

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. |
At what temperature is the Fahrenheit scale reading equal to (a) twice (b) half of Celsius ? |
Answer» Correct Answer - `-24.6^(@)C` |
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52. |
The normal temperature of the human body is `98.4^(@)F`.Calculcate this temperature on celsius scale and absolute scale. |
Answer» Correct Answer - `36.88^(@)F, 310.03K` |
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53. |
A new scale fo temperature is introduced. One degree temperature difference on the new scale is found to be equal to `(5)/(8).^(@)C`. Determine the temperature in Celsium scale when the new scale and the celsium scale shows the saer eading. The lower fixed point on the new scale is `33^(@)` degree. |
Answer» On new scale, One degree temperature difference=`5//8^(@)C` Lower fixed point (LFP) on new scale=33degrees find least count and then, ltBrgt Find temperatrue in new sclae=temperature on Celsium scale. |
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54. |
If 3360 J of heat is required to melt 10 g of ice, how many kilocalories of heat should be supplied to melt 1 kg ice? |
Answer» heat required to melt 10 of ice `=3360J=(3360)/(4.2)cal=800cal` heat required to melt 1 g of ice=80cal heat required tomelt 1000 g of ice=80,000 cal `therefore80` kcal of heat is required to melt 1 kg ice. |
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55. |
The temperature of an iron piece is heated from `30^@ C "to" 90^@ C`. What is the change in its temperature on the fahrenheit scale and on the kelvin scale?A. `108^(@)`, 60KB. `100^(@)`F, 55 KC. `100^(@)`F, 65KD. `60^(@)`F, 108 K |
Answer» a) `DeltaT_(c)=90^(@)-30^(@)`C=`60^(@)`C `DeltaT_(F) = 9/5DeltaT_(C) = 9/5(60^(@)) = 108^(@)`F `DeltaT = DeltaT_(c)=60K` |
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56. |
Celsius temperature scale. |
Answer» Celsius temperature scale: The temperature scale which has a lower limit of 0° C and a higher limit of 100° C and is divided into 100 parts. |
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57. |
Convert `55^(@)C` into Fahrenheit and Kelvin scale. |
Answer» `(55.0)/(100)=(F-32)/(180)` `f=131^(@)F` `(55-0)/(100)=(K-273)/(100)` `K=273+55=328K` |
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58. |
A: If roofs were weak, they could be lifted and blown away. R: The moving air creates high pressure. A) A & R are correct and R supports A B) A & R are correct but R does not support A C) A is correct but R is wrong D) R is correct but A is wrong |
Answer» The correct option is C) A is correct but R is wrong. |
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59. |
Which instrument is used to measure the maximum (highest) and minimum (lowest) temperatures of a place in a day? |
Answer» Six’s maximum and minimum thermometer is one of the Meteorological Instruments used to measure maximum (highest) and minimum (lowest) temperatures of a place in a day. |
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60. |
The pressure of air in the bulb of constant volume air thermometer is 75 cm of mercury at `0^(@)C, 100 cm` at `100^(@)C` and 80 cm at the room temperature. Calculate the room temperature. |
Answer» Correct Answer - `20^(@)C` |
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61. |
0.75 gram of petroleum was burnt in a bomb calorimeter which contains 2 kg of water and has a water equivalent 500 gram. The rise in temperature was `3^(@)C`. Determine the calorific value of petroleum. |
Answer» Correct Answer - `10^(4) cal g^(-1)` |
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62. |
Convert 212°F into Kelvin scale |
Answer» 212°F= 100°C. So 100 + 273 = 373 K. |
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63. |
Convert 25°C into Kelvin scale. |
Answer» 25°C = (273 + 25) K = 298 K |
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64. |
Why a bottle completely filled with water and closed with a tight cap break after freezing? |
Answer» 1. Density of ice is less than that of water. 2. This means water expands on freezing and converts into ice. 3. So, excess space is required to expand the water. 4. The bottle completely filled with water and closed with a tight cap has no excess space to expand water. 5. So, bottle breaks while freezing water in it. |
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65. |
A: The freezing point of alcohol is more than -100°C. R: Alcohol can be used to measure very low temperatures. A) A & R are correct and R supports A B) A & R are correct but R does not support A C) A is correct but R is wrong D) R is correct but A is wrong |
Answer» The correct option is D) R is correct but A is wrong. |
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66. |
A person weighing 60 kg takes in 2000 kcal diet in a day. If this energy were to be used in heating the person without any losses, his rise in temperature would be nearly (Given sp. Heat of human body is `0.83 cal g^(-1), .^@C^(-1)` |
Answer» Correct Answer - `40.16^(@)C` |
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67. |
What does Kelvin scale mean? |
Answer» Kelvin scale: A temperature scale that has a lower fixed point 273.15 K and an upper fixed point of 373.15 K is called the Kelvin scale. |
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68. |
What is the value of following temperatures in Kelvin scale? (a) 30° C b) 70° C |
