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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.
| 51. |
What is the change in potential energy (in calories) of a `10 kg` mass after `41.8 m` fall?A. `1000 cal`B. `800 cal`C. `1500 cal`D. `2000 cal` |
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Answer» Correct Answer - A Change in potential energy `Delta U = mgh = 10xx10xx41.8=4180J` If mechanical work//energy `(W)` in joule produces the same temperature as heat `(H)` , we write, `W=JH` where `J=4.18` is called mechanical equivalent of heat. `J` is expressed in joule//calorie. Hence equivalent calories will be `H=(Delta U)/(J) = (4180)/(4.18) = 1000 cal`. |
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| 52. |
The graph between two temperature scales `A` and `B` is shown in Fig. Between upper fixed point and lower fixed point there are `150` equal divisions on scales `A` and `100` on scale `B`. The relation between the temperature in two scales is given by_ |
| Answer» When `t_(B) = 0, t_(A) = 30^(@) :. (t_(A) - 30)/(150) = (t_(B) - 0)/(100)` | |
| 53. |
How much work can be done by `100` calories of heat?A. 100 JB. 420 JC. 42 JD. 4200 J |
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Answer» Correct Answer - B `1` calorie = 4.2 j `:.` Work done = obtained in clories `xx 4.2 j` or `W = 100 xx 4.2 = 420 j` |
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| 54. |
An accurate Celsius rhwemometer and a faulty Fahrenheit thermometer register `60^(@)` and `141^(@)` respectively when placed in the same constant temperature enclosure. What is the error in the Fahrenheit thermometer ? |
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Answer» From `(C)/(5) = (F - 32)/(9) rArr (60)/(5) = (F - 32)/(9) rArr F = 140^(0)F` Error = 140 - 140 = `1^(0)F`, Correction `= -1^(0)F` |
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| 55. |
An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature asA. B. C. D. |
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Answer» Correct Answer - C `(DeltaV)/(V) = (DeltaT)/(T) :. Lambda(1)/(L)` |
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| 56. |
Two identical long, solid cylinders are used to conduct heat from temp `T_(1)` to Temp `T_(2)`. Originally the cylinders are connected in series and the rate of heat transfer is H. If the cylinders are connected in parallel then the rate of heat transer would be:A. `H//4`B. `2H`C. `4H`D. `8H` |
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Answer» Correct Answer - C Initially effective resistance = `2R` . In parallel effective resistance =`R/2` . It has reduced by a factor of `1//4` so rate of heat transfer would be increased by a factor of 4, keeping other parameters same. |
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| 57. |
The graph shown gives the temperature along an `X` axis that extends directly through a wall consisting of three layers `A`, `B` and `C`. The air temperature on one side of the wall is `150^(@)C` and on the other side is `80^(@)C`. Thermal conduction through the wall is steady. out of the three layers `A` `B` and `C` thermal conductivity is greatest of the layer A. `A`B. `B`C. `C`D. Thermal conductivity of A = Thermal conductivity of B |
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Answer» Correct Answer - A Rate of heat transfer is same through all walls `(K_(1)*A*(10))/(40 cm) = (K_(2)A(20))/(20cm) = (K_(2)A(40))/(10 cm)` `rArr (K_1)/(4) = K_(2)=4K_(3) rArr K_(1) = 4K_(2) = 16K_(3)`. |
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| 58. |
The volume of an air bubble increases by `x%` as it raises from the bottom of a water lake to its surface. If the water barometer reads `H`, the depth of the lake is |
