A load of 20kg is suspended by a steel wire. Its frequency when rubbed with a resigned cloth is found to be 20 times its freqency when plucked. Find the area of cross section of the wire. Young's modulus of steel is `19.6 xx 10^(10)N//m^2`.

Explanation

To solve the problem step by step, we will use the relationship between the frequencies of the waves, tension in the wire, and Young's modulus.

### Step 1: Understand the relationship between frequencies
Given that the frequency when rubbed (F1) is 20 times the frequency when plucked (F2), we can write:
\[ F_1 = 20 \times F_2 \]

### Step 2: Determine the tension in the wire
The tension (T) in the wire due to the suspended load is given by:
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