GATE 2023 EC question paper PDF and answer key

The graph is even, non-negative, zero at $x = 0$ with a sharp dip, and decays on both sides.

• (B) and (D) are odd, so they go negative for $x < 0$.
• (C) grows without bound for $x \to -\infty$.
• Only $x^2 2^{-|x|}$ fits.

There are 200 pooled tests. In the worst case every positive person is in a different group, so 100 groups test positive and need $100 \times 5 = 500$ individual tests. The total is $200 + 500 = 700$.

• Accumulation ($V_g < V_{fb}$): no depletion, $W_{dep} = 0$.
• Depletion: $W_{dep}$ grows roughly as the square root of the gate overdrive.
• Inversion ($V_g > V_t$): the inversion charge screens further increases, so $W_{dep}$ saturates at $W_{max}$.

This is plot (B).

Work in mA, kΩ and V. Let $V$ be the left top node (joined to the node above the upward 1 mA source) and $V_C$ the node on the right of the 1 kΩ.

KCL at $V$ (the upward 1 mA source in and the rightward 1 mA source out cancel): $\frac{V - 2}{2} + \frac{V}{2} + \frac{V - V_C}{1} = 0 \Rightarrow 2V - V_C = 1$

KCL at $V_C$: $\frac{V_C}{0.2} = 1 + \frac{V - V_C}{1} \Rightarrow 6V_C = 1 + V$

Solving gives $V_C = \frac{3}{11}$ V, so $i = \frac{V_C}{0.2\ \text{k}\Omega} = \frac{15}{11} \approx 1.36\ \text{mA}$

$\lambda/2 = 30$ cm, so $\lambda = 60$ cm.

• $R_m = Z_0/\text{VSWR} = 25\ \Omega$ puts a voltage minimum at $R_m$, with minima every 30 cm. A minimum at 10 cm on the generator side needs $l_m = 20$ cm.
• $R_m = Z_0 \cdot \text{VSWR} = 100\ \Omega$ puts a maximum at $R_m$, with the nearest minimum 15 cm away. A minimum at 10 cm needs $l_m = 5$ cm.

Both (B) and (C) are valid.