An object is taken 1.0 , km deep in the sea. | vedclass

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

Question

(B) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.Since the mass $M = \rho V$ is constant,differentiating gives $\rho \Delta V + V

Vedclass · Accepted Answer

$0.64$

Vedclass · Answer

(B) The bulk modulus $B$ is defined as $B = -\frac{P}{\Delta V / V}$.Since the mass $M = ho V$ is constant,differentiating gives $ho \Delta V + V \Delta ho = 0$,which implies $\frac{\Delta ho}{ho} = -\frac{\Delta V}{V}$.Substituting this into the bulk modulus formula,we get $\frac{\Delta ho}{ho} = \frac{P}{B}$.The pressure at depth $h$ is $P = ho_{sea} g h$.Given $ho_{sea} = 1.025 	imes 10^3 \, kg/m^3$,$g = 10 \, m/s^2$,$h = 1000 \, m$,and $B = 1.6 	imes 10^9 \, Pa$ $(1.6 	imes 10^6 \, kPa = 1.6 	imes 10^9 \, Pa)$.$\frac{\Delta ho}{ho} = \frac{1.025 	imes 10^3 	imes 10 	imes 1000}{1.6 	imes 10^9} = \frac{1.025 	imes 10^7}{1.6 	imes 10^9} = 0.6406 	imes 10^{-2} \approx 0.0064$.Percentage change $= \frac{\Delta ho}{ho} 	imes 100 = 0.64 \%$.

An object is taken $1.0 \, km$ deep in the sea. The density of sea water is $1.025 \times 10^3 \, kg/m^3$ and the bulk modulus of the object is $1.6 \times 10^6 \, kPa$. Find the percentage change in the density of the object.

Similar Questions

The approximate depth of an ocean is $2700 \, m$. The compressibility of water is $45.4 \times 10^{-11} \, Pa^{-1}$ and the density of water is $10^3 \, kg/m^3$. What fractional compression of water will be obtained at the bottom of the ocean?

The isothermal bulk modulus of an ideal gas at pressure $P$ is:

The compressibility of water is $5 \times 10^{-10} \ m^2/N$. $A$ pressure of $15 \times 10^6 \ Pa$ is applied on $100 \ ml$ volume of water. The change in the volume of water is:

The density of a metal at normal pressure $P$ is $\varrho$. When it is subjected to an excess pressure $p$,the density becomes $\varrho^{\prime}$. If $K$ is the bulk modulus of the metal,then the ratio $\frac{\varrho^{\prime}}{\varrho}$ is

Determine the volume contraction of a solid copper cube,$10 \; cm$ on an edge,when subjected to a hydraulic pressure of $7.0 \times 10^{6} \; Pa$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module