiling point are respectively.A. 500 cal,`50^(@)C`B. 1000 cal, `100^(@)C`C. 1500 cal, `200^(@)C`D. 1000 cal, `200^(@)C` |
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Answer» Correct Answer - C From t =0 to t=1min `DeltaQ = msDeltaT = 50/100 xx 06 xx 1000 xx(50 xx 0) = 1500 cal` |
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| 47. |
A substance of mass M kg requires a power input of P wants to remain in the molten state at its melting point. When the power source is turned off, the sample completely solidifies in time t seconds. The latent heat of fusion of the substance is …….A. Pm/tB. Pt/mC. m/PtD. t/mc |
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Answer» Correct Answer - B Heat lost in t second = mL or heat lost per second = `(mL)/t` This must be the heat supplied for keeping the substance in matter state per second `therefore` `(mL)/t = P` or `(Pt)/m` |
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| 48. |
A bubble of 8 mole of helium is subnerged at a certain depth in water. The temperature of water increses by `30^(@)C`. How much hea is added approximately to helium during expansionA. 4000 JB. 3000 JC. 3500 JD. 5000 J |
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Answer» Correct Answer - D n= 8mole, `Deltat = 30^(@)`C Q = `nc_(p)DeltaT rArr Q= 8 xx 5/2 xx 8.31 xx 30~~5000`. |
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| 49. |
The temperature gradient in a rod of `0.5 m` length is `80^(@)C//m`. It the temperature of hotter end of the rod is `30^(@)C`, then the temperature of the cooler end isA. `0^(@)`CB. `-10^(@)`CC. `10^(@)`CD. `40^(@)`C |
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Answer» Correct Answer - B Temperature gradient = `(theta_(1)-theta_(2))/l` Here, `theta_(1) = 30^(@)`C rArr l =0.5 cm` `therefore 80=(30-theta_(2))/0.5` `theta_(2) = 30- 80 xx 0.5 rArr theta_92) = 30-40 = 10^(@)`C |
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