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.**

**Statement I**Two particles moving in the same direction do not lose all their energy in a completely in elastic collision.

**Statement II**Principle of conservation of momentum holds true for all kinds of collisions.?

**Q2.**A ball is dropped from a height h. If the coefficient of restitution be e, then to what height will it rise after jumping twice from the ground?

Solution

h

h

_{n}=he^{2n},if n=2 then h_{2}=he^{4}**Q3.**The work done in dragging a stone of mass 100 kg up an inclined plane 1 in 100 through a distance of 10 m is (take g=9.8 ms

^{-2})?

Solution

Slope of inclined plane, sinÎ¸= 1/100 Component o weight down the inclined plane F=mg sinÎ¸=100×9.8×1/100=9.8 N s= distance moved = 10 m W=F s=9.8×10=98 J

Slope of inclined plane, sinÎ¸= 1/100 Component o weight down the inclined plane F=mg sinÎ¸=100×9.8×1/100=9.8 N s= distance moved = 10 m W=F s=9.8×10=98 J

**Q4.**Four smooth steel balls of equal mass at rest are free to move along a straight line without friction. The first ball is given a velocity of 0.4 ms

^{-1}. It collides head on with the second one elastically, the second one similarly with the third and so on. The velocity of the last ball is?

Solution

If after the collision of two bodies, the total kinetic energy of the bodies remains the same as it was before the collision, and also momentum remains same, then it is a case of perfectly elastic collision. Momentum before collision= Momentum after collision Kinetic energy before collision =Kinetic energy after collision Also, u

If after the collision of two bodies, the total kinetic energy of the bodies remains the same as it was before the collision, and also momentum remains same, then it is a case of perfectly elastic collision. Momentum before collision= Momentum after collision Kinetic energy before collision =Kinetic energy after collision Also, u

_{1}-u_{2}=-(v_{1}-v_{2}) Where (u_{1}-u_{2}) is the relative velocity before the collision and (v_{1}-v_{2})is the relative velocity after the collision. Thus, in a perfectly elastic collision the relative velocity remains unchanged in magnitude, but is reserved in direction. Hence, velocity of the last ball is -0.4 ms^{-1}.**Q5.**A car of mass 1000 kg moves at a constant speed of 20 ms

^{-1}up an incline. Assume that the frictional force is 200 N and that sinÎ¸ = 1/20 where, Î¸ is the angel of the incline to the horizontal. The g=10 ms

^{-2}. Find the power developed by the engine?

Solution

P=(mg sinÎ¸+F)v =(1000×10×1/20+200)×20 =1400 W=14 kW

P=(mg sinÎ¸+F)v =(1000×10×1/20+200)×20 =1400 W=14 kW

**Q6.**Two springs A and B are identical but A is harder than B(k

_{A}>k

_{B}). Let W

_{A}and W

_{B}represent the work done when the springs are stretched through the same distance and W'

_{A}and W'

_{B}are the work done when these are stretched by equal forces, then which of the following is true?

Solution

k

k

_{A}>k_{B},x is the same ∴1/2 k_{A}x^{2}>1/2 k_{B}x^{2}⇒W_{A}>W_{B}Forces are the same k_{A}x_{A}=k_{B}x_{B},Ask_{A}>k_{B},x_{A}less than x_{B}W_{A}^{'}=1/2 (k_{A}x_{A}) x_{A}and W_{B}^{'}=1/2 (k_{B}x_{B}) x_{B}∴W_{A}^{'}less than W_{B}^{'}; ∴W_{A}>W_{B}but W_{A}^{'}less than W_{B}^{'}**Q7.**A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2 ms

^{-1}at a constant acceleration of 2 ms

^{-2}. Then rate of change of kinetic energy?

Solution

K= 1/2 mv

K= 1/2 mv

^{2}dK/dt=mv.dv/dt =(m dv/dt)v=(ma v=4v) As m=2 kg and a=2 ms^{-2}.**Q8.**A rubber ball is dropped from a height of 5 m on a planet, where the acceleration due to gravity is not known. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of ?

Solution

Potential energy=Kinetic energy Ie, mgh=1/2 mv

Potential energy=Kinetic energy Ie, mgh=1/2 mv

^{2}Or v=√2gh If h_{1}and h_{2}are initial and final heights, then v_{1}=√(2gh_{1}), v_{2}=√(2gh_{2}) Loss in velocity ∆v=v_{1}-v_{2}=√(2gh_{1})-√(2gh_{2}) ∴Fractional loss in velocity =∆v/v_{1}=(√(2gh_{1}) -√(2gh_{2}))/√(2gh_{1}) ∆v/v_{1}=1-√(h_{2}/h_{1}) =1-√(1.8/5) =1-√0.36=1-0.6=0.4=2/5**Q9.**If a body of mass 200 g falls from a height 200 m and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. of the body at the contact ( g=10 m/s

^{2})?

Solution

∆U=mgh=0.2×10×200=400J ∴ Gain in K.E. = decrease in P.E. =400 J

∆U=mgh=0.2×10×200=400J ∴ Gain in K.E. = decrease in P.E. =400 J

**Q10.**A machine which is 75 percent efficient, uses 12 joules of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in ms

^{-1})?:

Solution

Potential energy of a body =75% of 12 J mgh=9 J⇒h=9/(1×10)=0.9 m Now when this mass allow to fall then it acquire velocity v=√2gh=√(2×10×0.9)=√18 m/s

Potential energy of a body =75% of 12 J mgh=9 J⇒h=9/(1×10)=0.9 m Now when this mass allow to fall then it acquire velocity v=√2gh=√(2×10×0.9)=√18 m/s