Advertisements
Advertisements
प्रश्न
What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?
Advertisements
उत्तर १
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa
Density of water at the surface, ρ1 = 1.03 × 103 kg m–3
Let ρ2 be the density of water at the depth h.
Let V1 be the volume of water of mass m at the surface.
Let V2 be the volume of water of mass m at the depth h.
Let ΔV be the change in volume.
ΔV = V1 - V2
`=m(1/rho_1 - 1/rho_2)`
`:."Volumetric strain" = (triangle V)/V_1`
`= m(1/rho_1 - 1/rho_2) xx rho_1/m`
`:.triangleV/V_1 = 1 - rho_1/rho_2` ....(i)
Bulk modulus, `B = (rhoV_1)/(triangleV)`
`(triangleV)/V_1 = rho/B`
Compressibity of water = `1/B = 45.8 xx 10^(-11) Pa^(-1)`
`:.(triangleV)/V_1 = 80xx 1.013 xx 10^5 xx 45.8 xx 10^11 = 3.71 xx 10^(-3)` ...(ii)
For equation i and ii we get
`1 - rho_1/rho_2 = 3.71 xx 10^(-3)`
`rho_2 = (1.03 xx 10^3)/(1-(3.71xx10^(-3)))`
`=1.034 xx 10^3 kg m^(-3)`
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.
उत्तर २
Compressibility of water `k = 1/B = 45.8 xx 10^11 Pa^(-1)`
Change in pressure, `trianglep = 80 atm - 1 atm`
`= 79 atm = 79 xx 1.013 xx 10^5 Pa`
Density of water at the surface
`rho = 1.03 xx 10^3 kg m^(-3)`
As `B= (trianglep.V)/(triangleV) or (triangleV)/V = (trianglep)/B = trianglep xx 1/B = trianglep xx k`
or `triangleV/V = 79 xx 1.013 xx 10^5 xx 45.8 xx 10^(-11) = 3.665 xx 10^(-5)`
Now `(triangleV)/V = ((M/rho)-(M/(rho')))/(M/rho) = 1 - rho/rho'`
or `rho/(rho') = 1 - (triangleV)/V`
or `rho' = rho/(1-(triangleV/V))`
or `rho' = (1.03 xx 10^3)/(1-3.665 xx 10^(-3)) = (1.03 xx 10^3)/0.996`
`= 1.034 xx 10^3 "kg/m"^3`
संबंधित प्रश्न
Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
How much should the pressure on a litre of water be changed to compress it by 0.10%?
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 × 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk modulus of water = 2.1 × 109 N m−2.
The ratio of adiabatic bulk modulus and isothermal bulk modulus of gas is `("where" γ = "C"_"P"/"C"_"V")`
Bulk modulus of a perfectly rigid body is ______.
A ball falling in a lake of depth 300 m shows a decrease of 0.3% in its volume at the bottom. What is the bulk modulus of the material of the ball? (g = 10 m/s2)
For an ideal liquid ______.
- the bulk modulus is infinite.
- the bulk modulus is zero.
- the shear modulus is infinite.
- the shear modulus is zero.
For an ideal liquid ______.
- the bulk modulus is infinite.
- the bulk modulus is zero.
- the shear modulus is infinite.
- the shear modulus is zero.
To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. (The bulk modulus of rubber is 9.8 × 108 Nm–2; and the density of sea water is 103 kg m–3.)
A gas undergoes a process in which the pressure and volume are related by VPn = constant. The bulk modulus of the gas is ______.
Bulk modulus applies to ______.
Which of the following is an alternate name for bulk modulus?
Volume strain is calculated as ______.
Which of the following materials has the highest resistance to compression?
Bulk modulus is defined as the ratio of ______ to ______.