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Answer» `P_(1)V_(1) = P_(2)V_(2), (H + h) dgV_(1) = HdgV^(2)` `(H + h)V_(1) = HV_(2) , H + h = (HV_(2))/(V_(1))` `h = H((V_(2) - V_(1))/(V_(1))), h = H((DeltaV)/(V))` Here `(DeltaV)/(V) xx 100 = (x)/(100) :. h = (Hx)/(100)` |
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| 59. |
A vessel is filled with an ideal gas at a pressure of `10` atmospheres and temp `27^(0)C`. Half of the mass of the gas is removed from the vessel the temperature of the remaining gas is increased to `87^(0)C`. Then the pressure of the gas in the vessel will beA. `5 atm`B. `18 atm`C. `7 atm`D. `8 atm` |
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Answer» Correct Answer - B `P prop mT , (P_(1))/(P_(2)) = (m_(1))/(m_(2)) xx (T_(1))/(T_(2))` |
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| 60. |
An air bubble starts rising from the bottom of a lake. Its diameter is ` 3.6 mm` at the bottom and `4 mm` at the surface. The depth of the lake is `250 cm` and the temperature at the surface is `40^@ C`. What is the temperature at the bottom of the lake? Given atmospheric pressure = `76 cm of Hg and g = 980 cm//s^2`. |
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Answer» At the bottom of the lake, volume of the bubble `V_(1) = (4)/(3)pi (0.18)^(3)cm^(3)` Pressure on the bubble `P_(1) =` Atmospheric pressure `+` Pressure due to a column of `250cm` of water `= (76 xx 13.6 + 250)980 "dyne"//cm^(2)` At the surface of the lake, volume of the bubble `V_(2) = (4)/(3)pi(0.2)^(3)cm^(3)` Pressure on the bubble , `P_(2)` = atm. pressure `= (76 xx 13.6 xx 980)"dyne"//cm^(2)` `T_(2) = 273 + 40^(0)C = 313 K` Now `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` `rArr ((76 xx 13.6 + 250)680xx ((4)/(3))pi(0.18)^(3))/(T_(1))` `= ((76 xx 13.6 + 250)680xx ((4)/(3))pi(0.2)^(3))/(313)` or `T_(1) = 283.37 K , T_(1) = 283.37 - 273 = 10.37^(0)C` |
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| 61. |
The gas in a vessel is subjected to a pressure of `20` atmosphere at a temperature `27^(@)C`. The pressure of the gas in the vessel after one half of the gas is released from the vessel and the temperature of the remainder is raised by `50^(@)C` isA. `8.5 atm`B. `11.7 atm`C. `17 atm`D. `10.8 atm` |
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Answer» Correct Answer - B `PV = (m)/(M)RT, P prop mT , (P_(1))/(P_(2)) = (m_(1))/(m_(2)) xx (T_(1))/(T_(2))` |
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| 62. |
At the bottom of a lake where temperature is `7^(0)C` the pressure is `2.8` atmosphere. An air bubble of radius `1 cm` at the bottom rises to the surface. Where the temperature is `27^(0)C`. Radius of air bubble at the surface isA. `3^(1//3)`B. `4^(1//3)`C. `5^(1//3)`D. `6^(1//3)` |
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Answer» Correct Answer - A `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` |
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| 63. |
At what temperature is the Fahrenheit scale reading equal to (a) twice (b) half of Celsius ?A. `40^(@)C`B. `20^(@)C`C. `160^(@)C`D. `80^(@)C` |
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Answer» Correct Answer - C `((F - 32)/(18) = (C - 0)/(100)) , F = 2C` |
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| 64. |
A compound slab is made of two parallel plates of copper and brass of the same thickness and having thermal conductivities in the ratio `4:1`. The free face of copper is at `0^(@)C`. The temperature of the internal is `20^(@)C`. What is the temperature of the free face of brass?A. `0^(@)C`B. `20^(@)C`C. `40^(@)C`D. `100^(@)C` |
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Answer» Correct Answer - D `(Q)/(t) = (4KA(20-0))/(d) = (KA(theta-20))/(d)` or `80=theta-20 or theta = 100^(@)C`. |
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| 65. |
