Saved Bookmarks
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. |
A liquid having mass `m = 250` g is kept warm in a vessel by use of an electric heater. The liquid is maintained at `50^(@)C` when the power supplied by heater is 30 watt and surrounding temperature is `20^(@)C`. As the heater is switched off, the liquid starts cooling and it was observed that it took 10 second for temperature to fall down from `40^(@)C` to `39.9^(@)C`. Calculate the specific heat capacity of the liquid. Assume Newton’s law of cooling to be applicable. |
|
Answer» Correct Answer - `8000(J)/(kg^(@)C)` |
|
| 52. |
A metal ball of mass 1.0 kg is kept in a room at `15^(@)C`. It is heated using a heater. The heater supplies heat to the ball at a constant rate of 24 W. The temperature of the ball rises as shown in the graph. Assume that the rate of heat loss from the surface of the ball to the surrounding is proportional to the temperature difference between the ball and the surrounding. Calculate the rate of heat loss from the ball when it was at temperature of `20^(@)C`. |
|
Answer» Correct Answer - 4 Watt |
|
| 53. |
Majority of radiation from the Sun is in visible and near infra-red `(0.7 t o 4 mum)` region. What can you say about the composition of the radiation from the Earth? |
|
Answer» Correct Answer - Earth radiates almost `100%` in far infra-red region |
|
| 54. |
Figure shows a large tank of water at a constant temperature `theta_(0)` . and a small vessel containinng a mass m of water at an initial temperature `theta(lttheta_(0). A metal rod of length L, area of cross section A and thermal conductivity K connect the two vessels. Find the time taken for the temperature of the water in the smaller vessel to become `theta_(2)(theta_(1)lttheta_(2)lttheta_(0))` . Specific heat capacity of water is s and all other heat capacities are negligible. |
|
Answer» Suppose, the temperature of the water in the smaller vessel is theta at time t. In the next time interval dt, a heat (DeltaQ) is transferred to it where `DeltaQ=(KA)/(L)(theta_(0)-theta)dt` . This heat increases the temperature of the water of mass m to `theta+d theta where `DeltaQ=msd theta` . From (i) and (ii), (KA)/L(theta_(0)-theta)dt=msd theta` . or, `dt=(Lms)/(KA)(d theta)/(theta_(0)-theta` . or, `int_(0)^(T)dt=(Lms)/(KA)int_(theta_(1))^(theta^(2))(d theta)/(theta_(0)-theta)` . where T is the time requierd for the temperature of the water to become `theta_(2)` . Thus, `T=(Lms)/(KA) ln `(theta_(0)-theta_(1))/(theta_(0)-theta_(2))` . |
|
| 55. |
On a cold winter day, the atmosheric temperature is -theta (on celsium scale) which is below `0^(@)C. A cylindrical drum oof height h made of a bad conductor is completely filled with water at 0^(@)C and is kept outside without any lid. Calculate the time taken for the whole mass of water to freeze. Thermal conductivity of ice is K and its laternt heat of fusion is L. Neglect expansion of water on freezing. |
|
Answer» suppose, the ice starts forming at time `t=0` and a thickness x is formed at time t. The amount of heat flown from the water to the surrounding in the time interval t to `t+dt` is `DeltaQ=(KAtheta)/(x)dt` . The mass of the ice formed due to the loss of this amount of heat is `dm=(DeltaQ)/(L)=(KAtheta)/(xL)dt` . The thickness dx of ice formed in time dt is dx=(dm)/(Ap)=(Ktheta)/(pxL)dt` . or, `dt=(pL)/(Ktheta)xdX` . Thus, the time T taken for the whole mass of water to freeze is given by `int_(0)^(T)dt=(pL)/(Ktheta)int_(0)^(h)xdx` . or, `T=(pLh^(2))/(2Ktheta)` |
|
| 56. |
