InterviewSolution
Saved Bookmarks
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. |
A ball is dropped on a floor from a height of `2.0m`. After the collision it rises up to a height of `1.5m`. Assume that `40%` of the mechanical energy lost goes as thermal energy into the ball.Calculate the rise in the temperature of the ball in the collision.Heat capacity of the ball is `800J K^(-1)` |
|
Answer» Correct Answer - A::B::C `0.4 [mgDeltah] = msDeltatheta ` `=Deltatheta = (0.4gDeltah)/s = (0.4 xx 9.8 xx 0.5)/800` `=2.5 xx (10^-3)^@C`. |
|
| 2. |
A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B . Under thermal equilibrium, the temperature difference across the wall is `36^@C.` The temperature difference across the layer A isA. `6^@C`B. `12^@C`C. `18^@C`D. `24^@C` |
|
Answer» Correct Answer - C Thermal resistance, `R = l/(KA) prop 1/K rArr K_A = 2K_B` `:. R_A = (R_B)/2` So, let `R_A = R "then" R_B = 2R` Now, `H_A = H_B` `:. (TD)_A/R_A = (TD)_B/R_B or (TD)_A = (R_A/R_B)(TD)_B` `= (R/(2R))(36^@C) = 18^@C` . |
|
| 3. |
A rod is heated at one end as shown in figure. In steady state temperature of different sections becomes constant but not same. Why so? . |
|
Answer» Correct Answer - A::B::C::D |
|
| 4. |
Show that the SI units of thermal conductivity are `W//m-K`. |
|
Answer» Correct Answer - A `Q= ((KA(theta_1-theta_2)t)/l)` `:. K = (Ql)/(A(theta_1-theta_2)t) = (J -m)/(m^2K-s)` `=W/(m-K) (as J//s=W)` . |
|
| 5. |
Three discs, A, B and C having radii 2m, 4m and6m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are `300nm, 400nm` and `500 nm`, respectively. The power radiated by them are `Q_A, Q_B and Q_C` respectively (a) `Q_A` is maximum (b) `Q_B` is maximum (c) `Q_C` is maximum (d) `Q_A = Q_B = Q_C` |
|
Answer» Correct Answer - A (b) `Q prop AT^4 and lambda_mT= constant.` Hence, `Q prop A/(lambda_m)^4 or Q prop (r^2)/(lambda_m)^4` `Q_A:Q_B:Q_C = (2)^2/(3)^4:(4)^2/(4)^4: (6)^2/(5)^4 = 4/81 : 1/16 : 36/625 = 0.05 : 0.0625 : 0.0576` i.e.`Q_B` is maximum. |
|
| 6. |
A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be (a)225 (b)450 (c) 900 (d)1800 |
|
Answer» Correct Answer - A Power radiated = `e_rsigma T^4A` or Power radiated `prop` (surface area)`(T)^4`. The radius is halved, hence, surface area will become `1/4` times. Temperature is doubled, therefore, `T^4` becomes 16 times. New power `= (450)(1/4)(16)= 1800 W.` `:.` The correct option is (d). |
|
| 7. |
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 true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
|
Answer» Correct Answer - C If temperature of a normal body is more than a perfectly black body, it can radiate more energy. |
|
| 8. |
The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the North star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the sun and the north star isA. 1.46B. 0.69C. 1.21D. 0.83 |
|
Answer» Correct Answer - B `lambda_m prop 1/T` `:. (lambda_m)_1/(lambda_m)_2 = T_2/T_1 or T_1/T_2 = (lambda_m)_2/(lambda_m)_1` ` = 350/510 = 0.69` . |
|
| 9. |
Assertion : Emissivity of any body (e) is always less than its absorptive power (a). Reason : Both the quantities are dimensionless.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - D | |
| 10. |
Choose the correct optios.A. Good absorbers of a particular wavelength are good emitters of same wavelength. This statement was given by Kirchhoff.B. At low temperature of a body the rate of cooling is directly proportional to temperature of the body. This statement was given by the NewtonC. Emissive power of a perfectly black body is 1D. Absorptive power of a perfectly black body is 1 |
