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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.
| 251. |
The temperatures of equal masses of three different liquids A, D and C are `12^@C`,`19^@C` and `28^@C`, respectively. The temperature when A and B are mixed is `16^@C`, while when B and C are mixed, it is `23^@C`. What would be the temperature when A and C are mixed? |
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Answer» Let `m=`mass of each liquid, when A and B are mixed, Heat lost by `B=` heat gained by A `impliesmS_B(19-16)=mS_A(16-12)implies2S_B=4S_A` …(i) When B and C are mixed, Heat lost by `C=` heat gained by B `impliesmS_C=(28-23)=mS_B(23-19)implies5S_C=4S_B` .(ii) From eqs. (i) and (ii), we get `16S_A=12S_B=15S_C` ..(iii) When A and C are mixed, Let `theta=`final temperature Heat lost by `C=` heat gained by A `impliesmS_C=(28-theta)=mS_A(theta-12)` Using Eq. (iii), we get `implies15S_C(28-theta)=15S_A(theta-12)` `implies16S_A(28-theta)=15S_A(theta-12)` On solving for `theta`, we get `theta=(16xx28+12xx15)/(16+15)` `impliestheta=20.26^@C` |
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| 252. |
Three liquids `A,B` and `C` of masses `400 gm, 600 gm` and `800 gm` are at `30^@C, 40^@C` and `50^@C` respectively. When `A` and `B` are mixed resultant temperature is `36^@C` when `B` and `C` are mixed resultant temperature is `44^@C` Then ration of their specific heats areA. `2 : 1 : 1`B. `3 : 2 : 1`C. `2 : 2 : 1`D. `1 : 4: 9` |
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Answer» Correct Answer - C When `A,B` are mixed `m_(A)S_(A) (Delta theta)_(A) = m_(B)S_(B)(Delta theta)_(B)`….(i) When `B,C` are mixed `m_(B) S_(B)(Delta theta)_(B) = m_(C) S_(C)` …(ii) From (i) and (ii) we get relation between `S_(A)` and `S_(C)`. When `A` and `C` are mixed `m_(A)S_(A) (Delta theta)_(A) = m_(C) S_(C) (Delta theta)_(C)`. |
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| 253. |
Two sphere of radii in the ratio `1 : 2`, have specific heats in the ration `2 : 3`. The densities are in the ratio `3 : 4`. Find the ration of their thermal capacities. |
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Answer» Thermal capacity of a body `= mS` The ratio of thermal capacities `(m_(1)S_(1))/(m_(2)S_(2)) = (V_(1)rho_(1)S_(1))/(V_(2)rho_(2)S_(2)) = ((4)/(3)pi r_(1)^(3) rho_(1)S_(1))/((4)/(3) pi r_(2)^(3) rho_(2) S_(2)) = ((r_(1))/(r_(2)))^(3) ((rho_(1))/(rho_(2))) ((S_(1))/(S_(2)))` Here, `(r_(1))/(r_(2)) = (1)/(2) ,(S_(1))/(S_(2)) = (2)/(3) , (rho_(1))/(rho_(2)) =(3)/(4)` The ratio of thermal capacities =`((1)/(2))^(3) ((3)/(4))((2)/(3)) =(1)/(16)`. |
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| 254. |
Equal masses of `3` liquids `A,B` and `C` have temperatures `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` & `C` are mixed the temperature of the mixture isA. `16^@ C`B. `35^@C`C. `20^@ C`D. `25^@C` |
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Answer» Correct Answer - A When `A` & `B` are mixed, `mS_(A) (5) = mS_(B) (10)` :. `S_(A) = 2 S_(B)` When `B` & `C` are mixed, `mS_(B) (5) = mS_(C) (10)` :. `S_(B)=2S_(C)` So, `S_(A) = 4S_(C)` , When `A` & `B` are mixed `mS_(A) (theta -10) =mS_(C)(40-theta)`. |
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| 255. |
Two beakers `A` and `B` contain liquids of mases `300 g` and `420 g` respectively and specific heats `0.8 cal//g-.^(0) C` and `0.6 cal//g-.^(0) C`. The amount of heat on them is equal. If they are joined by a metal rodA. heat flows from the beaker `B` to `A`B. heat flows from `A` to `B`C. not heat flowsD. heat flows neither from `A` to `B` not `B` to `A`. |
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Answer» Correct Answer - B Quanity of heat on `A` = Quanity of heat on `B` `m_(A) xx S_(A) xx theta_(1) = m_(B) xx S_(B) xx theta_(2) rArr theta_(1) gt theta_(2)`. |
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| 256. |
Two liquids `A` and `B` of equal volumes have their specific heats in the ration`2 : 3`. If they have same thermal capacity, then the ratio of their densities isA. `1 : 1`B. `2 : 3`C. `3 : 2`D. `5 : 6` |
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Answer» Correct Answer - C `m_(1)S_(1)=m_(2)S_(2) rArr v_(1)rho__(1)S_(1)=v_(2)rho_(2)S_(2) (rho_(1))/(rho_(2)) =(S_(2))/(S_(1))` |
