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 stone of mass m tied to string of length l is rotating along a circular path with constant speed v. The torque on the stone is?

Solution

Torque acting on a body in circular motion is zero

Torque acting on a body in circular motion is zero

**Q2.**An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min. What torque does it deliver?

Solution

Ï‰=2Ï€n= (2Ï€×1800)/60=60Ï€ rad/s P=Ï„×Ï‰⇒Ï„=P/Ï‰=(100×10

Ï‰=2Ï€n= (2Ï€×1800)/60=60Ï€ rad/s P=Ï„×Ï‰⇒Ï„=P/Ï‰=(100×10

^{3})/60Ï€=531 N-m**Q3.**The flywheel is so constructed that the entire mass of it is concentrated at its rim, because?

Solution

It increases the moment of inertia

It increases the moment of inertia

**Q4.**A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml

^{2}/3, the initial angular acceleration of the rod will be?

Solution

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is I=(ml

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is I=(ml

^{2})/3 Where m is mass of rod and l is length. Torque (Ï„=IÎ±) acting on centre of gravity of rod is given by Ï„=mg 1/2 or IÎ±=mg 1/2 or (ml^{2})/3 Î±=mg 1/2 or Î±=3g/2l**Q5.**The centre of mass of a system of two particles divides the distance them?

Solution

m

m

_{1}r_{1}=m_{2}r_{2}r_{1}/r_{2}=m_{2}/m_{1}∴r∝1/m .**Q6.**From a disc of radius R,a concentric circular portion of radius r is cut out so as to leave an annular disc of mass M. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is?

Solution

The moment of inertia of this annular disc about the axis perpendicular to its plane will be 1/2 M(R

The moment of inertia of this annular disc about the axis perpendicular to its plane will be 1/2 M(R

^{2}+r^{2}).**Q7.**Two bodies of moment of inertia I

_{1}and I

_{2}(I

_{1}>I

_{2}) have equal angular momenta. If E

_{1},E

_{2}are their kinetic energies of rotation, then?

Solution

I

I

_{1}Ï‰_{1}=I_{2}Ï‰_{2}∴ Ï‰_{1}/Ï‰_{2}=I_{2}/I_{1}Now, E_{1}/E_{2}=(1/2 I_{1}Ï‰_{1}^{2})/(1/2 I_{2}Ï‰_{2}^{2})=I_{1}/I_{2}×(I_{2}/I_{1})^{2}=I_{2}/I_{1}As I_{1}>I_{2}∴E_{1}Less than E_{2}**Q8.**Moment of inertia along the diameter of a ring is?

Solution

1/2 MR

1/2 MR

^{2}**Q9.**Two wheels A and B are mounted on the same axle. Moment of inertia of A is 6 kgm

^{2}and it is rotating at 600 rpm when B is at rest. What is moment of inertia of B, if their combined speed is 400 rpm?

Solution

Applying the principle of conservation of angular momentum, (I

Applying the principle of conservation of angular momentum, (I

_{1}+I_{2})Ï‰=I_{1}Ï‰_{1}+I_{2}Ï‰_{2}(6+I_{2}) 400/60×2Ï€=6×600/60×2Ï€+I_{2}×0 Which gives, I_{2}=3 kg m^{2}**Q10.**A particle of mass m moving with a velocity (3i ̂+2j ̂)ms

^{-1}collides with a stationary body mass M and finally moves with a velocity (-2i ̂+j ̂)ms

^{-1}. If m/M=1/13, then ?

Solution

Impulsive received by m J =m(v

Impulsive received by m J =m(v

_{f}-v_{i}) =m(-2i ̂+j ̂-3i ̂-2j ̂) =m(-5i ̂-j ̂) And impulse received by M =-J =m(5i ̂+j ̂) (b) mv=m(5i ̂+j ̂) Or v=m/M (5i ̂+j ̂ )=1/13(5i ̂+j ̂) (c) e= (relative velocity of separation/relative velocity of approach) in the direction of -J =11/17