Answer» a) 30° C = 30 + 273 = 303 K b) 70° C = 70 + 273 = 343 K |
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69. |
What is internal energy? |
Answer» Internal energy : It is the energy possessed by the system by virtue of its molecular motion and molecular configuration. It is a stored energy. It depends on the temperature of the system. |
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70. |
Statement I : The internal energy of water decreases during freezing. Statement II : The heat energy given to the ice is totally utilized in breaking the bonds between the water molecules during melting.A) both statements are correct B) both statements are incorrect C) I is correct but II is incorrect D) I is incorrect but II is correct |
Answer» A) both statements are correct |
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71. |
A body of mass 25 kg is dragged on a horizontal rough road with a constant speed of `20 kmh^(-1)`. If the coefficient of friction is 0.5, find the heat generated in one hour. If 50% of the heat is absorbed by the body, find the rise in temperature. Specific heat of the material of the body is 0.1 cal `g^(-1).^(@)C^(-1)`. |
Answer» Correct Answer - `5.83 xx 10^(5) cal ,116.6^(@).^(@)C` |
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72. |
Internal energy of a system is A) vibrational energy B) linear kinetic energy C) rotational and potential energy D) all the above |
Answer» D) all the above |
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73. |
Write the formula for resultant temperatures of a mixture, when V1 ml of water at T1°C is mixed with V ml of water at T2° C. |
Answer» Resultant temperature T = \(\frac {V_1T_1+V_2T_2}{V_1+V_2}\) |
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74. |
When 10 g of coal is burnt, it raises the temperature of 2 litres of water from `20^(@)C` to `55^(@)C`. Calculate the heat of combustion of fuel. |
Answer» Correct Answer - `7000 cal g^(-1)` |
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75. |
Match the following.AB1. formation of ice bergsa.kelvin scale2.rise of mercury levelb.sudden3. automatic electric iron(c) celsius scale4. cracking of hot glass plate when touches by cold objectd.boiling pointe. bimetal stripf.thermal expansiong. anomalour expansion of water. |
Answer» 1. g 2. F 3. e 4. b |
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76. |
What is meant by internal energy? |
Answer» Combination of linear kinetic energy, rotational kinetic energy, vibrational energy, and potential energy of molecules is known as internal energy of the system. |
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77. |
When water is boiled at `2` atm pressure the latent heat of vaporization is `2.2xx10^(6) J//kg` and the boiling point is `120^(@)C` At `2` atm pressure `1` kg of water has a volume of `10^(-3)m^(-3)` and `1` kg of steam has volume of `0.824m^(3)` . The increase in internal energy of `1` kg of water when it is converted into steam at `2` atm pressure and `120^(@)C` is [`1` atm pressure `=1.013xx10^(5)N//m^(2)`] |
Answer» Correct Answer - `10.5^(@)C` |
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78. |
How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g) |
Answer» Answer is 10800 cal |
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79. |
Define latent heat of vaporization? |
Answer» The heat energy required to change one unit mass of liquid to gas at constant temperature is called latent heat of vaporization L = Q/m The value of latent heat of vapourization of water is 540 cal/gm. |
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80. |
Latent heat of vaporization. |
Answer» Latent heat of vaporization: The amount of heat required to convert one gram of liquid into vapor at a constant temperature. |
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81. |
Latent heat of vaporization? |
Answer» Heat energy is used to change the state of water from liquid to vapor (gas). This is called the latent heat of vaporization. Latent heat of vapourization (L) = Q/m Units in C.G.S – cal/gm. Units in S.l. – J/kg. The latent heat of vaporization of water is 540 cal/gm. |
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82. |
The specific latent heat of vaporization of water is 540 cal/g. Explain this statement. |
Answer» When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal. |
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83. |
Rewrite the following statements by selecting the correct options:i. The density of water is maximum at……….. (a) 0 °C (b) – 4 °C (c) 100 °C (d) 4 °Cii. ………..heat is needed to raise the temperature of 1 kg of water from 14.5 °C to 15.5 °C. (a) 4180 J (b) 103 J (c) 1 cal (d) 4180 caliii. ………..heat is needed to convert 1 g of water at 0 °C and at a pressure of one atmosphere into 1 g of steam under the same conditions. (a) 80 cal (b) 540 cal (c) 89 J (d) 540 Jiv. Water expands on reducing its temperature below………..°C.(a) 0 (b) 4 (c) 8 (d) 12 |
Answer» i. (d) 4 °C ii. (a) 4180 J iii. (b) 540 cal iv. (b) 4 |
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84. |
This mode of transfer of heat doesn’t require any media. A) Conduction B) Convection of heat C) Radiation of heat D) All |
Answer» The correct option is C) Radiation of heat. |
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85. |
How do we written degree of Celsius, degree of Fahrenheit and Kelvin? |