A spherical ball oof surface area `20cm^(@)` absorbs any radiation that falls on it. It is susoended in a closed box maintained at `57^(@)C` . (a) Find the amount of radiation falling on the ball per second. (b0 Find the net rate of heat flow to or from the ball at an instant when its temperature is `200^(@)C` . Stefan constant `=6.0xx10^(-s)Wm^(-2)K^(-4)` . |
|
Answer» Correct Answer - A::B::C::D `(a) A=20cm^2=20xx10^(-4)m^2` `T=57^(2)C=57+273=330K` `E=A sigma T^4` `=20xx10^(-4)xx6xx10^(-8)xx(330)^(4)` `=1.42J.` `(b)(E/t)=(A sigma e)(T_(1)^(4)-(T_(2)^(4)` `[A=20 cm^2=20xx10^(-4)m^2, sigma= 6xx10^(-8)` `t_1=473K, t_2=330 K` `=20xx10^(-4)xx6xx10^(-8)` `xx1[(473)^4-(330)^(4))]` `=20xx6xx10^(-12)` `[5.005xx10^(10)-1.185xx10^(10)] =20xx6xx3.82xx10^(-2)` `=4.58W from the ball.` |
|
| 57. |
Water at `506(@)C` is filled in a closed cylindercal vessel of height `10cm` and cross sectional area `10cm^(2)` The walls of the vessel are adiabatic but the flat parts are made of `1-mm` thick aluminium `(K=200Js^(-2)m^(-1)`^(@)C^(-1)` . Assume that the outside temperature is `20^(@)C` . The density of water is `1000kgm^(-3)` and the specify heat capacity of water `=4200Jkg^(-1)`^(@)C^(-1)` .Estimate the time taken for the temperature to fall by `1.0^(@)C` . Make any simplifying assumotion you need but speify them. |
|
Answer» Correct Answer - C `A=10 cm^(2): H=10 cm` `( Delta Q)/( Delta T)=(kA(T_1-T_2))/(l)` `=(200xx10xx10^(-4)xx30)/(1xx10^(-3))` `6000 J//s` Since heat goes out form both surfaces hence net heat coming out `( Delta Q)/( Delta T)=6000xx2=12000` `( Delta Q)/( Delta T)=ms ( Delta T)/( Delta l) `implies6000xx2=10^(-3)xx10^(-1)xx1000` `xx4200xx( Delta Q)/( Delta T).` `implies( Delta Q)/( Delta T)=12000/420=28.57` so in one sec 28.57^(@)C is dropped Hence time to drop of `1^(@)C=(1)/(28.57)sec.` `=0.035 sec is required . |
|
| 58. |
Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at `0^@C` and `90^@C`, respectively. The temperature of junction of the three rods will be (a) `45^@C` (b) `60^@C` (c) `30^@C` (d) ` 20^@C`. A. `45^(@)C`B. `60^(@)C`C. `30^(@)C`D. `20^(@)C` |
|
Answer» `(b)` Let `theta^(@)C` be the temperature at `B`. Let `Q` is the heat flowing per second from `A` to `B` on account of temperature difference by conductivity. `:. Q=(KA(90-theta))/(l)`…….`(i)` Where `K=` Thermal conducitivity of the rod `A=` Area of cross-section of the rod `l=` Length of the rod By symmetry, the same will be the case for heat flow from `C` to `B` `:.` The heat flowing per second from `B` to `D` will be `2Q=(KA(theta-0))/(l)`......`(ii)` Diveding equation `(ii)` by equation `(i)` `2=(theta)/(92-theta)` `implies theta=60^(@)` |
|
| 59. |
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`A. `Q_(a)` is maximumB. `Q_(b)` is maximumC. `Q_(C )` is maximumD. `Q_(a) = Q_(b) = Q_(c )` |
|
Answer» Correct Answer - C (3) `lambda_(1) T_(1) = lambda_(2) T_(2) = lambda_(3) T_(3)` `300 T_(1) = 400 T_(2) = 500 T_(3)` `3 T_(1) = 4 T_(2) = 5 T_(3) = k, k:` constant `Q = sigma A T^(4) = sigma. Pi r^(2) T^(4)` `Q prop r^(2) T^(4)` `Q_(a) : Q_(b) : Q_(c ) = (2)^(2) ((k)/(3))^(4) : (4)^(2) ((k)/(4))^(4) : (6)^(2) ((k)/(5))^(4)` `= (1)/(81) : (1)/(16) : (36)/(25)` `Q_(C ) gt Q_(b) gt Q_(a)` |
|
| 60. |
The radiancy of a black body is `M_(e)=3.0W//cm^(2)`. Find the wavelength corresponding to the maximum emmissive capacity of the body. `b=2.9xx10^(-3)m.k`, `sigma=5.67xx10^(-2)K^(-4)`. |
|
Answer» `M_(e)=sigmaT^(4)impliesT=4sqrt((M_(e))/(sigma))` Now `lambdaT=bimplieslambda=(b)/(T)=bxx4sqrt(sigma//M_(e))` `implieslambda=2.9xx10^(-3)4sqrt((5.67xx10^(-8))//3.0xx10^(4))=3.4xx10^(-6)m` |
|
| 61. |
Two bodies of masses `m_(1)` and `m_(2)` and specific heat capacities `S_(1)` and `S_(2)` are connected by a rod of length l, cross-ssection area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time `t=0` , the temperature of the fisrt body is `T_(1)` and the temperature of the second body is `T_(2)(T_(2)gtT_(1))` . Find the temperature difference between the two bodies at time t. |
