Question

A glass of water up to a height of $10{\text{cm}}$ has a bottom area of $5{\text{c}}{{\text{m}}^2}$ and top area of $10{\text{c}}{{\text{m}}^2}$. Find the downward force exerted by water on the bottom. (Take $g = 10{\text{m}}{{\text{s}}^{ - 2}}$, ${\rho _w} = {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$ and ${P_a} = 1.01 \times {10^5}{\text{N}}{{\text{m}}^{ - 2}}$ ).
A) $102{\text{N}}$
B) $120{\text{N}}$
C) $22{\text{N}}$
D) $51{\text{N}}$

Answer 1

Hint: The force exerted at the bottom of the glass by the water will be the product of the pressure and the area of the base of the glass. As the glass is kept open to the atmosphere, the pressure at the bottom of the glass will be the sum of the atmospheric pressure and the pressure for the height of the water in the glass.

Formula used:
The force exerted on a container is given by, $F = P \times A$ where $P$ is the pressure exerted and $A$ is the area of the container.

Complete step by step answer:
Step 1: List the parameters given in the question.
The height up to which water exists in the glass is given to be $h = 10{\text{cm}}$ .
The bottom and top area of the glass are given to be ${A_b} = 5{\text{c}}{{\text{m}}^2}$ and ${A_b} = 10{\text{c}}{{\text{m}}^2}$ respectively.
It is also given that the acceleration due to gravity is $g = 10{\text{m}}{{\text{s}}^{ - 2}}$, the density of water is ${\rho _w} = {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$ and the atmospheric pressure is ${P_a} = 1.01 \times {10^5}{\text{N}}{{\text{m}}^{ - 2}}$ .
Let $F$ be the force exerted at the bottom of the glass by the water in it.

Step 2: Express the force exerted downward at the bottom of the glass.
The total pressure $P$ experienced at the bottom of the glass will be the sum of the atmospheric pressure and the pressure for the height up to which water is present in the glass.
 i.e., $P = {P_a} + {\rho _w}gh$ -------- (1)
Then force exerted at the bottom of the glass can be expressed as $F = P \times {A_b}$ ---------- (2)
Substituting equation (1) in (2) we get, $F = \left( {{P_a} + {\rho _w}gh} \right){A_b}$ --------- (3)

Substituting for ${P_a} = 1.01 \times {10^5}{\text{N}}{{\text{m}}^{ - 2}}$, $g = 10{\text{m}}{{\text{s}}^{ - 2}}$, $h = 10{\text{cm}}$, ${\rho _w} = {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$ and ${A_b} = 5{\text{c}}{{\text{m}}^2}$ in equation (3) we get,
$F = \left( {1.01 \times {{10}^5} + {{10}^3} \times 10 \times 0.1} \right) \times 5 \times {10^{ - 4}}$
On calculating we get the downward force exerted by water as $F = 51{\text{N}}$

So the correct option is (D).

Note: While substituting values of the different physical quantities in equation (3) make sure that all the quantities are expressed in their respective S.I. units. If not, then the necessary conversion of units must be done. Here the height $h$ and the bottom area ${A_b}$ were not expressed in their S.I. units. So during substitution, we converted the height as $h = 0.1{\text{m}}$ and the area as ${A_b} = 5 \times {10^{ - 4}}{{\text{m}}^2}$ .