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| 257. |
Two liquids of masses `m` and `5 m` at temperatures `3 theta, 4 theta` are mixed, If their specific heats are `2S,3S` respectively, the resultant temperature of mixture isA. `(66)/(17) theta`B. `(55)/(17) theta`C. `(44)/(17)theta`D. `(33)/(17) theta` |
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Answer» Correct Answer - A From principle of calorimetry `theta = (m_(1)S_(1)theta_(1)+m_(2)S_(2)theta_(2))/(m_(1)S_(1)+m_2S_(2))`. |
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| 258. |
Two liquids are at `40^@ C` and `30^@ C`. When they are mixed in equal masses, the temperature of the mixture is `36^@ C`. Ratio of their specific heats isA. `3 : 2`B. `2 : 3`C. `4 : 3`D. `3 : 4` |
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Answer» Correct Answer - A Heat lost by `1^(st)` liquid = Heat gaind by `2^(nd)` liquid. `mS_(1)(40^@ -36^@) = mS_(2)(36^@ -30^@)`. |
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| 259. |
Two liquids `A` and `B` are at temperatures of `75^@ C` and `150^@ C` respectively. Their masses are in the ratio of `2 : 3` and specific heats are in the ratio `3 : 4`. The resultant temperature of the mixture, when the above liquids, are mixed (Neglect the water equivalent of container) isA. `125^@C`B. `100^@ C`C. `50^@ C`D. `150^@ C` |
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Answer» Correct Answer - A `m_(A) S_(A) (theta - 75) = m_(B)S_(B) (150 - theta)`. |
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| 260. |
Quantity of heat lost in condensation of `10 gm` of steam at `100^@ C` isA. `2.26 xx 10^(5) J`B. `2.26 xx 10^(4) J`C. `22.6 J`D. `44.52 xx 10^(4) J` |
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Answer» Correct Answer - B `Q = mL_(steam)`. |
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| 261. |
Two liquids at temperature `60^@ C` and `20^@ C` respectively have masses in the ratio `3:4` their specific heat in the ratio `4:5`. If the two liquids are mixed, the resulatant temperature is.A. `70^@ C`B. `50^@ C`C. `40^@C`D. `35^@ C` |
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Answer» Correct Answer - D Heat lost = Heat gained `theta = (m_(1)S_(1)theta_(1)+m_(1)S_(2)theta_(2))/(m_(1)S_(1)+m_(2)S_(2))`. |
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| 262. |
A spherical ball of radius 1cm coated with a metal having emissivity 0.3 is maintained at 1000 K temperature and suspended in a vacuum chamber whose walls are maintained at 300 K temperature. Find rate at which electrical energy is to be supplied to the ball to keep its temperature constant. |
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Answer» It is given that the temperature of ball is constant. This means that the rate at which it loosing heat by radiation must be equal to the rate at which heat is supplied to this the rate of heat loss by the ball is given as `(dQ)/(dt)=eAsigma(T^4-T_s^4)` `=0.3xx4pi(0.01)^2xx5.67xx10^-8xx[(1000)^4-(300)^4]` `=21.1W` Thus electrical energy must be supplied to the ball at a rate of 21.1 W. |
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| 263. |
When a hot body warms a cool one, are their temperature changes equal in magnitude? |
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Answer» No. `m_1s_1(T_1-T_2)=M_2s_2(T-T_2)` `(T_1-T)/(T-T_2)=(m_2s_2)/(m_2s_1)` If `m_1s_1nem_2s_2`, then, `T_1-TneT-T_2` |
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| 264. |
A copper calorimeter, blackened outside, is filled with some hot liquid and placed on a table. It is found to cool from `60^(@)C` to `50^(@)C` in 4 minutes and `40^(@)C` to `30^(@)C` in 8 minutes. What is the temperature of the surroundings? Why is blackened? |
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Answer» Correct Answer - `15^(@)C`, to promote radiation |
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| 265. |
A cube and a sphere of equal edge and radius, made of the same substance are allowed to cool under identical conditions. Determine which of the two will cool at a faster rate. |