|
Answer» Correct Answer - A::B::D `(Q/t)=(KA(T_1-T_2))/L` `T_2=(KA(T_1-T_2))/(Lms)` `Fall in temperature in T_1=(KA(T_1-T_2))/(Lm_1s_1)` `Final tempareture in T_1=T_1-=(KA(T_1-T_2))/(Lm_1 s_1)` `Final temprature in ` ` T_2=T_2+(KA(T_1-T_2))/(Lm_2 s_2)` `Change in temparature ` ` T_1-(KA(T_1-T_2))/(Lm_1 s_1)` `=(T_2+(KA(T_1-T_2))/(Lm_2 s_2)` `=(T_1-T_2)` `-[(KA(T_1-T_2))/(Lm_1 s_1)+(KA(T_1-T_2))/(Lm_2 s_2)]` `IndT=(KA)/(L)((M_2)(S_2) +(M_1)(S_1)/(M_1)(S_1)(M_2)(S_2))` So difference in tempareture `=(T_2-T_1)e^(-lambda t)` `where (lambda)=(KA)/(l)((m_1)(s_1)+(m_2)(s_2))/((m_1)(s_1)(m_2)(s_2))`. |
|
| 62. |
Consider the Sun to be a black body at temperature `T_(S) = 5780 K`. Radius of the Sun is `r_(S) = 6.96 xx 108 m`. The Earth - Sun distance is `R = 1.49 xx 1011m`. Assume that `30%` of the solar radiation that hits the earth is scattered back into space without absorption. (a) Calculate the steady state average temperature of the earth assuming it to be a black body. Take `(0.7 )^((1)/(4))= 0.91` (b) We know that average temperature of earth is `~=288K`. How does this value compare with that obtained in (a)? The difference is due to greenhouse effect. Comment on the following statement - “Emissivity of earth is reduced more than absorptivity due to green house effect.” |
|
Answer» Correct Answer - (a) 254 K |
|
| 63. |
The temperature of an isolated black body falls from `T_(1)` to `T_(2)` in time `t`. Let `c` be a constantA. `t = c [(1)/(T_(2)) - (1)/(T_(1))]`B. `t = c [(1)/(T_(2)^(3)) - (1)/(T_(1)^(2))]`C. `t = c [(1)/(T_(2)^(3)) - (1)/(T_(1)^(3))]`D. `t = c [(1)/(T_(2)^(4)) - (1)/(T_(1)^(4))]` |
|
Answer» Correct Answer - C (3) `ms (dT)/(dt) = - sigam A (T^(4) - 0)` `int_(T_(1))^(T_(2)) (dT)/(T^(4)) = - sigma A int_(0)^(t) dt` `[(1)/(T_(2)^(3)) - (1)/(T_(1)^(3))] = (3 sigma A) t` `t - (1)/((3 sigma A)) [(1)/(T_(2)^(3)) - (1)/(R_(1)^(3))]` `t = c [(1)/(T_(2)^(3)) - (1)/(T_(1)^(3))], c = (1)/(3 sigma A)` |
|
| 64. |
When two ends of a rod wrapped with cotton are maintained at differences and after some time every point of the rod attains a constant temperature, thenA. Condition of heat at different points of the rod stops becasuse the temperature is not increasingB. rod is bad conductor of heatC. heat is being radiated from each point of the rodD. each point of the rod is giving heat to its neighbour at the same rate at which it is receving heat |
| Answer» Correct Answer - D | |
| 65. |
The ends of a metre stick are maintained at `100^(@)C` and `0^(@)C`. One end of a rod is maintained at `25^(@)C`. Where should its other end be touched on the metre stick so that there is no heat current in the rod in steady state?A. `25 cm` form the hot endB. `40 cm` from the cold endC. `25 cm` from the cold endD. `60 cm` from the cold end |
|
Answer» Correct Answer - C `25 cm` from cold end is at `25^(@)C` (Since temperature gradient is `1^(@)C//cm)` `:.` rod shound be touched at this point to ensure no heat transfter. |
|
| 66. |
One end of a themally insulated copper rod of length `1 m` long and of `10 cm^(2)` in X-section is immersed in boiling water `(100^(@)C)` and the other in ice `(0^(@)C)` . If `K_(Cu) = 0.92 cal//cm.//s//.^(@)C` and `L_(ice) = 80 cal//g`, the ice that will melt per minute isA. `9.2xx10^(-3)kg`B. `8xx10^(-3)kg`C. `6.9xx10^(-3)kg`D. `5.4xx10^(-3)kg` |
|
Answer» `(dH)/(dt)=(kA(DeltaT))/(l)=(92xx10^(-3)xx100)/(1)=9.2cal//sec` in one minute `H=((dH)/(dt))xx60=9.2xx60` `9.2xx60=mxx8xx10^(4)` `m=(9.2xx6)/(8xx10^(3))` |
|
| 67. |
