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 solid sphere rolls down without slipping on an inclined plane at angle 60

^{°}over a distance of 10 m. The acceleration (in ms

^{-2}) is?

Solution

Here Î¸=60°,l=10 m,a=? a=(g sinÎ¸)/(1+K

Here Î¸=60°,l=10 m,a=? a=(g sinÎ¸)/(1+K

^{2}⁄R^{2}) For solid sphere K^{2}=2/5 R^{2}a=(9.8 sin60°)/(1+2/5)=5/7×9.8×√3/2=6.00 ms^{-2}**Q2.**A sphere of mass 50 g and diameter 20 cm rolls without slipping with a velocity of 5cm/sec. Its total kinetic energy is?

Solution

K

K

_{N}=1/2 mv^{2}(1+K^{2}/R^{2})=1/2×50×(5)^{2}×(1+2/5) =875erg**Q3.**Identify the correct statement for the rotational motion of a rigid body?

Solution

In rotational motion of a rigid body, the centre of mass of the body moves uniformly in a circular path.

In rotational motion of a rigid body, the centre of mass of the body moves uniformly in a circular path.

**Q4.**Three masses of 2 kg,4 kg and 4 kg are placed at the three points (1,0,0),(1,1,0) and (0,1,0) respectively. The position vector of its centre of mass is?

Solution

For centre of mass, x

For centre of mass, x

_{cm}=(2×1+4×1+4×0)/(2+4+4)=6/10=3/5 y_{cm}=(2×0+4×1+4×1)/(2+4+4)=8/10=4/5 ∴ Coordinate for cm=(3/5 i ̂,4/5 j ̂ ) Where i ̂ and j ̂ are unit vector along x and y axis**Q5.**The moment of inertia of a flywheel having kinetic energy 360 J and angular speed of 20 rads

^{-1}is?

Solution

Rotational kinetic energy of flywheel K=360 J Angular speed of flywheel (Ï‰)=20 rads

Rotational kinetic energy of flywheel K=360 J Angular speed of flywheel (Ï‰)=20 rads

^{-1}Rotational kinetic energy, K=1/2 IÏ‰^{2}∴ Moment of inertia, I=2K/Ï‰^{2}= (2×360)/(20)^{2}=1.8 kg-m^{2}**Q6.**Two spherical bodies of the same mass M are moving with velocities v

_{1}and v

_{2}. These collide perfectly inelastically?

Solution

Loss of kinetic energy =1/2 (m

Loss of kinetic energy =1/2 (m

_{1}m_{2})/(m_{1}+m_{2}) (v_{1}-v_{2})^{2}=1/2 (M×M)/((M+M)) (v_{1}-v_{2})^{2}**Q7.**A spring pong ball of mass m is floating in air by a jet of water emerging out of a nozzle. If the water strikes the ping pong ball with a speed v and just after collision water falls dead, the rate of flow of water in the nozzle is equal to?

Solution

The impact force F=∆p/∆t=v ∆m/∆t where ∆m/∆t= rate of flow of water in the nozzle and v the velocity of water jet. Since the ball is in equilibrium F=mg where m= mass of ping pong ball. ⇒ v ∆m/∆t=mg or rate of flow of water ∆m/∆t=mg/v

The impact force F=∆p/∆t=v ∆m/∆t where ∆m/∆t= rate of flow of water in the nozzle and v the velocity of water jet. Since the ball is in equilibrium F=mg where m= mass of ping pong ball. ⇒ v ∆m/∆t=mg or rate of flow of water ∆m/∆t=mg/v

**Q8.**A solid sphere and a hollow sphere of the same material and of a same size can be distinguished without weighing?

Solution

Time of descent will be less for solid sphere i.e. solid sphere will reach first at the bottom of inclined plane

Time of descent will be less for solid sphere i.e. solid sphere will reach first at the bottom of inclined plane

**Q9.**A door 1.6 m wide requires a force of 1 N to be applied at the free end to open or close it. The force that is required at a point 0.4 m distance from the hinges for opening or closing the door is ?

Solution

Here, torque Ï„=1.6×1=1.6 Nm So, when d=0.4 m, F=Ï„/d=1.6/0.4=4 N

Here, torque Ï„=1.6×1=1.6 Nm So, when d=0.4 m, F=Ï„/d=1.6/0.4=4 N

**Q10.**A cylinder rolls down an inclined plane of inclination 30°, the acceleration of cylinder is?

Solution

Let a plane be inclined at an angle Î¸ and a cylinder rolls down then the acceleration of the cylinder of mass m, radius R, and I I as moment of inertia is given by a=(g sinÎ¸)/((1+I/(mR

Let a plane be inclined at an angle Î¸ and a cylinder rolls down then the acceleration of the cylinder of mass m, radius R, and I I as moment of inertia is given by a=(g sinÎ¸)/((1+I/(mR

^{2})) ) Moment of inertia (I) of a cylinder =(mR^{2})/2 ∴a=(g sinÎ¸)/(((mR^{2})/(1+2/(mR^{2}))) )=2/3 g sin30° ⇒a=g/3