|
Answer» Correct Answer - A::D (b) If temperature difference is small, the rate of cooling is proportional to TD. |
|
| 11. |
A nuclear power plant generates 500 MW of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise by `10^@C`, what is the required flow rate in `kg//s`? |
|
Answer» Correct Answer - A::B::D `P = (dm)/(dt) s*Deltatheta` `:. (dm)/(dt) = P/(sDeltatheta) = (500xx10^6)/((4200)xx 10)` `=11904 kg//s` `=1.2 xx (10^4) kg//s` . |
|
| 12. |
As a physicist, you put heat into a 500 g solid sample at the rate of `10.0 kJ//min`, while recording its temperature as a function of time. You plot your data and obtain the graph shown in figure. . (a) What is the latent heat of fusion for this solid? (b) What is the specific heat of solid state of the material? |
|
Answer» Correct Answer - A::B::C::D (a) Between `t = 1 min` to` t = 3 min`, there is no rise in the temperature of substance. Therefore, solid melts in this time. `L = Q/m = (HT)/m = (10 xx2)/0.5` `=40kJ//kg` (b) From `Q = msDeltaT or s = Q/(mDeltaT)` Specific heat in solid state `s = (10 xx 1)/(0.5 xx 15) = 1.33 kJ//kg-^@C` |
|
| 13. |
A liquid takes 5 minutes to cool from `80^@C` to` 50^@C` . How much time will it take to cool from `60^@C` to `30^@C`? The temperature of surroundings is `20^@C`. |
|
Answer» Let us start with the assumption that `|(Deltatheta)/Delta| prop "Temperature difference"` `:. ((80-50)/5) = alpha [(80+50)/2 -20]` and `(60-30)/t = alpha [(60+30)/2 -20]` Solving these two equations, we get `t = 9 min` . |
|
| 14. |
Assertion : Warming a room by a heat blower is an example of forced convection. Reason : Natural convection takes place due to gravity.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - B | |
| 15. |
Two ends of rods of length L and radius R of the same material of kept at the same temperature. Which of the following rods conducts the maximum heat?A. L=50 cm, R = 1 cmB. L = 100cm, R = 2 cmC. L = 25 cm, R = 0.5 cmD. L = 75 cm , R = 1.5 cm |
|
Answer» Correct Answer - B Thermal distance `= l/(KA) = l/(K(piR^2)) prop l/R^2` The rod for which `l/(R^2)` is minimum will conduct maximum heat. |
|
| 16. |
In a container of negligible mass 30g of steam at `100^@C` is added to 200g of water that has a temperature of `40^@C` If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take `L_v = 539 cal//g and c_(water) = 1 cal//g-^@C.` |
|
Answer» Correct Answer - A::B::C::D Let Q be the heat required to convert 200 g of water at `40^@C "into" 100^@C`, then `Q= mcDeltaT = (200)(1.0)(100-40) = 12000 cal` Now, suppose `m_0` mass of steam converts into water to liberate this much amount of heat, then `m_0 = Q/L = 12000/539 = 22.26 g ` Since, it is less than 30g, the temperature of the mixture is `100^@C`. Mass of steam in the mixture `= 30- 22.26 = 7.74g ` and mass of water in the mixture `= 200 + 22.26 = 222.26g ` |
|
| 17. |
A bullet of mass 10 g moving with a speed of `20 m//s` hits an ice block of mass 990 g kept on a frictionless floor and gets stuck in it. How much ice will melt if `50%` of the lost kinetic energy goes to ice? (Temperature of ice block `= 0^@C`.) |
|
Answer» Correct Answer - C Velocity of bullet + ice block, `V = ((10g)xx(20 m//s))/1000 g = 0.2 m//s` Loss of KE `= 1/2 mv^2 - 1/2 (m+M)V^2` `= 1/2 [0.01 xx (20)^2 - 1 xx (0.2)^2]` `=1/2 [4- 0.04] = 1.98 J` `:.` Heat received by ice block `= (1.98)/(4.2 xx2) cal ` `= 0.24 cal` `:.` Mass of ice melted `= (0.24cal)/(80 cal//g)` `=0.003 g` . |
|
| 18. |
Equal masses of ice `(at 0^@C)` and water are in contact. Find the temperature of water needed to just melt the complete ice. |
