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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.
| 301. |
A piece of metal weighs 46 g in air and 30 g in lipuid of density `1.24 xx 10^3 kg m^(-3)` kept at `27^0C`. When the temperature of the liquid is raised to `42^0C` the metal piece weights 30.5 g . The density of the liqued at `42^0 C` is `1.20 xx 10^3 kg m^(-3)`. Calculate the coefficient of linear expandsion of the metal.A. `3.316 xx 10^(-5)//^(@)C`B. `2.316 xx 10^(-5)//^(@)C`C. `4.316 xx 10^(-5)//^(@)C`D. None of these |
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Answer» Correct Answer - B Loss of weight at `27^(@)C` is `=46-30 = 16 = V_(1) xx 1.24 rho_(l) xx g` ...(i) Loss of weight at `42^(@)C` is `=46 - 30.5 = 15.5 = V_(2) xx 1.2 rho_(l) xxg` ..(ii) Now dividing (i) by (ii) , we get , `(16)/(15.5) = (V_1)/(V_2) xx (1.24)/(1.2)` But `(V_2)/(V_1) = 1+ 3 alpha (t_(2)-t_(1))=(15.5xx1.24)/(16xx1.2) = 1.001042` `rArr 3 alpha(42^(@)-27^(@)) = 0.001042 rArr alpha = 2.316 xx 10^(-5)//^(@)C`. |
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| 302. |
Three liquids A, B and C having same sepcific heats have masses m,2m and 3m. Their temperaures are 0,20 and `3theta` respectively. ltBrgt Q. What is the temperature of mixture, when A and B are mixed?A. `(5)/(2)theta`B. `(5)/(3)theta`C. `(7)/(3)theta`D. `(13)/(5)theta` |
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Answer» Correct Answer - B |
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| 303. |
A "flow calorimeter" is used to measure the specific heat of a liquid. Heat is added at a known rate to a stream of the liquid as it passes through the calorimeter at known rate, then a measurement of the resulting temperature difference between the in flow and the out flow points of the liquid stream enables us to compute the specific heat of the liquid. A liquid of density 0.85 `g//cm^(3)` flows through a calorimeter at the rate of 8.0 `cm^(3)//s`. heat is added by means of a 250 watt electric heating coil, and a temperature difference of `15^(@)C` is established in steady state conditions between the in flow and out flow points. Q. Specific heat of the liquid isA. `0.59cal//g.^(@)C`B. `4cal//g.^(@)C`C. `2cal//g.^(@)C`D. `9cal//g.^(@)C` |
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Answer» Correct Answer - A |
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| 304. |
A "flow calorimeter" is used to measure the specific heat of a liquid. Heat is added at a known rate to a stream of the liquid as it passes through the calorimeter at known rate, then a measurement of the resulting temperature difference between the in flow and the out flow points of the liquid stream enables us to compute the specific heat of the liquid. A liquid of density 0.85 `g//cm^(3)` flows through a calorimeter at the rate of 8.0 `cm^(3)//s`. heat is added by means of a 250 watt electric heating coil, and a temperature difference of `15^(@)C` is established in steady state conditions between the in flow and out flow points. Q. Rate of heat absorption per unit specific heat capacity isA. 100g-`.^(@)C//s`B. 102g-`.^(@)C//s`C. 50g-`.^(@)C//s`D. `10.2g-.^(@)C//s` |
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Answer» Correct Answer - B |
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| 305. |
The water equivalent of a copper calorimeter is 4.5g. If the specific heat of copper is 0.09 cal `g^(-1).^(@)C^(-1)`. ThenA. mass of the calorimeter is 0.5 kgB. Thermal capacity of the calorimeter is 4.5 cal `cal^(@)C^(-1)`C. heat requried to raise the temperature of the calorimter by `8^(@)C` will be 36 calD. heat required to melt 15 gm of ice placed in the calorimeter in equilibrium with ice, will be 1200 cal. |
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Answer» Correct Answer - B::C::D |
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| 306. |
The thermal radiation from a hot body travels with a velocity ofA. `330 m s^(-1)`B. `2 xx 10^(8) m s^(-1)`C. `1200 m s^(-1)`D. `3 xx 10^(8) m s^(-1)` |
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Answer» Correct Answer - D The thermal radiation from a hot body travels with a velocity of light in vacuum i.e., `3 xx 10^(8) ms^(-1)`. |
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| 307. |
