Question

A glass of water upto a height $10$cm has bottom area $5c{m}^2$ and top area $10c{m}^2$. The downward force exerted by water on the bottom is? (Take $g=10m/s^2,{\rho }_w={10}^{3}kg/m^3,P_a=1.01\times {10}^{5}N{m}^{-2}$)

• A. $102$N
• B. $120$N
• C. $22$N
• D. $51$N

Answer 1

The correct option is D $51$N

Given, $p_a=1.01\times {10}^{5}N/m^2$
${\rho }_w={10}^{3}kg/m^3$
$g=10m/s^2$
$h=10cm=0.10$m
Area of base $=A_1=5c{m}^2=5\times {10}^{-4}{m}^2$
Area of top $=A_2=10c{m}^2={10}^{-3}{m}^2$
Force exerted by water on the bottom of the glass is
$F_1=(p_a+h{\rho }_{w}g)\cdot A_1$
$=(1.01\times {10}^5+0.10\times {10}^3\times 10)5\times {10}^{-4}$
$=(101000+1000)5\times {10}^{-4}$
$=102000\times 5\times {10}^{-4}=10.2\times 5$
$F_1=51$N.