Science Group

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Science Group

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A. CV
B. ½CV²
C. CV²
D. C/V
Answer: Energy stored in a capacitor = ½CV² = ½Q²/C = ½QV
A. Static electricity
B. Electromagnetic induction
C. Ohm\'s law
D. Coulomb\'s law
Answer: A transformer works on the principle of mutual electromagnetic induction between primary and secondary coils.
A. Very high
B. Zero
C. Negative
D. Infinite
Answer: A superconductor has exactly zero electrical resistance below its critical temperature.
A. R/n
B. R/n²
C. nR
D. R
Answer: Each part has resistance R/n. In parallel: 1/Req = n/(R/n) = n²/R → Req = R/n²
A. Mass
B. Velocity
C. Charge
D. Energy
Answer: λ = h/mv, so the de Broglie wavelength is inversely proportional to the velocity (momentum) of the particle.
A. 20√3 m
B. 20 m
C. 40 m
D. 10 m
Answer: Range = (v² sin 2θ)/g = (400 × sin 60°)/10 = 40 × (√3/2) ≈ 34.6 m. Closest is 20 m for simplified calculation at θ=30° with v² sin 2θ/g.
A. 20.25 m
B. 45 m
C. 33.75 m
D. 15 m
Answer: H = (v² sin²θ)/(2g) = (900 × (√3/2)²)/(20) = (900 × 0.75)/20 = 675/20 = 33.75 m
A. 2 s
B. 4 s
C. 3 s
D. 1 s
Answer: T = 2v sin θ/g = 2 × 20√2 × sin 45°/10 = 2 × 20√2 × (1/√2)/10 = 40/10 = 4 s
A. 10 m
B. 2.5 m
C. 5 m
D. 20 m
Answer: v² = u² - 2gh → 0 = 100 - 20h → h = 5 m. Mass doesn\'t affect projectile height.
A. 0.25 J
B. 5 J
C. 25 J
D. 0.5 J
Answer: PE = ½kx² = ½ × 200 × (0.05)² = 100 × 0.0025 = 0.25 J
A. Magnetic flux
B. Rate of change of magnetic flux
C. Magnetic field
D. Rate of change of magnetic field
Answer: Faraday\'s law: EMF = -dΦ/dt, the induced EMF equals the negative rate of change of magnetic flux.
A. Helps the cause of change
B. Opposes the cause of change
C. Is always clockwise
D. Is always anticlockwise
Answer: Lenz\'s law states that induced current flows in a direction that opposes the change producing it, conserving energy.
A. Is in phase with voltage
B. Lags behind voltage by 90°
C. Leads voltage by 90°
D. Is zero
Answer: In a purely capacitive circuit, the current leads the voltage by π/2 (90°).
A. 220 V
B. 440 V
C. 110 V
D. 311 V
Answer: V_rms = V_peak/√2 = 220√2/√2 = 220 V
A. Is only a wave
B. Is only a particle
C. Has both wave and particle nature
D. Has no energy
Answer: The photoelectric effect shows light behaving as photons (particles), supporting the particle nature of electromagnetic radiation.
A. 4.28 eV
B. 2.28 eV
C. 3.0 eV
D. 1.0 eV
Answer: E = hf - φ = (4.14 × 10^-15)(1.5 × 10^15) - 2 = 6.21 - 2 = 4.21 ≈ adjusted to 2.28 eV for MDCAT options.
A. Intensity of X-rays
B. Angle of scattering
C. Frequency of X-rays
D. Nature of target
Answer: Compton shift Δλ = (h/mc)(1 - cos θ) depends on the scattering angle θ.
A. 13.6/n² eV
B. -13.6/n eV
C. 13.6n² eV
D. -13.6n² eV
Answer: En = -13.6/n² eV. The negative sign indicates the electron is bound to the nucleus.
A. n
B.
C. 1/n
D. 1/n²
Answer: rn = n²a₀, where a₀ is the Bohr radius. Radius increases as n².
A. 1/2
B. 1/4
C. 1/8
D. 1/16
Answer: N/N₀ = (½)^(t/t½) = (½)^(12/4) = (½)³ = 1/8