Thermal Physics Quiz-14
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. Pick out the statement which is not true:
Solution
IR radiations arise due to inner electron transitions in atoms
IR radiations arise due to inner electron transitions in atoms
Q2. Dry ice is
Solution
We know that when solid carbondioxide is heated, it becomes vapour directly without passing through its liquid phase. Therefore it is called dry ice
We know that when solid carbondioxide is heated, it becomes vapour directly without passing through its liquid phase. Therefore it is called dry ice
Q3. The latent heat of vaporization of a substance is always
Solution
The latent heat of vaporization is always greater than latent heat of fusion because in liquid to vapour phase change there is a large increase in volume. Hence more heat is required as compared to solid to liquid phase change
The latent heat of vaporization is always greater than latent heat of fusion because in liquid to vapour phase change there is a large increase in volume. Hence more heat is required as compared to solid to liquid phase change
Q4. According to the experiment of IngenHausz the relation between the thermal conductivity of a metal rod is K and the length of the rod whenever the wax melts is
Solution
K/l2= constant
K/l2= constant
Q5. A conductor of area of cross-section 100 cm2 and length 1 cm has coefficient of thermal conductivity 0.76cals-1 m-1 K-1. If 30 cal of heat flows through the conductor per second. Find the temperature difference across the conductor.
Solution
Heat current H=KA∆θ/l ∴∆θ=Hl/KA = (30×1×10-2)/(0.76×100×10-4)=39.47 =40℃(approx.) .
Heat current H=KA∆θ/l ∴∆θ=Hl/KA = (30×1×10-2)/(0.76×100×10-4)=39.47 =40℃(approx.) .
Q6. At a certain temperature for given wave length, the ratio of emissive power of a body to emissive power of black body in same circumstances is known as
Solution
Emissivity
Emissivity
Q7. A body cools from 62℃ to 50℃ in 10 min and to 42℃ in the next 10 min. The temperature of the surrounding is
Solution
According to Newton’s law of cooling ((θ1-θ2))/t=K((θ1+θ2)/2-θ0 ) ∴ ((62-50))/10=K((62+50)/2-θ0 ) 12/10=K(56-θ0) …(i) For further cooling ((50-42))/10=K((50+42)/2-θ0 ) 8/10=K(46-θ0) …(ii) Dividing Eq (i) by Eq. (ii), we get, 12/8=((56-θ0))/((46-θ0)) 3(46-θ0 )=2(56-θ0) 138-3θ0=112-2θ0 θ0=26℃
According to Newton’s law of cooling ((θ1-θ2))/t=K((θ1+θ2)/2-θ0 ) ∴ ((62-50))/10=K((62+50)/2-θ0 ) 12/10=K(56-θ0) …(i) For further cooling ((50-42))/10=K((50+42)/2-θ0 ) 8/10=K(46-θ0) …(ii) Dividing Eq (i) by Eq. (ii), we get, 12/8=((56-θ0))/((46-θ0)) 3(46-θ0 )=2(56-θ0) 138-3θ0=112-2θ0 θ0=26℃
Q8. 5 g of ice at 0℃ is dropped in a beaker containing 20 g of water at 40℃. The final temperature will be
Solution
Let final temperature be θ Now heat taken by ice=m1 L+m1 c1 θ1 =5×80+5×1(θ-0) =400+5θ…(i) Heat given by water at 40℃ =m2 C2 θ2=20×1×(40°-θ) …(ii) Heat given=Heat taken 800-20θ=400+5θ or 25θ = 400, or θ=400/25=16℃
Let final temperature be θ Now heat taken by ice=m1 L+m1 c1 θ1 =5×80+5×1(θ-0) =400+5θ…(i) Heat given by water at 40℃ =m2 C2 θ2=20×1×(40°-θ) …(ii) Heat given=Heat taken 800-20θ=400+5θ or 25θ = 400, or θ=400/25=16℃
Q9. A red flower kept in green light will appear
Solution
Black
Black
Q10. Which of the prism is used to see infra-red spectrum of light:
Solution
Ordinary glass prism (crown, flint) absorbs the infrared radiation but rock salt prism transmit them. Hence it is used to obtain the spectrum of infrared radiation
Ordinary glass prism (crown, flint) absorbs the infrared radiation but rock salt prism transmit them. Hence it is used to obtain the spectrum of infrared radiation