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Answer» From equation, the rate of cooling of a body is given by `(-dT)/(dt)=(Aesigma)/(rhoVs)(T_4-T_0^4)` since, substance is same for both bodies, so `e//rho s=`constant. Finally, thay are allowed to cool under identical conditions so `(T^4-T_0^4)=` constant. `(-dT)/(dt)prop(A)/(V)` Let the edge of the cube or radius of the sphere be a, then for cube, `A=6a^2` and `V=a^3` so `(A)/(V)=6a` For the sphere, `A=4pia^2` and `V=(4//3)pia^3`, so `A//V=3//a` Evidently, the ratio `A//V` is more for cube, so the cube cools at a faster rate. Not the special technique used in this problem. |
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| 266. |
A sphere and a cube of same material and same volume are heated up to same temperature and allowed to cool in the same sorroundings. The radio of the amounts of radiations emitted in equal time intervals will beA. `1:1`B. `(4pi)/(3):(1)/(86400)//^@C 1`C. `((pi)/(6))^(1//3):1`D. `(1)/(2)((4pi)/(3))^(2//3):1` |
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Answer» Correct Answer - C `Q=sigmaAt(T^4-T_0^4)` If `T,T_0,sigma` and t are same for both bodies then `(Q_("sphere"))/(Q_("cube"))=(A_("sphere"))/(A_("cube"))=(4pir^3)/(6a^2)` ..(i) But according to problem, volume of sphere = volume of cube `implies(4)/(3)pir^3=a^3` `impliesa=((4)/(3)pi)^(1//3)r` substituting the value of a in Eq. (i) we get `(Q_("sphere"))/(Q_("cube"))=(4pir^2)/(6a^2)` `=(4pir^2)/(6{((4)/(3)pi)^(1//3)r}^2)=(4pir^2)/(6{((4)/(3)pi)^(2//3)r^2))` `=((pi)/(6))^(1//3):1` |
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| 267. |
Two identical objects A and B are at temperatures `T_A` and `T_B`. Respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature `T(T_AgtTgtT_B)`. The objects A and B attain the temperature T eventually. Select the correct statements from the following:A. A only emits radiation while B only obsorbs it until both attain the temperature TB. A loses more heat by radiation than it absorbs, while B absorbs more radiation than it emits, until they attain the temperature TC. Both A and B only absorb radiation, but do not emit it, until they attain the temperature TD. Each object continuous to emit and absorb radiation even after attaining the temperature T. |
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Answer» Correct Answer - B::D Every object emits and absorbs the radiation simulaneously. If energy emitted is more than energy absorbed, temperature falls and vice versa. |
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| 268. |
A bullet of mass `20g` enters into a fixed wooden block with a speed of `40 ms^(-1)` and stope in it .Find the change in internal energy during the process |
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Answer» Correct Answer - A Since the bullet enters and stops insides the block , the total K.E.change by the bolled is responsible for the internal changes of the block Change in internal energy K.E. of bollet `= (1)/(2) xx (20)/(1000) xx 40 xx 40 = 16J` |
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| 269. |
A 2 g bullet moving with a velocity of 200 m/s is brought to a sudden stoppage by an obstacle. The total heat produced goes to the bullet. If the specific heat of the bullet is `0.03 cal//g^@C`, the rise in its temperature will beA. `158.0^@C`B. `15.80^@C`C. `1.58^@C`D. `0.1580^@C` |
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Answer» Correct Answer - A The kinetic energy of the bullet will be utilized to melt the bullet `(1)/(2)mv^2=(msDeltatheta)J` `(1)/(2)xx2xx10^-3xx(200)^2=2xx0.03xxDeltathetaxx4.2` `Deltatheta=158^@C` |
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| 270. |
During the melting of a slab of ice at 273K at atmospheric pressure,A. positive work is done by the ice water system on the atmosphereB. positive work is done on the ice water system by the atmosphereC. the internal energy of the ice water system increasesD. the internal energy of the ice water system decreases. |
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Answer» Correct Answer - B::C Melting involves an increase in potential energy and hence an increase in internal energy. |
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| 271. |
The temperature drop through a two layer furnace wall is `900^@C`. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature `theta` of the interface?A. By increasing the thermal conductivity of outer layerB. By increasing the thermal conductivity of inner layerC. By increasing thickness of outer layerD. By increasing thickness of inner layer |
