Properties Of Solids And Liquids Question 301

Options:

A) $ \rho ={\rho _{0}}\left[ 1-\frac{{\rho _{0}}gy}{B} \right] $

B) $ \rho ={\rho _{0}}\left[ 1+\frac{{\rho _{0}}gy}{B} \right] $

C) $ \rho ={\rho _{0}}\left[ 1+\frac{Beta }{{\rho _{0}}hgy} \right] $

D) $ \rho ={\rho _{0}}\left[ 1-\frac{B}{{\rho _{0}}gy} \right] $

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Answer:

Correct Answer: B

Solution:

Bulk modulus, $ B=-V _{0}\frac{\Delta p}{\Delta V}\Rightarrow \Delta V=-V _{0}\frac{\Delta p}{B} $

therefore $ V=V _{0}\left[ 1-\frac{\Delta p}{B} \right] $

Density, $ \rho ={\rho _{0}}{{\left[ 1-\frac{\Delta p}{B} \right]}^{-1}}={\rho _{0}}\left[ 1+\frac{\Delta p}{B} \right] $

where, $ \Delta p=p-p _{0}=h{\rho _{0}}g $

= pressure difference between depth and surface of ocean $ \rho ={\rho _{0}}\left[ 1+\frac{{\rho _{0}}gy}{B} \right] $ (As h = y)



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