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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. |
Consider two rods of same length and different specific heats (`S_1` and `S_2`), conductivities `K_1` and `K_2` and area of cross section (`A_1` and `A_2`) and both having temperature `T_1` and `T_2` at their ends. If the rate of heat loss due to conduction is equal thenA. `K_(1)=K_(2)`B. `K_(1)S_(1)=K_(2)S_(2)`C. `(K_(1))/(A_(1)S_(1))=(K_(2))/(A_(2)S_(2))`D. `K_(1)A_(1)=K_(2)A_(2)` |
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Answer» Correct Answer - D Thermal resistance should be equal, to `l/(K_(1)A_(1))=l/(K_(2)A_(2))rArr K_(1)A_(1) =K_(2)A_(2)` |
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| 2. |
Two identical rectangular rods of metal are welded as shown in figure `(1)` and `20 J` of heat flows through the rods in `1 min`. How long would it take for `20 J` heat to flow through the rods if they are welded as shown in figure `(2)` |
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Answer» Let `R` be the thermal resistance of each rod. `:. ` In first case `(1)/(R_(1))=(1)/(R)+(1)/(R)` or `R_(1)=(R )/(2)` So the rate of flow of heat in this situation will be `(DeltaQ)/(Deltat)=(DeltaT)/(R_(1))=(100-0)/(R//2)=(20)/(60)` `R=600^(@)C//W` Now for case `(2)` `R_(2)=R+R=600+600=1200^(@)C//W` `:. (DeltaQ)/(Deltat)=(DeltaT)/(R_(2))` `(20)/(t)=(100)/(1200)` `t=240sec`. |
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| 3. |
Two identical square rods of metal are welded end to end as shown in figure (a). Assume that `10` cal of heat flows through thr rod in `2min`. Now the rods are welded as shown in figure (b). The time it would take for `10` cal to flow through the rods now, is A. `0.75 min`B. `0.5 min`C. `1.5 min`D. `1 min` |
| Answer» Correct Answer - B | |
| 4. |
A lake surface is exposed to an atmosphere where the temperature is `lt 0^(@)C`. If the thickness of the ice layer formed on the surface grows form `2cm` to `4cm` in `1` hour. The atmospheric temperature, `T_(a)` will be- (Thermal conductivity of ice `K = 4 xx 10^(-3) cal//cm//s//.^(@)C`, density of ice `= 0.9 gm//ice.` Latent heat of fustion of ice `= 80 cal//m`. Neglect the change of density during the state change. Assume that the water below the ice has `0^(@)` temperature every where)A. `-20^(@)C`B. `0^(@)C`C. `-30^(@)C`D. `-15^(@)C` |
| Answer» Correct Answer - C | |
| 5. |
A copper slab is 2 mm thick. It is protected by a 2 mm layer of stainless steel on both sides. The temperature on one side of this composite slab is `400^(@)C` and `200^(@)C` on the other side. Value of thermal conductivities are- `k_(cu)=400 Wm^(-1)K^(-1) ` and `k_(s)=16 Wm^(-1)K^(-1)` (a) Just by knowing that thermal conductivity of steel is much less than copper, find (approximately) the temperature of the copper slab. (b) Plot the variation of temperature across the thickness of the composite wall. |
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Answer» Correct Answer - (a) Temperature drop will take palace almost entirely in steel. The copper plate will have almost uniform temperature pf `(400+200)/(2)=300^(@)` |
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| 6. |
monatomic ideal gas is contained in a rigid container of volume V with walls of totsl inner surface area A, thickness x and thermal condctivity K. The gas is at an initial temperature `t_(0)` and pressure `P_(0)` . Find tjr pressure of the gas as a function of time if the temperature of the surrounding air is `T_(s)` . All temperature are in absolute scale. |
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Answer» As the volume of the gas is constant, a heat `DeltaQ given to the gas increases its temperature by `deltaT=DeltaQ//C_(v)` . Also, for a monatomic gas, C_(v)=3/2R` . If the temperature ot the gas at time t is T, the heat current into the gas is `(DeltaQ)/(Deltat)=(KA(T_(s)-T))/(x)` . or, `(DeltaT)/(Deltat)=(2KA)/(3xR)(T_(s)-T)` . or, `int_(to))^(T)(dT)/(T_(s)-T)=int_(0))^(t)(2KA)/(3xR)t` . or, `T_(s)-T(T_(s)-T_(0))e^(-(2KA)/(3xR)t` . or, `T=T_(s)(T_(s)-T_(0))e^(-(2KA)/(3xR)t` . As the volume remains constant, `P/T=p_(0)/T_(0)` . or, `p=p_(0)/T_(0)T` . `=p_(0)/(T_(0))[T_(s)-(T_(s)-T_(0))e^((2KA)/(3xR)t)]` . |
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| 7. |
A and B are two sphere made of same material. Radius of A is double that of B and initially they are at same temperature (T). Both of them are kept far apart in a room at temperature `T_(0)(lt T)`. Calculate the ratio of initial rate of cooling (i.e. rate of fall of temperature) of sphere A and B if (a) the spheres are solid, (b) the spheres are hollow made of thin sheets of same thickness |
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Answer» Correct Answer - (a) `(1)/(2)` (b) 1 |
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| 8. |
The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is `1.4 "kilowatt"//m^2`. The average earth-sun distance is `1.5 xx 10^11` meters. The mass lost by the sun per day is.A. `4.4 xx 10^(9) kg`B. `7.6 xx 10^(14) kg`C. `3.8 xx 10^(12) kg`D. `3.8 xx 10^(14) kg` |
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Answer» Correct Answer - D (4) `I = (u)/(4 pi r^(2))` `u = (I) (4 pi r^(2)) = mc^(2)` `m = (1.4 xx 10^(3) xx4 pi xx (1.5 xx 10^(11))^(2))/((3 xx 10^(8))^(2)) xx 86400` `= 3.8 xx 10^(14) kg` |
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| 9. |
Heat received by the Earth due to solar radiations is `1.35 KWm^(-2)`. It is also known that the temperature of the Earth’s crust increases `1^(@)C` for every 30 m of depth. The average thermal conductivity of the Earth’s crust is `K = 0.75 J (msK)^(-1)` and radius of the Earth is `R = 6400 km`. (i) Calculate rate of heat loss by the Earth’s core due to conduction. (ii) Assuming that the Earth is a perfect block body estimate the temperature of its surface. |
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Answer» Correct Answer - (i) `1.29xx10^(13)W` (ii) 276 K |
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| 10. |
The container `A` is constantly maintained at `100^(@)C` and insulated container `B` in the contains ice at `0^(@)C` Different rods are used to connect them For a rod made of copper, it takes 30 mintes for the ice to melt and for a rod of steel of same cross-section taken in different experiment it takes 60 minutes for ice to melt. When these rods are simultaneously connected in parallel, the ice melts is . |
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Answer» `Q=` it where `i=` heat flows rate `=(DeltaT)/(R )=(100)/(R )` For copper rod : `Q=((100)/(R_(1)))(30)implies R_(1)=((100)/(Q))xx30` Also for steel rod : `Q=((100)/(R_(2)))xx60implies R_(2)=((100)/(Q)xx60` Now, `Q=((1)/(R_(1))+(1)/(R_(2)))t_(reqd.)` `:. t_(reqd)=(Q)/(((Q)/(100))xx(1)/(30)+((Q)/(100))xx(1)/(60))=20min` |
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| 11. |
A black metal foil is warmed by radiation from a small sphere at temperature `T` and at a distance d it is found that the power received by the foil is `P` If both the temperature and the distance are doubled the power received by the foil will be .A. `16P`B. `4P`C. `2P`D. `P` |
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Answer» `P_(Recived)=IA` `I="Intensity"` `"Intensity"=("Power of same")/("Area of sphere")` `(P_(0))/(4pir^(2))=(sigmaeA_(1)T_(0)^(4))/(4pir^(2))` `P=(sigmaeA_(1)T_(0)^(4))/(4pir^(2))A` [`A=` Area of foil] `P_(2)=(sigmaeA_(1)(2T_(0))^(4)A)/(4pi(2r^(2)))=4P` `A_(1)=` area of source |
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| 12. |
A system `S` receives heat continuously from an electric heater of power `10 W`. The temperature of `S` becomes constant at `50^(@)C` when the surrounding temperature is `20^(@)C`. After the heater is switched off, `S` cools from `35.1^(@)C` to `34.9^(@)C` in `1 minute`. the heat capacity of `S` isA. `100 J//^(@)C`B. `300 J//^(@)C`C. `750 J//^(@)C`D. `1500 J//^(@)C` |
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Answer» Correct Answer - D (4) Rate of loss of heat `prop` difference in temperature with the surroundings At `50^(@)C, (dq)/(dt) = k_(0) (50 - 20) = 10`, where `k_(0)` constant `k_(0) = (1)/(3)` At an average temperature of `35^(@)C` `(dQ)/(dt) = (1)/(3) (35 - 20) = 5 J//s` Heat lost in min = `(dQ)/(dt) xx 60 xx 5 xx 60 = 300 J = Q` Fall in temperature `= 0.2^(@)C` `Q = (ms) (Delta theta)` Hetat capacity `ms = (Q)/(Delta theta) = (300)/(0.2) = 1500 J//^(@)C` |
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| 13. |
If the radius of a star is `R` and it acts as a black body, what would b the temperature of the star, in which the rate of energy production is `Q`?A. `Q//4 pi R^(2) sigma`B. `(Q//4 pi R^(2) sigma)^(-1//2)`C. `(4 pi R^(2) Q// sigma)^(1//4)`D. `(Q // 4 pi R^(2) sigma)^(1//4)` |
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Answer» Correct Answer - D (4) `Q = sigma A T^(4) = sigma 4 pi R^(2) T^(4)` `T = (Q // 4 pi R^(2) sigma)^(1//4)` |
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| 14. |
Stars `S_(1)` emits maximum radiation of wave length `420nm` and the star `S_(2)` emits maximum radiation of wavelength `560nm`, what is the ratio of the temperature of `S_(1)` and `S_(2)` :A. `4//3`B. `(4//3)^(1//4)`C. `3//4`D. `(3//4)^(1//2)` |
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Answer» Using wein displacement law `lambdaT=b` `lambda_(1)T_(1)=lambda_(2)T_(2)` `(T_(1))/(T_(2))=(lambda_(2))/(lambda_(1))=(560)/(420)=(4)/(3)` |
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| 15. |
The colour of a star indicates itsA. temperatureB. distanceC. velocityD. size |
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Answer» Correct Answer - A According to weins displacement law, `lambda_(m)T`=constant or temperature `prop 1/(lambda_(m))`. It represents that greater the temperature T of an emitted star, smaller the value of wave length `lambda`, we also know the wave length of ray depends upon its colour. hence, when the temperature of star increases, the wave length of star decreases. when temperature of star decreases, thewave length increases and star moves towards the red colour therefore, the colour of star indicates its temperature. |
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| 16. |
One end of a metal rod is kept in a furnace. In steady state, the temperature of the rodA. increases decreaseB. decrasesC. remains constantD. is non-uniform |
| Answer» Correct Answer - D | |
| 17. |
A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature `theta` along the length x of the bar from its hot end is best described by which of the following figures?A. B. C. D. |
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Answer» Correct Answer - A `(dQ)/(dt) =- kA (d theta)/(dx)` at steady state `(dQ)/(dt) = constant`. `d theta alpha - dx` `overset(theta)underset(theta_(0))int d theta == k overset(x)underset(0)int dx` `theta = theta_(0) - kx` |
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| 18. |
One end of a metal rod is kept in a furnce. In steady state, the temperature of the rodA. increasesB. decreasesC. remains constantD. is non uniform. |
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Answer» Correct Answer - D is nonuniform. |
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| 19. |
Mass m of a liquid A is kept in a cup and it is at a temperature of `90^(@)C`. When placed in a room having temperature of `20^(@)C`, it takes 5 min for the temperature of the liquid to drop to `30^(@)C`. Another liquid B has nearly same density as that of A and its sample of mass m kept in another identical cup at `50^(@)C` takes 5 min for its temperature to fall to `30^(@)C` when placed in room having temperature `20^(@)C`. If the two liquids at `90^(@)C` and `50^(@)C` are mixed in a calorimeter where no heat is allowed to leak, find the final temperature of the mixture. Assume that Newton’s law of cooling is applicable for given temperature ranges. |
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Answer» Correct Answer - `66^(@)` |
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| 20. |
A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquied and the room is plot will be very nearlyA. a straight lineB. a circular arcC. a parabolaD. an ellipse. |
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Answer» Correct Answer - A a straight line |
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| 21. |
A hot body is suspended inside a room that is maintained at a constant temperature. The temperature difference between the body and the surrounding becomes half in a time interval `t_(0)`. In how much time the temperature difference between the body and the surrounding will becomes `(1)/(4)` the original value? |
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Answer» Correct Answer - `t=2t_(0)` |
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| 22. |
A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time.The plot will be very nearlyA. a straight lineB. a circular areC. a parabolaD. an ellips |
| Answer» Correct Answer - A | |
| 23. |
The plots of intensity versus wavelength for three black bodies at temperatures `T_(1),T_(2)` and `T_(3)` respectively are shown in Their temperatures are shown in How their temperatures are related ? .A. `T_(1) gt T_(2) T_(3)`B. `T_(1) gt T_(3) T_(2)`C. `T_(2) gt T_(3) T_(1)`D. `T_(3) gt T_(2) T_(1)` |
| Answer» Correct Answer - B | |
| 24. |
What is the `S.I`. Unit of thermal conductivity? |
| Answer» Correct Answer - `S.I` unit of termal conductivity is `Jm^(-1)s^(-1)//.^(@)C`. | |
| 25. |
Five rods of same dimensions are arrranged as shown in the figure. They have thermal conductivities `K_(1), K_(2), K_(3), K_(4)` and `K_(5)`. When points `A` and `B` are maintained at different themperature, 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 (2) No heat will flow in in `CD` if `(R_(AC))/(R_(AD)) = (R_(CB))/(R_(DB)) implies (K_(3))/(K_(1)) = (K_(4))/(k_(2))` `K_(1) K_(4) = K_(2) K_(3)` |
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| 26. |
Two rods `A` and `B` are of equal lengths. Their ends of kept between the same temperature and their area of cross-section are `A_(1)` and `A_(2)` and thermal conductivities `K_(1)` and `K_(2)`. The rate of heat transmission in the two rods will be equal, ifA. `K_(1) A_(2) = K_(2) A_(1)`B. `K_(1) A_(1) = K_(2) A_(2)`C. `K_(1) = K_(2)`D. `K_(1) A_(1) = K_(2) A_(2)` |
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Answer» Correct Answer - B (2) `i = (theta_(1) - theta_(2))/(R )` `R_(A) = R_(B)` `(l)/(K_(1) A_(1)) = (l)/(K_(2) A_(2)) implies K_(1) A_(1) = K_(2) A_(2)` |
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| 27. |
A room has a `4mxx4mxx10cm` concrete roof `(K=1.26Wm^(-1)`^(@)C^(-1)` . At some instant, the temperature outside is `40(@)C` and that inside is 32(@)C` . (a) Neglecting converction, calculate the amount of heat flowing per second into the room through the roof. (b) Bricks `(K=0.65Wm^(-1)`^(@)C^(-1) of thickness `7.5cm` are loid down on the roof. Calculate the new rate of heat flow under the some temperature conditions. |
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Answer» The area of the roof `=4mxx 4m=16m^(2)` . The thickness `x=10cm=0.01m` . (a) The thermal resistance of the roof is R_(1)=(1)/(K)(x)/(A)=(1)/(1.26Wm^(-1)`^(@)C^(-1))(0.01m)/(16m^(2)` . `=4.96xx10_(-3)`^(@)CW_(-1)` . The heat current is `=(DeltaQ)/(Deltat)=(theta_(1)-theta_(2))/R_(1)=(46^(@)C-32^(@)C)/(4.96xx_(-3)`^(@)CW^(-1)` . `=2822W` . (b) The thermal resistance of the brick layer is `R_(2)=(1)/(K)(x)/(A)=(1)/(0.65Wm^(-1)`^(@)C^(-1))(7.5xx10^(-2)m)/(16m^(2)`. `=7.2xx10^(-3)`^(@)CW^(-2)` . The equivalent thermal resistance is `R=R_(1)+R_(2)=(4.96+7.2)xx(10)^(-3)`^(@)CW^(-1)`. `=1.216xx10^(-2)`^(@)CW^(-1)` . The heat current is `(DeltaQ)/(Deltat)=(theta_(1)-theta_(2))/(R)=(46^(@)C-32^(@)C)/(1.216xx10_(-2)`^(@)CW^(-1)` . `=1152W` . |