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Answer» Correct Answer - A::D `H=`rate of heat flow `=(900)/((l_i)/(K_iA)+(l_0)/(K_0A))` Now `1000-theta=(Hl_i)/(K_iA)` or , `theta=1000[(900)/((l_i)/(K_iA)+(l_0)/(K_0A)](l_i)/(K_iA)` `=100-(900)/(1+(l_0)/(K_iA)(K)/(l_i))` Now, we can see that `theta` can be decreased by increasing thermal conductivity of outer layer `(K_0)` and thikcness of inner layer `(l_i)` |
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| 272. |
About `5g` of water at `30^(@)C` and `5g` of ice at `-20^(@)C` are mixed together in a calorimeter. Calculate final temperature of the mixture. Water equivalent of the calorimeter is negligible. Specific heat of ice `=0.5"cal"g^(-1)C^(@)-1)` and latent heat of ice `=80"cal"g^(-1)`A. `0^@C`B. `10^@C`C. `-30^@C`D. `gt10^@C` |
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Answer» Correct Answer - A Here ice will absorb heat while hot water will release it. So if T is the final temperature of the mixture heat given by water `Q_1=mcDeltaT=5xx1xx(30-T)` and heat absorbed by ice `Q_2=5xx((1)/(2))[0-(-20)]+5xx80+5xx1(T-0)` ltbr gt So, by principle of calorimetry `Q_1=Q_2` i.e., `150-5T=450+5T` `T=-30^@C` Which is impossible as a body cannot be cooled to a temperature below the temperature of cooling body. the physical reason for this descrepancy is the heat remaining after changing the temperature of ice from `-20^@C` to `0^@C` with some ice left unmelted and we are taking it for granted that heat is transferred from water at `0^@C` to ice at `0^@C` so that temperature of system drops below `0^@C`. However, as heat cannot flow from one body (water) to the other (ice) at same temperature `(0^@C)`, the temperature of system will not fall below `0^@C` |
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| 273. |
About `5g` of water at `30^(@)C` and `5g` of ice at `-20^(@)C` are mixed together in a calorimeter. Calculate final temperature of the mixture. Water equivalent of the calorimeter is negligible. Specific heat of ice `=0.5"cal"g^(-1)C^(@)-1)` and latent heat of ice `=80"cal"g^(-1)` |
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Answer» Correct Answer - `3.75 kg` |
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| 274. |
`M` gram of ice at `0^@ C` is mixed with `3 M` gram of water at `80^@ C` then the final temperature is.A. `30^@ C`B. `40^@C`C. `50^@ C`D. `60^@ C` |
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Answer» Correct Answer - B `80 M + M xx 1 xx (theta^@ - 0) = M xx 1 xx (80^@ - theta^@)`. |
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| 275. |
`10` grams of stream at `100^@ C` is mixed with `50 gm` of ice at `0^@ C` then final temperature isA. `20^@ C`B. `50^@ C`C. `40^@ C`D. `100^@ C` |
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Answer» Correct Answer - C Heat lost = Heat gained `m_(ice) L_(ice) + m_(ice)S_(w)(theta)= m_(steam) L_(steam) + m_(steam) S_(w) (100-theta)`. |
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| 276. |
Victoria falls in Africa is 122 m in height. Calculate the rise in temperature of the water if all the potential energy lost in the fall is converted into heat. |
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Answer» Consider mass m of water falling. `mgy=mcDeltat` We express both sides in joules by noting `c=1(kcal)/(kg)`,`K=(4184J)/(kgK)` Then `9.8(122)=4184Deltat` and `Deltat=0.29K` |
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| 277. |
Equal volumes of three liquids of densities `rho_1`,`rho_2` and `rho_3`, specific heat capacities `c_1,c_2` and `c_3` and temperatures `t_1,t_2` and `t_3`, respectively are mixed together. What is the temperature of the mixture? Assume no changes in volume on mixing. |
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Answer» Let t be the temperature of the mixture. Net loss of heat by all `=(vrho_1)c_1(t_1-t)+(Vrho_2)c_2(t_2-t)+(Vrho_3)c_3(t_3-t)` but net loss`=0` `Vrho_1c_1(t_1-t)+Vrho_2c_2(t_2-t)+Vrho_3c_3(t_3-t)=0` or `t=(rho_1c_1t_1+rho_2c_2t_2+rho_3c_3t_3)/(rho_1c_1+rho_2c_2+rho_3c_3)` |
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| 278. |