A wall has two layers A and B each made of different materials. The thickness of both the layers is the same. The thermal conductivity of A, `K_(A) = 3K_(B)`. The temperature different across the wall is `20^(@)C` in thermal equilibriumA. The temperature difference across A is `15^(@)C`.B. Rate of heat transfer across A is more than across BC. Rate of heat transfer across both is same.D. Temperature difference aross A is `5^(@)C` |
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Answer» Correct Answer - D `(Q)/(At) =(K(theta_(1)-theta_(2)))/(d)`= constant `:. K_(A)((theta_(1)=theta)/(d)) = K_(B)((theta-theta_(2))/(d))` `(K_A)/(K_B) = (theta-theta_(2))/(theta_(1)-theta) or 3 = (theta-theta_(2))/(theta_(1)=theta)` or, `3theta_(1)+theta_(2) = 4 theta` ...(1) Given `theta_(1)-theta_(2) = 4 theta` ..(2) Solving (1) and (2) we have, `theta-theta_(2)=15^(@)C` `:. theta_(1) - theta = theta_(1)-theta_(2)+theta_(2) - theta` `=(theta_(1)-theta_(2)) - (theta-theta_(2))` `=20^(@)C - 15^(@)C = 5^(@)C`. |
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| 66. |
Figure shows in cross section a wall consisting of four layers with thermal conductivities `K_(1) = 0.06 W // mK, K_(3) = 0.04 W// mK and K_(4) = 0.10 W// mK`. The layer thickness are `L_(1) = 1.5 cm, L_(3) = 2.8 cm and L_(4) = 3.5 cm`. The temperature of interfaces is as shown in figure. energy transfer through the wall is in steady state. the temperature of the interface between layer `3 and 4` is: A. `-1^(@)C`B. `-3^(@)C`C. `2^(@)C`D. `0^(@)C` |
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Answer» Correct Answer - B In steady state `(Delta Q)/(DeltaT)|_("layer" 1)` = `(Delta Q)/(Delta t)|_("layer" 4)` `rArr (0.06xxAxx(30-25))/(1.5xx10^(-2)) = (0.10xxAxxDelta T)/(3.5xx10^(-2))` `rArr Delta T = 7^(@)C` `T_(3)=(-10+7)^(@)C = -(3)^(@)C`. |
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| 67. |
A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities `K_(1) and K_(2)`. The equivalent conductivity of the combination isA. `K_(1)+K_(2)`B. `(K_(1)+K_(2))/(2)`C. `(2K_(1)K_(2))/(K_(1)+K_(2))`D. `(K_(1)+K_(2))/(2K_(1)K_(2))` |
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Answer» Correct Answer - B |
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| 68. |
Assertion : A change in the temperature of a body cause change in dimentions. Reason : The dimentions of a body decrease due to the increase in its temperature.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is true and reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C Substance expand on heating and contract on cooling. |
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| 69. |
If a graph is plotted taking the temperature in Fahrenheit along the `Y`-axis and the corresponding temperature in Celsius along the `X`-axis, it will be a straight lineA. `5//9`B. `9//5`C. `4//5`D. `5//4` |
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Answer» Correct Answer - B `DeltaF = (9)/(5)DeltaC` |
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| 70. |
A wall has two layers A and B each made of different materials. The layer A is 10cm thick and B is 20 cm thick. The thermal conductivity of A is thrice that of B. Under thermal equilibrium temperature difference across the wall is `35^@C`. The difference of temperature across the layer A isA. `28^(@)C`B. `14^(@)C`C. `7^(@)C`D. `5^(@)C` |
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Answer» Correct Answer - D In the state, rate of heat across each layer of the wall is the same. `(K_(A)A(DeltaT))/(10) = (K_(B)A(35-DeltaT))/(20)` Using `K_(A) = 3K_(B)`, solving , `Delta T = 5^(@)C` |
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| 71. |
The Kelvin scale temperature is 0 K. What is the corresponding Celsius scale temperature ?A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is true and reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - B | |
| 72. |