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| 28. |
A solid body A of mass m and specific heat capacity ‘s’ has temperature `T_(1) = 400 K`. It is placed, at time `t = 0`, in atmosphere having temperature `T_(0) = 300 K`. It cools, following Newton’s law of cooling and its temperature was found to be `T_(2) = 350 K `at time `t_(0)`. At time `t_(0)`, the body A is connected to a large water bath maintained at atmospheric temperature `T_(0)`, using a conducting rod of length L, cross section A and thermal conductivity k. The cross sectional area A of the connecting rod is small compared to the overall surface area of body A. Find the temperature of A at time `t = 2t_(0)`. |
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Answer» Correct Answer - `T=300+50e^(-[(In2)/(t_(0))+(kA)/(msL)]t_(0)` |
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| 29. |
Solar constant, `I_(s)` is defined as intensity of solar radiation incident on the Earth. Its value is close to `1.4 kW//m^(2)`. Nearly `68%` of this energy is absorbed by the Earth. The average temperature of Earth is about 290 K. Radius of the Earth is `R_(e) = 6000 km` and that of the Sun is `R_(s) = 700,000 km`. Earth - Sun distance is `r = 1.5 xx 10_(8) km`. Assume Sun to be a black body. (a) Estimate the effective emissivity of earth. (b) Find the power of the sun. (c) Estimate the surface temperature of the Sun. |
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Answer» Correct Answer - (a) `0.6` (b) `3.96xx10^(26)` (c) 5800 K |
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| 30. |
The surface of a household radiator has an emissivity of `0.55` and an area of `1.5 m^(2)`. (a) At what rate is radiation emitted by the radiator when its temperature is `50^(@)C`? (b) At what rate is the radiation absorbed by the radiator when the walls of the room are at `22^(@)C`? (c ) What is the net rate of radiation from the radiator? (stefan constant `sigma = 6 xx 10^(-8)W//m^(2)-K^(4))` |
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Answer» Correct Answer - `539 W, 375W, 164 W` (a) `e sigma AT^(4) = 0.55 xx 6 xx 10^(-8) xx 1.5 xx (323)^(4) = 539W` (b) `e sigma AT_(surr)^(4) = 0.55 xx 6 xx 10^(-8) xx 15 xx (295)^(4) = 375W` (c ) `539 -375 = 164 W`. |
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| 31. |
A 3 mm diameter and 5 m long copper wire is insu- lated using a 2 mm thick plastic cover whose thermal con- ductivity is `K = 0.15 Wm^(-1)K^(-1)`. The wire has a potential difference of 10 V between its ends and the current through it is 8A. The outer surface of the wire is at `30^(@)C`. Neglect convection. (i) Calculate the temperature at the interface of the wire and the plastic cover. (ii) Determine whether doubling the thickness of the plastic cover will increase or decrease the interface temperature. [Given `ln(2.33)=0.85`] |
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Answer» Correct Answer - (i) `44.4^(@)C` (ii) increase |
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| 32. |
A solid copper sphere (density `=8900kgm^(-3)` and specific heat `C=390Jkg^(-1)K^(-1)`) of radius `r=10cm` is at an initial temperature `T_(1)=200K`. It is then suspended inside an chamber whose walls are at almost `OK`. Calculate the time required for the temperature of the sphere to drop to `T_(2)=100 K`, `sigma=5.67xx10^(-8)Wm^(-2)K^(-4)`. |
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Answer» Let `T` be the temperture at any time `t`. Then rate of loss of heat `=((4pi)/(3)r^(3)rho)c (-(dT)/dt)`. Also the rate of loss of heat `=sxx4pir^(2)(T^(2)-0)` `:.-(4pi)/(3)r^(3)rhoc(dT)/(dt)=4pr^(2)sT^(4)impliesdt=(rhorc)/(3sigma)(dT)/(dT^(4))` `implies t=(rhorc)/(3sigma)int_(T_(1))^(T_(2))(dT)/(T^(4))=(rhorc)/(9sigma)(T_(1)^(3)-T_(2)^(3))/(T_(1)^(3)T_(2)^(3))` `impliest=(10xx10^(-2)xx8900xx390)/(9xx5.67xx10^(-8))xx(200^(3)-100^(3))/(200^(3)xx100^(3))=595165s=165h19min` |
