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

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

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

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

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

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

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

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

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

$${L_z}$$

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

For a function g(t), it is given that
$$\int\limits_{ - \infty }^{ + \infty } {g\left( t \right){e^{ - j\omega t}}dt = \omega {e^{ - 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ - \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ - \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

-j

$$ - {j \over 2}$$

$${j \over 2}$$

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)

finite thermal resistance

zero thermal resistance

an infinite thermal resistance

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

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

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

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

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

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

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

total momentum must be zero

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

the total force on the system must be zero

the total torque on the system must be zero

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

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

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

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

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

The primitive translation vectors of the body centred cubic lattice are $$\overrightarrow {\bf{a}} = \frac{a}{2}\left( {{\bf{\hat x}} + {\bf{\hat y}} - {\bf{\hat z}}} \right),\,\overrightarrow {\bf{b}} = \frac{a}{2}\left( { - {\bf{\hat x}} + {\bf{\hat y}} + {\bf{\hat z}}} \right)$$ and $$\overrightarrow {\bf{c}} = \frac{a}{2}\left( {{\bf{\hat x}} - {\bf{\hat y}} + {\bf{\hat z}}} \right)$$ . The primitive translation vectors $$\overrightarrow {\bf{A}} ,\,\overrightarrow {\bf{B}} $$ and $$\overrightarrow {\bf{C}} $$ of the reciprocal lattice are

$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$

$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} - {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$

$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat z}}} \right)$$

Let $$\nabla \cdot \left( {{\text{f}}\overrightarrow {\text{v}} } \right) = {{\text{x}}^2}{\text{y}} + {{\text{y}}^2}{\text{z}} + {{\text{z}}^2}{\text{x}},$$ where f and v are scalar and vector fields respectively. If $$\overrightarrow {\text{v}} = {\text{y}}\overrightarrow {\text{i}} + {\text{z}}\overrightarrow {\text{j}} + {\text{x}}\overrightarrow {\text{k}} ,$$ then $$\overrightarrow {\text{v}} \cdot \nabla {\text{f}}$$ is

x2y + y2z + z2x

2xy + 2yz + 2zx

x + y + z

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

$$2{m_0}u\omega \hat \theta $$

$$ - 2m{u_0}\omega \hat \theta $$

$$4m{u_0}\omega \hat \theta $$

$$ - m{u_0}\omega \hat \theta $$

For two non-zero vectors $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$, if $$\overrightarrow {\text{A}} $$ + $$\overrightarrow {\text{B}} $$ is perpendicular to $$\overrightarrow {\text{A}} $$ - $$\overrightarrow {\text{B}} $$ then,

Magnitude of $$\overrightarrow {\text{A}} $$ is twice magnitude of $$\overrightarrow {\text{B}} $$

Magnitude of $$\overrightarrow {\text{A}} $$ is half the magnitude of $$\overrightarrow {\text{B}} $$

$$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$ cannot be orthogonal

the magnitudes of $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$ are equal