Assertion : While measuring the thermal conductivity of liquid experimentally, the upper layer is kept hot and the lower layer kept cold. Reason : This avoids heating of liquid by convection.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 - A We know that to measure thermal conductivity of liquids experimentally they must be heated from the top i.e. Upper layer is kipt hot cold, so as to prevent convection in liquids. |
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| 73. |
Assertion : The density of water remains constant as it is cooled from room temperature untill its temperature reached `4^(@)C`. Reason : Below `4^(@)C`, the density increases.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is true and reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D The volume of water decreases when water is cooled from room temperature, until its temperature reaches `4^(@)C` and hence density increases. Below `4^(@)C`, the volume increases and therefore, the density decreases. |
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| 74. |
Assertion : In change of state from solid to liquid the temperature decreases untill the entire amount of the solid subtance melts. Reason : The phenomenon of refreezing is called meltingA. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is true and reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D Both the solid and liquid states of the substances coexist in thermal equilibrium during the change of state from solid to liquid. The phenomenon of refreezing is called regelation. |
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| 75. |
Two circular disc `A` and `B` with equal radii are blackened. They are heated to same temperature and are cooled under identical conditions. What inference do your draw from their cooling curves? A. A and B specific heatB. Specific heat of A is lessC. Specific heat of B is lessD. Nothing can be said |
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Answer» Correct Answer - B When a body cools by radiation, the rate of cooling is given by: `(d theta)/(dt) = (eAsigma)/(ms) (theta^(4)-theta_(0)^(4))` `-ve` sign shown that temperature decreases, i.e., the body cools, is the specific heat of materail and `theta_(0)` is teh surrounding temperature or, `(d theta)/(dt) prop 1/s` i.e., rate of cooling `(R = d theta//dt)` is inversely proportional to the specific heat of material. for a, rate of cooling is large, therefore, specific heat of `A` is smaller. |
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| 76. |
Assertion : The density of water remains constant as it is cooled from room temperature untill its temperature reached `4^(@)C`. Reason : Below `4^(@)C`, the density increases.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 - D The volume of water decreases when water is coolied from temperature, until its temperature reaches `4^(@)C` and hance density increases. Below `4^(@)C` the volume icreases and therefire, the density decreases. |
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| 77. |
When a metal bar is cooled, then which one of these staments is correct.A. Length, density and mass remain same.B. Length decreases, density increases but mass remains sameC. Length and mass decrease but density remains the same.D. Length and density decrease but mass reamains the same. |
| Answer» Correct Answer - B | |
| 78. |
When a metal bar is heated, the increase in length is greater, ifA. the bar has large diameterB. The bar is long.C. the temperature rise smallD. Small diameter |
| Answer» Correct Answer - B | |
| 79. |
A sphere, a cube and a thin circular plate, all of same material and same mass are initially heated to same high temperture.A. Plate will cool fastet and cube the slowest.B. Sphere will cool fastest and cube the slowest.C. Plate will cool fastest and sphere the slowest.D. Cube will cool fastest and plate the slowest. |
