A rigid frictionless rod rotates anticlockwise in a vertical plane with angular velocity $$\overrightarrow \omega $$. A bead of mass m moves outward along the rod with constant velocity $$\overrightarrow {{u_0}} $$ . The bead will experience a coriolis force

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$ where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$ is the angular momentum. The Hamiltonian does not commute with

A

$$\overrightarrow {\bf{L}} + \overrightarrow {\bf{S}} $$

B

$$\overrightarrow {{{\bf{S}}^2}} $$

C

$${L_z}$$

D

$$\overrightarrow {{{\bf{L}}^2}} $$

A hollow rod has frictionless inner walls. Inside the rod are two small spheres that are kept on either side of the centre of rod. The rod is rotated about its central axis which is perpendicular to the plane of rotation. If the rod had been rotated by an impulsive torque which gave it an instantaneous angular velocity of 5rad/s, what will be the angular velocity after some finite but long time ‘t’? Assume no external forces act on rod after the impulse. Also state what will happen to the spheres in the centre of the rod. The rod has a mass =2kg & length =10cm. The two spheres have a mass =1kg each.

A

2.5 rad/s, spheres will stay where they are as there is no friction to move them

B

0 rad/s, spheres will move outwards and decrease velocity of rod to zero

C

1.25 rad/s, spheres will move to opposite ends of rod

D

2.5 rad/s, spheres will move to opposite ends of rod

A rigid body is rotating about its centre of mass; fixed at origin with an angular velocity $$\overrightarrow \omega $$ and angular acceleration $$\overrightarrow \alpha $$. If the torque acting on it is $$\overrightarrow \tau $$ and its angular momentum is $$\overrightarrow {\bf{L}} $$, then the rate of change of its kinetic energy is

A

$$\frac{1}{2}\overrightarrow \tau \cdot \overrightarrow \omega $$

B

$$\frac{1}{2}\overrightarrow {\bf{L}} \cdot \overrightarrow \omega $$

C

$$\frac{1}{2}\left( {\overrightarrow \tau \cdot \overrightarrow \omega + \overrightarrow {\bf{L}} \cdot \overrightarrow \alpha } \right)$$

D

$$\frac{1}{2}\overrightarrow {\bf{L}} \cdot \overrightarrow \alpha $$

In a cubic system with cell edge a, two phonons with wave vectors $${\overrightarrow {\bf{q}} _1}$$ and $${\overrightarrow {\bf{q}} _2}$$ collide and produce a third phonon with a wave. vector $${\overrightarrow {\bf{q}} _3}$$ such that $${\overrightarrow {\bf{q}} _1} + {\overrightarrow {\bf{q}} _2} = {\overrightarrow {\bf{q}} _3} + \overrightarrow {\bf{R}} $$ where, $$\overrightarrow {\bf{R}} $$ is a lattice vector. Such a collision process will lead to(a)

A

finite thermal resistance

B

zero thermal resistance

C

an infinite thermal resistance

D

a finite thermal resistance for certain $$\left| {\overrightarrow {\bf{R}} } \right|$$ only

A bead of mass m slides along a straight frictionless rigid wire rotating in a horizontal plane with a constant angular speed ω. The axis of rotation is perpendicular to the wire and passes through one end of the wire. If r is the distance of the mass from the axis of rotation and v is its speed, then the magnitude of the Coriolis force is

A

$$\frac{{m{v^2}}}{r}$$

B

$$\frac{{2m{v^2}}}{r}$$

C

mvω

D

2mvω

If for a system of N particles of different masses m₁, m₂, . . . m_N with position vectors $${\overrightarrow {\bf{r}} _1},\,{\overrightarrow {\bf{r}} _2},\,.\,.\,.\,{\overrightarrow {\bf{r}} _N}$$ and corresponding velocities $${\overrightarrow {\bf{v}} _1},\,{\overrightarrow {\bf{v}} _2},\,.\,.\,.\,{\overrightarrow {\bf{v}} _N}$$ respectively such that $$\sum\limits_i {\overrightarrow {{{\bf{v}}_i}} = 0,} $$ then

