Two dielectric materials A and B exhibit both ionic and orientational polarizabilities. The variation of their susceptibilities $$\chi \left( { = {\varepsilon _r} - 1} \right)$$ with temperature T is shown in the figure, where $${\varepsilon _r}$$ is the relative dielectric constant. It can be inferred from the figure that <img src="/images/question-image/engineering-physics/solid-state-physics/1689768234-two-dielectric-materials-a-and-b-exhibit.png" title="Solid State Physics mcq question image" alt="Solid State Physics mcq question image">

At the interface between two linear dielectrics (with dielectric constants $${{\varepsilon _1}}$$ and $${{\varepsilon _2}}$$), the electric field lines bend, as shown in the figure. Assume that there are no free charges at the interface. The ratio $$\frac{{{\varepsilon _1}}}{{{\varepsilon _2}}}$$ is

A

$$\frac{{\tan {\theta _1}}}{{\tan {\theta _2}}}$$

B

$$\frac{{\cos {\theta _1}}}{{\cos {\theta _2}}}$$

C

$$\frac{{\sin {\theta _1}}}{{\sin {\theta _2}}}$$

D

$$\frac{{\cot {\theta _1}}}{{\cot {\theta _2}}}$$

The dependences of the magnetic susceptibility $$\left( \chi \right)$$ of a material with temperature (T) can be represented by $$\chi \propto \frac{1}{{T - \theta }},$$ where θ is the Curie-Weiss temperature. The plot of magnetic susceptibility versus temperature is sketched in the figure, as curves P, Q and R with curve Q having θ = 0. Which one of the following statements is correct?

A

Curve R represents a paramagnet and Q a ferromagnetic field

B

Curve Q represents a ferromagnet and P an antiferromagnet

C

Curve R represents an antiferromagnet and Q a paramagnet

D

Curve R represents an antiferromagnet and Q a ferromagnet

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?

A

0

B

5

C

1

D

4

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?

A

0

B

1

C

4

D

1 + abcd

Variation of electrical resistivity $$\rho $$ with temperature T of three solids is sketched (on different scales) in the figure, as curves P, Q and R.
Which one of the following statements describes the variations most appropriately?

A

P is for a superconductor and R for a semiconductor

B

Q is for a superconductor and P for a conductor

C

Q is for a superconductor and R for a conductor

D

R is for a superconductor and P for a conductor

The temperature dependence of the electrical conductivity σ of two intrinsic semiconductors A and B is shown in the figure. If E_A and E_B are the band gaps of A and B respectively, which one of the following is true?

A

EA > EB

B

EA B

C

EA = EB

D

EA and EB both depend on temperature

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

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

The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$ = ?

A

0

B

3

C

$$\frac{1}{3}$$

D

2

$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$ $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$ $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$ is simplified to :

A

0

B

1

C

2

D

3

In a diatomic molecule, the inter-nuclear separation of the ground and first excited electronic state are the same as shown in the figure. If the molecule is initially in the lowest vibrational state of the ground state, then the absorption spectrum will appear as

A

B

C

D