Rotational Motion Quiz-18
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. A circular disc is to be made by using iron and aluminium, so that it acquires maximum moment of inertia about its geometrical axis. It is possible with?
Solution
By doing so the distribution of mass can be made away from the axis of rotation
By doing so the distribution of mass can be made away from the axis of rotation
Q2. The moment of inertia of semicircular ring about an axis which is perpendicular to the plane of the ring and passes through the centre?
Solution
MR2
MR2
Q3. If a ball is dropped from rest, its bounces from the floor. The coefficient of restitution is 0.5 and the speed just before the first bounce is 5ms-1. The total time taken by the ball to come to rest is?
Solution
Acceleration a=(v-u)/t Or a=(v-v0)/t Or g=(v-v0)/t ∴ v=0 Speed before first bounce v0=-5ms-1 ∴ t=(vB-vA)/g=(0(-5))/10=5/10=0.5 s
Acceleration a=(v-u)/t Or a=(v-v0)/t Or g=(v-v0)/t ∴ v=0 Speed before first bounce v0=-5ms-1 ∴ t=(vB-vA)/g=(0(-5))/10=5/10=0.5 s
Q4. Two negatively charges particles having charges e1 and e2 and masses m1 and m2 respectively are projected one after another into a region with equal initial velocity. The electric field E is along the y-axis, while the direction of projection makes an angle a with the y-axis. If the ranges of the two particles along the x-axis are equal then one can conclude that?
Solution
Here, effected gravitational acceleration is g'=(mg-qE)/m ∴ R=(v02 sin2α)/g' It means, g’ for both particles are same This is possible when m1=m2 and e1=e2
Here, effected gravitational acceleration is g'=(mg-qE)/m ∴ R=(v02 sin2α)/g' It means, g’ for both particles are same This is possible when m1=m2 and e1=e2
Q5. The acceleration of the centre of mass of a uniform solid disc rolling down an inclined plane of angle α is?
Solution
The acceleration of the body which is rolling down an inclined plane of angle α is
a= (g sin α)/(1+K2/R2 ) where K=radius of gyration, R=radius of body. Now, here the body is a uniform solid disc. So, K2/R2 =1/2 ∴a=(g sin α)/(1+1/2) or a=(g sinα)/(3⁄2) or a=(2g sinα)/3
The acceleration of the body which is rolling down an inclined plane of angle α is
a= (g sin α)/(1+K2/R2 ) where K=radius of gyration, R=radius of body. Now, here the body is a uniform solid disc. So, K2/R2 =1/2 ∴a=(g sin α)/(1+1/2) or a=(g sinα)/(3⁄2) or a=(2g sinα)/3
Q6. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω, as shown in the figure. At time, t=0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t=T and stops. The angular speed of the system remains ω throughout. The magnitude of the torque (|τ | on the system about O, as a function of time is best represented by which plot?
Solution
Angular momentum about rotational axis Lt=[I+m(vt)2]ω (dLt)/dt=2mv2 tω; Torque τ=(2mv2 ω)t
Angular momentum about rotational axis Lt=[I+m(vt)2]ω (dLt)/dt=2mv2 tω; Torque τ=(2mv2 ω)t
Q7. A small object of uniform density roll up a curved surface with an initial velocity v. If reaches up to a maximum height of 3v2/4g with respect to the initial position. The object is ?
Solution
1/2 mv2=1/2 I(v/R)2=mg((3v2)/4g) ∴I=1/2 mR2 ∴ Body is disc.
1/2 mv2=1/2 I(v/R)2=mg((3v2)/4g) ∴I=1/2 mR2 ∴ Body is disc.
Q8. Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm2 units will be?
Solution
Moment of inertia of the system about axis AX, =IA+IB+IC =mA (rA )2+mg(rB )2+mc(rC )2 =m(0)2+m(l)2+m(l cos60 )2 =ml2+(ml2)/4=(5ml2)/4
Moment of inertia of the system about axis AX, =IA+IB+IC =mA (rA )2+mg(rB )2+mc(rC )2 =m(0)2+m(l)2+m(l cos60 )2 =ml2+(ml2)/4=(5ml2)/4
Q9. Two spheres of equal masses, one of which is a thin spherical shell and the other a solid, have the same moment of inertia about their respective diameters. The ratio of their radii will be?
Solution
Let the radii of the thin spherical and the solid sphere are R1 and R2 respectively. Then the moment of inertia of the spherical shell about their diameter I=2/3 MR12 …(i) and the moment of inertia of the solid sphere is given by I=2/5 MR22 …(ii) Given that the masses and moment of inertia for both the bodies are equal, then from Eqs. (i) and (ii) 2/3 MR12=2/5 MR22⇒(R12)/(R22 )=3/5 ⇒R1/R2 =√(3/5)⇒R1:R2=√3:√5
Let the radii of the thin spherical and the solid sphere are R1 and R2 respectively. Then the moment of inertia of the spherical shell about their diameter I=2/3 MR12 …(i) and the moment of inertia of the solid sphere is given by I=2/5 MR22 …(ii) Given that the masses and moment of inertia for both the bodies are equal, then from Eqs. (i) and (ii) 2/3 MR12=2/5 MR22⇒(R12)/(R22 )=3/5 ⇒R1/R2 =√(3/5)⇒R1:R2=√3:√5
Q10. A reel of thread unrolls itself falling down under gravity. Neglecting mass of the thread, the acceleration of the reel is?
Solution
a= g/(1+K2/R2 ) [For solid cylinder K2/R2 =1/2] ∴a=g/(1+1/2)=2/3 g
a= g/(1+K2/R2 ) [For solid cylinder K2/R2 =1/2] ∴a=g/(1+1/2)=2/3 g
