EMI AC OPTICS are extremely important chapters for NEET Physics. Students should practice these questions seriously because they are concept-based and help you understand your real preparation level.
These questions are not ordinary questions. They test conceptual clarity, logical thinking, circuit understanding, electrostatics fundamentals, current electricity concepts, and NEET-level problem-solving ability.
First revise the chapter properly, then attempt these questions honestly. After solving, compare your answer with the official answer and solution given below each question.
If you can solve these questions correctly, you will get a clear idea of where your NEET Physics preparation actually stands.
If you are searching for a Physics Tutor, or if you are facing difficulty in understanding NEET Physics concepts, you may contact Kumar Sir for one-to-one online Physics classes.
EMI, AC and Optics Practice Quiz
NEET marking: +4, -1, 0
Q1
The current I, potential difference VL across the inductor and potential difference VC across the capacitor in the circuit shown are best represented vectorially as:
Correct Answer: Option 4
Official Solution:
VL leads current by 90° and VC lags current by 90°. Therefore the correct vector representation is option 4.
Q2
In an AC sub-circuit as shown in the figure, the resistance R = 0.2 Ω. At a certain instant VA - VB = 0.5 V, I = 0.5 A, and current is increasing at the rate of ΔI/Δt = 8 A/s. The inductance of the coil is:
Correct Answer: Option 3
Official Solution:
VA - VB = L(dI/dt) + IR.
0.5 = 8L + 0.5 × 0.2, so L = 0.05 H.
Q3
As shown in the figure, the magnet is fixed and a horizontal coil is dropped. The acceleration of the coil is a, then:
Correct Answer: Option 1
Official Solution:
The induced currents oppose the motion of the falling coil. Hence the downward acceleration becomes less than g, so a < g.
Q4
Two infinitely long conducting parallel rails are connected through a capacitor C as shown in the figure. A conductor of length l is moved with constant speed V. Which graph represents the variation of current through the conductor with time?
Correct Answer: Option 3
Official Solution:
ε = vBl. At t = 0, resistance due to the capacitor is zero. At t = ∞, resistance due to the capacitor is infinite. Thus the current decreases with time.
Q5
A copper rod ab of length L, pivoted at one end a, rotates at constant angular velocity ω at right angles to a uniform magnetic field of induction B. The emf developed between the mid point c of the rod and end b is:
Correct Answer: Option 1
Official Solution:
E = ∫L/2L vBdr = ωB∫L/2L r dr = 3BωL2/8.
Q6
L, C and R represent inductance, capacitance and resistance respectively. The combination which has the dimensions of frequency is:
Correct Answer: Option 1
Official Solution:
The combinations 1/RC and R/L both have the dimensions of frequency.
Q7
The effective value of current i = 2 sin 100πt + 3 sin(100πt + 60°) is:
Correct Answer: Option 3
Official Solution:
I1 = 2, I2 = 3. Net peak current I = √(22 + 32 + 2×2×3 cos 60°) = √19.
Therefore Irms = √(19/2).
Q8
If I1, I2 and I3 are the respective r.m.s. values of the time-varying current as shown in the three cases I, II and III, then:
Correct Answer: Option 1
Official Solution:
From the waveforms, I1 = I2 < I0, and for the square wave I3 = I0.
Q9
The best material for the core of a transformer is:
Correct Answer: Option 3
Official Solution:
Soft iron is the best material for the core of a transformer.
Q10
A step down transformer transforms a supply line voltage of 2200 V into 220 V. The primary coil has 4000 turns. The efficiency of the transformer is 80% and power output is 8 kW, then power supplied is:
Correct Answer: Option 1
Official Solution:
Let supplied power be x kW. Then (8/x)×100 = 80, so x = 10 kW.
Q11
In an LCR series circuit, the rms voltage across R, L and C are 20 V, 50 V and 50 V respectively. The rms voltage across the L-C combination will be:
Correct Answer: Option 1
Official Solution:
VL-C = 50 - 50 = 0.
Q12
In an AC circuit the applied potential difference and the current flowing are given by V = 100 sin 100πt V and I = 20 cos(100πt - π/2) ampere. The power consumption is equal to:
Correct Answer: Option 2
Official Solution:
V = 100 sin100πt = 100 cos(100πt - π/2) and I = 20 cos(100πt - π/2).
The charge on a plate of a capacitor varies as q = q0 cos 2πft. The separation between plates (d) is very small compared to area (A) of the plate. The peak value of displacement current through the capacitor is:
Correct Answer: Option 1
Official Solution:
I = dq/dt = q02πf sin2πft. For maximum value, sin2πft = 1, so peak current is 2πfq0.