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| 33. |
A body is in thermal equilibrium with surrounding. Absorptive power of the surface of the body is a `= 0.5`. E is the radiant energy incident in unit time on the surface of the body. How much energy propagates from its surface in unit time? |
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Answer» Correct Answer - E |
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| 34. |
Reflection and absorption coefficient of a given surface at `0^@C` for a fixed wavelength are 0.5 (each). At the same temperature and wavelength the transmission (coefficient) of surface will be-A. 0.5B. 1C. zeroD. in between zero and one |
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Answer» Correct Answer - C `a+r+t=1 rArr 0.5 +0.5 +t =1 rArr t=0` |
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| 35. |
The two ends of a rod of length `L` and a uniform cross-sectional area `A` are kept at two temperature `T_(1)` and `T_(2)` `(T_(1) gt T_(2))`. The rate of heat transfer. `(dQ)/(dt)`, through the rod in a steady state is given byA. `(dQ)/(dt)=(KL(T_(1)-T_(2)))/A`B. `(dQ)/(dt)=(K(T_(1)-T_(2)))/(LA)`C. `(dQ)/(dt)=KLA(T_(1)-T_(2))`D. `(dQ)/(dt)=(KA(T_(1)-T_(2)))/L` |
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Answer» Correct Answer - D For a rod of length L and area of cross-section A whose faces are maintained at temperature `T_(1)` and `T_(2)` respectively. Then in steady state the rate of heat flowing from one face to the other face in time t is given by `(dQ)/(dt)=(KA(T_(1)-T_(2)))/L` The curved surface of rod is kept insulated from surrounding to avoid leakage of heat |
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| 36. |
A sphere, a cube and a thin circular plate, all having the same mass and made of the same material are heated to the same temperature and them allowed to cool. Which of them cools fastest ?A. SphereB. CubeC. Circular plateD. All at same rate |
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Answer» Correct Answer - C (3) Thin circular plate has maximum surface area. Rate of cooling `prop` Surface area |
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| 37. |
A 20 mm diameter copper pipe is used to carry heated water. The external surface of the pipe is at `T = 80^(@)C` and its surrounding is at `T_(0) = 20^(@)C`. The outer surface of the pipe radiates like a black body and also loses heat due to convection. The convective heat loss per unit area per unit time is given by `h(T – T_(0))` where h = 6 W `(m^(2)K)^(-1)`. Calculate the total heat lost by the pipe in unit time for one meter of its length. |
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Answer» Correct Answer - `51.7 Wm^(-1)` |
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| 38. |
Newton’s law of cooling says that the rate of cooling of a body is proportional to the temperature difference between the body and its surrounding when the difference in temperature is small. (a) Will it be reasonable to assume that the rate of heating of a body is proportional to temperature difference between the surrounding and the body (for small difference in temperature) when the body is placed in a surrounding having higher temperature than the body? (b) Assuming that our assumption made in (a) is correct estimate the time required for a cup of cold coffee to gain temperature from `10^(@)C` to `15^(@)C` when it is kept in a room having temperature `25^(@)C`. It was observed that the temperature of the cup increases from `5^(@)C` to `10^(@)C` in 4 min |
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Answer» Correct Answer - (a) Yes (b) `5.6 min` |
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| 39. |
A star having radius R has a small planet revolving around it at a distance d `(gtgt R)`. The star and the planet both behave like black bodies and radiate maximum amount of energy at wavelength `lambda_(s)` and `lambda_(p)` respectively. (i) Find d in terms of other given parameters.(ii) Show that `lambda_(p)gtgtlambda_(s) ` |