The normal body-temperature of a person is `97(@)F` . Calculate the rate at which heat is flowing out of this body through the clothes asssuming the following values. Room temperature `=47^(@)F` , surface of the body under clothes `=1.6m^(-2)` , condutivity of the cloth `=0.04Js^(-1)m^(-1)`^(@)C^(-1)` , thickness of the cloth `=0.5cm. |
|
Answer» Correct Answer - A::C `K=0.04 J//m-S^(@)C,` `A=1.6 m^(2)` `(T_(1))=97^(@)F=36.10^(@)C` ``T_(2)=47^(@)F=8.330^(@)C` `l=0.5cm=0.005m` `(Q)/(T)=(kA((theta)_1-(theta)_2))/(l)` `(0.04xx1.6xx(27.7))/(0.005)=356 J//s` |
|
| 68. |
A black body calorimeter filled with hot water cools from `60^(@)C` to `50^(@)C` in `4 min` and `40^(@)C` to `30^(@)C` in `8min`. The approximate temperature of surrounding is :A. `10^(@)C`B. `15^(@)C`C. `20^(@)C`D. `25^(@)C` |
|
Answer» `(d theta)/(dt)=k(Deltatheta)=k[theta_(w)-theta_(s)]` `theta_(s)=` Temperature of surrounding `theta_(w)=(theta_(1)+theta_(2))/(2)` `(10)/(4)=k[(60+50)/(2)-theta_(s)]`……..`(i)` `(10)/(8)=k[(40+30)/(2)-theta_(s)]`……..`(ii)` equation `(i)` divided by `(ii)` gives `z=(55-theta)/(35-theta)` `70-20theta=55-theta` `theta=15^(@)` |
|
| 69. |
One end of rod of length `L` and cross-sectional area `A` is kept in a furance of temperature `T_(1)`. The other end of the rod is kept at at temperature `T_(2)`. The thermal conductivity of the material of the rod is `K` and emissivity of the rod is `e`. It is given that `T_(2)=T_(S)+DeltaT` where `DeltaT lt lt T_(S)`, `T_(S)` being the temperature of the surroundings. If `DeltaT prop (T_(1)-T_(S))`, find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is `T_(2)`. |
|
Answer» From the figure it is clear that emission takes place from the surface at temperture `T_(2)` (circular cross section). Heat conduction and radiation through laeral surface is zero. Heat conducted through rod is `Q=(KA(T_(1)-T_(2))Deltat)/(l)` Energy emitted by the surface of the rod in the same time `Deltat`, is `E=epsilonsigmaA(T_(2)^(4)-T_(S)^(4))Deltat` Since rod is at thermal equilibrium `:. E=Q` hence, `(KA(T_(1)-T_(2)))/(l)=epsilon sigma A(T_(2)^(4)-T_(S)^(4))Deltat` `implies T_(1)-T_(2)=(epsilon sigma(T_(2)^(4)-T_(S)^(4)))/(K)` Here `T_(1)-T_(2)=DeltaT` and `T_(s) gt gt Delta T` `T_(1)-(DeltaT+T_(S))=(epsilonsigmal)/(K)[(DeltaT+T_(s))^(2)-T_(S)^(4)]` `T_(1)-(DeltaT+T_(S))=(epsilonsigmal)/(K)xxT_(S)^(4)[(1+(DeltaT)/(T_(S)))^(4)-1]` `T_(1)-(DeltaT+T_(S))=(epsilonsigmal)/(K)xxT_(S)^(4)[(1+(4DeltaT)/(T_(S)))^(4)-1]` or `T_(1)-(T_(S)+DeltaT)=(4epsilonsigmal)/(K)xxT_(S)^(3)DeltaT` or `T-T_(s)=(4epsilonsigmalT_(s)^(3))/(K)DeltaT+DeltaT` or `T_(1)-T_(s)=((4epsilonsigmalT_(s)^(3))/(K)+1)DeltaT` `:. The proportionality constant`= `(1+(4epsilonsigmalT_(s)^(3))/(K))` |
|
| 70. |
According to the experiment of Ingen Hauze the relation between the thermal conductivity of a metal rod is `K` and the length of the rod whenever the wax melt isA. `K//l` = constantB. `K^(2)//l` = constantC. `K//l^(2)` = constantD. `K l` = constant |
| Answer» Correct Answer - C | |
| 71. |
Parallel rays of light of intensity `I=912 WM^-2` are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan-Boltzmann constant `sigma=5.7xx10^-8` `Wm^-2K^-4` and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toA. `330K`B. `660K`C. `990K`D. `1550 K` |
|
Answer» `IxxpiR^(2)=4piR^(2)sigma(T^(4)-300^(4))` `(912)/(4xx5.7)xx10^(9)+300^(4)=T^(4)` `implies4xx10^(9)+8.1xx10^(9)=T^(4)` `121xx10^(8)=T^(4)` `sqrt(11)xx10^(2)=T` `T=330K` |
|
| 72. |
Parallel rays of light of intensity `I=912 WM^-2` are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan-Boltzmann constant `sigma=5.7xx10^-8` `Wm^-2K^-4` and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toA. `330 K`B. `660 K`C. `990 K`D. `1550 K` |
|
