The figure shows the vertical cross-section of a vessel filled with a liquid of density $\rho$. The normal thrust per unit area on the walls of the vessel at the point $P$,as shown,will be

$A$ submarine experiences a pressure of $5.05 \times 10^6 \, Pa$ at a depth of $d_1$ in a sea. When it goes further to a depth of $d_2,$ it experiences a pressure of $8.08 \times 10^6 \, Pa.$ Then $d_2 - d_1$ is approximately ........ $m$ (density of water $= 10^3 \, kg/m^3$ and acceleration due to gravity $= 10 \, m/s^2$).

Explain how an open-ended tube manometer measures pressure.

The pressure acting on a submarine is $3 \times 10^{5} \; Pa$ at a certain depth. If the depth is doubled,the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is $1 \times 10^{5} \; Pa$,density of water is $10^{3} \; kg \; m^{-3}$,and $g = 10 \; m \; s^{-2}$)

Who devised for the first time,a method for measuring atmospheric pressure?

$A$ healthy adult of height $1.7 \,m$ has an average blood pressure $(BP)$ of $100 \,mm$ of $Hg$. The heart is typically at a height of $1.3 \,m$ from the foot. Take the density of blood to be $10^3 \,kg/m^3$ and note that $100 \,mm$ of $Hg$ is equivalent to $13.3 \,kPa$ (kilo pascals). The ratio of $BP$ in the foot region to that in the head region is close to: