A submarine experiences a pressure of 5.05 × 10 6 P a at a depth of d 1 in a sea. When it goes further to a depth of d 2 , it experiences a pressure of 8.08 × 10 6 P a . Then d 2 - d 1 is approximately (density of water = 10 3 k g / m 3 and acceleration due to gravity = 10 m s - 2 ):

Question

A submarine experiences a pressure of 5.05 × 10 6 P a at a depth of d 1 in a sea. When it goes further to a depth of d 2 , it experiences a pressure of 8.08 × 10 6 P a . Then d 2 - d 1 is approximately (density of water = 10 3 k g / m 3 and acceleration due to gravity = 10 m s - 2 ):, 600 m, 500 m, 300 m, 400 m

MARKS App · Accepted Answer

The pressure at a depth d is p = p 0 + ρ g d , here p 0 is the atmospheric pressure. ∴ p 1 = p 0 + ρ g d 1 and p 2 = p 0 + ρ g d 2 The difference in pressure is p 2 - p 1 = ρ g d 2 - d 1 Putting the values in the above equation 8.08 × 10 6 - 5.05 × 10 6 = 10 3 × 10 d 2 - d 1 d 2 - d 1 = 3.03 × 10 6 10 4 = 303 m

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