Let $$\overrightarrow {\bf{L}} $$ = (Lx, Ly, Lz) denotes the orbital angular momentum operators of a particle and let L+ = Lx + i Ly and L- = Lx - i Ly. The particle is in aneigen state of L2 and Lz eigen values $${\hbar ^2}\left( {l + 1} \right)$$ and $$\hbar l$$ respectively. The expectation value of L+L- in this state is

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

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

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

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

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

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

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

A

x2y + y2z + z2x

B

2xy + 2yz + 2zx

C

x + y + z

D

0

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

A

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

B

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

C

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

D

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

Consider two particles with position vectors $$\overrightarrow {{{\bf{r}}_1}} $$ and $$\overrightarrow {{{\bf{r}}_2}} $$ . The force exerted by particle 2 on particle 1 is $$\overrightarrow {\bf{F}} \left( {\overrightarrow {{{\bf{r}}_1}} ,\,\overrightarrow {{{\bf{r}}_2}} } \right) = \left( {{{{\bf{\dot r}}}_2} - {{{\bf{\dot r}}}_1}} \right)\left( {{r_2} - {r_1}} \right).$$ The force is

A

central and conservative

B

non-central and conservative

C

central and non-conservative

D

non-central and non-conservative

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

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

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

For a multi-electron atom $$l$$, L and S specify the one electron orbital angular momentum, total orbital angular momentum and total spin angular momentum respectively. The selection rules for electric dipole transition between the two electronic energy levels, specified by $$l$$, L and S are

A

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = 0, ±1

B

ΔL = 0, ±1; ΔS = 0; Δ$$l$$ = ±1

C

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = 0, ±1

D

ΔL = 0, ±1; ΔS = ±1; Δ$$l$$ = ±1