Answer» Correct Answer - A In steady state `I piR^(2) = sigma (T^(4) -T_(0)^(4)) 4pi R^(2)` `rArr I = sigma (T^(4)-T_(0)^(4)) 4 rArr T^(4) -T_(0)^(4) = 40 xx 10^(6)` `rArr T^(4) - 81 xx 10^(8) xx 40 xx 10^(8) rArr T^(4) = 121 xx 10^(8)` `rArr T ~~ 330 K` |
|
| 73. |
Assume that the total surface area of a human body is `1.6m^(2)` and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is `37^(@)C` . Stefan constant `sigma` is `6.0xx10^(-s)Wm^(-2)K^(-4)` . |
|
Answer» Correct Answer - `887 J` `A = 1.6m^(1) e = 1` `(dQ)/(dt) = p sigmaeA T^(4) =(6 xx 10^(-8)) (1) 1.6 m^(2) (310)^(4)` `= 887 J` |
|
| 74. |
The power radiated by a black body is `P` and it radiates maximum energy around the wavelength `lambda_(0)` If the temperature of the black body is now changed so that it radiates maximum energy around a wavelength `3lambda_(0)//4` the power radiated by it will increase by a factor of .A. `(4)/(3)`B. `(16)/(9)`C. `(64)/(27)`D. `(256)/(81)` |
|
Answer» Correct Answer - D (4) `lambda_(0) T_(1) = (3 lambda_(0))/(4) T_(2)` `(T_(2))/(T_(1)) = (4)/(3)` `(u_(2))/(u_(1)) = ((T_(2))/(T_(1)))^(4) = ((4)/(3))^(4) = (256)/(81)` |
|
| 75. |
A black body at `1227^(@)C` emits radiations with maximum intensity at a wavelength of `5000 Å`. If the temperature of the body is increased by `1000^(@)`, the maximum intensity will be observed atA. 4000 ÅB. 5000 ÅC. 6000 ÅD. 3000 Å |
|
Answer» Correct Answer - D `(lambda_(1))/(lambda_(2))=(T_(2))/(T_(1))` `(5000 Å)/(lambda_(2))=(2500 K)/(1500 K)` `lambda_(1)=3000 Å` |
|
| 76. |
The thermal radiation emited by a body is proportional to `T^(n)` where `T` is its absolute temperature. The value of `n` is exactly 4 forA. a blackbodyB. all bodiesC. bodies painted black onlyD. polished bodies only |
| Answer» Correct Answer - B | |
| 77. |
A black body does not (i) emit radiation (ii) absorb radiation (iii) reflect radiation (iv) refract radiationA. (i), (iii)B. (ii), (iii)C. (iii), (iv)D. (i), (ii) |
| Answer» Correct Answer - C | |
| 78. |
A uniform cylinder of length `L` and thermal conductivity `k` is placed on a metal plate of the same area `S` of mass `m` and infinite conductivity. The specific heat of the plate is `c`. The top of the cylinder is maintained at `T_(0)`. Find the time required for the temperature of the plate to rise from `T_(1)` to `T_(2)(T_(1) lt T_(2) lt T_(0))`. |
|
Answer» Correct Answer - `(mCL)/(KS)ln((T_(0)-T_(1))/(T_(0)-T_(2)))` The whole metal plate will always be at uniform temperature `(T)` since it has infinite conductivity. then, `-mc (dT)/(dt) = (ks(T_(0)-T))/(l)` `rArr -mc overset(T_(2))underset(T_(1))int (dt)/(T_(0)-T) = (ks)/(l) overset(t)underset(0)int dt` `ln ((T_(0)-T_(2))/(T_(0)-T_(1))) = (ks t)/(mcl)` `rArr t = (mcl)/(ks) ln |(T_(0)-T_(2))/(T_(0)-T_(1))|` |
|
| 79. |
The thermal conductivity of a rod depends onA. lengthB. massC. area of cross sectionD. material of the rod. |
|
Answer» Correct Answer - D material of the rod. |
|
| 80. |
A spherical black body of radius `r` radiates power `P`, and its rate of cooling is `R` (i)`P prop r` (ii)`P prop r^(2)` (iii)`R prop r^(2)` (iv)`R prop (1)/(r)` |
|
Answer» Correct Answer - D (4) `P = e sigma A (T^(4) - T_(0)^(4)) = e sigma . 4 pi r^(2) (T^(4) - T_(0)^(4))` `P prop r^(2)` `ms (dT)/(dt) = e sigma A (T^(4) - T_(0)^(4))` `(dT)/(dt) = (e sigma A (T^(4) - T_(0)^(4)))/(ms) = (e sigma 4 pi r^(2) (T_(4) - T_(0)^(4)))/(rho . (4)/(3) pi r^(3) s` Rate of cooling `R = ( = (dT)/(dt)) prop (1)/(r )` |
|
| 81. |
Calculate the quantity of heat conducted through `2m^(2)` of a brick wall `12cm` thick in `1 hour` if the temperature on one side is `8^(@)C` and on the other side is `28^(@)C`. (Thermal conductivity of brick `=0.13Wm^(-1)K^(-1)`) |
|
Answer» Temperature gradient `=(28-8)/(12xx10^(-2))Km^(-1)` and `t=3600s` `:. Q=kAtxx` temperature gradient `=0.13xx2xx3600xx(28-8)/(12xx10^(-2))` `=156000J` |
