If for a system of N particles of different masses m1, m2, . . . mN 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}}

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

Correct Answer: the total force on the system must be zero

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

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

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

$${L_z}$$

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

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

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

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

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

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

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

central and conservative

non-central and conservative

central and non-conservative

non-central and non-conservative

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

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

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

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

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

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

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

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

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

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

-1

For a vector potential $$\overrightarrow {\bf{A}} ,$$ the divergence of $$\overrightarrow {\bf{A}} $$ is $$\overrightarrow \nabla \cdot \overrightarrow {\bf{A}} = - \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{Q}{{{r^2}}},$$ where Q is a constant of appropriate dimension. The corresponding scalar potential $$\phi \left( {r,\,t} \right)$$ that makes $$\overrightarrow {\bf{A}} $$ and $$\phi $$ Lorentz gauge invariant is

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{Q}{r}$$

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{r}$$

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{Q}{{{r^2}}}$$

$$\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Qt}}{{{r^2}}}$$

If $$\overrightarrow {\text{a}} $$ and $$\overrightarrow {\text{b}} $$ are two arbitrary vectors with magnitudes a and b, respectively, $${\left| {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right|^2}$$ will be equal to

$${{\text{a}}^2}{{\text{b}}^2} - {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$

$${\text{ab}} - \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$

$${{\text{a}}^2}{{\text{b}}^2} + {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$

$${\text{ab}} + \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$