Three liquids with masses `m_1`,`m_2`,`m_3` are throughly mixed. If their specific heats are `c_1`,`c_2`,`c_3` and their temperature `T_1`,`T_2`,`T_3`, respectively, then the temperature of the mixture isA. `(c_1T_1+c_2T_2+c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)`B. `(m_1c_1 T_1+m_2c_2t_2+m_3c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)`C. `(m_1c_1T_1+m_2c_2T_2+m_3c_3T_3)/(m_1T_1+m_2T_2+m_3T_3)`D. `(m_1T_1+m_2T_2+m_3T_3)/(c_1T_1+c_2T_2+c_3T_3)` |
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Answer» Correct Answer - B Let the final temperature be `T^@C` Total heat supplied by the three liquids in coming down to `0^@C` `=m_1c_1T_1+m_2c_2T_2+m_3c_3T_3` ..(i) Total heat used by three liquids in raising temperature from `0^@C` to `T^@C` `=m_1c_1T+m_2c_2T+m_3c_3T` ..(ii) By equating Eqs. (i) and (ii) we get `(m_1c_1+m_2c_2+m_3c_3)T=m_1c_1T_1+m_2c_2T_2+m_3c_3T_3` `impliesT=(m_1c_1T_1+m_2c_2T_2+m_3c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)` |
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| 279. |
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)T`B. `(8//5)T`C. `(3//5)T`D. `(3//2)T` |
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Answer» Correct Answer - D Temperature of mixture is given by `T=(m_1c_1T_1+m_2c_2T_2)/(m_1c_1+m_2c_2)=(mc2T+(m)/(2)2cT)/(mc+(m)/(2)2c)=(3)/(2)T` |
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| 280. |
A thermocole cubical icebox of side 30 cm has a thickness of 5.0 cm if 4.0 kg of ice are put ini the box, estimate the amount of ice remaining after 6 h. The outside temperature is `45^@C` and coefficient of thermal conductivity of thermocole`=0.01 J//kg`. |
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Answer» Area of six faces of the box`=6l^2=6xx(0.30)^2=0.54m^2` and `L=5.0cm=0.05cm` time, `t=6hr=6xx3600s` `T_1-T_2=45-0=45^@C` Total heat entering into the box in 6 hr `Q=(KA(T_1-T_2))/(L)` `=(0.01xx0.54xx45xx6xx3600)/(0.05)=104976J` If m be the amount of ice melted then `Q=mL` or `m=(Q)/(L)=(104976)/(335xx10^3)=-.313kg` Mass of the ice left after `6h=(4-0.313)kg` `=3.687kg` |
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| 281. |
A cylinderical brass boiler of radius 15 cm and thickness 1.0 cm is filled with water and placed on an elerctric heater. If the If the water boils at the rate of `200 g//s`, estimate the temperature of the heater filament. Thermal conductivity of `brass=109 J//s//m^@C` and heat of vapourization of water `=2.256xx10^3 J//g`. |
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Answer» Rate of boiling of water `=200 g//s`. Since the heat of vapourization of water is `2.256xx10^3 J//g`, the amount of heat energy required to boil 200 g of water is `2.256xx10^3J//gxx200g=4.512xx10^5J` Since water is boiling at the rate of 200 g/s, the rate at which heat energy is supplied by the heater to water is `(Q)/(t)=4.512xx10^5J//s` Now, radius of the boiler `r=15cm=0.15m` `:.` Base area of the boiler `A=pir^2=3.142xx(0.15)^2=0.0707m^2` Thickness of brass `d=1.0cm=1.0xx10^-2m` Thermal conductivity of brass`k=109 J//s//m^(@)C` Temperature of boiling water `T_W=100^@C` If `T_f` is the temperature of the filament the, rate at which heat energy is transmitted through the base is given by `(Q)/(t)=(kA(T_f-100))/(1.0xx10^-2)=4.512xx10^5` or `T_f-100=585.5` or `T_f=685.5^@C` |
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| 282. |
A brass boiler has a base area of `0.15m^2` and thickness 1.0 cm it boils water at the rate of `6.0 kg//min`, When placed on a gas. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass`=109 J//s-.^@C` ) and heat of vapourization of water `=2256 J//g`. |
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Answer» Let `T_1` be the temperature of the part of the flame in contact with boiler. The amount of heat flows into water in 1 min `Q=KA(T_1-T_2)/(L)t` `=(109xx0.15xx(T_1-100)xx60)/(0.01)J` Mass of the water boiled in 1 min`=6kg=6000g` Heat required to boil the water `Q=mL=6000xx2256J` `(109xx0.15xx(T_1-100)xx60)/(0.01)=6000xx2256` or `T_1-100=138` or `T_1=238^@C` |
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| 283. |
Write down the approximate range of temperature at which water boils in a pressure cooker. |
