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 rods of equal length and area of cross-section are kept parallel and lagged between temperature 20℃ and 80℃. The ratio of the effective thermal conductivity to that of the first rod is [the ratio (K

_{1}/K

_{2})=3/4]:

Solution

For parallel combination of two rods of equal length and equal area of cross-section. K= (K

For parallel combination of two rods of equal length and equal area of cross-section. K= (K

_{1}+K_{2})/2= (K_{1}+(4K_{1})/3)/2 = (7K_{1})/6 Hence , K/K_{1}=7/6**Q2.**Two plates of same thickness, of coefficients of thermal conductivity K

_{1}and K

_{2}and areas of cross section A

_{1}and A

_{2}are connected as shown in figure. The common coefficient of thermal conductivity K will be

Solution

As is clear from figure. dQ/dt=(dQ

As is clear from figure. dQ/dt=(dQ

_{1})/dt+(dQ_{2})/dt K(A_{1}+A_{2})dT/dx=K_{1}A_{1}dT/dx+K_{2}A_{2}dT/dx K= (K_{1}A_{1}+K_{2}A_{2})/(A_{1}+A_{2})**Q3.**Calorie is defined as the amount of heat required to raise temperature of 1 g of water by 1 ℃ and it is defined under which of the following conditions?

Solution

1 calorie is the heat required to raise the temperature of 1 g of water from 14.5℃ to 15.5℃ at 760 mm of Hg

1 calorie is the heat required to raise the temperature of 1 g of water from 14.5℃ to 15.5℃ at 760 mm of Hg

**Q4.**The thermal capacity of a body is 80 cal, then its water equivalent is

Solution

We know that thermal capacity of a body expressed in calories is equal to water equivalent of the body expressed in grams

We know that thermal capacity of a body expressed in calories is equal to water equivalent of the body expressed in grams

**Q5.**Recently, the phenomenon of superconductivity has been observed at 95 K. This temperature is nearly equal to.

Solution

(F-32)/9=(K-273)/5⇒(F-32)/9=(95-273)/5⇒F=-288

(F-32)/9=(K-273)/5⇒(F-32)/9=(95-273)/5⇒F=-288

^{o}F**Q6.**Assuming the sun to be a spherical body of radius Rat a temperature of TK, evaluate the total radiant power, incident on earth, at a distance r from the sun Where r

^{0}is the radius of the earth and Ïƒ is stefan’s constant.v

Solution

From Stefan’s law, the rate at which energy is radiated by sun at its surface is P=Ïƒ×4Ï€r

From Stefan’s law, the rate at which energy is radiated by sun at its surface is P=Ïƒ×4Ï€r

^{2}T^{4}[Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e=1.] The intensity of this power at earth’s surface(under the assumption r>>r_{0}) is I=P/(4Ï€R^{2})=(Ïƒ×4Ï€r^{2}T^{4})/(4Ï€R^{2})=(ÏƒR^{2}ÏƒT^{4})/r^{2}The area of earth which receives this energy is only one-half of total surface area of earth, whose projection would be Ï€r_{0}^{2}. ∴ Total radiant power as received by earth = Ï€r_{0}^{2}×I = (Ï€r_{0}^{2}×ÏƒR^{2}T^{4})/r^{2}=(Ï€r_{0}^{2}R^{2}ÏƒT^{4})/r^{2}
Solution

The volume of matter in portion AB of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state

The volume of matter in portion AB of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state

**Q8.**The ends of two rods of different materials with their thermal conductivities, radii of cross-sections and lengths all are in the ratio 1:2 are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is 4 cal/s, that in the shorter rod in cal/s will be

Solution

dQ/dt=K(Ï€r

dQ/dt=K(Ï€r

^{2})dÎ¸/dt⇒(dQ/dt)_{s}/(dQ/dt)_{l}=(K_{s}×r_{s}^{2}×l_{1})/(K_{l}×r_{1}^{2}×l_{s})=1/2×1/4×2/1 ⇒(dQ/dt)_{s}=(dQ/dt)_{l}/4=4/4=1**Q9.**A hot body will radiate heat most rapidly if its surface is

Solution

Black and rough surfaces are good absorber that’s why they emit well. (Kirchhoff’s law)

Black and rough surfaces are good absorber that’s why they emit well. (Kirchhoff’s law)

**Q10.**It is hotter for the same distance over the top of a fire than it is in the side of it, mainly because:

Solution

Convection significantly transfer heat upwards (Gravity effect)

Convection significantly transfer heat upwards (Gravity effect)