1
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

STATEMENT 1

A vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil.

IIT-JEE 2007 Paper 2 Offline Physics - Alternating Current Question 8 English

Because

STATEMENT 2

In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
C
Statement-1 is True, Statement-2 is False.
D
Statement-1 is False, Statement-2 is True.
2
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-0

Match the following columns.

Column I Column II
(A) Dielectric ring uniformly charged. (P) Time independent electrostatic field out of system.
(B) Dielectric ring uniformly charged rotating with angular velocity $$\omega$$. (Q) Magnetic field.
(C) Constant current in ring $$io$$ (R) Induced electric field.
(D) $$i=i_0\cos\omega t$$ (S) Magnetic moment.

A
A $$\to$$ (P); B $$\to$$ (Q, S); C $$\to$$ (Q, S); D $$\to$$ (Q, R, S)
B
A $$\to$$ (P); B $$\to$$ (S); C $$\to$$ (Q, S); D $$\to$$ (R, S)
C
A $$\to$$ (P); B $$\to$$ (Q); C $$\to$$ (Q, S); D $$\to$$ (S)
D
A $$\to$$ (P); B $$\to$$ (S); C $$\to$$ (Q); D $$\to$$ (Q, R, S)
3
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-1

The capacitor of capacitance $C$ can be charged (with the help of a resistance R ) by a voltage source V , by closing switch $\mathrm{S}_1$ while keeping switch $\mathrm{S}_2$ open. The capacitor can be connected in series with an inductor L by closing switch $\mathrm{S}_2$ and opening $\mathrm{S}_1$.

IIT-JEE 2006 Physics - Alternating Current Question 3 English Comprehension

Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

A

after time interval $\tau$, charge on the capacitor is $\frac{\mathrm{CV}}{2}$.

B

after time interval $2 \tau$, charge on the capacitor is $\mathrm{CV}\left(1-e^{-2}\right)$.

C

the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.

D

after time interval $2 \tau$, charge on the capacitor is $\mathrm{CV}\left(1-e^{-1}\right)$.

4
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-1

The capacitor of capacitance $C$ can be charged (with the help of a resistance R ) by a voltage source V , by closing switch $\mathrm{S}_1$ while keeping switch $\mathrm{S}_2$ open. The capacitor can be connected in series with an inductor L by closing switch $\mathrm{S}_2$ and opening $\mathrm{S}_1$.

IIT-JEE 2006 Physics - Alternating Current Question 1 English Comprehension

After the capacitor gets fully charged, $\mathrm{S}_1$ is opened and $S_2$ is closed so that the inductor is connected in series with the capacitor. Then,

A

at $t=0$, the energy stored in the circuit is purely in the form of magnetic energy.

B

at any time $t>0$, the current in the circuit is in the same direction.

C

at $t>0$, there is no exchange of energy between the inductor and the capacitor.

D

at any time $t>0$, the instantaneous current in the circuit may $\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$.

JEE Advanced Subjects

Browse all chapters by subject