Question

A beaker of circular cross section of radius 4 cm is filled with mercury up to a height of 10 cm. find the force exerted by the mercury on the bottom of the beaker. The atmospheric pressure = ${10}^{5}N{m}^{-2}$. Density of mercury = $13600kg{m}^{-3}$. Take $g=10m{s}^{-2}$.

• A. 571 N
• B. 674 N
• C. 486 N
• D. 544 N

Answer 1

The correct option is A

571 N

The pressure at the surface= atmospheric pressure

=${10}^{5}N{m}^{-2}$

The pressure at the bottom

=${10}^{5}N{m}^{-2}+hpg$

=${10}^{5}N{m}^{-2}+\left(0.1m\right)\left(13600kg{m}^{-3}\right)\left(10m{s}^{-2}\right)$

=$1.136\times {10}^{5}N{m}^{-2}$

The force exerted by the mercury on the bottom

=$(1.136\times {10}^{5}N{m}^{-2})\times (3.14\times 0.04m\times 0.04m)$

=571N