Depth of a river is 100 m. Magnitude of comp | vedclass

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(B) Given: Depth $h = 100 \ m$,Compressibility $k = 0.5 	imes 10^{-9} \ N^{-1} \ m^2$,Density of water $ho = 10^3 \ kg/m^3$,Acceleration due to gra

Vedclass · Accepted Answer

$0.5 	imes 10^{-3}$

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(B) Given: Depth $h = 100 \ m$,Compressibility $k = 0.5 	imes 10^{-9} \ N^{-1} \ m^2$,Density of water $ho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$.Pressure at the bottom of the river is given by $P = ho gh$.$P = 10^3 	imes 10 	imes 100 = 10^6 \ N/m^2$.Compressibility $k$ is defined as the reciprocal of Bulk Modulus $B$,where $k = \frac{1}{B} = \frac{(\Delta V / V)}{P}$.Therefore,the fractional compression (fractional change in volume) is $\frac{\Delta V}{V} = k 	imes P$.$\frac{\Delta V}{V} = (0.5 	imes 10^{-9}) 	imes 10^6 = 0.5 	imes 10^{-3}$.

Depth of a river is $100 \ m$. Magnitude of compressibility of the water is $0.5 \times 10^{-9} \ N^{-1} \ m^2$. The fractional compression in water at the bottom of the river is (Acceleration due to gravity $= 10 \ m/s^2$)

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