Rotational Motion Quiz-14
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.
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Q1. The radius of a rotating disc is suddenly reduced to half without any change in its mass. Then its angular velocity will be?
Solution
L= 1/2 MR2 ω= constant ∴ω∝1/R2 [If m= constant] If radius is reduced to half then angular velocity will be four times
L= 1/2 MR2 ω= constant ∴ω∝1/R2 [If m= constant] If radius is reduced to half then angular velocity will be four times
Q2. A non-uniform thin rod of length L is placed along X-axis as such its one of ends is at the origin. The linear mass density of rod is l=l0 x. The distance of centre of mass of rod from the origin is?
Solution
The mass of considered element is
The mass of considered element is
Q3. In the given figure four identical spheres of equal mass m are suspended by wires of equal length l0, so that all spheres are almost touching to each other. If the sphere 1 is released from the horizontal position and all collisions are elastic, the velocity of sphere 4 just after collision is?
Solution
When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l0= kinetic energy at bottom Or mgl0=1/2 mv2 Or v=√(2gl0 ) Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to √(2gl0 )
When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l0= kinetic energy at bottom Or mgl0=1/2 mv2 Or v=√(2gl0 ) Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to √(2gl0 )
Q4. If all of a sudden the radius of the earth decreases, then?
Solution
If radius of earth decreases then its M.I. decreases As L=Iω ∴ω∝1/I [L= constant] i.e. angular velocity of the earth will increase
If radius of earth decreases then its M.I. decreases As L=Iω ∴ω∝1/I [L= constant] i.e. angular velocity of the earth will increase
Q5. The moment of inertia of a thin uniform rod length L and mass M about an axis passing through a point at a distance of 1/3 from one of its ends and perpendicular to the rod is ?
Solution
ICM=(ML2)/12 (about middle point)
∴I=ICM+Mx2 =(ML2)/12+M(L/6)2 I= (ML2)/9
ICM=(ML2)/12 (about middle point)
∴I=ICM+Mx2 =(ML2)/12+M(L/6)2 I= (ML2)/9
Q6. A disc of moment of inertia 5 kg- m2 is acted upon by a constant torque of 40 Nm. Starting from rest the time taken by it to acquire an angular velocity of 24 rads-1 is?
Solution
Torque is defined as rate of change of angular momentum ∴Ï„=dJ/dt=d(Iω)/dt Given, Ï„=40 Nm,I=5 kgm2,ω=24 rads-1 dt=d(Iω)/Ï„=(5×24)/40=3 s
Torque is defined as rate of change of angular momentum ∴Ï„=dJ/dt=d(Iω)/dt Given, Ï„=40 Nm,I=5 kgm2,ω=24 rads-1 dt=d(Iω)/Ï„=(5×24)/40=3 s
Q7. A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is?
Solution
The angular momentum of a disc of moment of inertia I1 and rotating about its axis with angular velocity ω is L1=I1 ω When a round disc of moment of inertia I2 is placed on first disc, then angular momentum of the combination is L2=(I1+I2)ω' In the absence of any external torque, angular momentum remains conserved, ie., L1=L2 I1 ω=(I1+I2)ω' ⇒ω'=(I1 ω)/(I1+I2 )
The angular momentum of a disc of moment of inertia I1 and rotating about its axis with angular velocity ω is L1=I1 ω When a round disc of moment of inertia I2 is placed on first disc, then angular momentum of the combination is L2=(I1+I2)ω' In the absence of any external torque, angular momentum remains conserved, ie., L1=L2 I1 ω=(I1+I2)ω' ⇒ω'=(I1 ω)/(I1+I2 )
Q8. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio?
Solution
L=√2IE. If E are equal then L1/L2 =√(I1/I2 )=√(I/2I)=1/√2
L=√2IE. If E are equal then L1/L2 =√(I1/I2 )=√(I/2I)=1/√2
Q9. A thin metal disc of radius 0.25 m and mass 2 kg starts from rest and rolls down an inclined plane. If its rotational kinetic energy is 4 J at the foot of the inclined plane, then its linear velocity at the same point is?
Solution
Here, r=0.5 m,m=2 kg Rotational KE=1/2 Iω2=1/2×(1/2 mr2 ) ω2 4= 1/4 mv2=1/4×2v2 v=√8=2√2 ms-1
Here, r=0.5 m,m=2 kg Rotational KE=1/2 Iω2=1/2×(1/2 mr2 ) ω2 4= 1/4 mv2=1/4×2v2 v=√8=2√2 ms-1
Q10. A uniform disk of mass M and radius R is mounted on a fixed horizontal axis. A block of mass m hangs from a mass less string that is wrapped around the rim of the disk. The magnitude of the acceleration of the falling block (m) is?
Solution
Using a=g/(1+I/(mR2 )) a=g/(1+(MR2)/(2mR2 )) [I=1/2 MR2 ] ⇒a=2m/(M+2m) g .
Using a=g/(1+I/(mR2 )) a=g/(1+(MR2)/(2mR2 )) [I=1/2 MR2 ] ⇒a=2m/(M+2m) g .