|
Answer» Correct Answer - C `mL = ms(theta -0^@)` `:. theta = L/s = 80/1 = 80^@C`. |
|
| 19. |
Two rods are of same material and having same length and area. If heat `DeltaQ` flows through them for 12 min when they are joined side by side. If now both the rods are joined in parallel, then the same amount of heat `DeltaQ` will flow inA. 24 minB. 3 minC. 12 minD. 6 min |
|
Answer» Correct Answer - B Let R = thermal resistance of each rod. In first case, `R_("net") = R + R = 2R` (in series) In second case, `R_("net") = R/2` (in parallel) In second case thermal resistance has become `1/4 th.` So, heat current will become 4 times. So time taken to flow same amount of heat will be reduced to `1/4th`. |
|
| 20. |
A layer of ice of thickness y is on the surface of a lake. The air is at a constant temperature `-theta^@C` and the ice water interface is at `0^@C`. Show that the rate at which the thickness increases is given by ` (dy)/(dt) = (Ktheta)/(Lrhoy)` where, K is the thermal conductivity of the ice, L the latent heat of fusion and `rho` is the density of the ice. |
|
Answer» Correct Answer - A See the extra points just before solved examples. Growth of ice on ponds. We have already derived that `t = 1/2 (rhoL)/(Ktheta) y^2` `:. (dt)/(dy) = (rhoLy)/(Ktheta)` `:. (dy)/(dt) = (Ktheta)/(Lrhoy)` . |
|
| 21. |
A 2 m long wire of resistance `4 Omega` and diameter 0.64 mm is coated with plastic insulation of thickness 0.66 mm. A current of 5A flows through the wire. Find the temperature difference across the insulation in the steady state. Thermal conductivity of plastic is `0.16 xx (10^-2) cal//s cm^@-C`. |
|
Answer» Correct Answer - B::C Thermal resistance of plastic coating`(R_t = l/(KA))` `R_t = t/(K(pid)l)` (d = diameter) `(0.6xx10^(3))/((0.16xx10^(-2)xx4.18xx10^(2))(pi)(0.64xx10^(-3))(2))` `=0.0223^@C-s//J` Now, `i^2R_e = (TD)/R_t` `:. TD = i^2R_eR_t` `=(5)^2(4)(0.0223)` `=2.23^@C` |
|
| 22. |
An electric heater is placed inside a room of total wall area `137 m^2` to maintain the temperature inside at `20^@C`. The outside temperature is `-10^@C`. The walls are made of three composite materials. The inner most layer is made of wood of thickness 2.5 cm the middle layer is of cement of thickness 1 cm and the exterior layer is of brick of thickness 2.5 cm. Find the power of electric heater assuming that there is no heat losses through the floor and ceiling. The thermal conductivities of wood, cement and brick are `0.125 W//m^@-C, 1.5W//m^@-C and 1.0 W//m^@-C` respectively. |
|
Answer» Correct Answer - A::D Three thermal resistance are in series. `(R_t = l/(KA))` `:. R = R_1 + R_2 + R_3` `= ((2.5 xx (10^-2))/(0.125 xx 137)) + ((1.0 xx (10^-2))/(1.5 xx 137)) + ((2.5 xx (10^-2))/(1.0 xx 137))` `=0.0017^@C-s//J` Now, heat current `H = ("Temperature difference")/("Net thermal resistance")` `= (30/(0.0017)) = 17647 W` . |
|
| 23. |
A kettle with 2 litre water at `27^@C` is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate `160 J//s`, when its lid is open. In how much time will water heated to `77^@C` with the lid open ? (specific heat of water = `4.2 kJ//^@C-kg)`A. 8 min 20 sB. 6 min 2 sC. 14 minD. 7 min |
|
Answer» Correct Answer - A Net power available per second, `P = 1000 - 160 = 840 J//s` Heat required, `Q = msDeltatheta = (2)(4200)(77-27)` `=420000 J` `:. `Time required `= Q/P = 500 s = 8 min 20s`. |
|
| 24. |
Ice at `0^@C` is added to 200 g of water initially at `70^@C` in a vacuum flask. When 50 g of ice has been added and has all melted the temperature of the flask and contents is `40^@C`. When a further 80 g of ice has been added and has all melted the temperature of the whole becomes `10^@C`. Find the latent heat of fusion of ice. |
