ISRO EC 2025 Q.11: solution and answer

ISRO 2025 · ECQ.11 · 1 mark · MCQ

A lossless transmission line is 80 cm long and operates at a frequency of 600 MHz. The line parameters are L = 0.25 μH/m and C = 100 pF/m. Find the phase velocity :

(A)
3 × 10^8 m/s
(B)
2 × 10^8 m/s
(C)
4 × 10^8 m/s
(D)
6 × 10^8 m/s

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Answer and solution

ANSWER

The phase velocity is $2 \times 1 0^{8} m/s$ , option (B).

Official answer: BSource: ISRO's final answer key of 15 May 2026

For a lossless line the phase constant is $β = ω L C$ , so the phase velocity is $v_{p} = \frac{ω}{β} = \frac{1}{L C} .$
$v_{p}$ depends only on the per-unit-length $L$ and $C$ . The line length ( $80 cm$ ) and the frequency ( $600 MHz$ ) are not needed.
$L C = (0.25 \times 1 0^{- 6} H/m) (100 \times 1 0^{- 12} F/m) = 25 \times 1 0^{- 18} s^{2} / m^{2}$ .
$L C = 5 \times 1 0^{- 9} s/m$ .
$v_{p} = \frac{1}{5 \times 1 0 ^{- 9}} = 2 \times 1 0^{8} m/s$
This is below $c = 3 \times 1 0^{8} m/s$ , as expected for a line filled with a dielectric (for $μ_{r} = 1$ it corresponds to $ε_{r} = (c / v_{p})^{2} = 2.25$ ). This is option (B).

KEY IDEA

On a lossless line, $v_{p} = 1/ L C$ using the per-metre inductance and capacitance. Length and frequency are distractors.

WATCH OUT FOR

Assuming the velocity is $3 \times 1 0^{8} m/s$ regardless of the line, or trying to use the $80 cm$ length and $600 MHz$ frequency, or slipping on the prefixes ( $μ$ and $p$ ) when multiplying $L$ and $C$ .

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 6 ISRO Transmission lines questions, with notes

ISRO EC 2025 Q.11: solution and answer

Answer and solution

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