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 R

Rate of cooling R

_{C}=dÎ¸/dt=(AÎµÏƒ(T^{4}-T_{0}^{4}))/mc ⇒dÎ¸/dt∝A/V∝r^{2}/r^{3}⇒dÎ¸/dt∝1/r**Q3.**A black body of surface area 10cm

^{2}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=AÎµÏƒ(T

Loss of heat ∆Q=AÎµÏƒ(T

^{4}-T_{0}^{4})t ⇒ Rate of loss of heat ∆Q/t=AÎµÏƒ(T^{4}-T_{0}^{4}) =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=ÏƒT

The Stefan’s law, E=ÏƒT

^{4}where Ïƒ is Stefan’s constant, Given, T_{1}=27℃=27+273=300 K T_{2}=84℃=273+84=357 K ∴E_{1}/E_{2}=(T_{1}^{4})/(T_{2}^{4}) = (300)^{4}/(357)^{4}=1/(1.19)^{4}Rate of increase of energy is E_{2}/E_{1}=(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 ∆Î¸ (∆Q/t)

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?

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-Î¸

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

Solution

Î±= ∆L/(L

Î±= ∆L/(L

_{0}(∆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?:

Solution

According to Stefan’s law, E∝T

According to Stefan’s law, E∝T

^{4}or E_{2}/E_{1}=(T_{2}/T_{1})^{4}or E_{2}/E=((T⁄2)/T)^{4}=(1/2)^{4}or E_{2}=E/16