IIT-JEE 2006 | Alternating Current Question 16 | Physics

1

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)

2

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 $\mathrm{CV}\left(1-e^{-1}\right)$.

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 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 $\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$.

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 2 English Comprehension

If the total charge stored in the LC circuit.is $\mathrm{Q}_0$, then for $t \geq 0$,

A

the charge on the capacitor is

$$ \mathrm{Q}=\mathrm{Q}_0 \cos \left(\frac{\pi}{2}+\frac{t}{\sqrt{\mathrm{LC}}}\right) $$

B

the charge on the capacitor is

$$ \mathrm{Q}=\mathrm{Q}_0 \cos \left(\frac{\pi}{2}-\frac{1}{\sqrt{\mathrm{LC}}}\right) . $$

C

the charge on the capacitor is

$$ \mathrm{Q}=-\mathrm{LC} \frac{d^2 \mathrm{Q}}{d t^2} . $$

D

the charge on the capacitor is

$$ \mathrm{Q}=-\frac{1}{\sqrt{\mathrm{LC}}} \frac{d^2 \mathrm{Q}}{d t^2} . $$

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