Two spheres of same material have radius 1 m and 4 m and temperature 4000 K and 2000 K respectively. The energy radiated per second by the first sphere isA. greater than that by the secondB. less than that by the secondC. equal in both casesD. the information is incomplete to draw any conclusion |
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Answer» Correct Answer - C The energy radiated per second E, by a body of surface area A, at temperature T K, is given by `E = esigma AT^(4)` Hence, `E_(1) = esigma4pi(1)^(2) xx (4000)^(4) = esigmapi xx 1024 xx 10^(12) J s^(-1)` and `E_(2) = esigma4pi(4)^(2)(2000)^(4) = esigmapi xx 1024 xx 10^(12) J s^(-1)` So, `E_(1) = E_(2)` |
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| 308. |
In figure which strip brass or steel have higher coefficient of linear expansion. A. brass stripB. steel stripC. both strips has same coefficient of linear expansionD. cannot be decided from given data |
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Answer» Correct Answer - A As the coeffiecient of thermal expansion of the brass is greater than steel. Hence, the length of brass strip will be more than steel strip. Therefore, brass strip will be on convex side. |
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| 309. |
The length of each steel rail is `10 m` in winter. The coefficient of linear expansion of steel in `0.000012//^(@)C` in summer. The gap to be left between the railsA. `0.0018m`B. `0.00120m`C. `0.0022m`D. `0.05m` |
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Answer» Correct Answer - A `l_(2) - l_(1) = l_(1)alpha(t_(2) - t_(1))` |
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| 310. |
The coefficient of linear expansion for a certain metal varies with temperature as `alpha(T)`. If `L_0` is the initial length of the metal and the temperature of metal is changed from `T_0` to `T(T_0gtT)`, thenA. `L=L_(0)int_(T_(0))^(T)alpha(T)dT`B. `L=L_(0)[1+int_(T_(0))^(T)alpha(T)]dT`C. `L=L_(0)[1+underset(T_(0))overset(T)(int)alpha(T)dT]`D. `L gtL_(0)` |
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Answer» Correct Answer - C `(dL)/(L_(0)) = -alpha(T)dt,` integrate the equation |
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| 311. |
A steel teape is placed around the earth at the equator. When the temperature is `0^(0)C` neglecting the expansion of the earth, the neglecting between the tape and the ground if the temperature of the tape rises to `30^(0)C`, is nearly `(alpha_(steel) = 11 xx 10^(-6) // K)`A. `1.1 km`B. `0.5 km`C. `6400 km`D. `2.1 km` |
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Answer» Correct Answer - D Increase in length of tape: `Deltal = lalphaDeltaT` |
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| 312. |
The ratio of lengths of two rods is `1 : 2` and the ratio of coefficient of expansions is `2 : 3`. The first rod is heated through `60^(@)C`. Find the temperautre through `60^(@)C`. Find the temperature through `60^(@)C`. Find the temperature through which the second rod is to be heated so that its expansion is twice that of first isA. `60^(@)C`B. `40^(@)C`C. `30^(@)C`D. `10^(@)C` |
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Answer» Correct Answer - B `Deltal = alphalDeltat rArr (Deltat_(1))/(Deltat_(2)) = (alpha_(2))/(alpha_(1)) xx (l_(2))/(l_(1))` |
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| 313. |
if `alpha_(c)` and `alpha_(f)` denote the numerical values of coefficient of linear expansions of the solid, expressed per ``^(0)C` and per Kelvin respectibely, then.A. `alpha_(c) gt alpha_(k)`B. `alpha_(c) lt alpha_(k)`C. `alpha_(c) = alpha_(k)`D. `alpha_(c) = 2alpha_(k)` |
| Answer» Correct Answer - C | |
| 314. |
Two substance of same size are made of same material but one is hollow and the other is solid. They are heated to same temperature, thenA. both spheres will expand equallyB. hollow sphere will expand more than solid oneC. solid sphere will expand more than hollow oneD. hollow sphere will expand double that of solid one |
| Answer» Correct Answer - A | |
| 315. |
A solid sphere and a hollow sphere of same material have same mass. When they are heated by` 50^(@)C`, increase in volume of solide sphere is `5 c.c.` the expansion of hollow sphere isA. `5 c.c`B. more than `5 c.c.`C. Less than `5 c.c.`D. None |
| Answer» Correct Answer - B | |
| 316. |
