1
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
Two spheres P and Q for equal radii have densities $$\rho$$1 and $$\rho$$2, respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities $$\sigma$$1 and $$\sigma$$2 and viscosities $${\eta _1}$$ and $${\eta _2}$$, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure) If sphere P along in L2 has terminal velocity vP and Q alone in L1 ha terminal velocity vQ, then

JEE Advanced 2015 Paper 2 Offline Physics - Properties of Matter Question 22 English
A
$${{|{v_P}|} \over {|{v_Q}|}} = {{{\eta _1}} \over {{\eta _2}}}$$
B
$${{|{v_P}|} \over {|{v_Q}|}} = {{{\eta _2}} \over {{\eta _1}}}$$
C
vP . vQ > 0
D
vP . vQ < 0
2
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
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. JEE Advanced 2015 Paper 2 Offline Physics - Geometrical Optics Question 36 English Comprehension
For two structures namely S1 with n1 = $${{\sqrt {45} } \over 4}$$ and n2 = $${3 \over 2}$$, and S2 with n1 = $${8 \over 5}$$ and n2 = $${7 \over 5}$$ and taking the refractive index of water to be $${4 \over 3}$$ and that to air to be 1, the correct options is/are :
A
NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index $${{16} \over {3\sqrt {15} }}$$
B
NA of S1 immersed in liquid of refractive index $${6 \over {\sqrt {15} }}$$ is the same as that of S2 immersed in water
C
NA of S1 placed in air is the same as that S2 immersed in liquid of refractive index $${4 \over {\sqrt {15} }}$$
D
NA of S1 placed in air is the same as that of S2 placed in water
3
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
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. JEE Advanced 2015 Paper 2 Offline Physics - Geometrical Optics Question 33 English Comprehension
If two structures of same cross-sectional area, but different numeral apertures NA1 and NA2 (NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is
A
$${{N{A_1}N{A_2}} \over {N{A_1} + N{A_2}}}$$
B
$$N{A_1} + N{A_2}$$
C
$$N{A_1}$$
D
$$N{A_2}$$
4
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-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 accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until 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.

JEE Advanced 2015 Paper 2 Offline Physics - Magnetism Question 31 English Comprehension
Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths 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 w1 = w2 and d1 = 2d2, then V2 = 2V1
B
If w1 = w2 and d1 = 2d2, then V2 = V1
C
If w1 = 2w2 and d1 = d2, then V2 = 2V1
D
If w1 = 2w2 and d1 = d2, then V2 = V1
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