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Answer» Correct Answer - (i) `d=(R)/(2)((lambda_(p))/(lambda_(s))^(2)` |
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| 40. |
A planet of radius `r_(0)` is at a distance r from the sun `(r gtgt r_(0))`. The sun has radius R. Temperature of the planet is `T_(0)`, and that of the surface of the sun is `T_(s)`. Calculate the temperature of another planet whose radius is `2r_(0)` and which is at a distance 2r from the sun. Assume that the sun and the planets are black bodies. |
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Answer» Correct Answer - `(T_(0))/(sqrt2)` |
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| 41. |
An ideal Black-body at room temperature is thrown into a furnace. It is observed thatA. initially it is the darkest body and at later times brightestB. it is the darkest body at all timesC. it cannot be distinguised at all timeD. initially it is the darkest body and at later times cannot be distinguished |
| Answer» Correct Answer - A | |
| 42. |
A uniform slab of dimension `10cmxx10cmxx1cm` is kept between two heat reservoir at temperatures `10^(@)C` and `90^(@)C` The larger surface areas touch the reservoirs. The thermal conductivity of the material is `0.80Wm^(-1).^(@)C^(-1)` . Find the amount of heat flowing through the slab per minute. |
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Answer» Correct Answer - `64 J` `A = 100 cm^(2)` `i_(H) = (90 -10)/((1xx10^(-2))) xx (0.8) (100 xx 10^(-4))` `l = 1cm` `(dq)/(dt) = i_(H) = 80 xx 0.8 = 64 J//s` `Q = 64 xx 1 = 64 J` |
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| 43. |
Stefan’s contain (s) derives from other known con- stant of nature, viz. Boltzmann constant, (k) planck’s con- stant (h) and speed of light in vacuum (c). Value of the constant is `sigma=5.67xx10^(-8)Js^(-1)m^(-2)K^(-4)` If speed of light were `2%` more than its present value, how much different (in percentage) the value of `sigma` would have been? |
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Answer» Correct Answer - `4%` less than its present value |
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| 44. |
Three copper rods and three steel rods each of length `l = 10 cm` and area of cross-section `1cm^(2)` are connected as shown If ends `A` and `E` are maintained at temperatures `125^(@)C` and `0^(@)C` respectively, calculate the amount of heat flowing per second from the hot of cold function. `[K_(Cu) = 400 W//m-K, K_(steel) = 50 W//m-K]` |
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Answer» `R_("steel") = (L)/(KA) = (10^(-1)m)/(50(W//m-^(@)C)xx10^(-4)m^(2)) = (1000)/(50).^(@)C//W`. Similarly `R_(Cu) = (1000)/(400).^(@)C//W` Junction `C` and `D` are identical in every respect and both will have same temperature. Consequently. The rod `CD` is in thermal equilibrium and no heat will flow through it. Hence it can be neglected in further analysis. Now rod `BC` and `CE` are in series their equilvalet resistance is `R_(1) = R_(s) +R_(Cu)` similarly rods `BD` and `DE` are in series with same equilvalent resistance `R_(1) = R_(s) +R_(Cu)` these two are in parallel giving an equilvalent resistance of `(R_(1))/(2) = (R_(S)+R_(Cu))/(2)` This resistance is connected in series with rod `AB`. Hence the net equilvalent of combination is `R = R_(steel) +(R_(1))/(2) (3R_(steel)+R_(Cu))/(2) = 500 ((3)/(50)+(1)/(400)).^(@)C//W` Now, `i = (T_(H)-T_(C))/(R) =(125^(@)C)/(500((3)/(50)+(1)/(400)).^(@)C//W)=4`watt. |
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| 45. |
A body at temperature `40^(@)C` is kept in a surrounding of constant temperature `20^(@)C`. It is observed that its temperature falls to `35^(@)C` in `10` minutes. Find how much more time will it taken for the body to attain a temperature of `30^(@)C`. |
