This question was previously asked in

ISRO (VSSC) Technical Assistant Electronics 2019 Official Paper

Option 2 : 1500 m

Subject Test 1: Electrical Basics

3012

20 Questions
80 Marks
30 Mins

__Concept__:

In the RADAR system, the range of the target is:

\(R = \frac{{C\; \times \;PRT}}{2}m\)

And,

PRT = (PW + RT) μ sec

Where

C = 3 × 10^{8} m/s = speed of light at which EMW travel

PRT = Pulse repetition time

PW = Pulse width

RT = rest time i.e. time interval between sending a pulse and return from target to the receiver.

__Calculation__**:**

Given that,

PW = 2 μ sec

RT = 10 μ sec

∴ PRT = 2 + 10 = 12 μ sec

\(R = \frac{{3\; \times \;{{10}^8}\; \times \;12\; \times \;{{10}^{ - 6}}}}{2}\)

\(R = 1800\;m\)

If we take an approximate value of PRT, we can write:

PRT = 2 + 10 ≈ 10 μ sec

\(R = \frac{{3\; \times \;{{10}^8}\; \times \;10\; \times \;{{10}^{ - 6}}}}{2}\)