Q14
In an electromagnetic wave, the dimensional formula of (1/2)ε0E2 is, where E is the electric field:
An electron microscope gives better resolution than an optical microscope because:
Correct Answer: Option 3
Official Solution:
Resolving power is inversely proportional to wavelength: R.P. ∝ 1/λ. The effective wavelength of electrons is small.
Q16
The false statement for the refraction through a prism is, where symbols have their usual meanings:
Correct Answer: Option 2
Official Solution:
The false statement among the given prism relations is δ = 2i - A as a general relation.
Q17
An astronomical telescope has a large aperture to:
Correct Answer: Option 2
Official Solution:
Resolving power is proportional to aperture. A large aperture gives high resolution.
Q18
A car is moving towards a plane mirror at a speed of 30 m/s. The relative speed of its image with respect to the car will be, in m/s:
Correct Answer: Option 1
Official Solution:
For a plane mirror, the image approaches the object with twice the object's speed. Relative speed = 2×30 = 60 m/s.
Q19
A mirror is inclined at an angle of 30° with the horizontal. If a ray of light is incident at an angle of 30° as shown, then the angle made by the reflected ray with the horizontal is:
Correct Answer: Option 1
Official Solution:
Using the law of reflection with the given geometry, the reflected ray becomes horizontal. Hence the angle made with the horizontal is zero.
Q20
The rising or setting sun appears red because of:
Correct Answer: Option 4
Official Solution:
The rising or setting sun appears red due to scattering of light in the atmosphere.
Q21
An object O is placed at 8 cm in front of a glass slab, whose one face is silvered as shown in the figure. The thickness of the slab is 9 cm. If the image is formed 10 cm behind the silvered face, the refractive index of the glass is:
Correct Answer: Option 1
Official Solution:
Let the effective distance from the surface be x.
8 + x = 10 + 9 - x, so x = 11/2.
n = real/apparent = 9/(11/2) = 18/11 = 1.63.
Q22
In the given arrangement, a lens of refractive index 1.5 is placed with media of refractive indices n1 and n2 on either side as shown. The values of n1 and n2 respectively with respect to the lens are:
Correct Answer: Option 1
Official Solution:
With respect to the lens, the relative refractive indices are n1 = n2 = 1.
Q23
When blue light is replaced by yellow light, then the resolving power of a microscope:
Correct Answer: Option 2
Official Solution:
Resolving power ∝ 1/λ. Yellow light has larger wavelength than blue light, so resolving power decreases.
Q24
A farsighted person cannot focus distinctly on objects closer than 100 cm. The lens that will permit him to read from a distance of 40 cm will have a focal length of:
Correct Answer: Option 3
Official Solution:
u = -40 cm, v = -100 cm.
1/v - 1/u = 1/f, so 1/f = -1/100 + 1/40 and f = 66.67 cm.
Q25
A convex lens of focal length 10 cm and a concave mirror of radius of curvature 20 cm are placed as shown in the figure. Find d if the final image is erect and formed on the object itself.
Correct Answer: Option 3
Official Solution:
For the lens, f = 10 cm and u = -25 cm.
1/v - 1/u = 1/f gives 1/v + 1/25 = 1/10, hence v = 50/3 cm.
Using the mirror condition from the figure, d = 50/3 + 20 = 36.67 cm.
Q26
A light ray passes from a liquid to air. Find the value of critical angle if velocities in air and liquid are 3×108 m/s and 1.5×108 m/s respectively.
Correct Answer: Option 3
Official Solution:
n = c/v = (3×108)/(1.5×108) = 2.
sin C = 1/n = 1/2, so C = 30°.
Q27
A vertical rod of length l is moved with constant velocity v towards East. The vertical component of Earth's magnetic field is B and angle of dip is θ. The induced e.m.f. in the rod is:
Correct Answer: Option 1
Official Solution:
Let Earth's magnetic field be B0. Since B0sinθ = B, B0 = B/sinθ.
Induced emf E = vB0cosθ l = (vB cosθ l)/sinθ = Blv cotθ.
Q28
A and B are two metallic rings placed at opposite sides of an infinitely long straight conducting wire as shown. If current in the wire is slowly decreased, the direction of induced current will be:
Correct Answer: Option 2
Official Solution:
By Lenz's law, the induced current opposes the decrease in magnetic flux. The direction is anticlockwise in A and clockwise in B.
Q29
A conducting ring of radius r with a conducting spoke is in pure rolling on a horizontal surface in a region having uniform magnetic field B as shown. v is the velocity of the centre of the ring. The potential difference between point O and A is:
Correct Answer: Option 1
Official Solution:
ε = (1/2)Bωr2. Since v = rω, ε = Bvr/2.