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Answer» Correct Answer - C The circular plate has maximum surface area and as such it cools the fastest. For a given volume (mass in this case), a sphere has the least surface area and accordingly cools the slowest. |
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| 80. |
The substance which ha negative coefficient of linear expension isA. leadB. aluminumC. ironD. invar steel |
| Answer» Correct Answer - A | |
| 81. |
A peiece of metal weight `45g` in air and `25g` `30^(@)C`. When the temperature of the liquide raised to `40^(@)` |
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Answer» `M_(1) =` mass of the metal piece in air `M_(2) =` mass of the metal piece in liquid at `30^(@)C` `M_(3) =` Mass of the metal piece in liequid at `40^(@)C` `rho_(1) =` density of liquid at `30^(@)C` `rho_(2) =` density of liquid at `40^(@)C` Mass expelled `Deltam_(1) = M_(1) - M_(2) = V_(1)rho_(1)` at `30^(@)C` `Deltam_(2) = M_(1) - M_(3) = V_(2)rho_(2)` at `40^(0)C, (20)/(18) = (V_(1)rho_(1))/(V_(2)rho_(2))` `(10)/(9) = (V_(1) xx 1.5 xx 10^(3))/(V_(1)(1 + gammaDeltat) xx 1.25 xx 10^(3)), 1 + gammaDeltat = (9 xx 1.2)/(10)` `= (10.8)/(10) = 1.08, gammaDeltat = 1.08 - 1, 3alpha = 0.08` `alpha = (0.08)/(3) = :. alpha = 2.6 xx 10^(-3//@)C` |
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| 82. |
Density of gold is `19.30 g//cm^(3)` at `20^(0)C`. Computer the density of gold at `90^(0)C` by adding steam to it. `(alpha = 14.2 xx 10^(-6 //0) C)` |
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Answer» `(rho_(2))/(rho_(1)) = (V_(1))/(V_(2)) = V_(1)/(V_(1)(1 + 3alphaDeltaT)) = (1 + 3alphaDelta)^(-1)` or `(rho_(2))/(rho_(1)) = (1 - 3alphaDeltaT)` or `rho_(2) = rho_(1)(1 - 3alphaDeltaT)` `= (19.30)[1 - 3(14.2 xx 10^(-6)(70)] = 19.24g // cm^(3)` |
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| 83. |
The temperature of 100 g 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.A. 20gB. 15gC. 12gD. 18g |
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Answer» Correct Answer - C |
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| 84. |
2gm of steam condenses when passed through 40 gm of water initially at `25^(@)C` . The condensation of steam raises the temperature of water to `54.3^(@)C` . What is the latent heat of steamA. 540 cal/gB. 536 cal/gC. 270 cal/gD. 480 cal/g |
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Answer» Correct Answer - A |
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| 85. |
Work done in converting 1 g of ice at `-10^@C` into steam at `100^@C` isA. 3.04 kJB. 6.05 kJC. 0.721 kJD. 0.616 kJ |
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Answer» Correct Answer - A |
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| 86. |
Two identical vessels are filled with equal amounts of ice. The vessels are made from different materials. If the ice melts in the two vessels in times `t_(1) and t_(2)` respectively, then their thermal conductivities are in the ratioA. `t_(1)/t_(2)`B. `t_(2)/t_(1)`C. `t_(2)^(2):t_(1)^(2)`D. `t_(1)^(2):t_(2)^(2)` |
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Answer» Correct Answer - B `mL=(KA(theta_(1)-theta_(2))t)/(d)` in the given problem, `Kt = consta nt` `:. (K_1)/(K_2)=(t_2)/(t_1)`. |
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| 87. |
80 g of water at `30^@C` are poured on a large block of ice at `0^@C`. The mass of ice that melts isA. 30 gmB. 80 gmC. 1600 gmD. 150 gm |
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Answer» Correct Answer - A |
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| 88. |
When hot water is poured on a glass plate, it breaks because ofA. unequal expansion of glassB. equal contraction of glassC. unequal contraction of glassD. glass is delicate |
