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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. Two weights w1 and w2 are suspended from the ends of a light string over a smooth fixed pulley. If the pulley is pulled up with acceleration g, the tension in the string will be:
•  (4w1 w2)/(w1+w2 )
•  (2w1 w2)/(w1+w2 )
•  (w1 -w2)/(w1+w2 )
•  (w1 w2)/(2(w1+w2))
Solution
For solving the problem, we assume that observer is situated in the frame of pulley (non-inertial reference frame) m1 g=w1 m2 g=w2 From force diagram,
T-m2 a0-w2=m2 a or T-m2 g-w2=m2 a (∵a0=g) or T-2w2=m2 a…(i) From force diagram, m1 a0+w1-T=m1 a or m1 a0+w1-T=m1 a or 2w1-T=m1 a…(ii) (∵a0=g) From Eqs. (i) and (ii), T=(4w1 w2)/(w1+w2 )

Q2. A block of weight 5N is pushed against a vertical wall by a force 12N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is
•  12 N
•  5 N
•  7.2 N
•  13 N
Solution
Wall applies 2 forces of the block (i) normal reaction, R=12 N, and (ii) frictional force, f2=mg=5 N tangentially upward
∴ Total force exerted by wall on block F=√(N2+fs2 )=√((12)2+(5)2 )=13N

Q3. A solid disc of mass M is just held in air horizontal by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity 6 ms-1. If the mass of each stone is 0.05 kg. What is the mass of the disc (g=10 ms-2)
•  1.2 Kg
•  0.5 Kg
•  20 Kg
•  3 Kg
Solution
Weight of the disc will be balanced by the force applied by the bullet on the disc in vertically upward direction F=nmv =40×0.05×6=Mg M=(40×0.05×6)/16 =1.2 kg

Q4. What is the maximum value of the force F such that the block shown in the arrangement, does not move?
•  20 N
•  10 N
•  12 N
•  15 N
Solution
Free body diagram (FBD) of the block (shown by a dot) is shown in figure.
For vertical equilibrium of the block, N=mg+F sin⁡60°=√3 g+√3 F/2 ….(i) For no motion, force of friction f≥F cos⁡60° or °μN≥F cos⁡60° or 1/(2√3) (√3 g+(√3 F)/2)≥F/2 or g≥F/2 or F≤2 gor 20 N Therefore, maximum value of F is 20 N.

Q5. A rocket of mass 100 kg burns 0.1 kg of fuel per sec. If velocity of exhaust gas is 1 km/sec, then it lifts with an acceleration of.
•  1000 ms-2
•  100 ms-2
•  10 ms-2
•  1 ms-2
Solution
dm/dt=0.1 kg / sec; Mass of the rocket = 100 kg v=1 km/sec=1000 m/sec F=(d(mv))/dt=m dv/dt-v dm/dt=0 as the mass is decreasing 100a-1000×0.1=0 a=+1 m/s2

Q6. A person used force (F), shown in figure move a load with constant velocity on give surface.

Identify the correct surface profile?
Solution
The correct surface profile will be (a), because slope of surface should change from one constant value (non-zero) in terms of sign because force is constant picewise.

Q7. A 60 kg man stands on a spring scale in a lift. At some instant he finds that the scale reading has changed from 60 kg to 50 kg for a while and then comes again to 60 kg mark. What should he conclude?
•  The lift was in constant motion upwards
•  The lift was in constant motion downwards
•  The lift while in motion suddenly stopped
•  The lift while in motion upwards suddenly stopped
Solution
When lift falls with acceleration (a)or rises with retardation (-a), then a person apparently loses weight. R-mg=-ma R=m(g-a) In the given case scale reading changes from 60 kg to 50 kg for a while and then comes back to 60 kg mark. It happens while the lift in motion upwards suddenly stops

Q8. The monkey B shown in figure is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it? (Take g=10 ms-2)
•  105 N
•  108 N
•  10.5 N
•  100 N
Solution
If A is climbing with constant velocity, then T'=5g+T and T=2g T'=5g+2g=7g=7×10N=70N Suppose A is climbing with acceleration a such that T=30 N That T=30 N T-2a=2a 30-2×10=2a or a=5 ms-2 Again, T'-T-5g=5a or T'=T+5g+5a or T'=(30+50+25) N= 105 N

Q9. In the above question, if the lift is moving upwards with a uniform velocity, then the frictional resistance offered by the body is
•  Mg
•  μMg
•  2μMg
•  Zero
Solution
When the lift is moving upward with constant velocity then, R=mg ∴F=μR= μmg

Q10. A passenger is travelling in a train moving at 72 kmh-1. His suitcase is kept on the berth. The driver of the train applies brakes such that the speed of the train decreases at a constant rate of 36 kmh-1 in 5 s. What should be the minimum coefficient of friction between the suitcase and the berth if the suitcase is not the slide during retardation of the train?:
•  10/49
•  10/98
•  28/49
•  30/98
Solution
Retardation of train = (36 kmh-1)/(5 s) =(35×5/18 ms-1)/5s=2 ms-2 It acts in the backward direction Fictitious force on suitcase = 2m N, Where m is the mass of suitcase It acts in the forward direction Due to this force, the suitcase has a tendency to slide forward. If suitcase is not to slide, then 2m= force f of friction or 2m=μmg or μ=2/g=2/9.8=20/98=10/49 ## Want to know more

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NLM Quiz 19
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