Q30
If a prism having refractive index √2 has angle of minimum deviation equal to the angle of the prism, then the angle of refraction of the prism is:
Correct Answer: Option 3
Official Solution:
Given δ = A. At minimum deviation, δ = 2i - A, so 2A = 2i and i = A.
n = sin i / sin(A/2) = sin A / sin(A/2) = 2cos(A/2) = √2.
Thus cos(A/2)=1/√2, so A/2=45° and A=90°.
Q31
The given lens is broken into four parts and rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is:
For an LCR series circuit with an AC source of angular frequency ω, the correct statement is:
Correct Answer: Option 3
Official Solution:
In a series LCR circuit, power factor is unity at resonance, when capacitive reactance equals inductive reactance.
Q33
If the readings of v1 and v3 are 100 V each, then the reading of v2 in volt is:
Correct Answer: Option 3
Official Solution:
As in resonance, VL = VC. Thus VL + VC = 0.
Therefore V1 + V2 + V3 = 200, giving V2 = 200 V.
Q34
There is an electrical element in the block shown. The voltage and current are given by v = v0 sin(157t + 30°) and i = I0 sin(157t + 120°). Then the block has:
Correct Answer: Option 3
Official Solution:
The voltage lags the current. Hence the block has capacitive nature, so it is only a capacitor.
Q35
If instantaneous current in a circuit is given by I = (2 + 3 sinωt) A, then the effective value of resulting current in the circuit is:
Correct Answer: Option 1
Official Solution:
The effective value is I = √(a2 + b2/2).
Here a=2 and b=3, so I = √(4 + 9/2) = √(17/2) A.
Q36
A capacitor of capacitance C has initial charge Q0 and is connected to an inductor of inductance L as shown. At t = 0, switch S is closed. The current through the inductor when energy in the capacitor is three times the energy of the inductor is:
Correct Answer: Option 1
Official Solution:
Total energy = Q02/(2C). Given capacitor energy is three times inductor energy, inductor energy is one-fourth of total energy.
(1/2)LI2 = (1/4)×Q02/(2C), so I = Q0/(2√LC).
Q37
A car moves on a plane road. Induced e.m.f. across the axle is maximum when it moves:
Correct Answer: Option 1
Official Solution:
At the poles, the perpendicular component of Earth's magnetic field is maximum, so induced e.m.f. is maximum.
Q38
A force of 10 N is required to move a conducting loop through a non-uniform magnetic field at 2 m/s. The rate of production of electrical energy in the loop, in watt, is:
Correct Answer: Option 3
Official Solution:
P = Fv = 10 × 2 = 20 W. Kinetic energy is converted into heat/electrical energy in the loop.
Q39
The plane surface of a plano-convex lens is silvered and it then acts like a concave mirror of focal length 30 cm. If μ for the lens is 1.5, the radius of curvature of the convex surface is:
Correct Answer: Option 2
Official Solution:
For the silvered plano-convex lens, 1/F = 2(μ-1)/R.
1/30 = 2(1.5-1)/R, hence R = 30 cm.
Q40
A thin prism P1 of angle 5° and refractive index 1.54 is combined with another thin prism P2 of refractive index 1.70 so that the ray undergoes no deviation. The angle of prism P2 is:
Correct Answer: Option 1
Official Solution:
For no deviation, (n1-1)A1 = (n2-1)A2.
A2 = (0.54×5°)/0.70 = 3.85°.
Q41
A lens of unknown nature and focal length 20 cm forms an erect image four times the size of an object. The nature of lens and distance of object from lens are:
Correct Answer: Option 1
Official Solution:
Let distance of object from lens be x. Then u=-x and for an erect magnified image, v=-4x.
1/v - 1/u = 1/20 gives -1/(4x)+1/x=1/20, so x=15 cm. The lens is convex.
Q42
Focal length of a lens for red colour is:
Correct Answer: Option 2
Official Solution:
Due to dispersion, refractive index is lower for red light than for violet light. Therefore the focal length for red colour is greater than that for violet.
Q43
In a compound microscope, the intermediate image is:
Correct Answer: Option 3
Official Solution:
In a compound microscope, the objective forms an intermediate image that is real, inverted and magnified.
Q44
A lens made up of different material forms two images of a point object O as shown in the figure. Then:
Correct Answer: Option 1
Official Solution:
Deviation produced by μ1 is more than that produced by μ2. Therefore μ1 > μ2.
Q45
The distance between an object and its doubly magnified image by a concave mirror is, where f is focal length:
Correct Answer: Option 1
Official Solution:
Let distance of the object from the mirror be x. Then u=-x, v=-2x, and for a concave mirror f=-f.
Using mirror formula 1/v + 1/u = 1/f: -1/(2x)-1/x=-1/f, so x=3f/2.
Distance between object and image is 2x - x = x = 3f/2.
