Question 153: Properties of Matter

Question

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 ms–2
) :

Accepted Answer

The pressure experienced by a submarine at a certain depth in the sea is given by the formula:
$P = ho g h$
where:
$P$ is the pressure
$ho$ is the density of the fluid (sea water in this case)
$g$ is the acceleration due to gravity
$h$ is the height (or depth in this case)
Given:
$ho = 10^3 \, kg/m^3$
$g = 10 \, m/s^2$
We are looking for the difference in depth, $d_2 - d_1$, which corresponds to the difference in pressure $\Delta P$:
$\Delta P = P_2 - P_1 = ho g (d_2 - d_1)$
Rearranging the above equation, we get:
$d_2 - d_1 = \frac{\Delta P}{ho g}$
Given:
$P_2 = 8.08 	imes 10^6 \, Pa$
$P_1 = 5.05 	imes 10^6 \, Pa$
$\Delta P = P_2 - P_1 = 8.08 	imes 10^6 \, Pa - 5.05 	imes 10^6 \, Pa = 3.03 	imes 10^6 \, Pa$
So,
$d_2 - d_1 = \frac{3.03 	imes 10^6 \, Pa}{10^3 \, kg/m^3 	imes 10 \, m/s^2} = 303 \, m$
The closest answer among the options provided is Option C, 300 m.

Answer

600 m

Answer

400 m

Answer

300 m

Answer

500 m

Properties of Matter

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 ms–2 ) :

Choose the correct answer: