1
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the piston moves out by a distance x.

Ignoring the friction between the piston and the cylinder, the correct statements is/are

A
If V2 = 2V1 and T2 = 3Tl, then the energy stored in the spring is $${1 \over 4}{P_1}{V_1}$$
B
If V2 = 2V1 and T2 = 3T1, then the change in internal energy is $$3{P_1}{V_1}$$
C
If V2 = 3V1 and T2 = 4T1, then the work done by the gas is $${7 \over 3}{P_1}{V_1}$$
D
If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is $${17 \over 6}{P_1}{V_1}$$
2
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
A fission reaction is given by $$_{92}^{236}U \to _{54}^{140}Xe + _{38}^{94}Sr + x + y$$, where x and y are two particles. Considering $$_{92}^{236}U$$ to be at rest, the kinetic energies of the products are denoted by $${K_{Xe}},{K_{Sr}},{K_x}(2MeV)$$ $$\text { and } \mathrm{K}_{\mathrm{y}}(2 \mathrm{MeV})$$, respectively. Let the binding energies per nucleon of $$_{92}^{236}U$$, $$_{54}^{140}Xe$$ and $$_{38}^{94}Sr$$ be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are
A
x = n, y = n, Ksr = 129 MeV, KXe = 86 MeV
B
x = p, y = e$$-$$, Ksr = 129 MeV, KXe = 86 MeV
C
x = p, y = n, Ksr = 129 MeV, KXe = 86 MeV
D
x = n, y = n, Ksr = 86 MeV, KXe = 129 MeV
3
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

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
4
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.
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
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