Answer» 1. Degree Celsius: Celsius is written as °C and read as degree Celsius. 2. Degree Fahrenheit: Fahrenheit is written as °F and read as degree fahrenheit. 3. Kelvin is written as K and read as Kelvin. |
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86. |
Two rods of same length and material transfer a given amount ofheat in 12 seconds, when they are joined end to end. But when theyare joined lengthwise, then they will transfer same heat in sameconditions inA. increasesB. decreasesC. remains the sameD. fluctuates |
Answer» Correct Answer - A | |
87. |
Two rods of equal length and of the same material but having different diameters, are heated through and equal rise in temperature. Of the two, thin and thick rods, which will experience a greater extension? Explain. |
Answer» `DeltaV_("real")=DeltaV_("apparent")+DeltaV_("container")` `rArr" "gamma_("real")=gamma_("apperent")+gamma_("container")` `gamma_("apparent")=(DeltaV)/(VDeltaV)=(0.5)/(50xx20)=5xx10^(-4)K^(-1)` `=50xx10^(-6)K^(-1)` `"Given "alpha_("container")=9xx10^(-6)K^(-1)` `thereforegamma_("container")=27xx10^(-6)K^(-1)[becausegamma=3alpha]` `thereforegamma_("real")=500xx10^(-6)+27xx10^(-6)` `=527xx10^(-6)K^(-1)` |
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88. |
When 50 ml of a liquid is heated through `20^(@)C`, its apparent expansion is 0.5 ml. If the coefficient of linear expansion of the container is `9xx10^(-6)K^(-1)`, find the coefficient of real expansion of the liquid. |
Answer» `Q=ms_("ice")(Deltat_("ice"))+mL_(F)+ms_(w)(Deltat)` `=(1xx2.095xx5xx10^(3))+(1xx336xx10^(3))+1xx4.2xx10^(3)xx5=367.475` kJ |
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89. |
Have you heard about temperature, when you suffer from fever? |
Answer» Yes, I have. |
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90. |
Calculate the amount of heat energy required to increase the temperature of 250 g of water from `27^(@)C` to `67^(@)C` (Specific heat capacity of water is 1 cal `g^([email protected])C^(-1)`) |
Answer» Given mass of water =250 g Change in temperature `Delta`=final temperature -intitial temperature =`67^(@)C-27^(@)C=40^(@)C=40^(@)C` specific heat capacity of water =1 cal `gal^(-1).^@C^(-1)` Heat energy supplied ,Q =`MsDelta t` Q=`250xx1xx40`=10000 cal =kCal |
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91. |
What are the units of heat used? |
Answer» SI unit of heat is Joule. The unit calorie is also used |
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92. |
Fill in the blanks: 1. We feel heat on our body when the _____ shines. 2. _____ energy can be generated by the burning of fuels like coal, wood, charcoal, gasoline etc.3. When _____ flows through a conductor, heat energy is produced. 4. _____ is a form of energy. 5. _____ determines the direction of flow of heat. 6. The coldest temperature in the world was measured in the _____ continent. 7. Temperature measures the _____ kinetic energy of molecules. |
Answer» 1. Sun 2. Heat 3. Electric current 4. Heat 5. Temperature 6. Antarctic 7. average |
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93. |
How heat energy is transferred? |
Answer» Heat energy flows from higher temperature to lower temperature. |
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94. |
List and describe the sources of heat. |
Answer» 1. Sun: It is the primary source of light and it gives us light and heat. 2. Combustion: Heat energy can be generated by the burning of fuels like wood, Kerosene, Coal, Charcoal, Petrol etc. 3. Friction: We can generate heat by rubbing two surfaces of some substances. In the past people used to rub two stones together to light fire. 4. Electricity: When Electric current flows through a conductor, heat energy is produced. Ex.: Iron box, Electric kettle. |
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95. |
Match the following :1. Source of heat(a). heat energy2. Electric current(b) Calorie3. Gasoline(c) 37°C4. Unit of heat(d) Electric kettle5. Human body temperature(e) sun |
Answer» 1. – e 2. – d 3. – a 4. – b 5. – c. |
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96. |
How can heat energy be generated? |
Answer» Heat energy can be generated by the burning of fuels like wood, Kerosene, Coal, charcoal, gasoline/petrol, oil, etc |
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97. |
Analogy:1. Movement of molecules : Heat. Heat energy: _____2. Expansion is length : Linear expansion. Expansion in volume : _______3. Ordinary glass : Glass tumbler. Pyrex glass: _______ |
Answer» 1. calorie 2. cubical expansion 3. Laboratory glassware. |
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98. |
Differentiate linear expansion and cubical expansion. |
Answer» Linear expansion: 1. The expansion is in length. 2. Ex.: Expansion of railway track rod. Cubical expansion: 1. The expansion is in volume. 2. Ex.: Expansion of metal ball. |
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99. |
What is thermal expansion? |
Answer» The expansion of a substance on heating is called the thermal expansion of that substance. |
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100. |
The volume of a metal sphere is increased by 1% of its original volume when it is heated from 320 K to 522 K. calculate the coefficients of linear, superficial and cubical expansion of the metal. |
Answer» Correct Answer - `1.67 xx 10^(-5).^(@)C^(-1), 3.34 xx 10^(-5).^(@)C^(-1)` |
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