When a rod is heated, its linear expansions depends on (a) initial length (b) area of cross section (c) mass (d) temperature riseA. only a is correctB. a & d are correctC. b & c are correctD. a & c are correct |
| Answer» Correct Answer - B | |
| 317. |
Two ideal gas thermometer `A` and `B` use oxygen and hydrogen respectively . The following observations are made: Temperature,Pressure therometer A,Pressure therometer B Triple point of water,`1.250xx10^(5)Pa`,`0.200xx10^(5)Pa` Normal melting point of sulphur, ` 1.797xx10^(5)Pa`,`0.287xx10^(5)Pa` (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers `A` and `B` ? (b) What do you think is the reason for the slightly different answers from `A` and `B` ? (The thermometers are not faulty). what further procedure is needed in the experiment to reduce the discrepancy between the two readings. |
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Answer» (a) Triple point of water, T = 273.16 K. At this temperature, pressure in thermometer `A, P_(A) = 1.250 xx 10^(5) `Pa Let `T_(1)` be the normal melting point of sulphur. At this temperature, pressure in thermometer` A, P_(1) = 1.797x× 10^(5)` Pa According to Charles’ law, we have the relation: `(P_(A))/T=(P_(1))/(T_(1))` `:.T_(1)=(P_(1)T)/(P_(A))=(1.797xx10^(5)xx273.16)/(1.250xx10^(5))` =392.69 K Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K. At triple point 273.16 K, the pressure in thermometer B, `P_(B) = 0.200 ×x 10^(5)` Pa At temperature T1, the pressure in thermometer `B, P_(2) = 0.287 x× 10^(5)` Pa According to Charles’ law, we can write the relation: `(P_(B))/T=(P_(1))/(T_(1))` `(0.200xx10^(5))/(273.16)=(0.287xx10^(5))/(T_(1))` `:. T_(1)=(0.287xx10^(5))/(0.200xx10^(5))xx273.16=391.98 K` Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K. (b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B. To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases. |
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| 318. |
Answer the following :(a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ? (b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number `0^(@)C" and "100^(@)C` respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ? (c ) The absolute temperature (Kelvin scale) T is related to the temperature `t_(c)` on the Celsius scale by `t_(c)=T-273.15` Why do we have 273.15 in this relation, and not 273.16 ? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ? |
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Answer» (a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature. (b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale. (c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature `0^(@)C` on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K. Hence, absolute temperature (Kelvin scale) T, is related to temperature `t_(c)`, on Celsius scale as: ` t_(c)=T-273.15` (d) Let `T_(F)` be the temperature on Fahrenheit scale and `T_(K)` be the temperature on absolute scale. Both the temperatures can be related as: `(T_(F)-32)/(180)=(T_(K)-273.15)/(100)`...(i) Let` T_(F1)` be the temperature on Fahrenheit scale and `T_(K1)` be the temperature on absolute scale. Both the temperatures can be related as: `(T_(F1)-32)/(180)=T_(K1)-273.15)/(100)`...(ii) It is given that: `T_(K1)-T_(K)=1K` Subtracting equation (i) from equation (ii), we get: `(T_(F1)-T_(F))/(180)=(T_(K1)-T_(K))/(100)=1/(100)` `T_(F1)-T_(F)=(1xx180)/(100)=9/5` Triple point of water = 273.16 K `:.`Triple point of water on absolute scale`=273.16xx9/(5=491.69)` |
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| 319. |
As part of an experiment to determine the latent heat of vaporisation of water, a student boils some water in a beaker using an electric heater as shown below. The student notes two sources of error. Error-1: thermal energy is lost from the sides of the beaker Error -2: As the water is boling water splashes out of the beaker Which of the following gives the correct effect of these two errors on the calculated value for the specific lalent heat?A. Error-1 = increases , Error-2 = DecreaseB. Error-1 = increases , Error-2 = increaseC. Error-1 = Decreases , Error-2 = increaseD. Error-1 = Decreases , Error-2 = Decrease |
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Answer» Correct Answer - A (i) some energy is lost through sides of the beaker. (ii) Water Splahes, so mass of water becomes less. |
