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.**The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collides is?

Solution

Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE. According to conservation of energy 1/2 kL

Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE. According to conservation of energy 1/2 kL

^{2}=1/2 Mv^{2}Or kL^{2}=(Mv)^{2}/M MKL^{2}=p^{2}(p=Mv) ∴p=L√MK**Q2.**In which case does the potential energy decrease?

Solution

In compression or extension of a spring work is done against restoring force In moving a body against gravity work is done against gravitational force of attraction It means in all three cases potential energy of the system increases But when the bubble rises in the direction of upthrust force then system works so the potential energy of the system decreases

In compression or extension of a spring work is done against restoring force In moving a body against gravity work is done against gravitational force of attraction It means in all three cases potential energy of the system increases But when the bubble rises in the direction of upthrust force then system works so the potential energy of the system decreases

**Q3.**Two bodies A and B have masses 2 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times t

_{A}and t

_{B}, then the ratio t

_{A}/t

_{B}is ?

Solution

According to question, 1/2 m

According to question, 1/2 m

_{A}v_{A}^{2}=1/2 m_{B}v_{B}^{2}⇒v_{A}/v_{B}=√(m_{B}/m_{A})=√(5/20)=1/2 Using Impulse Momentum (F∆t_{A})/(F∆t_{B})=(m_{A}∆v_{A})/(m_{B}∆v_{B})⇒(∆t_{A})/(∆t_{B})=20/5×1/2=2

**Q4.**If a lighter body (Mass M

_{1}and velocity V

_{1}) and a heavier body (mass M

_{2}and velocity V

_{2}) have the same kinetic energy, then?

Solution

P=√2ME. If kinetic energy are equal then P∝√m i.e., heavier body posses large momentum As M

P=√2ME. If kinetic energy are equal then P∝√m i.e., heavier body posses large momentum As M

_{1}less than M_{2}therefore M_{1}V_{1}less than M_{2}V_{2}**Q5.**A body moves a distance of 10 m along a straight line under action of 5 N force. If work done is 25 J, then angle between the force and direction of motion of the body will be?

Solution

Given W=25 J,F=5 N,∆s=10m Work=Force×displacement W=(F cosÎ¸)×∆s Or cosÎ¸=W/(F∙∆s) Or cosÎ¸=25/(5×10)=1/2 or Î¸=cos

Given W=25 J,F=5 N,∆s=10m Work=Force×displacement W=(F cosÎ¸)×∆s Or cosÎ¸=W/(F∙∆s) Or cosÎ¸=25/(5×10)=1/2 or Î¸=cos

^{-1})(1/2)=60° Hence, angle between force and direction of body is 60°.**Q6.**A body of mass 5 kg moving with a velocity 10 m/s collides with another body of the mass 20 kg at, rest and comes to rest. The velocity of the second body due to collision is?

Solution

Momentum conservation 5×10+20×0=5×0+20×v⇒v=2.5 m/s

Momentum conservation 5×10+20×0=5×0+20×v⇒v=2.5 m/s

**Q7.**An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel Î¸ with the vertical. Work done by the force F is?

**Q8.**A spring with spring constant k is extended from x=0to x=x

_{1}. The work done will be?

Solution

1/2 kx

1/2 kx

_{1}^{2}**Q9.**A ball of mass m falls vertically to the ground from a height h

_{1}and rebound to a height h

_{2}. The change in momentum of the ball on striking the ground is?

Solution

When ball falls vertically downward from height h

When ball falls vertically downward from height h

_{1}its velocity v_{1}=√(2gh_{1}) And its velocity after collision v_{2}=√(2gh_{2}) Change in momentum ∆P=m(v_{2}-v_{1})=m(√(2gh_{1})+√(2gh_{2})) [Because v_{1}and v_{2}are opposite in direction]**Q10.**A body of mass M

_{1}collides elastically with another mass M

_{2}at rest. There is maximum transfer of energy when?:

Solution

M

M

_{1}=M_{2}