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1.

In a thin rectangular metallic strip, a constant current I flows along the positive x-direction, as shown in the figure. The length,  width and thickness of the strip are l, w and d respectively. A uniform magnetic field  B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in the accumulation of charge carries on the surface PQRS and the appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues untill the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

Consider two different metallic strips (1  and 2)  of same dimensions (length l, width w and thickness d)  with carrier densities n1  and n2, respectively. Strip  1 is placed in magnetic field  B1  and strip 2 is placed in magnetic field B2, both along positive y- directions. Then V1  and V2 are the potential differences developed between K and M in strips  1 and 2, respectively. Assuming that the current I is the  same for both the strips, the correct options is/are

A) If $B_{1}=B_{2}$ and $n_{1}=2n_{2}$ , then $V_{2}=2V_{1}$

B) If $B_{1}=B_{2}$ and $n_{1}=2n_{2}$ , then $V_{2}=V_{1}$

C) If $B_{1}=2B_{2}$ and $n_{1}=n_{2}$ , then $V_{2}=0.5V_{1}$

D) If $B_{1}=2B_{2}$ and $n_{1}=n_{2}$ , then $V_{2}=V_{1}$

2.

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length,  width and thickness of the strip are l, w and d respectively. A uniform magnetic field  B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in the accumulation of charge carries on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues untill the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section  of the strip and carried by electrons.

Consider two different metallic strips (1 and 2 )  of the same material. Their lengths are the same , width  are w1   and w2  and thickness  are d1 and d2 , respectively. Two points K and M are symmetrically  located on the opposite  faces parallel to the x-y plane  (see figure) .V1 and V2  are the potential  differences  between K and M in strips 1 and 2 respectively. Then , for  a given  current I flowing through  them in a given  magnetic field  strength  B, the correct statements is/are

A) if $w_{1}=w_{2}$ and $d_{1}=2d_{2}$, then $V_{2}=2V_{1}$

B) If $w_{1}=w_{2}$ and $d_{1}=2d_{2}$, then $V_{2}=V_{1}$

C) If $w_{1}=2w_{2}$ and $d_{1}=d_{2}$, then $V_{2}=2V_{1}$

D) If $w_{1}=2w_{2}$ and $d_{1}=d_{2}$, then $V_{2}=V_{1}$

3.

Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1  surrounded by a medium of lower refractive index n2. The light guidance in  the structure takes place due to successive total internal reflections at the interface of the media n1  and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1 . The numerical aperture  (NA) of the structure is defined as sin im.

If two structures of same cross-sectional area, but different numerical  apertures NA1  and NA2  (NA<NA1) are joined  longitudinally, the numerical aperture  of the combined structure is

A) $\frac{NA_{1}NA_{2}}{NA_{1}+NA_{2}}$

B) ${NA_{1}+NA_{2}}$

C) $NA_{1}$

D) $NA_{2}$

4.

Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1  surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1  and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture  (NA) of the structure is defined as sin I'm.

For two structures namely S1   with $n_{1}=\frac{\sqrt{45}}{4}$  and   $n_{2}=\frac{3}{2}$   and S2    with $n_{1}=\frac{8}{5}$  and $n_{2}=\frac{7}{5}$  and taking the  refractive index of water to be $\frac{4}{3}$  and that to air to be 1. the correct options is/are

A) NA of $S_{1}$ immersed in water is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$

B) NA of $S_{1}$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_{2}$ immersed in water.

C) NA of $S_{1}$ placed in air is the same as that $S_{2}$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$

D) NA of $S_{1}$ placed in air is the same as that of $S_{2}$ placed in water

5.

In terms of potential difference V, electric current i, Permittivity  $\epsilon_{0}$ , permeability  $\mu_{0}$ and speed of light c, the dimensionally correct equations is /are

A) $\mu_{0}I^{2}=\epsilon_{0}V^{2}$

B) $\epsilon_{0}I=\mu_{0}I$

C) $I=\epsilon_{0}cV$

D) $\mu_{0}cI=\epsilon_{0}V$

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