Question

A glass full of water has a bottom of area 20 cm², top of area 20 cm², height 20 cm and volume half a liter. Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on water. Atmospheric pressure = 105 N/m², density of water = 1000 kg/m³ and g = 10 m/s².

• A. 204 N upwards
• B. 205 N upwards
• C. 1 N upwards
• D. 205 N downwards

Answer 1

The correct option is C

1 N upwards

Consider the free body diagram of water contained,$F_b$ = Force by the beaker on water

For the contained water to be in equilibrium, the net force acting on it by various external means must be zero.

Let $F_g$ = Force of gravitational attraction = ρVg

Given,$V=\frac{1}{2}liters=500c{m}^3$

$1liter=1000c{m}^{3}approx$

$\Rightarrow F_g=\frac{1000\times 500\times 10}{(100{)}^3}=5Nupwards$

$F_{atm}$ = Force of atmospheric pressure from top

$=P_{atm}\times A=\frac{{10}^5\times 20}{100\times 100}=200N$

From Newton's 2nd law,

$F_{atm}+F_g=F_b$

$\Rightarrow F_b=205N$

Note: Now this force also includes the force from the bottom of the container. By subtracting this force from the total force, you will get the force due to sides of the glass container.
Let's calculate the force from the bottom = F'

F'=(Net~ pressure~at~the~bottom)$\times$(Area)

F'= 204N Upwards

Now force from sides =$F_b-F^{\prime }$

= (205 - 204)N

= 1 N upwards