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.**A coil of resistance R and inductance L is connected to a battery of emf E volt. The final current in the coil is

Solution

Final current in constant and L plays no role at that instant. Therefore, i=E/R.

Final current in constant and L plays no role at that instant. Therefore, i=E/R.

**Q2.**An alternating e.m.f. is applied to purely capacitive circuit. The phase relation between e.m.f. and current flowing in the circuit is or In a circuit containing capacitance only

Solution

For purely capacitive circuit e=e

For purely capacitive circuit e=e

_{0}sinÏ‰t i=i_{0}sin(Ï‰t+Ï€/2),i.e., current is ahead of emf by Ï€/2**Q3.**What is the value of inductance L for which the current is a maximum in a series L-C-R circuit with C=10Î¼Fand Ï‰=1000 s^(-1)?

Solution

Current in a L-C-R series circuit,

Where V is rms value of current, R is resistance, X

Current in a L-C-R series circuit,

_{L}is inductive reactance and X_{C}is capacitive reactance.For current to be maximum, denominator should be minimum which can be done, if

X

_{L}=X_{C}This happens in resonance state of the circuit

i.e, Ï‰L=1/Ï‰C

or L=1/(Ï‰

^{2}C) ...(i)Given, Ï‰=1000s

^{-1},C=10Î¼F=10×10^{-6}FHence, L=1/(1000

^{2}×10×10^{-6})=0.1 H=100 mH

**Q4.**What will be the self inductance of a coil, to be connected in a series with a resistance of Ï€√3 Î© such that the phase difference between the emf and the current at 50 Hz frequency is 30°

**Q5.**When an AC source of emf e=E

_{0}sin(100t) is connected across a circuit, the phase difference between the emfe and the current i in the circuit is observed to beÏ€/4 , as shown in the diagram. If the circuit consists possibly only of R – C or R – L or L – C in series, find the relationship between the two elements

Solution

As the current i leads the emf e by Ï€/4, it is an R – C circuit

As the current i leads the emf e by Ï€/4, it is an R – C circuit

tan Ï• =X

_{C}/Ror tan Ï€/4 =(1/Ï‰C)/R

∴ Ï‰CR=1
As Ï‰=100 rads

^{-1}The product of C – R should be 1/100 s

^{-1}.**Q6.**The current in series LCR circuit will be maximum when Ï‰ is

Solution

At resonant frequency current in series LCR circuit is maximum

At resonant frequency current in series LCR circuit is maximum

**Q7.**Three identical ring move with the same speed on a horizontal surface in a uniform horizontal magnetic field normal to the planes of the rings. The first (A) slips without rolling, the second (B) rolls without slipping, and the third rolls with slipping

Solution

The same emf is induced in all the three rings because emf is only due to linear motion and does not depend on spin.

The same emf is induced in all the three rings because emf is only due to linear motion and does not depend on spin.

**Q8.**An alternating voltage e=200 sin 100 t is applied to a series combination R=30 Î© and an inductor of 400 mH. The power factor of the circuit is

**Q9.**An alternating e.m.f. of angular frequency Ï‰ is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency

Solution

The instantaneous values of emf and current in inductive circuit are given by E=E

The instantaneous values of emf and current in inductive circuit are given by E=E

_{0}sin Ï‰t and i=i_{0}sin(Ï‰t-Ï€/2) respectivelySo, P

_{inst}=Ei=E_{0}sin Ï‰t × i_{0}sin(Ï‰t-Ï€/2)=E

_{0}i_{0}sin Ï‰t(sin Ï‰t cos Ï€/2-cos Ï‰t sinÏ€/2)=E

_{0}i_{0}sinÏ‰t cosÏ‰t=1/2 E

_{0}i_{0}sin 2Ï‰t(sin 2Ï‰t = 2sin Ï‰t cos Ï‰t)Hence, angular frequency of instantaneous power is 2Ï‰

**Q10.**The peak value of an alternating emf E given by E=E

_{0}cosÏ‰t is 10 V and its frequency is 50 Hz. At a time t=1/100 s, the instantaneous value of the emf is

Solution

E=E

E=E

_{0}cos Ï‰t= 10 cos (2Ï€ × ft) =10 cos(2Ï€ × 50 × 1/600) =10 cos(Ï€/6)=10 × √3/2 = 5√3 V