1
JEE Advanced 2017 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
Change Language
A circular insulated copper wire loop is twisted to form two loops of area $$A$$ and $$2A$$ as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $$\overrightarrow B $$ points into the plane of the paper. At $$t=0,$$ the loop starts rotating about the common diameter as axis with a constant angular velocity $$\omega $$ in the magnetic field.

Which of the following options is/are correct?

JEE Advanced 2017 Paper 1 Offline Physics - Electromagnetic Induction Question 15 English
A
The emf induced in the loop is proportional to the sum of the areas of the two loops
B
The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude if maximum emf induced in the smaller loop alone
C
The net emf induced due to both the loops is proportional to $$\cos \,\omega t$$
D
The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper
2
JEE Advanced 2017 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
Change Language
In the circuit shown, $$L = 1\,\mu H,C = 1\,\mu F\,$$ and $$R = 1\,k\Omega .$$ They are connected in series with an a.c. source $$V = {V_0}\sin \omega t$$ as shown.

Which of the following options is/are correct?

JEE Advanced 2017 Paper 1 Offline Physics - Alternating Current Question 13 English
A
The current will be in phase with the voltage if $$\omega = {10^4}$$ $$rad.{s^{ - 1}}$$
B
The frequency at which the current will be in phase with the voltage is independent of $$R$$
C
At $$\omega \sim 0$$ the current flowing through the circuit becomes nearly zero
D
At $$\omega > > {10^6}rad.{s^{ - 1}},$$ the circuit behaves like a capacitor
3
JEE Advanced 2017 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
Change Language
For an isosceles prism of angle $$A$$ and refractive index $$\mu $$, it is found that the angle of minimum deviation $${\delta _m} = A.$$

Which of the following options is/are correct?
A
For the angle of incidence $${i_1} = A,$$ the ray inside the prism is parallel to the base of the prism
B
For this prism, the refractive index $$\mu $$ and the angle of prism $$A$$ are related as

$$A = {1 \over 2}{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$
C
At minimum deviation, the incident angle $${i_1}$$ and the refracting angle $${r_1}$$ at the first refracting surface are related by $${r_1} = \left( {{i_1}/2} \right)$$
D
For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is

$${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A} \right]$$
4
JEE Advanced 2017 Paper 1 Offline
Numerical
+3
-0
Change Language
An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $${n_i}$$ to another with quantum number $${n_f}$$. $${V_i}$$ and $${V_f}$$ are respectively the initial and final potential energies of the electron. If $${{{V_i}} \over {{V_f}}} = 6.25$$, then the smallest possible $${n_f}$$ is
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