| Answer» Correct Answer - A | |
| 89. |
What length of brass and iron at `0^(0)C` must be used if the difference between their lengths is always `0.2m`? The values of `alpha` for brass and iron and `18 xx 10^(-6)//0 C` and `12 xx 10^(-6)//0 C` respectively, |
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Answer» `l_(1)alpha_(1) = l_(2)alpha_(2)` and `l_(2) - l_(1) = x , l_(1) = (l_(2)alpha_(2))/(alpha_(1))` `l_(2) - l_(1) = x = l_(2) - (l_(2)alpha_(2))/(alpha_(1)) = l_(2)((alpha_(1) - alpha_(2))/(alpha_(1)))` `l_(2) = (xalpha_(1))/(alpha_(1) - alpha_(2)) = (0.2 xx 12 xx 10^(+6))/((18 - 12) xx 10^(-6)) = 0.40m` `l_(1) = (xalpha_(2))/(alpha_(1) - alpha_(2)) = (0.2 xx 18 xx 10^(-6))/((18 - 12) xx 10^(-6)) = 0.60m` |
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| 90. |
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the ring are `5.243 m` and `5.231 m` respectively at `27^(@)C`. To what temperature should the ring be heated so as to fit the rim of the wheel ? Coefficient of linear expansion of iron =` 1.20 xx 10^(-5) K^(-1)` |
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Answer» Given` t_(1) = 27^(0)C , l_(1) = 5.231 m, l_(2) = 5.243 m` `l_(2) = l_(1) [1 + alpha_(1)(t_(2) - t_(1)]` `5.243 = 5.23 l [1 + 1.20 xx 10^(-5)(t_(2) - 27)]` or `t_(2) = 218^(@)C` |
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| 91. |
If the cold junction of thermocouple is kept at `0^(@)C` and the hot junction is kept at `T^(@)C`, then the relation between neutral temperature `(T_(n))` and temperature of inversion `(T_(i))` isA. `T_(n) = (T_(i))/(2)`B. `T_(n) = 2T_(i)`C. `T_(n) = T_(i) - T`D. `T_(n) = T_(i) + T` |
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Answer» Correct Answer - A We know temperature of inversion `(T_i)` is as much above the neutral temperature `(T_n)` as neutral temperature is above the temperature of the cold junction `(T)`, ie., `T_(i)-T_(n)=T_(n)-T or T_(i)=2T_(n)-T` But, here the cold junction is kept at `0^(@)C`, hence `T=0` Thus, `T_(i) = 2 T_(n) or T_(n) = (T_i)/(2)`. |
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| 92. |
A hot and a cold body are kept in vacuum separated from each other. Which of the following cause decrease in temperature of the hot bodyA. RadiationB. ConvectionC. ConductionD. Temperature remains same |
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Answer» Correct Answer - A |
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| 93. |
A block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a room free from air currents. Which of the following curves correctly represents the rate of coolingA. B. C. D. |
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Answer» Correct Answer - B |
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| 94. |
An 850 W consumer coffee maker can make 10 cups (1.75 litres) of `80^(@)C` coffee from `20^(@)C` tap water in 10 minutes. What percentage of the electrical energy consumed actually makes it to the coffee? |
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Answer» The energy needed for heating of water `Q_(out)=mcDeltaT=(1.75kg)(4200J//kg*K)(80-20^(@)C)=441,000J` The, energy consumed by the device (electrical energy) `W_("in")=Pt=(850W)(10"min")(60s//"min")=510,000J` The ratio of these two quantities is the efficiency of the device `eta=Q_("out")//W_("in")=(441,000J)//(510,000J)xx100=86.5%`. |
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| 95. |
The plots of intensity versus wavelength for three black bodies at temperatures `T_(1),T_(2)` and `T_(3)` respectively are shown in Their temperatures are shown in How their temperatures are related ? .A. `T_(1) gt T_(2) gt T_(3)`B. `T_(1)gtT_(3)gtT_(2)`C. `T_(2) gt T_(3) gt T_(1)`D. `T_(3) gt T_(2) gt T_(1)` |
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Answer» Correct Answer - B |
