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.**A black body radiates at the rate of W watts at a temperature T. If the temperature of the body is reduced to T/3, it will radiate at the rate of (in Watts):

Solution

P=(Q/t)∝T

P=(Q/t)∝T

^{4}⇒W/P_{2}=(T/(T/3))^{4}⇒P_{2}=W/81**Q2.**For a small temperature difference between the body and the surroundings the relation between the rate of loss heat R and the temperature of the body is depicted by

Solution

Rate of loss of heat (R)∝ temperature difference ⇒R∝(Î¸-Î¸

Rate of loss of heat (R)∝ temperature difference ⇒R∝(Î¸-Î¸

_{0})⇒R=k(Î¸-Î¸_{0})=kÎ¸-kÎ¸_{0}[k= constant] On comparing it with y=mx+c it is observed that, the graph between R and Î¸ will be straight line with slope =k and intercept =-kÎ¸_{0}**Q3.**There are two identical vessels filled with equal amounts of ice. The vessels are of different metals., If the ice melts in the two vessels in 20 and 35 minutes respectively, the ratio of the coefficients of thermal conductivity of the two metals is

Solution

Q= (KA(Î¸

Q= (KA(Î¸

_{1}-Î¸_{2}))/l t⇒K_{1}t_{1}=K_{2}t_{2}⇒K_{1}/K_{2}=t_{2}/t_{1}=35/20=7/4 [As Q,l,A and (Î¸_{1}-Î¸_{2}) are same]

**Q4.**The dimensions of thermal resistance are

Solution

Thermal resistance =l/KA=[L/(MLT

Thermal resistance =l/KA=[L/(MLT

^{-3}) K^{-1}×L^{2})]=[M^{-1}L^{-2}T^{3}K]**Q5.**Five identical rods are joined as shown in figure. Point A and C are maintained at temperature 120℃ and

20℃ respectively. The temperature of junction B will be.

**Q6.**While measuring the thermal conductivity of a liquid, we keep the upper part hot and lower part cool, so that?

Solution

Convection may be stopped.

Convection may be stopped.

**Q7.**A lead bullet of 10 g travelling at 300 m/s strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (specific heat of lead =150J/kg,K)

Solution

Since specific heat of lead is given in Joules, hence use W=Q instead of W=JQ. ⇒1/2×(1/2 mv

Since specific heat of lead is given in Joules, hence use W=Q instead of W=JQ. ⇒1/2×(1/2 mv

^{2})=m.c.∆Î¸⇒∆Î¸=v^{2}/4c=(300)^{2}/(4×150)=150℃**Q8.**A liquid cools down from 70℃ to 60℃ in 5 minutes. The time taken to cool it from 60℃ to 50℃ will be

Solution

According to Newton’s law of cooling Rate of cooling ∝ mean temperature difference Initially, mean temperature difference =((70+60)/2-Î¸

According to Newton’s law of cooling Rate of cooling ∝ mean temperature difference Initially, mean temperature difference =((70+60)/2-Î¸

_{0})=(65-Î¸_{0}) Finally, mean temperature difference =((60+50)/2-Î¸_{0})=(55-Î¸_{0}) In second case mean temperature difference decreases, so rate of fall of temperature decreases, so it takes more time to cool through the same range**Q9.**Two identical bodies have temperatures 277℃ and 67℃ . If the surroundings temperature is 27℃, the ratio of loss of heats of the two bodies during the same interval of time is(approximately)

Solution

T

T

_{1}=277℃=277+273=550 K T_{2}=67℃=67+273=340 K Temperature of surrounding T=27℃=27+273=300 K Ratio of loss of heat=(T_{1}^{4}-T^{4})/(T_{2}^{4}-T^{4}) =((T_{1}/T)^{4}-1)/((T_{2}/T)^{4}-1)=((550/300)^{4}-1)/((340/300)^{4}-1)=9.5/0.5=19/1**Q10.**The absolute zero is the temperature at which:

Solution

At absolute zero (i.e.,0 K)v

At absolute zero (i.e.,0 K)v

_{rms}becomes zero