A submarine experiences a pressure of 5.05 ti | vedclass

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(D) The pressure at a depth $d$ in a fluid is given by $P = P_{atm} + \rho g d.$For two different depths $d_1$ and $d_2,$ the pressure difference is g

Vedclass · Accepted Answer

$303$

Vedclass · Answer

(D) The pressure at a depth $d$ in a fluid is given by $P = P_{atm} + ho g d.$For two different depths $d_1$ and $d_2,$ the pressure difference is given by $\Delta P = P_2 - P_1 = ho g (d_2 - d_1).$Given $P_1 = 5.05 	imes 10^6 \, Pa$ and $P_2 = 8.08 	imes 10^6 \, Pa.$The difference in pressure is $\Delta P = (8.08 - 5.05) 	imes 10^6 \, Pa = 3.03 	imes 10^6 \, Pa.$Using the formula $\Delta P = ho g (d_2 - d_1)$ where $ho = 10^3 \, kg/m^3$ and $g = 10 \, m/s^2$:$3.03 	imes 10^6 = 10^3 	imes 10 	imes (d_2 - d_1)$$3.03 	imes 10^6 = 10^4 	imes (d_2 - d_1)$$d_2 - d_1 = \frac{3.03 	imes 10^6}{10^4} = 3.03 	imes 10^2 = 303 \, m.$Therefore,the value of $d_2 - d_1$ is $303 \, m.$

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