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| 320. |
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 `E = sigma xx` area pf `T^(4)` , `T` increase by a factor `3/2` Area increase by a factor `1/4`. |
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| 321. |
The volume of a block of a metal changes by `0.12%` when it is heated through `20^(@)C` . The coefficient of linear expansion of the metal isA. `4 xx 10^(-5)"^(@) C^(-1)`B. `2 xx 10^(-5)"^(@) C^(-1)`C. `0.5 xx 10^(-5)"^(@) C^(-1)`D. `4 xx 10^(-4)"^(@) C^(-1)` |
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Answer» Correct Answer - B Here, `(DeltaV)/(V) = 0.12%, = (0.12)/(100)` `DeltaT = 20 .^(@)C` Now, `gamma = (DeltaV)/(VDeltaT) = (0.12)/(100 xx 20) = 6 xx 10^(-5) .^(@)C^(-1)` `:. alpha = (gamma)/(3) = 2 xx 10^(-5) .^(@)C^(-1)` |
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| 322. |
The coefficient of volume expansion of glycerine is `49 xx 10^(-5)//^(@)C`. What is the fractional change in its density (approx.) for `30^(@)C` rise in temperature?A. `1.5 xx 10^(-2)`B. `2.5 xx 10^(-2)`C. `2.0 xx 10^(-2)`D. `2.8 xx 10^(-2)` |
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Answer» Correct Answer - A Coefficient of volume expansion of glycerine, `gamma = 49xx10^(-5)//^(@)C` Rise in temperature, `Delta T = 30^(@)C` As, `gamma = 1/(Delta T) ((Delta V)/(V)) rArr (Delta V)/(V) = gamma xx Delta T` or `(Delta V)/(V) = (49xx10^(-5))xx30 = 0.0147` Since mass remains constant, Fractional change in density = fractional change in volume = `0.0147 = 1.5 xx 10^(-2)`. |
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| 323. |
Two gases A and B having the same temperature T, same pressure P and same volume V are mixed. If the mixture is at the same temperature and occupies a volume V. The pressure of the mixture isA. `2P`B. `P`C. `P//2`D. `4P` |
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Answer» Correct Answer - A `(PV)/(T) = (P_(1)V_(1))/(T_(1)) + (P_(2)V_(2))/(T_(2)) , (P^(1)V)/(T) = (2PV)/(T)` |
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| 324. |
The height of the mercury column in a barometer provided with a brass scale corrercted at `0^(0)C` is observed to be `74.9 cm` at `15^(0)C`. Find the true height of the column at `0^(0)C`. `alpha_(b) = 20 xx 10^(-6 //0)C` and `gamma_(Hg) = 175 xx 10^(-6 //^(0)C)`A. `74.82cm`B. `79.92cm`C. `74.12cm`D. `72.64 cm` |
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Answer» Correct Answer - A `H_(0)rho_(0)g = H_(t)rho_(t)g , H_(0)rho_(0) = H(1 + alphat)(rho_(0))/((1 + gammat))` `:. H_(0) = H[1 - (gamma - alpha)t]` |
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| 325. |
A steel scale measures the length of a copper wire as `80.0 cm` when both area at `20^(@)C` (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at `40^(@)C`? (Given `alpha_("steel") = 11 xx 10^(-6) per ^(@)C` and `alpha_("copper") = 17 xx 10^(-6) per^(@)C`)A. `80.1126 cm`B. `80.2136 cm`C. `80.0096 cm`D. `80.1006 cm` |
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Answer» Correct Answer - C 1cm length of steel scale at `40^(@)C` `=1+1 xx (11 xx 10^(-6)) xx 940-20) == 1.00022 cm` Length of copper rod at `40^(@)C` `=80+(80xx17xx10^(-6))xx(40-20)=80.0272 cm` Number of division on the steel scale `=(80.0272)/(1.00022) = 80.0096` Length of the rod = `80.0096 cm`. |
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| 326. |
When a metasl sphere is heated maximum percentage increase occurs in itsA. density and moment of inertia increaseB. surface areaC. radiusD. volume |
| Answer» Correct Answer - D | |
| 327. |
A metal rod lenth `L_(0)` whose coefficient of linear expansion `alpha = 10^(-3) ^(0)C^(-1)` is heated such that its temperature changes by `1000K` assum-ing `alpha` is constant during the temperature change `(e =2.7)`A. Final length of the rod is greater than `2L_(0)`B. Final length of the rod is greater than `2.5L_(0)`C. Final length of the rod is greater than `3L_(0)`D. increase in length of rod is `L_(0)` |
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Answer» Correct Answer - A::B `dL = alphaLdt` `int_(L_(0))^(L)(dL)/(L)=int_(0)^(T)alphadt` `L = L_(0)e^(alphaT) = 2.7L_(0)` |
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| 328. |
The temperature gradient in a rod of 0.5 m long is `80^(@)C//m`. If the temperature of hotter end of the rod is `30^(@)C`, then the temperature of the colder end isA. `40^(@)C`B. `-10^(@)C`C. `10^(@)C`D. `0^(@)C` |
