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. Expansion during heating:
Solution
Solids, liquids and gases all expand on being heated, as a result density (= mass/volume) decreases
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
Solution
Rate of cooling RC=dθ/dt=(Aεσ(T4-T04))/mc ⇒dθ/dt∝A/V∝r2/r3 ⇒dθ/dt∝1/r
Rate of cooling RC=dθ/dt=(Aεσ(T4-T04))/mc ⇒dθ/dt∝A/V∝r2/r3 ⇒dθ/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
Solution
Loss of heat ∆Q=Aεσ(T4-T04 )t ⇒ Rate of loss of heat ∆Q/t=Aεσ(T4-T04) =10×10-4×1×5.67×10-8 {(273+127)4-(273+27)4} =0.99 W
Loss of heat ∆Q=Aεσ(T4-T04 )t ⇒ Rate of loss of heat ∆Q/t=Aεσ(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?
Solution
The Stefan’s law, E=σT4 where σ 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
The Stefan’s law, E=σT4 where σ 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.
Solution
Compensating leads
Compensating leads
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℃?
Solution
Rate of loss of heat (∆Q/t)∝ temperature difference ∆θ (∆Q/t)1/(∆Q/t)2 =(∆θ2)/(∆θ1 )⇒60/(∆Q/t)2 =(80-20)/(40-20)⇒(∆Q/t)2=20cal/s
Rate of loss of heat (∆Q/t)∝ temperature difference ∆θ (∆Q/t)1/(∆Q/t)2 =(∆θ2)/(∆θ1 )⇒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?
Solution
Warming of glass of bulb due to filament
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
Solution
In first case (50-40)/5=K[(50+40)/2-θ0 ]…(i) In second case (40-33.33)/5=K[(40+33.33)/2-θ0 ] …(ii) By solving θ0=20℃
In first case (50-40)/5=K[(50+40)/2-θ0 ]…(i) In second case (40-33.33)/5=K[(40+33.33)/2-θ0 ] …(ii) By solving θ0=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
Solution
α= ∆L/(L0 (∆θ) )=0.19/100(100-0) =1.9×10-5/℃ Now γ=3α=3×1.9×10-5/℃=5.7×10-5/℃
α= ∆L/(L0 (∆θ) )=0.19/100(100-0) =1.9×10-5/℃ Now γ=3α=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?:
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
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