Ncert Exercises & Solutions

Question 9.13:
What is the density of water at a depth where pressure is 80.0 atm, given that its density
at the surface is $1.03 \times 10^{3} {kgm}^{- 3}$ ?

Pressure at the depth, P = 80.0 atm = 80 × 1.01 × 10 $^{5}$ Pa
The density of water at the surface, $ρ_{1}$ = 1.03 × 10 $^{3}$ kg m $^{- 3}$
Let ρ $_{2}$ be the density of water at the depth h.
Let ΔV be the change in volume.

Bulk modulus, B= $\frac{{PV}_{1}}{∆ V}$ $\Rightarrow \frac{∆ V}{V_{1}} = \frac{P}{B}$

Compressibility of water = $\frac{1}{B} = 45.8 \times 10^{- 11} {Pa}^{- 1}$

$∴ \frac{∆ V}{V_{1}} = 80 \times 1.013 \times 10^{5} \times 45.8 \times 10^{- 11} = 3.71 \times 10^{- 3}$

Change in volume is given by:

$∆ V = V_{1} - V_{2} = m (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}}) ∴ Volumetric strain = \frac{∆ V}{V_{1}} = m (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}}) \times \frac{ρ_{1}}{m} ∴ \frac{∆ V}{V_{1}} = 1 - \frac{ρ_{1}}{ρ_{2}}$

$\Rightarrow 1 - \frac{ρ_{1}}{ρ_{2}} = 3.71 \times 10^{- 3} \Rightarrow ρ_{2} = \frac{1.03 \times 10^{3}}{1 - 3.71 \times 10^{- 3}} = 1.034 \times 10^{3} {Kgm}^{- 3}$

Therefore, the density of water at the given depth is $1.034 \times 10^{3} {Kgm}^{- 3} .$

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