A rectangular conducting loop ABCD with sides a and b is rotated with angular velocity ω in a uniform magnetic field B perpendicular to the plane of rotation. The induced EMF is:
Answer: B
When the loop rotates, the magnetic flux through it varies as Φ = BA·cosωt. The induced EMF = -dΦ/dt = BA·ω·sinωt = Bab·ω·sinωt
Q.222Medium
Two magnets are placed with their north poles facing each other. The force between them varies with distance r as:
Answer: D
Two magnetic dipoles interact with force F ∝ 1/r⁴ when aligned along the same axis. This is because the magnetic field of a dipole varies as 1/r³, and force on a dipole is proportional to the field gradient.
Q.223Medium
A conducting rod of length L moves with velocity v perpendicular to its length in a magnetic field B. The motional EMF induced is maximum when:
Answer: B
Motional EMF = B·L·v·sinθ, where θ is the angle between v and B. EMF is maximum when sinθ = 1, i.e., when v is perpendicular to B.
Q.224Medium
The magnetic susceptibility of a paramagnetic material is:
Answer: B
Paramagnetic materials have positive but small magnetic susceptibility (χ > 0, typically 10⁻⁵ to 10⁻³). Diamagnetic materials have small negative susceptibility, and ferromagnetic materials have large positive susceptibility.
Q.225Medium
A charged particle with charge q and mass m is moving with speed v in a circular path of radius r in a magnetic field. The magnetic field strength is:
Answer: A
From qvB = mv²/r (centripetal force equals magnetic force), we get B = mv/(qr). This is the relationship between field strength, particle properties, and circular path radius.
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Q.226Medium
The permeability of free space μ₀ has the value:
Answer: B
The permeability of free space μ₀ = 4π × 10⁻⁷ T·m/A. Option A is permittivity ε₀, option C is speed of light, and option D is Planck's constant.
Q.227Medium
A long straight wire carrying current I produces a magnetic field at distance r. If the current is doubled and distance is halved, the magnetic field becomes:
Answer: B
B = μ₀I/(2πr). If I → 2I and r → r/2, then B_new = μ₀(2I)/(2π(r/2)) = 4·μ₀I/(2πr) = 4B_initial
Q.228Medium
The SI unit of magnetic flux density (magnetic field) is:
Answer: B
The SI unit of magnetic field (flux density) is Tesla (T). 1 T = 1 Wb/m² = 1 kg/(A·s²). Weber is the unit of magnetic flux, Gauss is CGS unit, and Henry is unit of inductance.
Q.229Medium
Two slits separated by 0.5 mm are illuminated by light of wavelength 500 nm. The distance to the screen is 1 m. Find the fringe width.
Answer: B
Fringe width β = λD/d = (500×10⁻⁹ × 1)/(0.5×10⁻³) = 1×10⁻³ m = 1 mm
Q.230Medium
A concave mirror has a focal length of 20 cm. An object is placed at the center of curvature. Where will the image form?
Answer: A
When object is at center of curvature (u = R = 2f), image forms at the same position (v = R). Magnification = -1
Q.231Medium
A ray of light is incident on a glass slab of thickness 5 cm and refractive index 1.6. If the incident angle is 45°, find the lateral displacement.
Resolving power = 1/(1.22λ/2NA). It depends on wavelength and numerical aperture (NA = n×sin(θ))
Q.233Medium
When white light passes through a prism, violet light deviates more than red light. This is because:
Answer: D
Higher frequency → higher refractive index → slower speed in medium → greater deviation. All statements are correct.
Q.234Medium
An object is placed 10 cm from a concave lens of focal length 20 cm. Find the magnification.
Answer: C
Using lens formula: 1/(-20) = 101 + 1/v. v = -6.67 cm. Magnification m = -v/u = 6.1067 = 0.67
Q.235Medium
Polarization of light proves that light is:
Answer: B
Only transverse waves can be polarized. Polarization demonstrates the transverse nature of electromagnetic waves.
Q.236Medium
The intensity at a point in the interference pattern of two coherent sources is I₁ and I₂. The resultant intensity is maximum when the phase difference is:
Answer: B
Maximum intensity occurs for constructive interference when phase difference = 0 or 2π. I_max = (√I₁ + √I₂)²
Q.237Medium
An object is placed at distance u from a convex lens of focal length f. If the magnification is -2, what is the relationship between u and f?
Answer: A
Magnification m = -v/u = -2, so v = 2u. Using lens equation: 1/f = 1/u + 1/v = 1/u + 1/(2u) = 3/(2u). Therefore u = 3f/2.
Q.238Medium
In Young's double-slit experiment with slit separation d = 1 mm and distance to screen D = 1 m, if the 5th bright fringe is at 2.5 mm from the center, what is the wavelength of light?
Answer: A
For bright fringes: y = (m·λ·D)/d. For 5th bright fringe: 2.5 × 10⁻³ = (5 × λ × 1)/(1 × 10⁻³). Therefore λ = 500 nm.
Q.239Medium
A ray undergoes total internal reflection at a critical angle θc. If the refractive index of the denser medium is √2, what is θc?
Answer: B
At critical angle: sin(θc) = 1/n = 1/√2. Therefore θc = 45°. This occurs when light travels from denser to less dense (rarer) medium.
Q.240Medium
A concave lens of focal length -20 cm is used to form an image of an object placed 10 cm from it. What is the nature of the image?
Answer: B
For concave lens, images are always virtual, erect, and diminished regardless of object position. Using 1/v = 1/f - 1/u = -201 - 101 = -203, v = -320 ≈ -6.67 cm (virtual).
