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A submarine experiences a pressure of 5.05 × 10⁶ Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 10⁶ Pa. Then d2 − d1 is approximately (density of water = 10³ kg/m³ and acceleration due to gravity = 10 m s⁻²):
- 600 m
- 400 m
- 300 m
- 500 m
Correct answer: 300 m
Solution
Pressure difference = rho*g*(d2-d1), so d2-d1 = (8.08x10^6 - 5.05x10^6)/(10^3 * 10) = 3.03x10^6/10^4 = 303 m, i.e. about 300 m.
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