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Answer» In a pressure cooker, the water boils at about 120oC to 125oC. |
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| 284. |
How is the boiling point of water affected when some salt is added to it? |
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Answer» The boiling point of water increases on adding salt. |
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| 285. |
How is the volume of water affected when it boils at 100℃? |
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Answer» Volume of water wills increases when it boils at 100oC. |
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| 286. |
The rectangular surface of area `8 cm xx 4 cm` of a black body at temperature `127^(@)C` emits energy `E` per section if length and breadth are reduced to half of the initial value and the temperature is raised to `327^(@)C`, the ratio of emission of energy becomesA. `(3)/(8)E`B. `(81)/(16)E`C. `(9)/(16)E`D. `(81)/(64)E` |
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Answer» Correct Answer - D Energy radiated by body per second `(Q)/(t)Asigma T^4` or `(Q)/(t)proplxxbxxT^4` (Area `= l xx b`) `:. (E_2)/(E_1)=(l_2)/(l_1)xx(b_2)/(b_1)xx((T_2)/(T_1))^4=((l_(1)//2))/(l_(1))xx((b_(1)//2))/(b_(1))xx((600)/(400))^4` `=(1)/(2)xx(1)/(2)xx((3)/(2))^4impliesE_2=(81)/(64)E` |
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| 287. |
One end of a rod of length L and crosssectional area A is kept in a furnace at temperature `T_(1)`. The other end of the rod is kept at a temperature `T_(2)`. The thermal conductivity of the matrieal of the rod is `K` and emissivity of the rod is e. It is gives that `T_(!)=T_(s)+DeltaT`, where `DeltaTltlt T_(s),T_(s)` is the temperature of the surroundings. If `DeltaTprop(T_(1)-T_(2))` find the proportional constant, consider that heat is lost only by rediation at the end where the temperature of the rod is `T_(1)`. |
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Answer» Rate of heat conduction through rod=rate of the heat lost from right end of the rod. `(KA(T_(1)-T_(2)))/(L)`=`eAsigma(T_(2)^(4)-T_(s)^(4))` (i) Given that `T_(1)`=`T_(s)+DeltaT` `T_(2)^(4)`=`(T_(s)+DeltaT)^(4)`=`T_(5)^(4)(1+(DeltaT)/(T_(s)))` Using binomial expansion, we have `T_(2)^(4)`=`T_(s)^(4)(1+4(DeltaT)/(T_(s)))` (as `DeltaTltlt T_(s))` `T_(2)^(4)`=`-T_(s)^(4)`=`4(DeltaT)(T_(s)^(3))` Subsituting in Eq. (i), we have `(K(T_(1)-T_(S)-DeltaT))/(L)`=`4esigma T_(s)^(3)DeltaT` `(K(T_(1)-T_(s)))/(L)`=`(4esigmaT_(s)^(3)+(K)/(L))DeltaT` `DeltaT`=`(K(T_(1)-T_(s)))/((4esigma LT_(s)^(3)+K))` Comparing with the given relation, proportionality constant =`(K)/(4esigmaLT_(s)^(3)+K)` |
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| 288. |
An iron tyre is to be fitted onto a wooden wheel 1.0 m in diameter. The diameter of the tyre is 6 mm smaller than that of wheel the tyre should be heated so that its temperature increases by a minimum of (coefficient of volume expansion of iron is `3.6xx10^-5//^@C`)A. `167^@C`B. `334^@C`C. `500^@C`D. `1000^@C` |
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Answer» Correct Answer - C Initial diameter of tyre`=(100-6)mm`,`=994mm`, so initial radius of tyre `R=(994)/(2)=497mm` and change in diameter `DeltaD=6mm`, so `DeltaR=(6)/(2)=3mm` Given that after increasing temperature by `DeltaT` tyre will fit onto wheel Increment in the length (circumference) of the iron tyre `DeltaL=LxxalphaxxDeltaT=Lxx(gamma)/(3)xxDeltaT` `(Asalpha=(gamma)/(3))` `implies2piR=2piR((gamma)/(3))DeltaT` `impliesDeltaT=(3)/(gamma)(DeltaR)/(R )=(3xx3)/(3.6xx10^-5xx497)` [As `Delta R=3` mm and `R = 497` mm] `impliesDelta=500^@C` |
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| 289. |
A clock with a metallic pendulum gains 5 s each day at a temperature of `15^@C` and loses 10 s each day at a temperature of `30^@C`. Find the coefficient of thermal expansion of the pendulum metal.A. `(1)/(86400)//^@C`B. `(1)/(43200)//^@C`C. `(1)/(14400)//^@C`D. `(1)/(28800)//^@C` |
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Answer» Correct Answer - A Loss of time due to heating a pendulum is given as `DeltaT=(1)/(2)alphaDeltathetaT` `implies12.5=(1)/(2)xxalphaxx(25-0)^@Cxx86400` `implies alpha=(1)/(86400)//^(@)C ` |
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| 290. |
