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.**A light and a heavy body have equal kinetic energy. Which one has a greater momentum

(b) P=√2mE if Eare equal then P∝√m i.e., heavier body will possess greater momentum

**Q2.**Two bodies A and B have masses 20 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 ration t

^{A}/t

^{B}is

(b) a

_{A}=F/m

_{A}=(4×10)/20=2ms

^{-2}a

_{B}=F/m

_{B}=(4×10)/5=8ms

^{-2}Given that, K

_{A}=K

_{B}i.e., 1/2 m

_{A}v

_{A}

^{2}=1/2 m

_{B}v

_{B}

^{2}Or m

_{A}(u+a

_{A}t

_{A})

^{2}=m

_{B}(u+a

_{B}t

_{B})

^{2}(∵v=u+at) Or m

_{A}a

_{A}

^{2}t

_{A}

^{2}=m

_{B}a

_{B}

^{2}t

_{B}

^{2}(∵u=0) Or t

_{A}/t

_{B}=√(m

_{B}/m

_{A}×(a

_{B}

^{2})/(a

_{A}

^{2})) =√(5/20×(8)

^{2}/(2)

^{2})=√((5×64)/(20×4))=2

**Q3.**A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12kg.The velocity of 18 kg mass is 6 ms

^{-1}.The kinetic energy of the other mass is

(b) The linear momentum of exploding part will remain conserved. Applying conservation of linear momentum, We write, m1 u1=m2 u2 Here,m1=18kg,m2=12kg u1=6ms

^{-1},u2=? ∴18×6=12 u2 ⟹u2=(18×6)/12 9ms

^{-1}Thus, kinetic energy of 12 kg mass k2=1/2 m2 u2

^{2}=1/2×12×(9)

^{2}=6×81 =486 J

**Q4.**The graph between √E and 1/p is (E=kinetic energy and p= momentum)

(c) P=√2mE it is clear that P∝√E So the graph between P and √E will be straight line But graph between 1/P and √E will be hyperbola

**Q5.**A bag (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined (bag + bullet) system

(d) Initial momentum =mv Final momentum =(m+M)V By conservation of momentum mv=(m+M)V ∴Velocity of (bag + bullet) system V=mv/(M+m) ∴ Kinetic energy =1/2 (m+M) V

^{2}=1/2 (m+M) (mv/(M+m))

^{2}=1/2 (m

^{2}v

^{2})/(M+m)

**Q6.**The kinetic energy k of a particle moving along a circle of radius R depends upon the distance sas k=as

^{2}. The force acting on the particle is

(b) Here k=1/2 mv^2=as^2 ∴ mv

^{2}=2as

^{2}Differentiating w.r.t. time t 2mv dv/dt=4as ds/dt=4asv,m dv/dt=2as This is the tangential force, Ft=2as Centripetal force Fc=(mv

^{2})/R=(2as

^{2})/R ∴ Force acting on the particle F=√(Ft

^{2}+Fc

^{2})=√((2as)

^{2}+(2as/R)

^{2})=2as√(1+s

^{2}/R

^{2})

**Q7.**If the linear momentum is increased by 50%, then kinetic energy will be increased by

(c) The relation between linear momentum and kinetic energy is p

^{2}=2mk ….(i) But linear momentum is increased by 50%, then p'=150/100 p p'=3/2 p Hence, p'2=2mk' Or (3/2 p)

^{2}=2mk' Or 9/4 p

^{2}=2mk' ….(ii) On putting the value of p^2 from Eq. (i) in Eq. (ii) 9/4×2mk=2mk' Or K'=9/4 k So, the increase in kinetic energy is ∆K=9/4 k-k=5/4 k Hence, percent increase in kinetic energy =((5/4)K)/K×100% =5/4×100%=125%

**Q8.**A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kg of energy. The initial velocity of the shell is

(c) Mass of the shell =m1=0.2 kg Mass of the gun =m2=4kg Let energy of shell =E1, energy of gun =E2 Total energy liberated =E1+E2=1050 Joule …(i) As E=P

^{2}/2m ∴E1/E2 =m2/m1 =4/0.2=20⇒E2=E1/20 …(ii) From equation (i) and (ii) we get E1=1000 Joule ∴ Kinetic energy of the shell =1/2 m1 v1

^{2}=1000 ⇒1/2 (0.2) v1

^{2}=1000⇒v1=√10000=100 m/s

**Q9.**A block(B) is attached to two unstretched springs S_1 and S_2 with springs constants k and 4k,representively (see Fig. I)The other ends are attached to identical supports M

_{1}and M

_{2}not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall I by small distance x(Fig II) and released. The block returns and moves a maximum distance y towards wall 2.Displacements x and y are measured with respect to the equilibrium position of the block B The ratio y/x is

(c) From energy conservation, 1/2 kx

^{2}=1/2 (4k) y

^{2}y/x=1/2

**Q10.**A body of mass m moving with velocity vcollides head on another body of mass 2m which is initially at rest. The ratio of KE of colliding body before and after collision body before and after collision will be

(d) KE of colliding body before collision=1/2 mv^2 After collision its velocity becomes V’=((m1-m2)/(m1+m2 ))v=m/3m v=v/3 KE after collision=1/2 mv

^{2}2=1/2 m(v/3)

^{2}=1/2 (mv

^{2})/9 Ratio of kinetic energy=(KEbefore)/(KEafter ) =(1/2 mv

^{2})/(1/2 (mv

^{2})/9)=9/1