Rotational Motion Quiz 12

Rotational Motion Quiz-12

Dear Readers,

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. 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?

•  mlv
•  mv/l
•  mv²/l
•  Zero

Solution
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?

•  350 N-m
•  440 N-m
•  531 N-m
•  628 N-m

Solution
ω=2πn= (2π×1800)/60=60π rad/s P=τ×ω⇒τ=P/ω=(100×10³)/60π=531 N-m

Q3. The flywheel is so constructed that the entire mass of it is concentrated at its rim, because?

•  It increases the power
•  It increases the speed
•  It increases the moment of inertia
•  It saves the flywheel fan from breakage

Solution
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 ml²/3, the initial angular acceleration of the rod will be?

•  2g/3l
•  mg l/2
•  3/2 gl
•  3g/2l

Solution
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is I=(ml²)/3 Where m is mass of rod and l is length. Torque (τ=Iα) acting on centre of gravity of rod is given by τ=mg 1/2 or Iα=mg 1/2 or (ml²)/3 α=mg 1/2 or α=3g/2l

Q5. The centre of mass of a system of two particles divides the distance them?

•  In inverse ratio of square of masses of particles
•  In direct ratio of square of masses of particles
•  In inverse ratio of masses of particles
•  In direct ratio of masses of particles

Solution
 m₁ r₁=m₂ r₂ r₁/r₂ =m₂/m₁ ∴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?

•  1/2 M (R² +r² )
•  1/2 M(R²-r² )
•  1/2 M(R⁴+r⁴ )
•  1/2 M'(R⁴-r⁴ )

Solution
The moment of inertia of this annular disc about the axis perpendicular to its plane will be 1/2 M(R²+r²).

Q7. Two bodies of moment of inertia I₁ and I₂ (I₁>I₂) have equal angular momenta. If E₁,E₂ are their kinetic energies of rotation, then?

•  E₁>E₂
•  E₁=E₂
•  E₁ Less than E₂
•  Cannot be said

Solution
I₁ ω₁=I₂ ω₂ ∴ ω₁/ω₂ =I₂/I₁ Now, E₁/E₂ =(1/2 I₁ ω₁²)/(1/2 I₂ ω₂² )=I₁/I₂ ×(I₂/I₁ )²=I₂/I₁ As I₁>I₂ ∴E₁ Less than E₂

Q8. Moment of inertia along the diameter of a ring is?

•  3/2 MR²
•  1/2 MR²
•  MR²
•  2 MR²

Solution
1/2 MR²

Q9. Two wheels A and B are mounted on the same axle. Moment of inertia of A is 6 kgm² 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?

•  8 kg m²
•  4 kg m²
•  3 kg m²
•  5 kg m²

Solution
Applying the principle of conservation of angular momentum, (I₁+I₂ )ω=I₁ ω₁+I₂ ω₂ (6+I₂ ) 400/60×2π=6×600/60×2π+I₂×0 Which gives, I₂=3 kg m²

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 ?

•  The impulse received by each is ,m(5i ̂+j ̂)
•  The velocity of the M is 1/13(5i ̂+j ̂)
•  The coefficient of restitutions 11/7
•  All the above are correct

Solution
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