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Answer» Correct Answer - B |
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| 329. |
Assertion : It is hotter over the top of a fire than at the same distacne of the side. Reason : Air surrounding the fire conducts more heat upwardA. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A It is hotter over the top of a fire than at the same distance on the sides, because not air i.e., thermal radiation being light moves up. |
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| 330. |
Assertion : Temperature near the sea-coast are moderate. Reason : Water has a high thermal conductivity.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Temperature near seacoast are moderate becacuse thermal conduction never takes place at sea-cost causing moderature temperature. Water has high thermal condcutivity. |
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| 331. |
A cylindrical rod having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1) cal//sec`. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat `Q_(2)` will beA. `4 Q_(1)`B. `2 Q_(1)`C. `Q_(1)/4`D. `Q_(1)/2` |
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Answer» Correct Answer - B Rate of heat flow `(Q/t) = (kpir^(2)(theta_(1)-theta_(2)))/(L) prop (r^2)/(L)` `:. (Q_1)/(Q_2) = ((r_1)/(r_2))^(2)((l_2)/(l_1)) = (1/2)^(2)xx(2/1) = 1/2` `rArr Q_(2) = 2Q_(1)`. |
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| 332. |
The coefficient of thermal conductivity of copper, mercury and glass are respectively `K_(c), K_(m)` and `K_(g)` that `K_(c) gt K_(m) gt K_(g)`. If the same quantity of heat is to flow per second per unit of each and corresponding temperature gradients are `X_(c), X_(m)` and `X_(g)`, thenA. `X_(c) = X_(m) = X_(g)`B. `X_(c) lt X_(m) gt X_(g)`C. `X_(c) lt X_(m) lt X_(g)`D. `X_(m) lt X_(c) lt X_(g)` |
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Answer» Correct Answer - C `(Q)/(At) = K (Delta theta)/(l) rArr K(Delta theta)/(l) =` consta nt `rArr (Delta theta)/(l) prop (l)/(K)` Hence, if `K_(c) gt K_(m) gt K_(g)`, then `((Delta theta)/(l))_(c) lt ((Delta theta)/(l))_(m) lt ((Delta theta)/(l))_(g) rArr X_(c) lt X_(m) lt X_(g)` Because higher `K` implies lower value of the temerature gradient. |
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| 333. |
Ice has formed on a shallow pond, and a steady state has been reached, with the air above the ice at `-5.0^(@)C` and the bottom of the pond at `40^(@)C`. If the total depth of ice + water is `1.4 m`, (Assume that the thermal conductivities of ice and water are `0.40` and `0.12 cal//m C^(@)s`, respectively.) The thickness of ice layer isA. `1.1 m`B. `0.4 m`C. `2.1 m`D. `3.6 m` |
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Answer» Correct Answer - A The top surface of the ice is at `t_(C) =-5.0^(@)C` and teh bottom of the body of water is at `T_(H) = 4.0^(@)C` and the interface between the ice and the water is at `T_(X) = 0.0^(@)C`. The rate of heat flow through both layers should be equal `(k_("water")A(T_(H)-T_(X)))/(L-L_("ice")) = (k_(ice)A(T_(X)-T_(C)))/(L_("ice"))` We cancel the area `A` and solve for thickness of the ice layer:`L_("ice") = 1.1 m`, |
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| 334. |
The coefficient of thermal conductivity of copper, mercury and glass are respectively `K_(c), K_(m)` and `K_(g)` that `K_(c) gt K_(m) gt K_(g)`. If the same quantity of heat is to flow per second per unit of each and corresponding temperature gradients are `X_(c), X_(m)` and `X_(g)`, thenA. `X_(c) = X_(m) = X_(g)`B. `X_(c) gt X_(m) gt X_(g)`C. `X_(c) lt X_(m) lt X_(g)`D. `X_(m) lt X_(c) lt X_(g)` |
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Answer» Correct Answer - C use `(d theta)/(dt) prop 1/K`. |
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| 335. |
The temperature coefficient of resistance of a wire is `0.00125` per `""^(@)C`. At 300 K, its resistance is `1Omega`. The resistance of the wire will be `2Omega` atA. `1154 K`B. `1100 K`C. `1400 K`D. `1127 K` |
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Answer» Correct Answer - C `alpha = (R_(2) - R_(1))/(R_(1)t_(2) - R_(2)t_(1))` |
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