A clock with a metallic pendulum gains 5 s each day at a temperature of `15^@C` and loses 10 s each day at a temperature of `30^@C`. Find the coefficient of thermal expansion of the pendulum metal. |
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Answer» Loss or gain per day `=dT=(1)/(2)alphadtxx86400` Since `T=86400s` for each day At `15^@C`,`5=(1)/(2)alpha(t-15)xx86400` At `30^@C`,`10=(1)/(2)alpha(30-t)xx86400` `(30-t)/(t-15)=2implies3t=60^@Cimpliest=20^@C` `alpha=(10)/((t-15)xx86400)=(10)/(5xx86400)=0.000023` `2.3xx(10^-5^@)/(C)=2` |
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| 291. |
A cylindrical container has a cross sectional area of `A_(0) = 1 cm^(2)` at `0^(@)C`. A scale has been marked on vertical surface of the container which shows correct reading at `0^(@)C`. A liquid is poured in the container. When the liquid and container is heated to `100^(@)C`, the scale shows the height of the liquid as 83.33 cm. The coefficient of volume expansion for the liquid is `gamma = 0.001^(@)C^(-1)` and the coefficient of linear expansion of the material of cylindrical container is `alpha = 0.0005^(@)C^(-1)`. A beaker has `300 cm^(3)` of same liquid at `0^(@)C`. The two liquids are mixed. Find the final temperature of the mixture assuming that heat exchange takes place between the liquids only, and its specific heat capacity is independent of temperature |
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Answer» Correct Answer - `25^(@)C` |
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| 292. |
The temperature of three different liquids `A,B` and `C` are `14^(@)C, 24^(@)C` and `34^(@)C` respectively. When equal masses of `A` and `B` are mixed, the temperature of the mixture is `20^(@)C`. When equal masses of `B` and `C` are mixed, the temperature of the mixture is `31^(@)C`. Supposing equal masses of `A` and `C` were mixed, what would be the temperature of the mixture? |
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Answer» Correct Answer - `29.6^(@)C` |
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| 293. |
The temperature of samples of three liquids A, B and C are `12^(@)C`, `19^(@)C` and `28^(@)C` respectively. The temperature when A and B are mixed is `16^(@)C` and when B and C are mixed it is `23^(@)C`. (i) What should be the temperature when A and C are mixed? (ii) What is final temperature if all the three liquids are mixed? Assume no heat loss to the surrounding. |
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Answer» Correct Answer - (i) `20.26^(@)C` , (ii) `19.76^(@)C` |
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| 294. |
The temperature of equal masses of three different liquied A ,B and C are `12^(@),18^(@),19^(@) and 28^(@)`respectively. The temperature when A and B are mixed is`16^(@)` and when Band C are mixed it is `23^(@)` what will be the temperaturewhen A and c are mixed? |
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Answer» Correct Answer - B::C Given , The temperature of `A = 12^(@)C` The temperature of `B = 19^(@)C` The temperature of `C= 28^(@)C` `rArr` The temperature of `A + B= 16^(@)C` `rArr`The temperature of `B + C = 23^(@)C` In accordance with the principal of calorimetry, when A and B are mixed `M_(CA) (16- 23) = M_(CB)(19 - 16)` `rArr 4M_(CA)=3M_(CB)` `rArr M_(CA)=((3)/(4))M_(CB)` and when B and C are mixed `M_(CB) (23 - 19) = M_(OC) (28 - 23)` `rArr 4M_(CB)=5M_(OC)` `rArr M_(OC)=((4)/(5))M_(CB)` When A and C mixed, If T is the common tempetature of the mixture `M_(CA) (T - 12) = M_(OC) (28 - T)` `rArr ((3)/(4))M_(CB)= (T - 12) = ((4)/(5))M_(CB)(28 - T)` `rArr ((3)/(4))(T - 12) = ((4)/(5))(28 - T)` `rArr (3 xx 5) (T - 12) = (4 xx 4) (28 - T)` `rArr 15T =T - 180= 448 - 16T` `rArr 31T = 628` `rArr T = (628)/(31) = 20.253^(@)C` `= 20.3^(@)C` |
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| 295. |
The ends of a metal rod are kept at temperatures `theta_1` and `theta_2` with `theta_2gttheta_1`. The rate of flow of heat along the rod is directly proportional toA. the length of the rodB. the diameter of the rodC. the cross sectional area of the rodD. the temperature difference `(theta_2-theta_1)` between the ends of the rod |
