## Thermal Physics Quiz-20

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. Expansion during heating:
•  Occurs only in solids
•  Increases the weight of a material
•  Decreases the density of a material
•  Occurs at the same rate for all liquids and solids
Solution
Solids, liquids and gases all expand on being heated, as a result density (= mass/volume) decreases

Q2. A hot metallic sphere of radius r radiates heat. It’s rate of cooling is
•  Independent of r
•  Proportional to r
•  Proportional to r2
•  Proportional to 1/r
Solution
Rate of cooling RC=dÎ¸/dt=(AÎµÏƒ(T4-T04))/mc ⇒dÎ¸/dt∝A/V∝r2/r3 ⇒dÎ¸/dt∝1/r

Q3. A black body of surface area 10cm2 is heated to 127℃ and is suspended in a room at temperature 27℃. The initial rate of loss of heat from the body at the room temperature will be
•  2.99 W
•  1.89 W
•  1.18 W
•  0.99 W
Solution
Loss of heat ∆Q=AÎµÏƒ(T4-T04 )t ⇒ Rate of loss of heat ∆Q/t=AÎµÏƒ(T4-T04) =10×10-4×1×5.67×10-8 {(273+127)4-(273+27)4} =0.99 W

Q4. The temperature of a metal block is increased from 27℃ to 84℃. The rate of the radiated energy from the block will increase approximately?
•  2 times
•  4 times
•  8 times
•  16 times
Solution
The Stefan’s law, E=ÏƒT4 where Ïƒ is Stefan’s constant, Given, T1=27℃=27+273=300 K T2=84℃=273+84=357 K ∴E1/E2 =(T14)/(T24 ) = (300)4/(357)4 =1/(1.19)4 Rate of increase of energy is E2/E1 =(1.19)4=2

Q5. ‘Stem Correction’ in platinum resistance thermometers are eliminated by the use of.
•  Cells
•  Electrodes
•  None of the above
Solution

Q6. Liquid is filled in a vessel which is kept in a room with temperature 20℃. When the temperature of the liquid is 80℃, then it loses heat at the rate of 60 cal/s. What will be the rate of loss of heat when the temperature of the liquid is 40℃?
•  180 cal/s
•  40 cal/s
•  30 cal/s
•  20 cal/s
Solution
Rate of loss of heat (∆Q/t)∝ temperature difference ∆Î¸ (∆Q/t)1/(∆Q/t)2 =(∆Î¸2)/(∆Î¸1 )⇒60/(∆Q/t)2 =(80-20)/(40-20)⇒(∆Q/t)2=20cal/s

Q7. In which of the following process convection does not take place primarily?
•  Sea and land breeze
•  Boiling of water
•  Warming of glass of bulb due to filament
•  Heating air around a furnace
Solution
Warming of glass of bulb due to filament

Q8. A body takes 5 minutes for cooling from 50℃ to 40℃. Its temperature comes down to 33.33℃ in next 5 minutes. Temperature of surroundings is
•  15℃
•  20℃
•  25℃
•  10℃
Solution
In first case (50-40)/5=K[(50+40)/2-Î¸0 ]…(i) In second case (40-33.33)/5=K[(40+33.33)/2-Î¸0 ] …(ii) By solving Î¸0=20℃

Q9. A rod of silver at 0℃ is heated to 100℃. It’s length is increased by 0.19 cm. Coefficient of cubical expansion of the silver rod is
•  5.7×10-5/℃
•  0.63×10-5/℃
•  1.9×10-5/℃
•  16.1×10-5/℃
Solution
Î±= ∆L/(L0 (∆Î¸) )=0.19/100(100-0) =1.9×10-5/℃ Now Î³=3Î±=3×1.9×10-5/℃=5.7×10-5/℃

Q10. A black body radiates energy at the rate of E Wm-2 at a high temperature T K. When the temperature is reduced to (T/2) K, the radiant energy is?:
•  E/2
•  2E
•  E/4
•  E/16
Solution
According to Stefan’s law, E∝T4 or E2/E1 =(T2/T1 )4 or E2/E=((T⁄2)/T)4=(1/2)4 or E2=E/16

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