As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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**Q1.**The average power dissipated in a pure inductor of inductance L when an ac current is passing through it, is (Inductance of the coil L and current I)

Solution

It is a fact.

It is a fact.

**Q2.**The time taken by AC of 50 Hz in reaching from zero to the maximum value is

Solution

The time taken by AC in reaching from zero to maximum value is t=T/4=1/(4 f)

The time taken by AC in reaching from zero to maximum value is t=T/4=1/(4 f)

= 1/(50×4)=5×10

^{-3}s**Q3.**A 100 V, AC source of frequency 500 Hz is connected to an L-C-R circuit with L=8.1 mH, C=12.5 Î¼F, R=10 Î© all connected in series as shown in figure. What is the quality factor of circuit?

Solution

We know that Q - factor of series resonant circuit is given as Q=(Ï‰

We know that Q - factor of series resonant circuit is given as Q=(Ï‰

_{r}L)/RHere, L=8.1 mH,C=12.5 Î¼F,R=10Î©,f=500 Hz

∴ Q=(Ï‰

_{r}L)/R=2Ï€fL/R =(2×Ï€×500×8.1×10^{-3})/10=8.1Ï€/10

=2.5434

**Q4.**Two coils A and B have coefficient of mutual inductance M = 2H. The magnetic flux passing through coil A changes by 4 Wb in 10 s due to change in current in B. Then

Solution

Here, M=2H,dÏ•=4 Wb,dt=10 s

Here, M=2H,dÏ•=4 Wb,dt=10 s

As Ï• =M i

dÏ•=M di

Or di=dÏ•/M=4/2=2 A

Also, dÏ•=M (di)=2(1)

= 2 Wb

**Q5.**In an LCRcircuit R=100 ohm. When capacitance C is removed, the current lags behind the voltage by Ï€/3. When inductance L is removed, the current leads the voltage by Ï€/3. The impedance of the circuit is

Solution

When C is removed circuit becomes RL circuit

When C is removed circuit becomes RL circuit

hence
tan Ï€/3=X

_{L}/R…(i)When L is removed circuit becomes RC circuit hence
tan Ï€/3=X

_{C}/R …(ii)From equation (i) and (ii) we obtain X

_{L}=X_{C}. This is the condition of resonance and in resonance Z=R=100Î©**Q6.**The instantaneous values of current and emf in an ac circuit are I=1/√2 sin 314 t amp and E=√2 sin (314 t-Ï€/6) V respectively. The phase difference between E and I will be

Solution

Phase difference relative to the current

Phase difference relative to the current

Ï•=(314t-Ï€/6)-(314 t)=-Ï€/6

**Q7.**In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state are

Solution

In steady state current through the branch having capacitor is zero.

In steady state current through the branch having capacitor is zero.

∴ 1/R=1/1+1/2+1/3

1/R=(6+3+2)/6

R=6/11

As V=iR

∴ 6=i×6/11

Current through the battery i=11A

Charge on the capacitor q=CV
⇒ q=0.5×10

^{-6}×6q=3Î¼C

**Q8.**In a L – R circuit, the value of L is (0.4/Ï€)H and the value of Ris 30 Î©. If in the circuit, an alternating emf of 200 V at 50 cycle/s is connected, the impedance of the circuit and current will be

Solution

Z^2=R^2+(2Ï€fL)

Z^2=R^2+(2Ï€fL)

^{2}=(30)^2+(2Ï€×50×0.4/Ï€)

^{2}=(900+1600)=2500

or Z=50 Î©

Also, I=V/Z=200/50=4 A

Solution

Time constant of R – C circuit is Ï„=RC

Time constant of R – C circuit is Ï„=RC

Here effective resistance of the circuit

= (2R×3R)/(2R+3R)=6R/5

∴ Ï„=6R/5×C=6RC/5

**Q10.**A bulb and a capacitor are in series with an ac source. On increasing frequency how will glow of the bulb change

Solution

This is because, when frequency v is increased, the capacitive reactance X

This is because, when frequency v is increased, the capacitive reactance X

_{C}=1/2Ï€vC decreases and hence the current through the bulb increases