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 ball of mass 0.2 kg is thrown vertically upward by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 m height further, find the magnitude of the force. Consider g=10ms

^{-2}:

Solution

The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B, let acceleration of ball during PA is ams

.

The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B, let acceleration of ball during PA is ams

^{-2}[assumed to be constant] in upward direction and velocity of ball at A is vms^{-1}. Then for PA , v^{2}=0^{2}+2a×0.2 For AB , 0=v^{2}-2×g×2 ⟹v^{2}=2g×2 From above equation, a=10g=100 ms^{-2}Then for PA, FBD of ball is F-mg=ma [F is the force exerted by hand on ball] ⟹F=m(g+a)=0.2 (11g) =22 N.

**Q2.**Which one of the following is not a contact force

Solution

Magnetc Force

Magnetc Force

**Q3.**A block of mass 50 kg slides over a horizontal distance of 1m. If the coefficient of friction between their surface is 0.2, then work done against friction is

Solution

W=Î¼mgS=0.2×50×9.8×1=98 J

W=Î¼mgS=0.2×50×9.8×1=98 J

**Q4.**A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension in the thread will be

Solution

T= (2m

T= (2m

_{1}m_{2})/(m_{1}+m_{2}) g=(2×10×6)/(10+6)×9.8=73.5 N**Q5.**A machine gun fires a bullet of mass 40 g with a velocity 1200 ms

^{-1}. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most

Solution

u= velocity of bullet dm/dt= Mass of bullet fired per second by the gun dm/dt= Mass of one bullet (m

u= velocity of bullet dm/dt= Mass of bullet fired per second by the gun dm/dt= Mass of one bullet (m

_{B})× Bullets fired per sec (N) Maximum force that man can exert F=u(dm/dt) ∴F=u×m_{B}×N ⇒N=F/(m_{B}×u)=144/(40×10^{-3}×1200)=3**Q6.**A body shown in figure is accelerating downward with acceleration 2 m/s

^{2}. The tension in the string is

**Q7.**A wooden wedge of mass M and inclination angle Î± rests on a smooth floor. A block of mass m is kept on wedge. A force P ⃗ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force P ⃗ is

**Q8.**A force-time graph for a linear motion of a body is shown in the figure. The change in linear momentum between 0 and 7 s is

Solution

Since,F=∆p/∆t or ∆p=F ∆t We can say that momentum between 0 to 7 s is equal to the vector area enclosed by the force-time graph from 0 to 7 s. So, Change in linear momentum = vector area of triangle OAB + vector area of square BCDE+ vector area of triangle EFG + vector area of square GHIJ + vector area of triangle JKL =[1/2×1×(-1) ]+[2×2]+[1/2×2×(-2) ]+[1×1]+[1/2×1×(-1) ] =-1/2+4-2+1-1/2=2 Ns

Since,F=∆p/∆t or ∆p=F ∆t We can say that momentum between 0 to 7 s is equal to the vector area enclosed by the force-time graph from 0 to 7 s. So, Change in linear momentum = vector area of triangle OAB + vector area of square BCDE+ vector area of triangle EFG + vector area of square GHIJ + vector area of triangle JKL =[1/2×1×(-1) ]+[2×2]+[1/2×2×(-2) ]+[1×1]+[1/2×1×(-1) ] =-1/2+4-2+1-1/2=2 Ns

**Q9.**A piece of wire is bent in the shape of a parabola y=kx

^{2}(y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is

**Q10.**A gun fires bullet each of mass 1 g with velocity of 10 ms

^{-1}by exerting a constant force of 5 g weight. Then the number of bullets fired per second is (Take g=10 ms

^{-2}) :

Solution

Mass of each bullet(m)=1 g=0.001 kg Velocity of bullet (v)=10 ms

Mass of each bullet(m)=1 g=0.001 kg Velocity of bullet (v)=10 ms

^{-1}Applied force (F)=5 g-wt. =5/1000×10 N =0.05 N Let n bullets are fired per second, then Force= rate of change of linear momentum ie,F=n×mv ∴ Number of bullets fired per second n=F/mv =0.05/(0.001×10)=5