Processing math: 100%

Introduction to gpsurya blog

Dear Friends, In this blog you will find around 320 solved Geophysics GATE and CSIR-NET and other competitive exams solutions with a better explanation. Please follow and share if you like it. Thanks, gpsurya and group

GATE-2020 (22)

 22. Skin Depth in homogeneous media of resistivity \rho_{1} and \rho_{2} are 100 m and 200 m respectively, at 1000Hz frequency. The ratio \frac{\rho_{1}}{\rho_{2}} will be __________.

 (Thanks to Pragnath, AU)

Solution:

Skin depth formula  D = 503.8 * \sqrt{(\frac{\rho}{f})}

Given that,

 D_{1}= 100m~,D_{2}= 200 m;         f=1000~Hz

\frac{D_{1} = 503.8 * \sqrt{(\frac{\rho_{1}}{f})}}{D_{2} = 503.8 * \sqrt{(\frac{\rho_{2}}{f})}}

\frac{100}{200}=\frac{503.8 * \sqrt{(\frac{\rho_{1}}{1000})}}{ 503.8 * \sqrt{(\frac{\rho_{2}}{1000})}}

\frac{1}{2}=\frac{\sqrt{(\frac{\rho_{1}}{1000})}}{\sqrt{(\frac{\rho_{2}}{1000})}}

\frac{1}{2}=\frac{\sqrt{(\rho_{1}}}{\sqrt{(\rho_{2}})}

Squaring on both sides we get

\frac{1}{4}=\frac{{\rho_{1}}}{\rho_{2}}

\frac{{\rho_{1}}}{\rho_{2}}=0.25

Related Posts

Post a Comment

Subscribe Our Newsletter