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 backside of a truck is open and a box of 40 kg is placed 5m away from the rear end. The coefficient of friction of the box with the surface of the truck is 0.15. The truck starts from rest with 2 m/s

^{2}acceleration. Calculate the distance covered by the truck when the box falls off

Solution

Here, Mass of the box, M=40 kg Acceleration of the truck, a=2 ms

As the truck moves in forward direction with the acceleration a=2ms

Here, Mass of the box, M=40 kg Acceleration of the truck, a=2 ms

^{-2}Distance of the box from the rear end , d=5m Coefficient of friction between the box and the surface below it, Î¼=0.15 The various forces acting on the block are as shown in the figureAs the truck moves in forward direction with the acceleration a=2ms

^{-2}, the box experiences a force F in the backward direction and it is given by f=Ma=(40 kg)×(2 ms^{-2})=80 N in backward direction Under the action of this force, the box will tend to move toward the rear end of the truck . As it does so, its motion will be opposed by the force of friction which acts in the forward direction and it is given by f=Î¼N=Î¼Mg=0.15×40×10=60 N The acceleration of the box relative to the truck toward the rear end is, a_{1}=(F-f)/M=(80 N-60 N)/(40 kg)=0.5 ms^{-1}Let t be the time taken for the box to fall off the truck Using, S=ut+1/2 at^{2}, we get , d=0×t+1/2 a_{1}t^{2}[∵u=0] 5= 1/2×0.5×t^{2},t=√((2×5)/0.5)=√20 s During this time, the truck covers a distance x Using S=ut+1/2 at^{2}We get x=0×t+1/2×2×(√20)^{2}[∵u=0] x=20 m**Q2.**In the first second of its flight, rocket ejects 1/60 of its mass with a velocity of 2400 ms

^{-1}. The acceleration of the rocket is

Solution

Acceleration a=1/m (-dm)/dt) v

Acceleration a=1/m (-dm)/dt) v

_{r}=1/1 (1/60)×2400=40 ms^{-2}**Q3.**A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with velocity of 3 i ̂ms

^{-1}and the other with a velocity of 4 j ̂ ms

^{.-1}. If the explosion occurs in 10

^{-4}s, the force acting on the third piece in newtons is

Solution

By law of conservation of linear momentum. m

By law of conservation of linear momentum. m

_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}=0 Here m_{1}=m_{2}=m_{3}=1 kg, v_{1}=3 i ̇ ̂,v_{2}=4 j ̇ ̂ ∴ 3 i ̇ ̂+4 j ̇ ̂+u_{3}=0 The average force acting on the third piece is F=(mv_{3})/t =(1×-(3 i ̇ ̂+4 j ̇ ̂ ))/10^{4}N =-(3 i ̇ ̂+4 j ̇ ̂ )×10^{4}N

**Q4.**Two bodies A and B of masses 10 kg and 15 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. If T represents the tension in the spring when a horizontal force F=500 N is applied to A (as shown in figure 1) and T'' be the tension when it is applied to B (figure 2), then which of the following is true

**Q5.**A bullet of mass 10 g moving with 300 ms

^{-1}hits a block of ice of mass 5 kg and drops dead. The velocity of ice is

Solution

Applying law of conservation of liner momentum, ie, m

Applying law of conservation of liner momentum, ie, m

_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}Here,m_{1}=10 g=10^{-2}. kg,m_{2}=5 kg u_{1}=300 ms^{-1},u_{2}=0 v_{1}=0,v_{2}=? ∴ 10^{-2}×300+5×0=10^{-2}×0+5v^{2}or 5v_{2}=3 or v_{2}=3/2 ms^{-1}=60 cms^{-1}**Q6.**A force of 98 N is required to just start moving a body of mass 100 kg over ice. The coefficient of static friction is

Solution

Î¼= F/R=F/mg=98/(100×9.8)=1/10=0.1

Î¼= F/R=F/mg=98/(100×9.8)=1/10=0.1

**Q7.**A boy having a mass equal to 40 kilograms is standing in an elevator. The force felt by the feet of the boy will be greatest when the elevator g=9.8 metres/ sec

^{2}

Solution

Accelerates upward with an acceleration equal to 4 metres/sec

Accelerates upward with an acceleration equal to 4 metres/sec

^{2}**Q8.**The motion of a rocket is based on the principle of conservation of

Solution

Linear momentum

Linear momentum

**Q9.**A body of mass 2 kg is being dragged with uniform velocity of 2 m/s on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.20. The amount of heart generated in 5 sec is (J=4.2 joule/calandg=9.8 m/s

^{2})

Solution

Work done = Force × displacement =Î¼mg×(v×t) W=(0.2)×2×9.8×2×5 joule Heat generated Q=W/J=(0.2×2×9.8×2×5)/4.2=9.33 cal

Work done = Force × displacement =Î¼mg×(v×t) W=(0.2)×2×9.8×2×5 joule Heat generated Q=W/J=(0.2×2×9.8×2×5)/4.2=9.33 cal

**Q10.**The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of their resultant is 12 N. If the resultant is at 90° with the smaller force, the magnitude of the forces in N are Replace Diagram

Solution

Let smaller force be F

∴ R

Let smaller force be F

_{1}. Resultant R of the forces is at 90° to AB,∴ R

^{2}+F_{1}^{2}=F_{2}^{2}in ∆ABC or (12)^{2}=F^{2}^{2}-F_{1}^{2}……(i) or 144=(F_{2}-F_{1})(F_{2}+F_{1}) but F_{1}+F_{2}=18 N (given) ….(ii) ∴ F_{2}-F_{1}=144/18=8 …..(iii) From Eqs. (ii) and (iii), F_{1}=5, F_{2}=13 Hence, forces are 5 N and 13 N.