Same energy in two cylinders (same length, second has half diameter). Electric field amplitude:
A50 sin(ωt−kx)
B400 sin(ωt−kx)
C200 sin(ωt−kx)
D100 sin(ωt−kx)
✅ Correct: C
Energy density \(\propto E_0^2\). Same energy, area ratio \(4:1\). Energy \(= u\times V = \frac{\varepsilon_0 E_0^2}{2}\times Al\). Same energy: \(E_0^2 A = \text{const}\). If \(A\) reduces to \(1/4\), \(E_0\) doubles. If original is 100, new \(= \mathbf{200}\).
Q3
Capacitor area 16 cm², current 6 A. Displacement current through area 3.2 cm² (mA):
A600
B1200
C300
D2400
✅ Correct: B
Displacement current density is uniform. \(I_D(3.2) = I \times (3.2/16) = 6\times0.2 = 1.2\text{ A} = \mathbf{1200\text{ mA}}\).
Q4
EM wave: \(E = 300\sin(5\pi\times10^3 x - 3\pi\times10^{11}t)\) V/m. Magnetic field amplitude:
Correct relation between \(E_0\) and \(B_0\) of EM wave:
A\(E_0B_0 = \omega k\)
B\(E_0 = kB_0\)
C\(kE_0 = \omega B_0\)
D\(E_0 = \omega B_0\)
✅ Correct: C
From Maxwell: \(E_0/B_0 = \omega/k = c\). So \(kE_0 = \omega B_0\). \(\mathbf{kE_0 = \omega B_0}\).
Q7
Order of EM spectrum from lowest to highest frequency:
ARadio→Micro→IR→Visible→UV→X-ray→Gamma
BGamma→X-ray→UV→Visible→IR→Micro→Radio
CRadio→IR→Micro→Visible→UV→X-ray→Gamma
DMicro→Radio→IR→Visible→UV→X-ray→Gamma
✅ Correct: A
Correct order (lowest to highest frequency): Radio waves → Microwaves → Infrared → Visible → Ultraviolet → X-rays → Gamma rays. Gamma rays have highest frequency and energy.
Q8
Which EM wave is used in radar?
AX-rays
BInfrared
CMicrowaves
DGamma rays
✅ Correct: C
Radar (Radio Detection And Ranging) uses microwaves (wavelength 1 mm to 30 cm). Microwaves penetrate clouds and rain, reflect off metallic objects, enabling detection of aircraft and ships.
Q9
Displacement current arises due to:
AChanging magnetic field
BChanging electric flux
CMoving charges
DStatic charges
✅ Correct: B
Maxwell's displacement current: \(I_D = \varepsilon_0 d\Phi_E/dt\). It arises due to changing electric flux (not actual charge flow). This completes Ampere's law and predicts EM waves.
Q10
Speed of EM waves in vacuum:
A\(3\times10^6\) m/s
B\(3\times10^8\) m/s
C\(3\times10^{10}\) m/s
D\(3\times10^4\) m/s
✅ Correct: B
Speed of light \(c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} = \mathbf{3\times10^8\text{ m/s}}\). This is the same for all EM waves in vacuum regardless of wavelength. In medium: \(v = c/n\).
Q11
Wavelength of microwave oven (~2.45 GHz):
A12.2 cm
B1.22 cm
C1.22 m
D0.122 mm
✅ Correct: A
\(\lambda = c/f = 3\times10^8/(2.45\times10^9) = 0.122\text{ m} \approx \mathbf{12.2\text{ cm}}\). Microwave ovens use this frequency to resonate with water molecules.
Q12
Energy of photon with wavelength \(\lambda = 600\) nm (\(h = 6.6\times10^{-34}\) J·s):
Which EM wave is used for satellite communication?
ARadio waves
BX-rays
CMicrowaves
DInfrared
✅ Correct: C
Satellite communication uses microwaves (GHz range). They can penetrate the ionosphere (unlike longer radio waves) and travel to satellites. Radio waves reflect off ionosphere — useful for long-distance terrestrial communication.
Q14
Intensity of EM wave is proportional to:
A\(E_0\)
B\(B_0\)
C\(E_0^2\)
D\(E_0 B_0\)
✅ Correct: C
Intensity \(I = \frac{1}{2}\varepsilon_0 cE_0^2 = \frac{cB_0^2}{2\mu_0}\). Both \(E^2\) and \(B^2\) are proportional to intensity. Answer: \(\mathbf{E_0^2}\) (proportional, same as \(B_0^2\)).
Q15
Which EM wave has wavelength range 0.01–10 nm?
AGamma rays
BX-rays
CUV rays
DMicrowaves
✅ Correct: B
X-rays: wavelength \(\approx 0.01\) nm to 10 nm. Gamma rays: \(< 0.01\) nm. UV: 10–400 nm. Visible: 400–700 nm. X-rays are used in medical imaging and material analysis. \(\mathbf{\text{X-rays}}\).
Q16
Poynting vector represents:
AElectric field strength
BMagnetic field strength
CRate of energy flow per unit area
DMomentum of EM wave
✅ Correct: C
Poynting vector \(\vec{S} = \dfrac{1}{\mu_0}\vec{E}\times\vec{B}\). Its magnitude represents rate of energy flow per unit area (W/m²) — same as intensity. Direction gives direction of wave propagation.
Q17
Gamma rays are produced in:
ARadioactive nuclei transitions
BElectron transitions in atoms
CVibrations of molecules
DAcceleration of electrons
✅ Correct: A
Gamma rays originate from radioactive nuclear transitions (nucleus drops to lower energy state). X-rays: inner shell electron transitions. UV/Visible/IR: outer electron transitions. Microwaves: molecular vibrations/rotations.
Q18
EM wave traveling in +x direction has \(\vec{E}\) in +y direction. \(\vec{B}\) direction:
A+y
B+z
C−z
D+x
✅ Correct: B
\(\vec{S} = \vec{E}\times\vec{B} \propto +\hat{x}\). \(\hat{y}\times\vec{B} = +\hat{x}\). \(\hat{y}\times\hat{z} = \hat{x}\). So \(\vec{B}\) is in \(\mathbf{+z}\) direction.
Q19
In free space, ratio of electric to magnetic energy density of EM wave:
Ac
B1/c
C1
Dc²
✅ Correct: C
Electric energy density \(u_E = \varepsilon_0 E^2/2\). Magnetic energy density \(u_B = B^2/2\mu_0\). Since \(E = cB\) and \(c = 1/\sqrt{\mu_0\varepsilon_0}\): \(u_E = \varepsilon_0 c^2 B^2/2 = B^2/2\mu_0 = u_B\). Ratio \(= \mathbf{1}\).
Q20
Sun emits maximum radiation at \(\lambda_{max} \approx 500\) nm. This corresponds to:
AInfrared
BUltraviolet
CVisible (green)
DX-rays
✅ Correct: C
500 nm is in the visible region (green) of the EM spectrum. Wien's displacement law: \(\lambda_{max} T = 2.9\times10^{-3}\) m·K. Sun surface \(T \approx 5800\) K → \(\lambda_{max} \approx 500\) nm.
R
Roshan
Expert · 5 Years Experience
Top 20 questions from PYQ analysis. NeetJeeRankers pe free expert-level practice milti hai.