What would be the absolute pressure at depth $1 \mathrm{~km}$ below the ocean? [Given : density of water…

What would be the absolute pressure at depth $1 \mathrm{~km}$ below the ocean? [Given : density of water $=10^{3} \mathrm{~kg} / \mathrm{m}^{3}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}, 1$ atmospheric pressure $=$ $\left.1 \cdot 01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right]$
  1. $10 \cdot 101 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}$
  2. $10 \cdot 101 \times 10^{7}$ dyne $/ \mathrm{cm}^{2}$
  3. $10 \cdot 101 \times 10^{6}$ dyne $/ \mathrm{cm}^{2}$
  4. $10 \cdot 101 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}$

Solution

Given: $\mathrm{h}=1000 \mathrm{~m}, \mathrm{~d}=1.03 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ The absolute pressure is given by: absolute pressure $=$ pressure of water $+$ atmospheric pressure $\mathrm{P}=\mathrm{~hdg}+1 \mathrm{~atm}=1000 \times 1.03 \times 10^{3} \times 10+1\mathrm{~atm}$ $=1.03 \times 10^{7} \mathrm{~Pa}+1 \mathrm{~atm}$ $P=103 \mathrm{~atm}+1 \mathrm{~atm}=104 \mathrm{~atm}$ :

Asked in: MHT CET 2020 (20 Oct Shift 2)

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