First slide

Standing waves

Question

A load of 20 kg is suspended by a steel wire. Its frequency when rubbed with a resined cloth is found to be 20 times its frequency when plucked. If y of the wire is 19 .6 x 1o1r dyne/square cm, what is its area of cross-section ?

Moderate

A

$2 \times 10^{- 3} {cm}^{2}$
B

$3 \times 10^{- 3} {cm}^{2}$
C

$4 \times 10^{- 3} {c m}^{2}$
D

$5 \times 10^{- 3} {cm}^{2}$

Solution

When the wire is rubbed with resin cloth, its frequency is given by
$n_{1} = \frac{1}{2 l} \sqrt{(\frac{Y}{ρ})} for a loaded wire n_{2} = \frac{1}{2 l} \sqrt{(\frac{T}{aρ})} ∴ \frac{n_{1}}{n_{2}} = \sqrt{\frac{Ya}{T}} or 20 = \sqrt{(\frac{Ya}{T})} or a = \frac{400 T}{Y} = \frac{400 \times 20 \times 1000 \times 980}{19 \cdot 6 \times 10^{11}} = 4 \times 10^{- 3} {cm}^{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App