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 bomb of mass 9 kg explodes into two parts. One part of mass 3 kg moves with velocity 16 m/s ,then the KE of the other part is?

Solution

From the law of conservation of momentum 3×16+6×v=9×0 Or v=-8 ms

From the law of conservation of momentum 3×16+6×v=9×0 Or v=-8 ms

^{-1}⇒ v=8ms^{-1}(numerically) Therefore, its kinetic energy k=1/2×6×8^{2}=192J**Q2.**A body of mass ‘M’ collides against a wall with a velocity v and retraces its path with the same speed. The change in momentum is (take initial direction of velocity as positive)?

Solution

Change in momentum =mv

Change in momentum =mv

_{2}-mv_{1}=-mv-mv=-2mv**Q3.**The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x=1mto x=5m will be?

**Q4.**A block C of mass mis moving with velocity v

_{0}and collides elastically with block A of mass m and connected to another block B of mass 2m through spring constant k.What is k if x

_{0}is compression of spring when velocity of A and B is same ?

Solution

Using conservation of linear momentum, we have mv

Using conservation of linear momentum, we have mv

_{0}=mv+2mv Or v=v_{0}/3 Using conservation of energy, we have 1/2 mv_{0}^{2}=1/2 kx_{0}^{2}+1/2 (3m) v^{2}Where x_{0}=compression in the spring , ∴mv_{0}^{2}=kx_{0}^{2}+(3m) (v_{0}^{2})/9 Or kx_{0}^{2}=mv_{0}^{2}-(mv_{0}^{2})/3 Or kx_{0}^{2}=(2mv_{0}^{2})/3 ∴k=(2mv_{0}^{2})/(3x_{0}^{2})**Q5.**A canon ball is fired with a velocity 200 m/sec at an angle of 60° with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m/sec, the second one falling vertically downwards with a velocity 100 m/sec. The third fragment will be moving with a velocity ?

Solution

Momentum of ball (mass m) before explosion at the highest point =mvi ̂=mu cos60°i ̂ =m×200×1/2 i ̂=100 mi ̂ kgms

Let the velocity of third part after explosion is V After explosion momentum of system =P

Momentum of ball (mass m) before explosion at the highest point =mvi ̂=mu cos60°i ̂ =m×200×1/2 i ̂=100 mi ̂ kgms

^{-1}Let the velocity of third part after explosion is V After explosion momentum of system =P

_{1}+P_{2}+P_{3}=m/3×100j ̂-m/3×100j ̂+m/3×Vi ̂ By comparing momentum of system before and after the explosion m/3×100j ̂-m/3×100j ̂+m/3 Vi ̂=100mi⇒V=300m/s**Q6.**If w

_{1},w

_{2}and W

_{3}represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between w

_{1},w

_{2}and w

_{3}?

Solution

Gravitational field is a conservative force field. In a conservative force field work done is path independent. ∴ W

Gravitational field is a conservative force field. In a conservative force field work done is path independent. ∴ W

_{1}=W_{2}=W_{3}**Q7.**A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is?

Solution

R=u√(2h/g)⇒20=V

R=u√(2h/g)⇒20=V

_{1}√((2×5)/10) and 100=V_{2}√((2×5)/10) ⇒V_{1}=20 m/s,V_{2}=100 m/s Applying momentum conservation just before and just after the collision (0.01)(V)=(0.2)(20)+(0.01)(100) V=500 m/s**Q8.**A gun of mass 20 kg has bullet of mass 0.1 kg in it. The gun is free to recoil 804 J of recoil energy are released on firing the gun. The speed of bullet (ms

^{-1}) is?

Solution

Here, m

Here, m

_{1}=20 kg, m_{2}=0.1 kg, v_{1}= velocity of recoil of gun, v_{2}= velocity of bullet As m_{1}v_{1}=m_{2}v_{2}v_{1}=m_{2}/m_{1}v_{2}=0.1/20 v_{2}=v_{2}/200 Recoil energy of gun=1/2 m_{1}v_{1}^{2}=1/2×20(v_{2}/200)^{2}804= (10v_{2}^{2})/(4×10^{4})=(v_{2}^{2})/(4×10^{3}) v_{2}=√(804×4×10^{3}) ms^{-1}**Q9.**A particle is placed at the origin and a force F=kx is acting on it (where k is positive constant). If U(0)=0, the graph of U(x) versus x will be (where U is the potential energy function)?

**Q10.**A wire of length L suspended vertically from a rigid support is made to suffer extension l in its length by applying a force F. The work is?