|
| 82. |
Calculate the amount of heat radiated per second by a body of surface area `12cm^(2)` kept in thermal equilibrium in a room at temperature `20^(@)C` The emissivity of the surface `=0.80` and `sigma=6.0xx10^(-s)Wm^(-2)K^(-4)` . |
|
Answer» Correct Answer - B::D `A=12 cm^(2)=12xx10^(-4)m^(2),` `T=20^(@)C=293^(@)K,` `sigma=6xx10^(-8)W//m^(2)-K^(4)` `Q=AeST^4` ltbrge12xx10^(-4)xx0.8xx6xx10^(-8)xx(293)^4` `=0.4245=0.42`. |
|
| 83. |
The tungsten filament of an electric lamp has a length of `0.5m` and a diameter `6xx10^(-5)m`. The power rating of the lamp is `60 W`. Assuming the radiation from the filament is equivalent to `80 %` that of a perfect black body radiator at the same temperature, estimate the steady temperature of the filament.(Stefan constant `=5.7xx10^(-8)Wm^(-2)K^(-4)`) |
|
Answer» When the temperature is steady, power radiated from filament `=` power received `=60W` `:. 0.8xx5.7xx10^(-8)xx2pixx3xx10^(-5)xx0.5xxT^(4)=60` since surface area of cylindrical wire is `2pirh` with the usual notation. `:. T=((60)/(0.4xx5.7xx10^(-8)xx2pixx3xx10^(-5)))^(1//4)=1933K` |
|
| 84. |
A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^@C`. Initial temeprature of cooling is valid. The body cools to `38^@C` in 10 min The temperature of the body In `.^@C` denoted by `theta`. The veriation of `theta` versus time t is best denoted asA. B. C. D. |
| Answer» Correct Answer - A | |
| 85. |
A bulb made of tungsten filament of surface are a`0.5cm^(2)` is heated to a temperature `3000 K` when operatred at `220V`. The emissivity of the filament is `e=0.35` and take `sigma=5.7xx10^(-8)mks units`. Then calculate the wattage of the bulb (in watt) |
|
Answer» The emissive power `"watt"//m^(2)` is `E=esT^(4)` Therefore the power of the bulb is `P=exx`area(Watts) `:. P=e A sigma T^(4)` `:. P=0.35xx0.5xx10^(-4)xx5`, `7xx10^(-8)xx(3000)^(4)` `implies P=80.8 W` |
|
| 86. |
A body cools from `50^(@)C` to `40^(@)C` in 5 min. The surroundings temperature is `20^(@)C`. In what further times (in minutes) will it cool to `30^(@)C` ?A. 5B. `15//4`C. `25//3`D. 10 |
|
Answer» Correct Answer - C (3) `(Delta theta)/(Delta t) = - k (bar theta - theta_(0))` `(50 - 40)/(5) = - k ((50 + 40)/(2) - 20)` `2 = - k xx 25 implies K = - (2)/(25)` `(40 - 30)/(t) = - k ((40 + 30)/(2) - 20)` `(10)/(t) = (2)/(25) (15) implies t = (25)/(3)` min |
|
| 87. |
A body cools down from `50^(@)C` to `45^(@)C` in 5 minutes and to `40^(@)C` in another 8 minutes. Find the temperature of the surrounding. |
|
Answer» Correct Answer - C::D `V = 100CC, Deltatheta = 5^(0) C, T = 5 min` for water `(msDeltatheta)/(dt) = (KA)/l` for kerosene `(ms)/(dt) = (KA)/(l) 1/(Deltatheta)` `implies (100 xx 10^(-3) xx 800 xx 2100)/(t)` `= (100 xx 10^(-3) xx 1000 xx 4200)/(5)` `implies t = (5 xx 800 xx 2100)/(1000 xx 4200) = (2000)/(1000)` |
|
| 88. |
A liquid cools from `70^(@)C` to `60^(@)C` in `5` minutes. Calculate the time taken by the liquid to cool from `60^(@)C` to `50^(@)C` , If the temperature of the surrounding is constant at `30^(@)C` . |
|
Answer» Correct Answer - `7 minutes` `(70 -60)/(5) = K [(70 +60)/(2)-30]` `rArr (10)/(5) = K [65 - 30] ….(i)` Now `(60 -50)/(t) =K [65 - 30] …(ii)` Dividing equation (i) and (ii) `(t)/(5) = (35)/(25)` `t = (7)/(5) xx5 = 7 min` |
|
| 89. |
A metal ball cools from `62^(@)C` to `50^(@)C` in `10min` and to `42^(@)C` the next ten minutes. What will be its temperature at the end of next ten minutes ? |
|