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| 96. |
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of `57^(@)C` is drunk. You can take body (tooth) temperature to be `37^(@)C` and `alpha_(Cu) = 1.7 xx 10^(-5)//^(@)C` bulk modulus for copper `B_(Cu) = 140 xx 10^(9) N//m^(2)`.A. `42 xx 10^(6) N//m^(2)`B. `22 xx 10^(6) N//m^(2)`C. `36 xx 10^(6) N//m^(2)`D. `18 xx 10^(6) N//m^(2)` |
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Answer» Correct Answer - A Given , decrease in temperature `(Delta t) = 57-37 = 20^(@)C` Coefficient of cubical expansion of copper `gamma_(Cu) = 3 alpha_(Cu) = 5.1xx10^(-5)//^(@)C` Let initial volume of the cavity be `V` and its volume increases by `Delta V` due to increase in temperature `:. Delta V = gamma V Delta t` `rArr (Delta V)/(V) = gamma Delta t` Thermal stress produced = `B xx` Volumetric strain `=B xx (Delta V)/(V) = B xx gamma Delta t` `=140xx10^(9)xx(1.5xx10^(-5)xx20) = 42 xx 10^(6)N//m^(2)` This is about `10^(3)` times of atmospheric pressure. |
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| 97. |
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of `57^(@)C` is drunk. You can take body (tooth) temperature to be `37^(@)C` and `alpha_(Cu) = 1.7 xx 10^(-5)//^(@)C` bulk modulus for copper `B_(Cu) = 140 xx 10^(9) N//m^(2)`.A. `1.43 xx 10^(8) Nm^(-2)`B. `4.13 xx 10^(8) Nm^(-2)`C. `2.12 xx 10^(4) Nm^(-2)`D. `3.12 xx 10^(4) Nm^(-2)` |
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Answer» Correct Answer - A Volumetric strain in tooth cavity `= (V)/(V)` Let `gamma` be the coefficient of volume expansion with change in temperature `DeltaT .^(@)C`. Change in volume is `DeltaV = gammaVDeltaT` or `(DeltaV)/(V) = gammaDeltaT` Thermal stress in tooth cavity `= beta xx "volumetric strain" = beta xx gamma DeltaT` `= beta xx 3alpha DeltaT` `= 140 xx 10^(9) xx 3 xx 1.7 xx 10^(-5) xx (57^(@)C - 37^(@)C)` ` = 1.43 xx 10^(8) N m^(-2)`. |
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| 98. |
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. `1.46`B. `0.69`C. `1.21`D. `0.83` |
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Answer» Correct Answer - B `(T_S)/(T_N) = ((lambda_N)_(max))/((lambda_S)_(max)) = (350)/(510) = 0.69`. |
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| 99. |
Two rods (one semi-circular and other straight) of same material and of same cross-sectional area are joined as shown in the figure. The point A and B are maintained at different temperature. Find the ratio of the heat transferred through a cross-section of a semi-circular rod to the heat transferred through a cross section of the straight rod in a given time. A. `(2)/(pi)`B. `(1)/(pi)`C. `pi`D. `(pi)/(2)` |
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Answer» Correct Answer - A `(dQ)/(dt) = (KA Delta theta)/(l) `, for both rods `K, A` and `Delta theta` are same `rArr (dQ)/(dt) prop 1/l ` so `((dQ//dt)_("semicircular"))/((dQ//dt)_("straight"))` `=(l_("straight"))/(l_("semicircular")) = (2r)/(pi r) = 2/(pi)`. |
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| 100. |
Two rods of lengths `l_(1) and l_(2)` are made of materials having coefficients of linear expansion `alpha_(1) and alpha_(2)` respectively. What could be the relation between above values, if the difference in the lengths of the two rods does not depends on temperature variation? |
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Answer» The difference in length is `l=l_(1)-l_(2)` (assuming that `l_(1)gtl_(2)`) As l is independent of temperature `(Deltal)/(DeltaT)=0` `implies(Deltal_(1))/(DeltaT)-(Deltal_(2))/(DeltaT)=0` As we know that `Deltal=lalphaDeltaT` `implies(Deltal)/(DeltaT)=lalpha` So, we get `l_(1)alpha_(1)-l_(2)alpha_(2)=0` or `l_(1)alpha_(1)=l_(2)alpha_(2)` this is the desired result. |
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