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.**Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets melted in 20 minutes and 30 minutes. The ratio of their thermal conductivities will be:

Solution

Q= (KA(Î¸

Q= (KA(Î¸

_{1}-Î¸_{2})t)/l; in both the cases, A,l and (Î¸_{1}-Î¸_{2}) are same so Kt= constant ⇒K_{1}/K_{2}=t_{2}/t_{1}=30/20=3/2=1.5 .**Q2.**Ice formed over lakes has

Solution

Very low conductivity and retards further formation of ice

Very low conductivity and retards further formation of ice

**Q3.**On heating a liquid of coefficient of cubical expansion Î³ in a container having coefficient of linear expansion Î³/3, the level of liquid in the container will

Solution

As coefficient of cubical expansion of liquid equals coefficient of cubical expansion of vessel, the level of liquid will not change on heating

As coefficient of cubical expansion of liquid equals coefficient of cubical expansion of vessel, the level of liquid will not change on heating

**Q4.**Which of the following has maximum specific heat

Solution

Water has maximum specific heat

Water has maximum specific heat

**Q5.**When the room temperature becomes equal to the dew point, the relative humidity of the room is

Solution

If the room temperature becomes equal to the dew point, the relative humidity of the room is 100%.

If the room temperature becomes equal to the dew point, the relative humidity of the room is 100%.

**Q6.**A body cools from 60℃ to 50℃ in 10 min. if the room temperature is 25℃ and assuming Newton’s law of cooling to hold good, the temperature of the body at the end of the next 10 min will be

Solution

From Newton’s law of cooling when a hot body is cooled in air, the rate of loss of heat by the body is proportional to the temperature difference between the body and its surroundings. Given, Î¸

From Newton’s law of cooling when a hot body is cooled in air, the rate of loss of heat by the body is proportional to the temperature difference between the body and its surroundings. Given, Î¸

_{1}=60℃, Î¸_{2}=50℃, Î¸=25℃ ∴Rate of loss of heat=K (Mean temp.-Atmosphere temp.) Where K is coefficient of thermal conductivity (Î¸_{1}-Î¸_{2})/t=K((Î¸_{1}+Î¸_{2})/2-Î¸) (60-50)/10=K((60+50)/2-25) ⇒K=1/30 Also putting the value of K, we have (50-Î¸_{3})/10=1/30 ((50+Î¸_{3})/2-25) ⇒Î¸_{3}=42.85℃**Q7.**A metallic solid sphere is routing about its diameter as axis of rotation. If the temperature is increased by 200℃, the percentage in its moment of inertia is (Coefficient of linear expansion of the metal=10

^{-5}℃

^{-1})

Solution

The moment of inertia of a solid sphere about the axis along its diameter is I=2/5 mR

The moment of inertia of a solid sphere about the axis along its diameter is I=2/5 mR

^{2}⇒I∝R^{2}∴∆I/I×100=2[∆R/R]100 But Î±=∆R/(R×∆t ) ⇒ ∆R/R= Î±∆t ∴∆I/I×100=2(Î±)(∆t)100 = 2(10^{-5})(200)(100) = 0.4%**Q8.**Three discs, A,B and C having radii 2 m, 4 m and 6 m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensitios are 300 nm, 400 nm and 500 nm respectively. The power radiated by them are Q

_{A},Q

_{B}and Q

_{C}respectively

Solution

Power radiated, Q∝AT

Power radiated, Q∝AT

^{4}andÎ»_{m}T=constant.Hence,Q∝A/(Î»_{m})^{4}Q∝A/(Î»_{m})^{4}or Q∝r^{2}/(Î»_{m})^{4}Q_{A}:Q_{B}:Q_{C}=(2)^{2}/(3)^{4}∶(4)^{2}/(4)^{4}∶(6)^{2}/(5)^{4}=4/81:1/16:36/625 = 0.05:0.0625:0.0576 ie.,Q_{B}is maximum. .**Q9.**By increasing the temperature of a liquid its

Solution

When we increase the temperature of a liquid, the liquid will expand. So, the volume of the liquid will increase and hence, the density of the liquid will decrease.

When we increase the temperature of a liquid, the liquid will expand. So, the volume of the liquid will increase and hence, the density of the liquid will decrease.

**Q10.**The spectrum of a black body at two temperatures 27℃ and 327℃ is shown in the figure. Let A

_{1}and A

_{2}be the areas under the two curves respectively. The value of A

_{2}/A

_{1}is:

Solution

Area under given curve represents emissive power and emissive power ∝T

Area under given curve represents emissive power and emissive power ∝T

^{4}⇒A∝T^{4}⇒A_{2}/A_{1}=(T_{2}^{4})/(T_{1}^{4})=(273+327)^{4}/(273+27)^{4}=(600/300)^{4}=16/1