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Answer» Correct Answer - C::D The rate of flow of heat is given by `(Q)/(t)=(KA(theta_2-theta_1))/(l)` Where K is the thermal conductivity. A is the cross sectional area and l the length of the rod. Hence, the correct choices are (c ) and (d). |
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| 296. |
A glass vessel is filled up to 3/5th of its volume by mercury. If the volume expansivities of glass and mercury be `9xx 10^(-6)//^(@)C` and `18xx10^(-5)//^(@)C` respectively, then the coefficient of apparent expansion of mercury isA. `17.1xx10^(-5)//^(@)C`B. `9.9xx10^(-5)//^(@)C`C. `17.46xx10^(-5)//^(@)C`D. `16.5xx10^(-5)//^(@)C` |
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Answer» Correct Answer - D Let V be the volume of the glass vessel. Then volume of mercury will be `(3//(5)` V. Expansion of mercury`=(3//5)Vxxgamma_mDeltaT` And that of glass`=Vgamma_gDeltaT` Net (effective or apparent) expansion of mercury `DeltaV=V((3gamma_m)/(5)-gamma_g)DeltaT` `:.` Coefficient of apparent expansion will be `(DeltaV)/((3//5)VxxDeltaT)=((3)/(5)gamma_mgamma_g)(5)/(3)` |
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| 297. |
A vessel is partly filled with liquid. When the vessel is cooled to a lower temperature, the space in the vessel unoccupied by the liquid remains constant. Then the volume of the liquid `(V_L)` volume of the vessel `(V_V)` the coefficient of cubical expansion of the material of the vessel `(gamma_v)` and of the solid `(gamma_L)` are related asA. `gamma_L gt gamma_V`B. `gamma_L lt gamma_V`C. `(gamma_V)/(gamma_L)=(V_V)/(V_L)`D. `(gamma_V)/(gamma_L)=(V_L)/(V_V)` |
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Answer» Correct Answer - A::D `DeltaV_L=DeltaV_V` `Y_LV_L=Y_(V)V_V` or `(Y_L)/(T_V)=(V_V)/(V_L)` `V_VgtV_LimpliesT_LgtY_V` |
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| 298. |
Figure. Shows two air filled bulbs connected by a U-tube partly filled with alcohol. What happens to the levels of alcohol in the limbs X and Y when an electric bulb placed midway between the bulbs is lighted?A. The level of alcohol in limb X falls while that in limb Y risesB. The level of alcohol in limb X rises while that in limb Y fallsC. The level of alcohol falls in both limbsD. There is no change in the levels of alcohol in either of the two limbs |
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Answer» Correct Answer - A Black bulb absorbs more heat in comparison with painted bulb. So air in black bulb expands more. Hence the level of alcohol in limb X falls while that in limb Y rises. |
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| 299. |
The coefficient of apparent expansion of mercury in a glass vessel is `153xx10^(-6)//^(@)C` and in a steel vessel is `114xx10^(-6)//^(@)C`. If `alpha` for steel is `12xx10^(-6)//^(@)C`, then that of glass isA. `9xx10^(-6)//^(@)C`B. `6xx10^(-6)//^(@)C`C. `36xx10^(-6)//^(@)C`D. `27xx(10^(-6)//^(@)C` |
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Answer» Correct Answer - A `gamma_("real")=gamma_("app")+gamma_("vessel")` `implies153xx10^-6+(gamma_("vessel"))_("glass")=144xx10^-6+(gamma_("vessel"))_("steel")` Further `(gamma_("vessel"))_("steel")=3alpha=3xx(12xx10^-6)` `=36xx(10^(-6)//^@C` `153xx10^-6+(gamma_("vessel"))_("glass")=144xx10^-6+36xx10^-6` `(gamma_("vessel"))_("glass")=27xx10^(-6)//^@C` or `alpha=(gamma_("glass"))/(3)9xx10^(-6)//^@C` |
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| 300. |
A vessel is partly filled with a liquid. Coefficients of cubical expansion of material of the vessel and liquid are `gamma_v` unoccupied by the liquid will necessarilyA. remain unchanged if `gamma_v=gamma_L`B. increase if `gamma_v=gamma_L`C. decrease if `gamma_v=gamma_L`D. none of the above |
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Answer» Correct Answer - B Since the vessel is partly filled, volume of the vessel is greater that of the liquid. When a body having volume V is heated through `Deltatheta`, then increase in its volume is given by `DeltaV=VgammaDeltatheta` Since, `gamma_V=gamma_L`, therefore `DeltaVpropV`. Hence on heating expansion of vessel will be greater than that of liquid. It means unoccupied volume will necessarily increase. So, option (b) is correct. |
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