Rotational Motion Quiz 16

Rotational Motion Quiz-16

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 solid cylinder on moving with constant speed v₀ reaches the bottom of an incline of 30°. A hollow cylinder of same mass and radius moving with the same constant speed v₀ reaches the bottom of a different incline of θ. There is no slipping and both of them go through the same distance in the same time ; θ is then equal to?

•  37°
•  30°
•  42°
•  45°

Solution
For solid cylinder, θ=30°,K²=1/2 R² For hollow cylinder, θ=?,K²=R² Using we find, ((1+1/2))/sin⁡30° =(1+1)/sin⁡θ ∴ sin⁡θ=2/3=0.6667 θ=42°

Q2. A uniform rod of length 2L is placed with one end in contact with the horizontal and is then inclined at an angle α to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be?

•  ω=√((3g sin⁡α)/2L)
•  ω=√(2L/(3g sin⁡α ))
•  ω=√((6g sin⁡α)/2L)
•  ω=√(2L/(g sin⁡α ))

Solution
By the conservation of energy P.E. of rod = Rotational K.E.

mg l/2 sin⁡α=1/2 Iω²⇒mg l/2 sin⁡α=1/2 (ml²)/3 ω² ⇒ω=√((3g sin⁡α)/l) But in the problem length of the rod 2L is given ⇒ω=√((3g sin⁡α)/2L)

Q3. A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. For the combined system of bag and bullet, the correct option is?

•  Momentum is mMv/((m+M))
•  Kinetic energy is 1/1 Mv²
•  Momentum is mv
•  Kinetic energy is 1/2 (m² v²)/((M+m))

Solution
As F _ext=0, hence momentum remains conserved and final momentum = initial momentum = mv

Q4. A ball strikes a horizontal floor at an angle θ=45°. The coefficient of restitution between the ball and the floor is e=1/2. The fraction of its kinetic energy lost in collision in?

•  5/8
•  3/8
•  3/4
•  1/4

Solution
Let ball strikes at a speed u the K₁=1/2 mu² Due to collision tangential component of velocity remains unchanged at u sin 45°, but the normal component of velocity change to u sin45°=1/2 u cos⁡45° ∴ Final velocity of ball after collision v=√((u sin⁡45° )²+(1/2 u cos⁡45° )² ) =√((u/√2)²+(u/(2√2))² )=√(5/3) u. Hence final kinetic energy K₂=1/2 mv²=5/16 mu². ∴ Fractional loss in KE =(K₁-K₂)/K₁ =(1/2 mu²-5/16 mu²)/(1/2 mu² )=3/8

Q5. A marble and a cube have the same mass starting from rest, the marble rolls and the cube slides down a frictionless ramp. When they arrive at the bottom, the ratio of speed of the cube to the centre of mass and speed of the marble is ?

•  7 : 5
•  √7:√5
•  √2:1
•  5 : 2

Solution
 If h is height of the ramp, then in rolling of marble, speed v=√(2gh/(1+K²/R² )) The speed of the cube to the centre of mass v^'=√2gh ∴ v^'/v=√(1+K²/R² ) For marble sphere, K²=2/5 R² ∴ v^'/v=√(1+2/5)=√(7/5)=√7:√5

Q6. From a circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is?

•  4 MR²
•  40/9 MR²
•  10 MR²
•  37/9 MR²

Solution
I_remaining=I_whole-I_removed or I=1/2 (9M)(R² )-[1/2 m(R/3)²+1/2 m(2R/3)² ] …(i) Here, m=9M/(πR² )×π(R/3)²=M Substituting in Eq. (i), we have I=4MR²

Q7. A solid sphere (mass 2 M) and a thin hollow spherical shell (mass M) both of the same size, roll down an inclined plane, then?

•  Solid sphere will reach the bottom first
•  Hollow spherical shell will reach the bottom first
•  Both will reach at the same time
•  None of these

Solution
Time of descent t=1/sin⁡θ √(2h/g (1+K²/R² ) ) For solid sphere K²/R² =2/5 For hollow sphere K²/R² =2/3 As (K²/R² )_Hollow>(K²/R² )_Solid i.e. solid sphere will take less time so it will reach the bottom first

Q8. Two particles of masses 1 kg and 2 kg are located at x₁=0,y₁=0 and x₂=1,y₂=0 respectively. The centre of mass of the system is at?

•  x=1,y=2
•  x=2,y=1
•  x= 1/3,y=2/3
•  x= 2/3,y=0

Solution
x= 2/3,y=0

Q9. Two spherical bodies of masses M and 5M in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is?

•  2.5 R
•  4.5 R
•  7.5 R
•  1.5 R

Solution
Distance between the centre of spheres = 12R ∴ Distance between their surfaces =12R-(2R+R)=9R Since there is no external force, hence centre of mass must remain unchanged and hence ⇒ m₁ r₁=m₂ r₂⇒Mx=5M(9R-x)⇒x=7.5R

Q10. A flywheel rotating about a fixed axis has a kinetic energy of 360 joule when its angular speed is 30 rad/sec. The moment of inertia of the wheel about the axis of rotation is?

•  0.6 kg×m²
•   0.15 kg×m²
•  0.8 kg×m²
•  0.75 kg×m²

Solution
1/2 Iω²=360⇒I=(2×360)/(30)² =(2×360)/(30×30)=0.8kg×m².