Thermal Physics Quiz-12
Dear Readers,
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.
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Q1. In a pressure cooker, cooking is faster because the increase of vapour pressure:
Solution
Increase of vapour pressure increases the boiling point of water.
Increase of vapour pressure increases the boiling point of water.
Q2. The adjoining diagram shows the spectral energy density distribution Eλ of a black body at two different temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature T is
Solution
AT/A2000 =16/1 [Given] Area under eλ-λ curve represents the emissive power of body and emissive power ∝T4 [Hence area under eλ-λ curve) ∝T4 ⇒AT/A2000 =(T/2000)4⇒16/1=(T/2000)4⇒T =4000K
AT/A2000 =16/1 [Given] Area under eλ-λ curve represents the emissive power of body and emissive power ∝T4 [Hence area under eλ-λ curve) ∝T4 ⇒AT/A2000 =(T/2000)4⇒16/1=(T/2000)4⇒T =4000K
Q3. Shown below are the black body radiation curves at temperatures T1 and T2 (T2>T1). Which of the following plots is corre
Solution
According to Wien’s displacement law λm∝1/T⇒λm2λm1 [∵T1T2] There fore I-λ graph for T2 has lesser wavelength (λm) and so curve for T2 will shift towards left side
According to Wien’s displacement law λm∝1/T⇒λm2λm1 [∵T1T2] There fore I-λ graph for T2 has lesser wavelength (λm) and so curve for T2 will shift towards left side
Q4. In supplying 400 calories of heat to a system, the work done will be
Solution
W=JQ=4.18×400=1672 joule
W=JQ=4.18×400=1672 joule
Q5. Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is
Solution
In steady state energy absorbed by middle plate is equal to energy released by middle plate σA(3T)4-σA(T'')4=σA(T'')4-σA(2T)4 (3T)4-(T'' )4=(T'')4-(2T)4 2(T'')4-(16+81) T4 T''=(97/2)1/4 T
In steady state energy absorbed by middle plate is equal to energy released by middle plate σA(3T)4-σA(T'')4=σA(T'')4-σA(2T)4 (3T)4-(T'' )4=(T'')4-(2T)4 2(T'')4-(16+81) T4 T''=(97/2)1/4 T
Q6. The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar show in figure, what will be the temperature at the junction of copper ad steel?
Solution
Let the temperature of junction be θ. (∆Q/d1 )copper =(∆Q/∆T)steel K1 A= ((100-θ))/18=(K2 A(θ-0))/6 9K2 ((100-θ))/3=K2 θ 3θ=900-9θ 12θ=900 θ=75℃
Let the temperature of junction be θ. (∆Q/d1 )copper =(∆Q/∆T)steel K1 A= ((100-θ))/18=(K2 A(θ-0))/6 9K2 ((100-θ))/3=K2 θ 3θ=900-9θ 12θ=900 θ=75℃
Q7. 300 gm of water at 25℃ is added to 100 g of ice at 0℃.The final temperature of the mixture is
Solution
Which is not possible. Hence θmix=0℃
Which is not possible. Hence θmix=0℃
Q8. Surface of the lake is at 2℃. Find the temperature of the bottom of the lake
Solution
The densest layer of water will be at bottom. The density of water is maximum at 4℃. So the temperature of bottom of lake will be 4℃.
The densest layer of water will be at bottom. The density of water is maximum at 4℃. So the temperature of bottom of lake will be 4℃.
Q9. The temperature on Celsius scale is 25℃. What is the corresponding temperature on the Fahrenheit scale
Solution
C/5=(F-32)/9⇒25/5=(F-32)/9=F=77o F
C/5=(F-32)/9⇒25/5=(F-32)/9=F=77o F
Q10. A beaker is completely filled with water at 4℃. It will overflow:
Solution
Water will overflow, both when heated or cooled because water has maximum density at 4℃ or minimum volume at 4℃
Water will overflow, both when heated or cooled because water has maximum density at 4℃ or minimum volume at 4℃