Answer» Use `(theta_(1)-theta_(2))/(t)=K((theta_(1)-theta_(2))/(2)theta_(0))` ………`(1)` to get the following equations `(62-50)/(10)=K((62+50)/(2)-theta_(0))` and `(50-42)/(10)=((50+42)/(2)-theta_(0))`……..`(2)` Divide `2` by `1` and solve to get `theta_(0)` `(8)/(12)=(46-theta_(0))/(56-theta_(0))` `implies theta_(0)=26`.......`(3)` Let after the next `10min` the temperture falls to then `(42-theta)/(10)=K((42+theta)/(2)-26)`......`(4)` From `(3)` using value of `theta_(0)=26` we get `(12)/(10)-k(56-26)` Divide `(4)` by `(5)` to get `(42-theta)/(12)=(42+theta-52)/(2(56-26))` `implies theta=36.7^(@)C` |
|
| 90. |
At `1600K` maximum radiation is emitted at a wavelength of `2mu_(M)`. Then what will be corresponding wavelength at `2000K` ? |
|
Answer» Using `lambda_(m_(1))T_(1)=lambda_(m_(2))T_(2)` `:. Lambda_(m_(2))=(lambda_(m_(1))T_(1))/(T_(2))` `:. Lambda_(m_(2))=(2xx10^(-6)xx1600)/(2000)` `implies lambda_(m_(2))=1.6mu m` |
|
| 91. |
In example `3, K_(1) = 0.125 W//m-.^(@)C, K_(2) = 5K_(1) = 0.625 W//m-.^(@)C` and thermal conductivity of the unknown material is `K = 0.25 W//m^(@)C L_(1) = 4cm, L_(2) = 5L_(1) = 20cm`. If the house consists of a single room of total wall area of `100 m^(2)`. then find the power of the electric heater being used in the room. |
|
Answer» `ul(I^(ulst)"method") R_(1) =R_(2) = ((4xx10^(-2)m))/((0.125w//m-^(@)C)(100m^(2))) =32 xx 10^(-4).^(@)C//w` `:. (25 -20)/(R_(1)) = (20-T_(3))/(R )` `R = (L)/(KA) = 112 xx 10^(-4).^(@)C//W` the equivalent thermal resistance of the entire wall `= R_(1) +R_(2) +2R = 288 xx 10^(-4).^(@)C//W` `:.` Net heat current i.e. amount of heat flowing out of the house per second `= (T_(H)-T_(C))/(R)` `=(25^(@)C-(-20^(@)C))/(288xx10^(4).^(@)C//w) =(45xx10^(4))/(288)`watt `= 1.56 K`watt Hence the heater must supply `1.56 kW` to compenstate for the outflow of heat. `ul(II^(ulnd)method)` `i = (T_(1)-T_(2))/(R_(1)) = (25 -20)/(32 xx 10^(-4)) = 1.56 K`watt |
|
| 92. |
A solid cylinder and a sphere of same material are suspended in a room turn by turn, after heating them to the same temperature. The cylinder and the sphere have same radius and same surface area. (a) Find the ratio of initial rate of cooling of the sphere to that of the cylinder. (b) Will the ratio change if both the sphere and the cylinder are painted with a thin layer of lamp black? |
|
Answer» Correct Answer - (a) `(3)/(4)` (b) No |
|
| 93. |
An iron ball is heated to `727^(@)C` and it appears bright red. The plot of energy density distribution versus wavelength is as shown. The graph encloses an area `A_(0)` under it. Now the ball is heated further and it appears bright yellow. Find the area (A) of the energy density graph now. If the given that wavelengths for red and yellow light are 8000 Å and 6000 Å respectively. |
|
Answer» Correct Answer - `A=((4)/(3))^(4)A_(0)` |
|
| 94. |
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 propertional toA. `T_(2)-T_(1)`B. `T_(2)^(2)-T_(1)^(2)`C. `T_(2)^(3)-T_(1)^(3)`D. `T_(2)^(2)-T_(1)^(4)` . |
|
Answer» Correct Answer - D `T_(2)^(2)-T_(1)^(4)` . |
|
| 95. |
A copper sphere is suspended in an evacuated chamber maintained at 300K. The sphere is maitained at a constant temperature of 500K by heating it electrically. A total of 210W is electric power is needed to do it. When the surface of the copper sphere is completely blackned, 700W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper. |
|
Answer» Correct Answer - C `Q=eA rho(T_(1)^(4)-T_(2)^(4))` `For any body ,` `210= eA rho[(500)^(4)-(300)^(4)]` `For black body,` `700=1xxAxx rho [(500)^(4)-(300)^(4)]` `dividing ` `(210)/(700)=e/1` `e=0.3`. |
|
| 96. |