|
Answer» Correct Answer - A::C Let heat capacity of flask is C and latent heat of fusion of ice is L. Then, `C(70-40)+200 xx 1 xx (70-40) = 50L` `+50 xx 1xx (40 -0)` or `3C - 5L = -400 ……(i)` Further, `C(40-10)+250 xx 1 xx (40-10)` `=80L + 80 xx 1 xx (10 -0)` or `3C - 8L = -670` .......(ii) Solving Eq. (i) and (ii), we have `L = 90 cal//g` |
|
| 25. |
A certain amount of ice is supplied heat at a constant rate for 7 minutes. For the first one minute the temperature rises uniformly with time. Then, it remains constant for the next 4 minute and again the temperature rises at uniform rate for the last two minutes. Calculate the final temperature at the end of seven minutes. (Given, L of ice `= 336 xx (10^3) J//kg` and specific heat of water `= 4200 J//kg-K`). |
|
Answer» Correct Answer - C::D Let heat is supplied at a constant rate of `(alpha) J/(min)`. In 4 minutes, (when temperature remains constant) ice will be melting. `:. } mL = (4) alpha` `:. alpha/m = L/4 = (336 xx (10^3))/4` `=84 xx (10^3) J//min-kg` Now in last two minutes, `Q = msDeltatheta` `:. (alpha)(2) = (m)(4200)(theta - 0^@C)` Substituting the value of `alpha/m` we get, `theta = 40^@C` . |
|
| 26. |
Two plates eachb of area A, thickness `L_1 and L_2` thermal conductivities `K_1 and K_2` respectively are joined to form a single plate of thickness `(L_1+L_2)`. If the temperatures of the free surfaces are `T_1 and T_2`. Calculate. (a) rate of flow of heat (b) temperature of interface and (c) equivalent thermal conductivity. |
|
Answer» Correct Answer - A::B (a) If the thermal resistance of the two plates are `R_1 and R_2` respectively. Plates are in series. `R_s = R_1 + R_2 = (L_1/(AK_1)) + (L_2/(AK_2))` as `R = L/(KA)` and so `H = (dQ)/(dt) = (DeltaT)/R` `= ((T_1-T_2)/(R_1+R_2)) = (A(T_1 -T_2))/[(L_1/K_1) + (L_2/K_2)]` (b) If T is the common temperature of interface, then as in series rate of flow of heat remains same, i.e. `H = H_1 (=H_2)` `((T_1 - T_2)/(R_1 + R_2)) = (T_1 - T)/R_1` i.e. `T = ((T_1R_2 + T_2R_2)/(R_1 + R_2))` or ` T = ( [T_1 L_2/K_2] + T_2[L_1/K_1])/([L_1/K_1 + L_2/K_2] ) (as R = L/(KA))` (c) If K is the equivalent conductivity of composite slab, i.e. slab of thickness `L_1 + L_2` and cross - sectional area A, then as in series. `R_S = R_1 + R_2 or (L_1+L_2)/(AK_(eq)) = R_1 + R_2` i.e. `K_(eq) = (L_1 + L_2)/(A(R_1 + R_2)) = (L_1 + L_2)/([L_1/K_1 + L_2/K_2])` (as `R = L/(KA)` ). |
|
| 27. |
For an enclosure maintained at 2000K, the maximum radiation occurs at wavelength `lambda_m`. If the temperature is raised to 3000K, the peak will shift toA. `0.5 lambda_m`B. `lambda_m`C. `2/3 lambda_m`D. `3/2lambda_m` |
|
Answer» Correct Answer - C `lambda_m prop 1/T` `:. ((lambda_m)_2)/((lambda_m)_1) = T_1/T_2 = 2000/3000 = 2/3` `:. (lambda_m)_2 = 2/3(lambda_m)_1`. |
|
| 28. |
15 g ice at `0^@C` is mixed with 10 g water at `40^@C`. Find the temperature of mixture. Also, find mass of water and ice in the mixture. |
|
Answer» Correct Answer - A::C Heat liberated by 10g water `(at 40^@C)` when it converts into water at `0^@C`. `Q = m_1sDeltatheta` `=10xx1xx40` =400cal Mass of ice melted by this heat, `m_2 = Q/L = 400/80 = 5g lt 15g` Therefore, whole ice is not melted. Temperature of mixture is `0^@C`. Mass of water `=m_1 + m_2 = 15g` Mass of ice = 15-`m_2` = 10g. |
|
| 29. |