A

total momentum must be zero

B

total angular momentum must be independent of the choice of the origin

C

the total force on the system must be zero

D

the total torque on the system must be zero

A rod of length L with uniform charge density $$\lambda $$ per unit length is in the XY-plane and rotating about Z-axis passing through one of its edge with an angularvelocity $$\overrightarrow \omega $$ as shown in the figure below. $$\left( {{\bf{\hat r}},\,\hat \phi ,\,{\bf{\hat z}}} \right)$$ refer to the unit vectors at Q, $$\overrightarrow {\bf{A}} $$ is the vector potential at a distance d from the origin O along Z-axis for d ≪ L and $$\overrightarrow {\bf{J}} $$ is the current density due to the motion of the rod. Which one of the following statements is correct?

A

$$\overrightarrow {\bf{J}} {\text{ along }}{\bf{\hat r}};\overrightarrow {\bf{A}} {\text{ along }}{\bf{\hat z}};\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{d}$$

B

$$\overrightarrow {\bf{J}} {\text{ along }}\hat \phi ;\overrightarrow {\bf{A}} {\text{ along }}\hat \phi ;\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{{{d^2}}}$$

C

$$\overrightarrow {\bf{J}} {\text{ along }}{\bf{\hat r}};\overrightarrow {\bf{A}} {\text{ along }}{\bf{\hat z}};\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{{{d^2}}}$$

D

$$\overrightarrow {\bf{J}} {\text{ along }}\hat \phi ;\overrightarrow {\bf{A}} {\text{ along }}\hat \phi ;\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{d}$$

An atom with net magnetic moment $$\overrightarrow \mu $$ and net angular momentum $$\overrightarrow {\bf{L}} \left( {\overrightarrow \mu = \gamma \overrightarrow {\bf{L}} } \right)$$ is kept in a uniform magnetic induction $$\overrightarrow {\bf{B}} = {B_0}{\bf{\hat k}}.$$ The magnetic moment $$\overrightarrow \mu \left( { = {\mu _x}} \right)$$ is

A

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + \gamma {B_0}{\mu _x} = 0$$

B

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + \gamma {B_0}\frac{{d{\mu _x}}}{{dt}} + {\mu _x} = 0$$

C

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + {\gamma ^2}B_0^2{\mu _x} = 0$$

D

$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + 2\gamma {B_0}{\mu _x} = 0$$

Given $$\overrightarrow {\text{F}} = \left( {{{\text{x}}^2} - 2{\text{y}}} \right)\overrightarrow {\text{i}} - 4{\text{yz}}\overrightarrow {\text{j}} + 4{\text{x}}{{\text{z}}^2}\overrightarrow {\text{k}} ,$$ the value of the line integral $$\int\limits_{\text{c}} {\overrightarrow {\text{F}} \cdot d\overrightarrow l } $$ along the straight line c from (0, 0, 0) to (1,1,1) is

A

$$\frac{3}{{16}}$$

B

0

C

$$\frac{{ - 5}}{{12}}$$

D

-1

Two monochromatic waves having frequencies $$\omega $$ and $$\omega + \Delta \omega \left( {\Delta \omega \ll \omega } \right)$$ and corresponding wavelengths $$\lambda $$ and $$\lambda - \Delta \lambda \left( {\Delta \lambda \ll \lambda } \right)$$ of same polarization, travelling along X-axis are superimposed on each other. The phase velocity and group velocity of the resultant wave are respectively given by

A

$$\frac{{\omega \lambda }}{{2\pi }},\,\frac{{\Delta \omega {\lambda ^2}}}{{2\pi \Delta \lambda }}$$

B

$$\omega \lambda ,\,\frac{{\Delta \omega {\lambda ^2}}}{{\Delta \lambda }}$$

C

$$\frac{{\omega \Delta \lambda }}{{2\pi }},\,\frac{{\Delta \omega \Delta \lambda }}{{2\pi }}$$

D

$$\omega \Delta \lambda ,\,\omega \Delta \lambda $$