At what depth below the surface of water total pressure would be equal to $1.5$ atmosphere?
$1 a t m o s p h e r e = 10^{5} N m^{- 2}$ .

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Solution

Total pressure at depth $h$ below the water surface
$= a t m o s p h e r i c p r e s s u r e + w a t e r p r e s s u r e$ .
Let atmospheric pressure be $P$ . Then, total pressure $= 1.5 P$
$∴ 1.5 P = P + h d g \Rightarrow h d g = 0.5 P$
$\Rightarrow h = \frac{0.5 P}{d g} = \frac{0.5 \times 10^{5} N m^{- 2}}{(10^{3} k g m^{- 2}) \times 9.8 m s^{- 2}}$
$= \frac{5 \times 10^{5}}{10^{3} \times 98} m = \frac{500}{98} m = 5.1 m$

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At what depth below the surface of water total pressure would be equal to 1.5 atmosphere?1atmosphere=105Nm−2.

Total pressure at depth h below the water surface=atmosphericpressure+waterpressure.Let atmospheric pressure be P. Then, total pressure =1.5P∴1.5P=P+hdg⇒hdg=0.5P⇒h=0.5Pdg=0.5×105Nm−2(103kgm−2)×9.8ms−2=5×105103×98m=50098m=5.1m

At what depth below the surface of water total pressure would be equal to $1.5$ atmosphere?
$1 a t m o s p h e r e = 10^{5} N m^{- 2}$ .