Given: sigma is the compressibility of water, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The pressure at a depth $h$ in a lake is given by the hydrostatic pressure formula: $P = h\rho g$.This pressure acts as the stress on the water.Th

Vedclass · Accepted Answer

$\frac{1}{2}\sigma (h\rho g)^2$

Vedclass · Answer

(A) The pressure at a depth $h$ in a lake is given by the hydrostatic pressure formula: $P = hho g$.This pressure acts as the stress on the water.The energy density $u$ stored in a material under stress is given by the formula: $u = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.Since the bulk modulus $K = \frac{	ext{stress}}{	ext{strain}}$,we have $	ext{strain} = \frac{	ext{stress}}{K}$.Substituting this into the energy density formula: $u = \frac{1}{2} 	imes 	ext{stress} 	imes \frac{	ext{stress}}{K} = \frac{1}{2} \frac{(	ext{stress})^2}{K}$.Given that compressibility $\sigma = \frac{1}{K}$,we can rewrite the expression as: $u = \frac{1}{2} \sigma (	ext{stress})^2$.Substituting the stress $P = hho g$,we get: $u = \frac{1}{2} \sigma (hho g)^2$.

Given: $\sigma$ is the compressibility of water,$\rho$ is the density of water,and $K$ is the bulk modulus of water. What is the energy density of water at the bottom of a lake $h$ metre deep?

Similar Questions

$A$ uniform cubical block is subjected to volumetric compression,which decreases its each side by $2 \%$. The Bulk strain produced in it is ............

How much should the pressure on a litre of water be changed to compress it by $0.10\%?$

The isothermal bulk modulus of a perfect gas at normal pressure is

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at a temperature of $57\,^{\circ}C$ is drunk. You can take the body (tooth) temperature to be $37\,^{\circ}C$ and $\alpha = 1.7 \times 10^{-5}/^{\circ}C$,bulk modulus for copper $= 140 \times 10^9\, N/m^2$.

The density of a metal at normal pressure is $\rho$. Its density when it is subjected to an excess pressure $p$ is $\rho^{\prime}$. If $B$ is the Bulk modulus of the metal,the ratio $\frac{\rho^{\prime}}{\rho}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module