NEET · Physics · STD 11 - 9.1 fluid mechanics
A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface ?
(Consider the density of water \(=1000 kg m ^{-3}, 1 atm=1 \times 10^5 Pa\) and gravitational acceleration \(g=10 m / s ^2\) )
- A 990 m
- B 9900 m
- C 99 m
- D 9000 m
Answer & Solution
Correct Answer
(A) 990 m
Step-by-step Solution
Detailed explanation
(A) 990 m
The absolute pressure at a depth \(h\) below the water surface is given by \(P=P_0+\rho g h\), where \(P_0\) is the atmospheric pressure.
Given:
\(P=100 atm=100 \times 10^5 Pa\)
\(P_0=1 atm=1 \times 10^5 Pa\)
\(\rho=1000 kg m^{-3}\)
\(g=10 m / s^2\)
Substituting the values into the equation:
\(100 \times 10^5=1 \times 10^5+1000 \times 10 \times h\)
\(99 \times 10^5=10^4 \times h\)
\(h=\frac{99 \times 10^5}{10^4}\)
\(h=990 m\)
The absolute pressure at a depth \(h\) below the water surface is given by \(P=P_0+\rho g h\), where \(P_0\) is the atmospheric pressure.
Given:
\(P=100 atm=100 \times 10^5 Pa\)
\(P_0=1 atm=1 \times 10^5 Pa\)
\(\rho=1000 kg m^{-3}\)
\(g=10 m / s^2\)
Substituting the values into the equation:
\(100 \times 10^5=1 \times 10^5+1000 \times 10 \times h\)
\(99 \times 10^5=10^4 \times h\)
\(h=\frac{99 \times 10^5}{10^4}\)
\(h=990 m\)
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