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}

central and conservative

non-central and conservative

central and non-conservative

non-central and non-conservative

Correct Answer: non-central and non-conservative

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

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

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

$${L_z}$$

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

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

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

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

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

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

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

$$R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$$ the formula represent?

Series combination

Parallel combination

Bridge combination

Linear combination

The wavefunctions of two identical particles in stated n and s are given by $${\phi _n}\left( {{r_1}} \right)$$ and $${\phi _s}\left( {{r_2}} \right)$$ , respectively. The particles obey Maxwell-Boltzmann statistics. The state of the combined two particles system is expressed as

$${\phi _n}\left( {{r_1}} \right) + {\phi _s}\left( {{r_2}} \right)$$

$$\frac{1}{{\sqrt 2 }}\left$$

$${\phi _n}\left( {{r_1}} \right){\phi _s}\left( {{r_2}} \right)$$

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?

1 + abcd

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

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

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

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

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

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