Figure shows a copper rod joined to a steel rod. The rods have equal length and equal cross-sectional area. The free end of the copper rod is kept at `0^@C` and that of the steel rod is kept at `100^@C`. Find the temperature `theta` at the junction of the rods. Conductivity of copper `=390 W//m-^@C` and that of steel `= 46 W//m-^@C`. . |
|
Answer» Correct Answer - A::C `H_(steel) = H_(copper)` (two rods are in series) `:. ((TD)_S/(l//KA)_S) = ((TD)_C/(l//KA)_C)` or `K_S (TD)_S = K_C (TD)_C` or ` 46(100-theta) = 390 (theta-0)` Solving we get, `theta = 10.6^@C`. |
|
| 30. |
One end of the rod of length l, thermal conductivity K and area of cross-section A is maintained at a constant temperature of `100^@C`. At the other end large quantity of ice is kept at `0^@C` . Due to temperature difference, heat flows from left end to right end of the rod. Due to this heat ice will start melting. Neglecting the radiation losses find the expression of rate of melting of ice. . |
|
Answer» Correct Answer - A::D Putting, `((dQ)/dt)_("conduction")=((dQ)/(dt))_("calorimetry")` `:. (TD)/R = L* (dm)/(dt)` `rArr (dm)/(dt) = (TD)/(RL)` So, this is the desired expression of `(dm)/(dt)`. In the above expression, `TD = 100^@C, R = l/(KA)` and L = latent heat of fusion. |
|
| 31. |
The temperature of the two outer surfaces of a composite slab consisting of two materials having coefficient of thermal conductivity K and 2K and thickness x and 4x respectively are `T_2` and `T_1(T_2gtT_1)`. The rate of heat transfer through the slab in steady state is `((AK(T_2 -T_1))/x)f`. where, f is equal to .A. 1B. `1//2`C. `2//3`D. `1//3` |
|
Answer» Correct Answer - D `R_("net") = R_1 + R_2 = (X/(KA)) + (4X)/(2KA) = (3X)/(KA)` Now, `H = (dQ)/(dt) = (TD)/(R_("net"))` `=((T_2 -T_1))/((3X//KA)) = [(KA(T_2-T_1))/x] (1/3)` `:. f = 1/3` . |
|
| 32. |
`0 g` ice at `0^@C` is converted into steam at `100^@C`. Find total heat required . `(L_f = 80 cal//g, S_w = 1cal//g-^@C, l_v = 540 cal//g)` |
|
Answer» Correct Answer - A::B::C `Q = mL_f+ms_w Deltatheta + mL_f` `= m(L_f + s_wDeltatheta + L_f)` `= 10[80 + 1 xx 100 + 540]` `=7200 cal` |
|
| 33. |
The temperature of 100g of water is to be raised from `24^@C` to `90^@C` by adding steam to it. Calculate the mass of the steam required for this purpose. |
|
Answer» Correct Answer - A::B Let m be the mass of the steam required to raise the temperature of 100 g of water from `24^@C` to `90^@C`. Heat lost by steam = Heat gained by water. `:. m(L + sDeltatheta_1) = 100sDeltatheta_2` or `m=((100)(s)(Deltatheta_2))/(L + s(Deltatheta_1))` Here, s = specific heat of water `= 1 cal//g-^@C`, L = latent heat of vaporization `=540 cal//g` `Deltatheta_1 = (100-90) = 10^@C` and `Deltatheta_2 = (90-24) = 66^@C` Substituting the values, we have `m=((100)(1)(66))/((540) + (1)(10)) = 12g ` `:. m = 12g` . |
|
| 34. |
A steam cylindrical pipe of radius 5 cm carries steam at `100^@C` . The pipe is covered by a jacket of insulating material 2 cm thick having a thermal conductivity 0.07 `W//m-K` . If the temperature at the outer wall of the pipe jacket is `20^@C` , how much heat is lost through the jacket per metre length in an hours? . |
|
Answer» Correct Answer - A::C Thermal resistance per metre length of an element at distance r of thickness dr is `dR = (dr)/(K(2pir))" " (R = l/(Ka))` `:.` Total resistance `R = int _(r_1= 5cm)^(r_2 = 7cm) dR` `=1/(2piK) int_(7.0 xx 10^-2 m)^(5.0 xx 10^-2 m) (dr)/r` `= 1/(2piK) In (7/5)` `=1/((2pi)(0.07)) In (1.4)` = 0.765 `K//W` Heat current , `H = ("Temperature difference")/("thermal resistance")` `(100 -20)/(0.765) = 104.6W` `:." Heat lost in one hour" = "Heat current" xx "time"` `= (104.6)(3600)J` `= 3.76 xx (10^5)J`. |
|