A scuba diver is exploring a freshwater lake at a depth of $10$ m. What is the ratio of the absolute pressure to the gauge pressure experienced by the diver at this depth?
(Take the density of water as $1000$ kg m $^{-3}$ , atmospheric pressure as $1 \times 10^5$ Pa, and $g = 10$ m/s $^2$ )

Question

A scuba diver is exploring a freshwater lake at a depth of

10

m. What is the ratio of the absolute pressure to the gauge pressure experienced by the diver at this depth?
(Take the density of water as

1000

kg m

^{-3}

, atmospheric pressure as

1 \times 10^5

Pa, and

g = 10

m/s

^2

)

Quantrex Academy · Accepted Answer

Correct answer: B. 2:1

The gauge pressure at a depth h is given by P_ gauge = g h . Given: h = 10 m = 1000 kg m ^ -3 g = 10 m/s ^2 Substituting the values: P_ gauge = 1000 × 10 × 10 = 1 × 10^5 Pa The absolute pressure is the sum of atmospheric pressure and gauge pressure: P_ abs = P_ atm + P_ gauge Given P_ atm = 1 × 10^5 Pa: P_ abs = 1 × 10^5 + 1 × 10^5 = 2 × 10^5 Pa The required ratio is: P_ abs P_ gauge = (2 × 10^5)/(1 × 10^5) = (2)/(1) The ratio is 2:1 . Answer: 2:1

A scuba diver is exploring a freshwater lake at a depth of 10 m....

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