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.
.

**Q1.**The energy which an e

^{-}acquires when accelerated through a potential difference of 1 volt is called?

Solution

1 eV

1 eV

**Q2.**A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of 60° with the vertical with the same initial speed. At highest points of their journey, the ratio of their potential energies will be?

**Q3.**If the K.E. of a body is increased by 300%, its momentum will increase by?

Solution

Let initial kinetic energy, E

Let initial kinetic energy, E

_{1}=E Final kinetic energy, E_{2}=E+300% of E=4E As P∝√E⇒P_{2}/P_{1}=√(E_{2}/E_{1})=√(4E/E)=2⇒P_{2}=2P_{1}⇒P_{2}=P_{1}+100% of P_{1}i.e., Momentum will increase by 100%

**Q4.**A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The initial velocity of its projection is?

Solution

Let ball is projected vertically downward with velocity v from height h Total energy at point A=1/2 mv

50%(1/2 mv

Let ball is projected vertically downward with velocity v from height h Total energy at point A=1/2 mv

^{2}+mgh During collision loss of energy is 50% and the ball rises up to same height. It means it possess only potential energy at same level50%(1/2 mv

^{2}+mgh)=mgh 1/2 (1/2 mv^{2}+mgh)=mgh v=√2gh=√(2×10×20) ∴v=20 m/s**Q5.**An engine of power 7500 W makes a train move on a horizontal surface with constant velocity of 20 ms

^{-1}. The force involved in the problem is?

Solution

Power = 7500, W = 7500 Js

Power = 7500, W = 7500 Js

^{-1}, velocity v=20 ms^{-1}P=Fv or F=P/v=(7500 Js^{-1})/(20 ms^{-1})=375 N**Q6.**Identify the wrong statement?

Solution

If a body has momentum, it must have kinetic energy also, (a) is the wrong statement If the energy is totally potential, it need not have momentum (b) is correct (c) and (d) are also correct

If a body has momentum, it must have kinetic energy also, (a) is the wrong statement If the energy is totally potential, it need not have momentum (b) is correct (c) and (d) are also correct

**Q7.**If a shell fired from a cannon, explodes in mid air, then?

Solution

Its total kinetic energy increases

Its total kinetic energy increases

**Q8.**A body of mass 2 kg is thrown up vertically with kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of its original value is?

Solution

Potential energy at the required height =490/2=245 J Again, 245=2×10×h or h=245/20 m=12.25 m

Potential energy at the required height =490/2=245 J Again, 245=2×10×h or h=245/20 m=12.25 m

**Q9.**A particle is moving under the influence of a force given by F=kx,where k is a constant and x is the distance moved. The energy (in joule )gained by the particle in moving from x=0 to x=3 is?

Solution

The energy gained by the particle U=1/2 k(x

The energy gained by the particle U=1/2 k(x

_{2}^{2}-x_{1}^{2}) =1/2 k(3^{2}-0^{2})=9/2 k 4.5k**Q10.**The kinetic energy acquired by a mass m in travelling a certain distance d starting from rest under the action of a constant force is directly proportional to?:

Solution

Kinetic energy acquired by the body = Force applied on it × distance covered by the body K.E. =F×d If F and d both are same then K. E. acquired by the body will be same

Kinetic energy acquired by the body = Force applied on it × distance covered by the body K.E. =F×d If F and d both are same then K. E. acquired by the body will be same