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 copper block of mass 4 kg is heated in a furnance to a temperature 425℃ and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper=500 J kg

^{-1}℃

^{-1}and heat of fusion of ice=336 k J kg

^{-1}):

Solution

Fall in temperature of copper block when it is placed on the ice block = ∆T=425-0=425℃. Heat lost by copper block when it is placed on the ice block. Q

Fall in temperature of copper block when it is placed on the ice block = ∆T=425-0=425℃. Heat lost by copper block when it is placed on the ice block. Q

_{1}=m_{1}s∆T = 4×500×425=850kJ Heat gained by ice in melting into m_{2}kg of water. Q_{2}=m_{2}L =m_{2}×336 =336m_{2}kJ According to Calorimetry principle, Heat lost=Heat gained ie,850=336m_{2}∴m_{2}=850/336=2.5kg**Q2.**Which of the substance A,B or C has the highest specific heat? The temperature vs time graph is shown

Solution

Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that t

Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that t

_{C}>t_{B}>t_{A}⇒C_{C}>C_{B}>C_{A}**Q3.**A spherical black body with a radius of 12 cm radiates 440 W power at 500K. If the radius were halved and the temperature doubled, the power radiated in watt would be

Solution

Radiated power by blackbody P=Q/t=AÏƒT

Radiated power by blackbody P=Q/t=AÏƒT

^{4}⇒P∝AT^{4}∝r^{2}T^{4}⇒P_{1}/P_{2}=(r_{1}/r_{2})^{2}(T_{1}/T_{2})^{4}⇒440/P_{2}=(12/6)^{2}(500/1000)^{4}⇒P_{2}=1760 W≈1800 W

**Q4.**How many grams of a liquid of specific heat 0.2 at a temperature 40℃ must be mixed with 100 gm of a liquid of specific heat of 0.5 at a temperature 20℃, so that the final temperature of the mixture becomes 32℃

Solution

Temperature of mixture Î¸=(m

Temperature of mixture Î¸=(m

_{1}c_{1}Î¸_{1}+m_{2}c_{2}Î¸_{2})/(m_{1}c_{1}+m_{2}c_{2}) ⇒32=(m_{1}×0.2×40+100×0.5×20)/(m_{1}×0.2+100×0.5)⇒m_{1}=375gm**Q5.**Equal masses of two liquids are filled in two similar calorimeters. The rate of cooling will.

Solution

Liquid having more specific heat has slow rate of cooling because for equal masses rate of cooling dÎ¸/dt∝1/c

Liquid having more specific heat has slow rate of cooling because for equal masses rate of cooling dÎ¸/dt∝1/c

**Q6.**The wavelength of the radiation emitted by a body depends upon?

Solution

According to Wien’s law, Î»∝1/T ie.,it depends on the temperature of the surface.

According to Wien’s law, Î»∝1/T ie.,it depends on the temperature of the surface.

**Q7.**The freezer in a refrigerator is located at the top section so that

Solution

The entire of the refrigerator is cooled quickly due to convection

The entire of the refrigerator is cooled quickly due to convection

**Q8.**A black body of mass 34.38 g and surface area 19.2 cm

^{2}is at an initial temperature of 400 K. It is allowed to cool inside an evacuated enclosure kept at constant temperature 300 K. The rate of cooling is 0.04℃s

^{-1}. The specific heat of the body in J kg

^{-1}K

^{-1}is (Stefan’s constant Ïƒ=5.73×10

^{-8}Wm

^{-2}K

^{-4})

Solution

According toNewtons’s law of cooling dÎ¸/dt=(ÏƒA(T

According toNewtons’s law of cooling dÎ¸/dt=(ÏƒA(T

^{4}-T_{0}^{4}))/ms ∴ Specific heat s= (ÏƒA(T^{4}-T_{0}^{4}))/m(dÎ¸/dt) Substituting the values ∴ s=((5.73×10^{-8}) )(19.2×10^{-4})[4^{4}-3^{4}]×10^{8})/((34.38×10^{-3})(4×10^{-2}))=1400**Q9.**Two slabs A and B of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab A is twice that of B. The coefficient of thermal conductivity of slab A is twice that of B. The first surface of slab A is maintained at 100℃, while the second surface of slab B is maintained at 25℃. The temperature at contact of their surfaces is

Solution

The temperature at the contact of the surface =(K

The temperature at the contact of the surface =(K

_{1}d_{2}Î¸_{1}+K_{2}d_{1}Î¸_{2})/(K_{1}d_{2}+K_{2}d_{1}) = (2K_{2}d_{2}×100+2d_{2}×K_{2}×25)/(2K_{2}d_{2}+K_{2}2d_{2}) =(200+50)/4=62.6℃**Q10.**Thermoelectric thermometer is based on:

Solution

Thermoelectric thermometer is based on Seebeck Effect

Thermoelectric thermometer is based on Seebeck Effect