MHT CET · Physics · Mechanical Properties of Fluids
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 \(=\) \(\left.1 \cdot 01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right]\)
- A \(10 \cdot 101 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\)
- B \(10 \cdot 101 \times 10^{7}\) dyne \(/ \mathrm{cm}^{2}\)
- C \(10 \cdot 101 \times 10^{6}\) dyne \(/ \mathrm{cm}^{2}\)
- D \(10 \cdot 101 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\)
Answer & Solution
Correct Answer
(D) \(10 \cdot 101 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\)
Step-by-step Solution
Detailed explanation
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
\(\mathrm{P}=\mathrm{~hdg}+1 \mathrm{~atm}=1000 \times 1.03 \times 10^{3} \times 10+1\mathrm{~atm}\) \(=1.03 \times 10^{7} \mathrm{~Pa}+1 \mathrm{~atm}\)
\(P=103 \mathrm{~atm}+1 \mathrm{~atm}=104 \mathrm{~atm}\)
The absolute pressure is given by: absolute pressure \(=\) pressure of water \(+\) atmospheric pressure
\(\mathrm{P}=\mathrm{~hdg}+1 \mathrm{~atm}=1000 \times 1.03 \times 10^{3} \times 10+1\mathrm{~atm}\) \(=1.03 \times 10^{7} \mathrm{~Pa}+1 \mathrm{~atm}\)
\(P=103 \mathrm{~atm}+1 \mathrm{~atm}=104 \mathrm{~atm}\)
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