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Answer» from equation `(14.3)` `Delta theta_(1) = Delta theta_(i)e^(-kt)` for the interval in which temperature falls from `40 to 35^(@)C` `(30 -20) = (40 -20)e^(-k.10)` `rArr e^(-10k) = (3)/(4)` `rArr k =("ln"(4)/(3))/(10)` for the next interval `(30 - 20) =(35 - 20)e^(-kt)` `rArr e^(-kt) = (2)/(3) rArr kt = "ln"(3)/(2)` `rArr (("ln"(4)/(3))t)/(10) ="ln"(3)/(2)` `rArr t = 10 (("ln"(4)/(3)))/(("ln"(4)/(3)))` minute `- 14.096 min` Aliter: (by approximate method) for the interval in whcih temperature falls form `40 to 35^(@)C` `lt theta gt = (40 +35)/(2) = 37.5^(@)C` from equation `(14.4) ltlt(d theta)/(dt)gtgt =-k(lt thetagt -theta_(0))` `rArr ((35^(@)C-40^(@)C))/(10(min)) =- K(37.5^(@)C-20^(@)C)` `rArr K = (1)/(35) (min^(-1))` for the interval in which temperature falls from `35^(@) to 30^(@)C` `lt theta gt = (35 +30)/(2) = 32.5^(@)C` from equation `(14.4)` `((30^(@)C-35^(@)C))/(t) =- (32.5^(@)C-20^(@)C)K` `rArr` required time, `t = (5)/(12.5) xx 35 min = 14 min` |
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| 46. |
A body of emissivity `(e = 0.75)`, surface area of `300cm^(2)` and temperature `227^(@)C` is kept in a room at temperature `27^(@)C`. Calculate the initial value of net power emitted by the body. |
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Answer» Using equation `(13.3) P = e sigma A(T^(4)-T_(0)^(4))` `= (0.75) (5.67 xx 10^(-8) W//m^(2) -K^(4)) (300 xx 10^(-4)m^(2)) xx {(500 K)^(4) -(300K)^(4)}` `= 69.4` Watt. |
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| 47. |
A man, the surface area of whose skin is `2m^(2)` , is sitting in a room where air temperature is `20^(@)C` if his skin temperature is `28^(@)C` and emissivity of his skin equals 0.97, find the rate at which his body loses heat. |
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Answer» Correct Answer - `92.2 W` `e sigma A(T^(4)-T_("surroundings"))` `= 0.97 xx 5.67 xx 10^(-8) 2 (301^(4) - 293^(4))` `= 92.2 W` |
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| 48. |
If the ratio of coefficient of thermal conductivity of silver and copper is `10 : 9`, then the ratio of the length upto which wax will melt in Ingen Hauze experiment will beA. `6 : 10`B. `sqrt 10 : 3`C. `100 : 81`D. `81 : 100` |
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Answer» Correct Answer - B (2) `K prop l^(2)` `(K_(1))/(K_(2)) = ((l_(1))/(l_(2)))^(2) implies (l_(1))/(l_(2)) = sqrt((K_(1))/(K_(2))) = sqrt((10)/(5)) = (sqrt(10))/(3)` |
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| 49. |
For a temperature difference `DeltaT=20.0^(@)C`, one slab of material conducts `10.0W//m^(2)` , another of the same shape conduct `20.0W//m^(2)`. What is the rate of heat flow `per m^(2)` of surface area whenthe slabs are placed side by side with `DeltaT_(tot)=20.0^(@)C` ? |
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Answer» `(dQ)/(dt)=(KADeltaT)/(l)=(DeltaT)/(RT_(1))` `10=(20)/(R_(1))impliesR_(1)=Q`unit Similarly `R_(2)=1` unit when placed side by side `implies` series combination `R_(new)=R_(1)+R_(2)` `implies (dQ)/(dt)=(DeltaT)/(R_(New))=(20)/(3unit)=(20)/(3)w//m^(2)` |
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| 50. |
(i) A cylindrical pipe of length L has inner and outer radii as a and b respectively. The inner surface of the pipe is at a temperature `T_(1)` and the outer surface is at a lower temperature of `T_(2)`. Calculate the radial heat current if conductivity of the mate- rial is K. (ii) A cylindrical pipe of length L has two layers of material of conduc- tivity `K_(1)` and `K_(2)`. (see figure). If the inner wall of the cylinder is maintained at `T_(1)` and outer surface is at `T_(2) (lt T_(1))`, calcu- late the radial rate of heat flow. |
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Answer» Correct Answer - (i) `((T_(1)-T_(2))2piKL)/(ln((b)/(c)))` (ii) `(T_(1)-T_(2))/((1)/(2piK_(1)L)ln((r^(2))/(r^(1)))+(1)/(2piK_(2)L)ln((r_(3))/(r_(2))))` |
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