JEE Physics - MCQ Practice Questions

The process transconductance parameter k_n = μ_n × C_ox determines how efficiently the MOSFET converts gate voltage to drain current. Both μ_n and C_ox are independent device parameters.

FinFETs employ multiple gates (typically 3 gates) wrapping around a thin silicon fin, providing superior electrostatic control of the channel, reducing short-channel effects and subthreshold swing.

Due to lower hole mobility in PMOS (~2-3 times lower than electron mobility in NMOS), the PMOS width must be 2-3 times larger to match switching speeds and minimize inverter delay.

The depletion width W = √(2εε₀(V_bi + V_R)/(eN_D N_A/(N_D + N_A))) depends on applied voltage, doping concentrations, and built-in potential. Both doping levels and reverse bias voltage are critical factors.

In photoconductive (reverse-bias) mode, the depletion region widens significantly, reducing junction capacitance and transit time, resulting in faster response (MHz to GHz range) and shot noise is suppressed due to the reverse field.

The Early effect occurs because increased reverse bias widens the depletion region, reducing the effective base width. This decreases recombination in the base and increases collector current. The Early voltage V_A is inversely proportional to base doping concentration.

Silicon has dominated modern rectifiers because of significantly lower reverse saturation current (leakage ~nA at room temperature vs ~µA for Ge), better temperature stability, and superior thermal properties. The extra 0.4V drop is offset by reduced losses in high-power applications.

Using v = fλ, where v = 300 m/s and f = 150 Hz, λ = v/f = $\frac{300}{150}$ = 2 m

In transverse waves, particle oscillation is perpendicular to the direction of wave propagation, such as in electromagnetic waves and ripples on water

Frequency remains constant during refraction. λ = v/f. Since speed increases from 330 to 1500 m/s, wavelength increases proportionally

In constructive interference, phase difference is 0° or 2πn. Resultant amplitude = A₁ + A₂

Doppler effect: When source approaches observer, f' = f(v)/(v-vs), which is greater than f

For a string fixed at both ends, fₙ = nf₁ where n is the harmonic number. Second harmonic (n=2) has f₂ = 2f₁

For destructive interference, the path difference must be odd multiples of λ/2: (2n+1)λ/2 or (n+$\frac{1}{2}$)λ

From y = Asin(2πx/λ - 2πt/T), A = 5 m, and f = 1/T = $\frac{1}{2}$π × 2π = 1 Hz

Standing waves result from the superposition of two coherent waves traveling in opposite directions with the same frequency

Intensity ∝ A²f². New intensity = (2A)² × (f/2)² × I/(A²f²) = 4 × ($\frac{1}{4}$) × I = 2I

For closed pipe, f₁ = v/(4L) = 340/(4×0.5) = $\frac{340}{2}$ = 170 Hz

Since I ∝ A², the ratio A₁/A₂ = √(I₁/I₂)

Upon reflection from a denser medium (fixed end), there is a phase change of π radians or 180°