Two parallel plates `A` and `B` are joined together to form a compound plate . The thicknesses of the plate are `4.0cm` and `2.5cm` respectively and the area of cross section is `100cm^(2)` for each plate. The thermal conductivities are `K_(A)=200Wm^(-1).^(@)C^(-1)` for the plate `A` and `K_(B)=400Wm^(-1).^(@)C^(-1)` for the plate `B`. The outer surface of the plate `A` is maintained at `100^(@)C`and the outer surface of the plate A is maintained at `100^(@)C` . and the outer surface of the plate B is maintained at `0^(@)C`. Find(a) the rate of heat flow through any cross section,(b) the temperature at the interface and(c) the equivalent thermal conductivity of the compound plate. |
|
Answer» `(a)` Let the temperature of the interface be `T`. The area of cross-section of each pate is `A=100cm^(2)=0.01m^(2)`. The thickness are `x_(A)=0.04m` and `x_(B)=0.025m` The thermal resistance of the plate `A` is `R_(A)=(x_(A))/(K_(A)A)` and that of the plate `B` is `R_(B)=(x_(B))/(K_(B)A)` The equivalent thermal resistance is `R_(eq)=R_(A)+R_(B)=(1)/(A)((x_(A))/(K_(A))+(x_(B))/(K_(B)))`.........`(i)` Thus, `(DeltaQ)/(Deltat)=(T_(1)-T_(2))/(R_(eq))=(A(T_(1)-T_(2)))/(x_(A)//K_(A)+x_(B)//K_(B))` `=((0.01m^(2))(100^(@)C))/((0.04m)//(200W//m-^(@)C)+(0.025m)//(400W//m-^(@)C))=3810W`. `(b)` We have `(DeltaQ)/(Deltat)=(A(T-T_(2)))/(x_(B)//K_(B))` or, `3810W=((0.01m^(2))(T-0^(@)C))/((0.025m)//(400W//m-^(@)C))` or , `T=25^(@)C` `(c )` If `K` is the equivalent thermal conductivity of the compound plate, its thermal resistance is `R_(eq)=(1)/(A)(x_(A)+x_(B))/(K_(eq))` Comparing with `(i)`, `(x_(A)+x_(B))/(K_(eq))=(x_(A))/(K_(A))+(x_(B))/(K_(B))` or, `K_(eq)=(x_(A)+x_(B))/(x_(A)//K_(A)+x_(B)//K_(B))=248W//m-^(@)C` |
|
| 97. |
A spherical tungsten pices of radius `1.0cm` is suspended in an evacuated chamber maintained at `300K. The pices is maintained at 1000K by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is `0.30` and the Stefan constant sigma is `6.0xx10^(-s)Wm^(-2)K^(-4)` . |
|
Answer» Correct Answer - B `r=1cm =1xx10^(-2)m,` `A =4 pi (10^(-2))^(2)` `=4 pi xx10^(-4)m^(2)` `e=0.3, sigma =6xx10^(-8)` `(E/t)e sigma A (T_(2)^(4)-T_(1)^(4))` `=0.3xx6xx10^(-8)xx4 pixx10^(-4)` `xx[(1000)^4-(300)^4]` `=0.3xx6xx4 pi xx10^(-12)` `xx[1-0.0081]xx1012` `=0.3xx64xx3.14xx9919xx10^(-5)` `=22.4=22W`. |
|
| 98. |
A spherical tungsten piece of radius `1.0cm` is suspended in an evacuated chamber maintained at `300K`. The piece is maintained at `1000K` by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is `0.30` and the Stefan constant sigma is `6.0xx10^(-s)Wm^(-2)K^(-4)` . |
|
Answer» Correct Answer - 22 `E = sigma eA lfloor(100)^(4) -(300)^(4)rfloor` where `A = 4pi r^(2) & r = 0.01 m`. |
|
| 99. |
One end of a `2.35m` long and `2.0cm` radius aluminium rod `(K=235W.m^(-1)K^(-1))` is held at `20^(@)C`. The other end of the rod is in contact with a block of ice at its melting point. The rate in `kg.s^(-1)` at which ice melts is [Take latent heat of fusion for ice as `(10)/(3)xx10^(5)J.kg^(-1)`]A. `48pixx10^(-6)`B. `24pixx10^(-6)`C. `2.4pixx10^(-6)`D. `4.8pixx10^(-6)` |
|
Answer» `l=2.35m`, `r=2cm` `K=235`, `T=20^(@)C`, `DeltaT=20^(@)C` `(dH)/(dt)=((dm)/(dt))L_(f)` `((dm)/(dt))=((dH)/(dt))_(L_(f))^(bot)=(KA)/(l)(DeltaT)(1)/(L_(f))` `((dm)/(dt))=(235xxpi(2xx10^(-2))^(2)xx20)/(2.35xx(10)/(3)xx105)` `=(100pixx4xx10^(-4)xx20)/(10xx105)xx3` `=2.4pixx10^(-6)ks//sec` |
|
| 100. |
Two identical adiabatic containers of negligible heat capacity are connected by conducting rod of length L and cross sectional area A. Thermal conductivity of the rod is k and its curved cylindrical surface is well insulated from the surrounding. Heat capacity of the rod is also negligible. One container is filled with n moles of helium at temperature `T_(1)` and the other one is filled with equal number of moles of hydrogen at temperature `T_(2) (lt T_(1))`. Calculate the time after which the temperature difference between two gases will becomes half the initial difference. |
|
Answer» Correct Answer - `(15nRL)/(16kA)ln2` |
|